Chemical Equilibrium

1
Chapter 6

• Chemical Equilibrium

2
Equilibrium

• The state in which the forward and reverse rates
  of all reactions are equal, so that the
  concentrations of all species remain constant

3
Equilibrium Constant

• A value that describes whether a reaction is
  favored
• aA bB ? cC dD
• K
• A reaction is favored if Kgt1
• The larger the K value the more favored the
  reaction
• C - represents the ratio of a species to its
  concentration in its standard state
• Standard state - for solutes (1M), for gases
  (1bar)

CcDd
AaBb
4
Equilibrium Constant

• K is a dimensionless constant
• If discussing solutes (the minor components of a
  solution), all units must be expressed as mols/L
• If discussing gasses, all units must be expressed
  in terms of bars
• If discussing pure solids and liquids, ratio of
  the species 1, so these are omitted from the
  equilibrium expression

5
Equilibrium Constant

• If the direction of a reaction is reversed, then
  the K for that reaction is simply the reciprocal
  of the forward reaction (K 1/K)
• The K for two reactions that are added together,
  is the product of the K for each of the
  individual reactions (K3 K1K2)

6
Equilibrium Constant

• Write the Equilibrium Constant for the following
  reactions
• A. HA ? H A-
• H C ? CH
• HA C ? A- CH
• B. H2O ? H OH- Kw 1 x
  10-14
• NH3(aq) H2O ? NH4 OH-
  K(NH3) 1.8 x 10-5

7
Equilibrium Constant

• Write the Equilibrium Constant for the following
  reactions
• C. CuN3(s) ? Cu N3- K 4.9 x 10-9
• HN3 ? H N3- K 2.2 x 10-5
• Cu HN3 ? CuN3(s) H

8
Equilibrium and Thermodynamics

• Equilibrium constants are derived from heat
  changes (enthalpy) and the disorder (entropy)
  associated with the reactants and products

9
Enthalpy (?H)

• Enthalpy describes the heat given off or absorbed
  during a reaction
• ?H is expressed as kJ/mol at 25oC
• - ?H - indicates that heat is released and hence
  is exothermic
• ?H - indicates that heat is absorbed during the
  reaction and hence the reaction is endothermic
• ?H (reaction) ?H (products) - ?H (reactants)

10
Entropy (?S)

• Entropy describes the disorder associated with
  the reaction
• ?S is expressed as J/(K . mol) at 25oC
• - ?S - the products are less disordered than the
  reactants
• ?S - the products are more disordered than the
  reactants
• ?S (reaction) ?S (products) - ?S (reactants)

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Will the Reaction Proceed?

• -?H and ?S - reaction is favored
• ?H and -?S - reaction is unfavored
• ?H and ?S - Gibbs Free Energy
• -?H and -?S - must be calculated
• to determine the favorability of the
  reaction

12
Gibbs Free Energy (?G)

• ?G ?H - T?S
• ?G has the unit of kJ/mol
• Reaction is favored if ?G is negative
• Gibbs Free Energy relates to the Equilibrium
  Constant in that
• K e- ?G/RT
• the more negative ?G, the larger the equilibrium
  constant

13
Gibbs Free Energy (?G) Example

• For the reaction
• Mg2 Cu(s) ? Mg(s) Cu2
• K 1 x 10-92 ?S 18 J/(K . mol)
• Under standard conditions, is ?G positive or
  negative? The term standard conditions means
  that reactants and products are in their standard
  states.
• Under standard conditions, is the reaction
  endothermic or exothermic?

14
Le Châteliers Principle

• If a system at equilibrium is disturbed, the
  direction in which it proceeds back to
  equilibrium is such that the disturbance is
  partially offset
• See-Saw
• Reaction Quotient (Q) - verifies Le Châteliers
  Principle algebraically
• Calculation is exactly the same as that for the
  equilibrium constant

15
Le Châteliers Principle

• QltK - the reaction proceeds toward the products
  until QK
• QgtK - the reaction proceeds toward the reactants
  until QK
• If a change in temperature occurs, use the
  following equation to predict the effect observed
• K e- ?G/RT e-(?H - T?S)/RT
• e(-?H/RT ?S/R)
• e-?H/RT . e?S/R

16
Equilibrium Constants

• There are several types of equilibrium
  expressions that are derived depending on the
  system being discussed
• Kw
• Ksp
• Ka
• Kb

17
Solubility Product (Ksp)

• The equilibrium constant for the dissociation of
  a solid salt to give its ions in solution
• Appendix F lists values for some solubility
  products

18
Solubility Product (Ksp)

• Example
• Hg2Cl2(s) ? Hg22 2Cl-
• Ksp Hg22Cl-2 1.2 x 10-18
• Meaning
• Hg2Cl2(s) will dissolve until the solution
  reaches a concentration 1.2 x 10-18 M in
  Hg2Cl2(s). At this point, the solution is said
  to be saturated
• Any additional solid Hg2Cl2(s) will remain intact
  in solution
• Ksp value does not tell the entire story of the
  solution since pure solids are omitted from the
  equilibrium constant

19
Solubility Product (Ksp)

• Example
• Hg2Cl2(s) ? Hg22 2Cl-
• Ksp Hg22Cl-2 1.2 x 10-18
• Determine the concentration of Hg22 in a
  saturated solution

20
The Common Ion Effect

• Occurs when a salt is dissolved in a solution
  containing one of the ions in the salt. The salt
  is less soluble than it would be otherwise. (Le
  Châteliers Principle)

21
The Common Ion Effect

• Example
• Hg2Cl2(s) ? Hg22 2Cl-
• Ksp Hg22Cl-2 1.2 x 10-18
• .030 M NaCl
• Determine the concentration of Hg22 in this
  solution

22
Separation by Precipitation

• Utilizes the common ion effect and Ksp values for
  the solids to determine whether complete
  separation of two ions can be accomplished.
• Extremely important when doing an experiment in
  which one ion might interfere with the desired
  effect caused by the other

23
Separation by Precipitation

• Example
• Hg2I2(s) ? Hg22 2I- Ksp
  Hg22I-2 1.1 x 10-28
• PbI2(s) ? Pb2 2I- Ksp Pb2I-2 7.9 x
  10-9
• Consider a solution that is 0.010 M in Pb2 and
  Hg22, given the Ksp values above, can the iodide
  ion be used to completely separate the two
  cations?

24
Separation by Precipitation Example

• Is it possible to separate 99.90 of 0.020 M Mg2
  from 0.10 M Ca2 without precipitation of Ca(OH)2
  by addition of NaOH?

25
Separation by Precipitation

• Too simple
• Coprecipitation a substance whose solubility is
  not exceeded precipitates along with another
  substance whose solubility is exceeded
• Complex Formation
• More than one form of the complex exists causing
  some of the desired cation complex to redissolve
  in solution

26
Complex Formation

• Terminology
• Complex ions an ion that consists of two or
  more simple ions bonded to each other
• Al(OH)4- CuCl3-
• Ligand an atom or group of atoms attached to
  the species of interest

27
Complex Formation

• Terminology
• Lewis Acid accepts a pair of electrons (cation
  of interest)
• Lewis Base donates a pair of electrons (anion)
• Adduct product of the reaction between a Lewis
  acid and Lewis base
• Dative or Coordinate covalent bond bond formed
  between a Lewis acid and Lewis base

28
Complex Ion Formation Effects on Solubility

• Example
• PbI2
• At low I- concentrations, formation of PbI2
  governs the concentration of Pb in solution
• At higher I- concentrations, the total
  concentration of dissolved Pb is considerably
  greater due to the formation of PbI, PbI3-,
  PbI42-, PbI2(aq)
• Pbtotal Pb2 PbI PbI3- PbI42-
  PbI2(aq)

29
Complex Ion Formation Effects on Solubility

• Note
• Because all equilibrium conditions are satisfied
  simultaneously, the concentration of the species
  of interest (Pb2) that satisfies any one
  equilibria, must satisfy all of the equilibria
  because there can only be one concentration of
  the desired species in solution

30
Complex Ion Formation Effects on Solubility

• Example
• PbI2(s) ? Pb2 2I- Ksp Pb2I-2
  7.9 x 10-9
• Pb2 I- ? PbI K1 PbI/Pb2I-
  1 x 102
• Pb2 2I- ? PbI2(aq) ?2
  PbI2(aq)/Pb2I-2 1.4 x 103
• Pb2 3I- ? PbI3- ?3
  PbI3-/Pb2I-3 8.3 x 103
• Pb2 4I- ? PbI42- ?4
  PbI3-/Pb2I-4 3.0 x 104
• Determine the total lead concentration assuming a
  dissolved I- concentration of 0.0010 M and 1.0 M.

31
Complex Ion Formation Effects on Solubility

• Using the previous equations for Pb formations,
  calculate the concentrations of Pb2, PbI,
  PbI3-, PbI42-, and PbI2(aq) in a solution whose
  total I- concentration is somehow fixed at
  0.050M.

32
Complex Formation - Example

• Consider the following equilibria
• AgCl(s) ? Ag Cl- Ksp 1.8 x 10-10
• AgCl(s) Cl- ? AgCl2- K2 1.5 x 10-2
• AgCl2- Cl- ?AgCl32- K3 0.49
• Find the total concentration of silver-containing
  species in a silver-saturated, aqueous solution
  containing the following concentrations of Cl-
  (a) 0.010 M, (b) 0.20 M, (c) 2.0 M.

33
Acids and Bases

• Definitions of Acids
• Lewis Acid Accepts a pair of electrons
• Aqueous Acid a substance that increases the
  hydronium ion concentration
• Brønsted-Lowry Acid proton donors
• Definitions of Bases
• Lewis Base donates a pair of electrons
• Aqueous Base decreasing the concentration of
  hydronium or increasing the hydroxide
  concentration
• Brønsted-Lowry Base proton acceptors

34
Acids and Bases

• Unlike the definition for an aqueous acid or
  base, Brønsted-Lowry Acids and Bases do not
  require the formation of hydronium so these
  definitions can be extended to nonaqueous
  solutions
• This class will refer to acids and bases using
  the Brønsted-Lowry definitions

35
Identification of Brønsted-Lowry Acids and Bases

• Identify the Brønsted-Lowry Acids on both sides
  of the reaction
• NaHSO3 NaOH ? Na2SO3 H2O

36
Acids and Bases

• Terminology
• Neutralization The process in which a
  stoichiometric equivalent of acid is added to a
  base
• Salt the formation of an ionic solid as a
  product of an acid-base reaction. Salts are
  composed of anions and cations that when
  dissolved become strong electrolytes.

37
Acids and Bases

• Conjugate Acid-Base pairs An acid and a base
  that differ only through the gain or loss of a
  single proton.
• The acid reactant loses a proton and becomes the
  conjugate base product.
• The base reactant gains a proton and becomes the
  conjugate acid product

38
Conjugate Acids-Base Pairs

• Identify the conjugate acid-base pairs in the
  reaction
• H2NCH2CH2NH2 H2O ? H3NCH2CH2NH2 OH-

39
Acids and Bases

• Water
• Can be an acid or a base
• H commonly written in chemical equations
  although H3O is meant
• H-O-H CH3-O- ? H-O- CH3-O-H
• H2O HBr ? H3O Br-
• Undergoes autoprotolysis self-ionization
• H2O H2O ? H3O OH-
• H2O ? H OH-

40
Acids and Bases

• Protic Solvents contain an acidic proton -
  undergo autoprotolysis
• Aprotic Solvents - no acidic protons

41
pH

• pH - ? - log H
• The H concentration is the same as the H
  concentration indicated in an equilibrium
  constant
• Water
• H2O ? H OH-
• Kw 1 x 10-14
• Ideally - H 1 x 10-7 and OH- 1 x 10 7
• pH pOH 14
• Solution Acidic if pH lt 7 and basic if pH gt 7

42
pH

• Calculate the concentration of H and the pH of
• 0.0010 M HCLO4
• 0.050 M HBr
• 0.050 M LiOH
• 3.0 M NaOH
• 0.0050 M (CH3CH2)4NOH

43
Acid / Base Strengths

• The strength of an acid or base is determined by
  the extent it dissociates in water to form H or
  OH- respectively
• Table 6-2 lists some common strong acids and
  bases most often discussed in this class
• Most other acids and bases for this class will be
  assumed to be weak

44
Strong Acids / Bases

• Strong acids dissociate entirely in water to give
  H anion
• Therefore the H concentration is equal to the
  strength of the acid
• Strong bases dissociate entirely in water to give
  OH- cation
• Therefore the OH- concentration is equal to the
  strength of the base

45
Weak Acid

• Weak acids only partially dissociate to give H
  in H2O
• Equilibrium Constant
• HA H2O ? H3O A-
• Ka HA- / HA
• Ka acid dissociation constant
• The smaller the Ka value, the weaker the acid

46
Weak Base

• Weak bases react with water by abstracting a
  proton
• Equilibrium Constant
• B H2O ? BH OH-
• Kb BHOH- / B
• Kb base hydrolysis constant
• Hydrolysis reaction with water
• The smaller the Kb value, the weaker the base

47
Weak Base Strength

• Which is the stronger base cyclohexylamine or
  imidazole? Write the base hydrolysis reaction of
  each. In the case of imidazole, the hydrogen
  atom without a hydrogen is the one that accepts
  H.

48
Common Weak Acids / Bases

• Acids
• Carboxylic acid - -COOH
• Ammonium ions - -NH4, R3NH, R2NH2
• Bases
• Carboxylate anion - - COO-
• Amines - RNH2, R2NH, R3N

49
Polyprotic Acids and Bases

• Polyprotic the ability to accept or donate more
  than one proton (e.g., H2SO4, C2H2O4)
• Equilibrium Notation
• Acids K1, K2, K3 (commonly used instead of Ka1,
  Ka2, Ka3 unless necessary for clarity)
• Bases Kb1, Kb2, Kb3
• K1 and Kb1 - indicating acidic or basic species
  with the most protons

50
Relationship Between Ka and Kb

• The relationship for the acid and its conjugate
  base is
• Ka . Kb Kw
• Same for polyprotic
• Ka . Kb Kw except you must be aware of which
  acid / base pair discussed

51
Relationship Between Ka and Kb

• Carbonic Acid Example
• H2CO3 ? HCO3- H K1 4.41 x 10-7
• HCO3- ? CO32- H K2 4.69 x 10-11
• Corresponding Kbs
• K1 corresponds to Kb2 K1 . Kb2 Kw
• K2 corresponds to Kb1 K2 . Kb1 Kw

52
Identifying the Equilibrium Constant

• Write the Ka reaction for formic acid, HCO2H, and
  for methylammonium ion, CH3NH3.
• Write the Kb reactions for piperidine and
  benzoate.
• Write the Ka and kb reactions of K2HPO4.

53
Identifying the Equilibrium Constant

• Write the stepwise acid-base reactions for
  piperazine and the phthalate ion in water. Write
  the correct symbol (e.g., Kb1) for the
  equilibrium constant for each reaction.
• Write the Kb reaction of hypochlorite, OCl-.
  Given that the Ka value for HOCl is 3.0 x 10-8,
  calculate the Kb for OCl-.

54
Identifying the Equilibrium Constant

• Write the Ka2 reaction of H2SO4 and the Kb2
  reaction for the trisodium salt.
• Citric Acid has the following values 7.44 x 10-4,
  1.73 x 10-5, 4.02 x 10-7 for Ka1, Ka2, and Ka3
  respectively. Write the reactions, identify the
  Ka and Kb values and calculate the Kb values for
  Citric Acid.

55
Chapter 6 - Homework

• Problems 5, 6, 7, 9, 12, 15, 17, 21, 25, 26,
  28, 30, 34, 39, 40, 41, 44, 45, 48, 49, 52, 53, 54