First, system reaches equilibrium

1
Chemical Equilibrium

• Introduction
• 1.) Equilibria govern diverse phenomena
• Protein folding, acid rain action on minerals to
  aqueous reactions
• 2.) Chemical equilibrium applies to reactions
  that can occur in both directions
• reactants are constantly forming products and
  vice-versa
• At the beginning of the reaction, the rate that
  the reactants are changing into the products is
  higher than the rate that the products are
  changing into the reactants.
• When the net change of the products and reactants
  is zero the reaction has reached equilibrium.

Then, system continually exchanges products and
reactants, while maintaining equilibrium
distribution.
First, system reaches equilibrium
Product
Reactants
At equilibrium the amount of reactants and
products are constant, but not necessarily equal
2
Chemical Equilibrium

• Equilibrium Constant
• 1.) The relative concentration of products and
  reactants at equilibrium is a constant.
• 2.) Equilibrium constant (K)
• For a general chemical reaction

Equilibrium constant
Where - small superscript letters are the
stoichiometry coefficients - A concentration
chemical species A relative to standard state
3
Chemical Equilibrium

• Equilibrium Constant
• 2.) Equilibrium constant (K)
• A reaction is favored when K gt 1
• K has no units, dimensionless
• - Concentration of solutes should be expressed as
  moles per liter (M).
• - Concentrations of gases should be expressed in
  bars.
• ? express gas as Pgas, emphasize pressure
  instead of concentration
• ? 1 bar 105 Pa 1 atm 1.01325 bar
• - Concentrations of pure solids, pure liquids and
  solvents are omitted
• ? are unity
• ? standard state is the pure liquid or solid
• 3.) Manipulating Equilibrium Constants

Consider the following reaction
Reversing the reaction results in a reciprocal
equilibrium reaction
4
Chemical Equilibrium

• Equilibrium Constant
• 3.) Manipulating Equilibrium Constants

If two reactions are added, the new K is the
product of the two individual K values
K1
K2
K3
5
Chemical Equilibrium

• Equilibrium Constant
• 3.) Manipulating Equilibrium Constants
• Example

Given the reactions and equilibrium constants
Kw 1.0 x 10-14
KNH3 1.8 x 10-5
Find the equilibrium constant for the reaction
Solution
K1 Kw
K21/KNH3
K3Kw1/KNH35.6x10-10
6
Chemical Equilibrium

• Equilibrium and Thermodynamics
• 1.) Equilibrium constant derived from the
  thermodynamics of a chemical
• reaction.
• deals with the relationships and conversions
  between heat and other forms of energy
• 2.) Enthalpy
• DH is the heat absorbed or released when the
  reaction takes place under constant applied
  pressure
• DH Hproducts Hreactants
• Standard enthalpy change (DHo)
• all reactants and products are
• in their standard state.
• DHo negative ? heat released
• - Exothermic
• - Solution gets hot
• DHo positive ? heat absorbed

7
Chemical Equilibrium

• Equilibrium and Thermodynamics
• 3.) Entropy
• Measure of a substances disorder
• Greater disorder ? Greater Entropy
• - Relative disorder Gas gt Liquid gt solid
• DS Sproducts Sreactants
• DSo change in entropy when all species are in
  standard state.
• - positive?
• product more disorder
• - negative ?
• product less disorder

DSo 76.4 J/(K.mol) at 25oC More disorder for
aqueous ions than solid
8
Chemical Equilibrium

• Equilibrium and Thermodynamics
• 3.) Entropy
• Increase in temperature results in an increase in
  Entropy (S)
• Increase occurs for all products and reactants

9
Chemical Equilibrium

• Equilibrium and Thermodynamics
• 4.) Free Energy
• Systems at constant temperature and pressure have
  a tendency toward lower enthalpy and higher
  entropy
• Chemical reaction is favored if
• - DH is negative ? heat given off
• and
• - DS is positive ? more disorder
• Chemical reaction is not favored if
• - DH is positive and DS is negative
• Gibbs Free Energy (DG) determines if a reaction
  is favored or not when both DH and DS are
  positive or negative
• - A reaction is favored if DG is negative

Free energy DG DH -TDS
10
Chemical Equilibrium

• Equilibrium and Thermodynamics
• 4.) Free Energy
• Example

Is the following reaction favored at 25oC?
DHo -74.85 x 103 J/mol DSo -130.4 J/K.mol
Free energy DG DH TDS (-74.85x103 J/mol)
(298.15K)(-130.4 J/K.mol) DG -35.97 kJ/mol ?
DG negative ? reaction favored
Favorable influence of enthalpy is greater than
unfavorable influence of entropy
11
Chemical Equilibrium

• Equilibrium and Thermodynamics
• 5.) Free Energy and Equilibrium
• Relate Equilibrium constant to the energetics (DH
  DS) of a reaction
• Equilibrium constant depends on DG
• where
• R (gas constant) 8.314472 J/(K.mol)
• T temperature in kelvins
• The more negative DG ? larger equilibrium
  constant
• Example

DG -35.97
Because K is very large, HCl is very soluble in
water and nearly completely ionized
12
Chemical Equilibrium

• Equilibrium and Thermodynamics
• 5.) Free Energy and Equilibrium
• If DGo is negative or K gt1 the reaction is
  spontaneous
• Reaction occurs by just combining the reactants
• If DGo is positive or K lt 1, the reaction is not
  spontaneous
• - Reaction requires external energy or process
  to proceed

Gas flows towards a vacuum. spontaneous
A vacuum does not naturally form. nonspontaneous
13
Chemical Equilibrium

• Le Châteliers Principal
• 1.) What Happens When a System at Equilibrium
  is Perturbed?
• Change concentration, temperature, pressure or
  add other chemicals
• Equilibrium is re-established
• Reaction accommodates the change in products,
  reactants, temperature, pressure, etc.
• Rates of forward and reverse reactions
  re-equilibrate

14
Chemical Equilibrium

• Le Châteliers Principal
• 1.) What Happens When a System at Equilibrium
  is Perturbed?
• Le Châteliers Principal
• - the direction in which the system proceeds
  back to equilibrium is such that the change is
  partially offset.

Consider this reaction
At equilibrium
To return to equilibrium (balance), some (not
all) CO and H2 are converted to CH3OH
Add excess CO(g)
If all added CO was converted to CH3OH, then
reaction would be unbalanced by the amount of
product
15
Chemical Equilibrium

• Le Châteliers Principal
• 2.) Example

Consider this reaction
At one equilibrium state
16
Chemical Equilibrium

• Le Châteliers Principal
• 2.) Example

What happens when
According to Le Châteliers Principal, reaction
should go back to left to off-set dichormate on
right
Use reaction quotient (Q), Same form of
equilibrium equation, but not at equilibrium
17
Chemical Equilibrium

• Le Châteliers Principal
• 2.) Example

Because Q gt K, the reaction must go to the left
to decrease numerator and increase
denominator. Continues until Q K 1. If the
reaction is at equilibrium and products are added
(or reactants removed), the reaction goes
to the left 2. If the reaction is at
equilibrium and reactants are added ( or products
removed), the reaction goes to the right
18
Chemical Equilibrium

• Le Châteliers Principal
• 3.) Affect of Temperature on Equilibrium

Combine Gibbs free energy and Equilibrium
Equations
Only Enthalpy term is temperature dependent
19
Chemical Equilibrium

• Le Châteliers Principal
• 3.) Affect of Temperature on Equilibrium

1. Equilibrium constant of an endothermic
reaction (DHo ) increases if the
temperature is raised. 2. Equilibrium
constant of an exothermic reaction (DHo
-)decreases if the temperature is
raised.
D
DH
D
DH -
20
Chemical Equilibrium

• Le Châteliers Principal
• 4.) Thermodynamics vs. Kinetics
• Thermodynamics predicts if a reaction will occur
  
• - determines the state at equilibrium
• Thermodynamics does not determine the rate of a
  reaction
• - Will the reaction occur instantly, in minutes,
  hours, days or years?

DG - spontaneous
Diamonds
Graphite
21
Chemical Equilibrium

• Solubility Product
• 1.) Equilibrium constant for the reaction
  which a solid salt dissolves to give its
  constituent ions in solution
• Solid omitted from equilibrium constant because
  it is in a standard state
• Example

22
Chemical Equilibrium

• Solubility Product
• 1.) Saturated Solution contains excess,
  undissolved solid
• Solution contains all the solid capable of
  dissolving under the current conditions
• Example
• Find Cu2 in a solution saturated with
  Cu4(OH)6(SO4) if OH- is fixed at 1.0x10-6M.
  Note that Cu4(OH)6(SO4) gives 1 mol of SO42- for
  4 mol of Cu2?

23
Chemical Equilibrium

• Solubility Product
• 2.) If an aqueous solution is left in contact
  with excess solid, the solid will dissolve until
  the condition of Ksp is satisfied
• Amount of undissolved solid remains constant
• Excess solid is required to guarantee ion
  concentration is consistent with Ksp
• 3.) If ions are mixed together such that the
  concentrations exceed Ksp, the solid will
  precipitate.
• 4.) Solubility product only describes part of the
  solubility of a salt
• Only includes dissociated ions
• Ignores solubility of solid salt

24
Chemical Equilibrium
Common ion effect a salt will be less soluble
if one of its constituent ions is already present
in the solution.
Decrease in the solubility of MgF2 by the
addition of NaF
PbCl2 precipitate because the ion product is
greater than Ksp.
25
Chemical Equilibrium

• Common Ion Effect
• 1.) Affect of Adding a Second Source of an Ion
  on Salt Solubility
• Equilibrium re-obtained following Le Châteliers
  Principal
• Reaction moves away from the added ion

Find Cu2 in a solution saturated with
Cu4(OH)6(SO4) if OH- is fixed at 1.0x10-6M and
0.10M Na2SO4 is added to the solution.
26
Chemical Equilibrium

• Complex Formation
• 1.) High concentration of an ion may
  redissolve a solid
• Ion first causes precipitation
• Forms complex ions, consists of two or more
  simple ions bonded to each other

ppt. formation
Complex forms and redissolves solid
27
Chemical Equilibrium

• Complex Formation
• 2.) Lewis Acids and Bases
• M acts as a Lewis acid ? accepts a pair of
  electrons
• X- acts as a Lewis base ? donates a pair of
  electrons
• Bond is a coordinate covalent bond

adduct
ligand
Lewis base
Lewis acid
28
Chemical Equilibrium

• Complex Formation
• 3.) Affect on Solubility
• Formation of adducts increase solubility
• Solubility equation becomes a complex mixture of
  reactions
• - dont need to use all equations to determine
  the concentration of any species

Ksp
Implies low Pb2 solubility
Only one concentration of Pb2 in solution
Concentration of Pb2 that satisfies any one of
the equilibria must satisfy all of the equilibria
All equilibrium conditions are satisfied
simultaneously
29
Chemical Equilibrium

• Complex Formation
• 3.) Affect on Solubility
• Total concentration is dependent on each
  individual complex species

Total solubility of lead depends on I- and the
solubility of each individual complex formation.
30
Chemical Equilibrium

• Complex Formation
• 3.) Affect on Solubility
• Example

Given the following equilibria, calculate the
concentration of each zinc-containing species in
a solution saturated with Zn(OH)2(s) and
containing OH- at a fixed concentration of
3.2x10-7M. Zn(OH)2 (s) Ksp
3.0x10-16 Zn(OH) b1 2.5 x104 Zn(OH)3- b3
7.2x1015 Zn(OH)42- b4 2.8x1015
31
Chemical Equilibrium

• Acids and Bases
• 1.) Protic Acids and Bases transfer of H
  (proton) from one molecule to
• another
• Hydronium ion (H3O) combination of H with
  water (H2O)
• Acid is a substance that increases the
  concentration of H3O
• Base is a substance that decreases the
  concentration of H3O
• - base also causes an increase in the
  concentration of OH- in aqueous solutions
• 2.) Brønsted-Lowry definition does not require
  the formation of H3O
• Extended to non-aqueous solutions or gas phase

acid
acid
base
salt
32
Chemical Equilibrium

• Acids and Bases
• 3.) Salts product of an acid-base reaction
• Any ionic solid
• Acid and base neutralize each other and form a
  salt
• Most salts with a single positive and negative
  charge dissociate completely into ions in water
• 4.) Conjugate Acids and Bases

Products of acid-base reaction are also acids and
bases
A conjugate acid and its base or a conjugate base
and its acid in an aqueous system are related to
each other by the gain or loss of H
33
Chemical Equilibrium

• Acids and Bases
• 5.) Autoprotolysis acts as both an acid and
  base
• Extent of these reactions are very small

water

• - H3O is the conjugate acid of water
• - OH- is the conjugate base of water
• Kw is the equilibrium constant for the
  dissociation of water

Acetic acid
34
Chemical Equilibrium

• Acids and Bases
• 6.) pH negative logarithm of H
  concentration
• Ignores distinction between concentration and
  activities (discussed later)
• A solution is acidic if H gt OH-
• A solution is basic if H lt OH-

35
Chemical Equilibrium

• Acids and Bases
• 6.) pH
• pH values for some common samples

36
Chemical Equilibrium

• Acids and Bases
• 6.) pH
• Example

What is the pH of a solution containing 1x10-6 M
H?
What is OH- of a solution containing 1x10-6 M
H?
37
Chemical Equilibrium

• Acids and Bases
• 7.) Strengths of Acids and Bases
• Depends on whether the compound react nearly
  completely or partially to produce H or OH-
• strong acid or base completely dissociate in
  aqueous solution
• - equilibrium constants are large
• - everything else termed weak

Strong ? no undissociated HCl or KOH
38
Chemical Equilibrium

• Acids and Bases
• 7.) Strengths of Acids and Bases
• weak acids react with water by donating a proton
• - only partially dissociated in water
• - equilibrium constants are called Ka acid
  dissociation constant
• - Ka is small
• weak bases react with water by removing a proton
• - only partially dissociated in water
• - equilibrium constants are called Kb base
  dissociation constant
• - Kb is small

Ka
Equivalent
Ka
Kb
Equivalent
Kb
39
Chemical Equilibrium
Some Common Weak Acids (carboxylic acids)
40
Chemical Equilibrium
Some Common Weak Acids (Metals cations)
41
Chemical Equilibrium
Some Common Weak Bases (amines)

• The Ka or Kb of an acid or base may also be
  written in terms of pKa or pKb
• As Ka or Kb increase ? pKa or pKb decrease
• - a strong acid/base has a high Ka or Kb and a
  low pKa or pkb

42
Chemical Equilibrium

• Acids and Bases
• 8.) Polyprotic Acids and Bases can donate or
  accept more than one proton
• Ka or Kb are sequentially numbered
• - Ka1,Ka2,Ka3 Kb1,Kb2,Kb3

43
Chemical Equilibrium

• Acids and Bases
• 8.) Relationship Between Ka and Kb

44
Chemical Equilibrium

• Acids and Bases
• 8.) Relationship Between Ka and Kb
• Example

Write the Kb reaction of CN-. Given that the Ka
value for HCN is 6.2x10-10, calculate Kb for CN-.