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					 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
   First, and reactants, equilibrium
productssystem reacheswhile maintaining
equilibrium distribution.


                                                     Reactants                       Product


              At equilibrium the amount of reactants and products are constant,
              but not necessarily equal
 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




                                            c       d
                                      [C ] [ D]
     Equilibrium constant:    K
                                      [ A]a [ B]b
            Where:
                      - small superscript letters are the stoichiometry coefficients
                      - [A] concentration chemical species A relative to standard state
 Chemical Equilibrium
Equilibrium Constant

   2.) Equilibrium constant (K):
           A reaction is favored when K > 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:
                                                                        [ H  ][ A ]
                                                                   K1 
                                                                           [ HA]
      Reversing the reaction results in a reciprocal equilibrium reaction:

                                                                            [ HA]
                                                                   K'                       1 / K1
                                                                    1                  
                                                                          [ H ][ A ]
 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


          [ H  ][ A ]                     [CH  ]                [ A ][CH  ]
     K1                           K2                        K3 
             [ HA]                        [ H  ][C ]                [ HA][C ]

                       [ H  ][ A ] [CH  ] [ A ][CH  ]
         K 3  K1K 2                          
                          [ HA]                  [ HA][C ]
                                      [ H ][C ]
  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

                                                  K2=1/KNH3

                                                  K3=Kw*1/KNH3=5.6x10-10
   Chemical Equilibrium
Le Châtelier’s 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
   Chemical Equilibrium
Le Châtelier’s Principal

    1.) What Happens When a System at Equilibrium is Perturbed?
           Le Châtelier’s 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
   Chemical Equilibrium
Le Châtelier’s Principal

    2.) Example:

  Consider this reaction:



                        [ Br - ][Cr2O7 - ][ H  ]8
                                     2
                   K                                 1  10 11 at 25 o C
                            [ BrO3 ][Cr 3 ]2
                                 -


   At one equilibrium state:


            [H  ]  5.0 M [Cr2O7 - ]  0.10 M [Cr 3 ]  0.0030 M
                                2

            [Br  ]  1.0 M [BrO 3 ]  0.043 M
                                 -
   Chemical Equilibrium
Le Châtelier’s Principal

    2.) Example:

  What happens when:
                     2
                [Cr2O7 - ] increased from 0.10 M to 0.20 M
  According to Le Châtelier’s 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:


       Q
            [ Br - ][Cr2O7 - ][ H  ]8
                         2
                                         
                                           1.0 0.20 5.0 8    2  10 11  K
               [ BrO3 ][Cr 3 ]2
                    -
                                           0.0430.0030 2
   Chemical Equilibrium
Le Châtelier’s Principal

    2.) Example:

  Because Q > 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
   Chemical Equilibrium
Le Châtelier’s Principal

    3.) Affect of Temperature on Equilibrium

         1. Equilibrium constant of an endothermic reaction (DHo = +) increases if
            the temperature is raised.
                                            D
                                                                     DH = +

         2. Equilibrium constant of an exothermic reaction (DHo = -)decreases if the
            temperature is raised.

                                     D
                                                                     DH = -
 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:




                         K sp  [ Hg 2 ][Cl- ]2  1.2  10 18
                                     2
  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+?

                     K sp  2.3  10 69
 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
  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
                                                    PbCl2 precipitate because the
      the addition of NaF
                                                    ion product is greater than Ksp.
 Chemical Equilibrium

Common Ion Effect

   1.)   Affect of Adding a Second Source of an Ion on Salt Solubility
            Equilibrium re-obtained following Le Châtelier’s Principal
            Reaction moves away from the added ion
            EXAMPLE:

     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.
   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
 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




                                ligand                adduct


             Lewis acid                     Lewis base
  Chemical Equilibrium
Complex Formation

      3.)   Affect on Solubility
                Formation of adducts increase solubility
                                       Ksp
 Implies low Pb2+ solubility:                                       K sp  [ Pb2  ][ I - ]2  7.9  10 9

                Solubility equation becomes a complex mixture of reactions
                 - don’t need to use all equations to determine the concentration of any species


                                         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
 Chemical Equilibrium
Complex Formation

     3.)   Affect on Solubility
                Total concentration is dependent on each individual complex species




 Pbtotal  Pb2   PbI   PbI2 ( aq )  PbI3  PbI4  
                                                             2




Total solubility of lead depends on [I-] and the
solubility of each individual complex formation.
 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
 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


                        acid

   2.) Brønsted-Lowry – definition does not require the formation of H3O+
            Extended to non-aqueous solutions or gas phase

            Acid – proton donor

            Base – proton acceptor

                       acid         base                      salt
   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+
    Chemical Equilibrium
  Acids and Bases

       5.)    Autoprotolysis – acts as both an acid and base
                 Extent of these reactions are very small




     water
                                                                 K w  [ H  ][OH - ]  1.0  10 14

                   - 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                                                                K  3.5  10 15
 Chemical Equilibrium
Acids and Bases

   6.)   pH – negative logarithm of H+ concentration
             Ignores distinction between concentration and activities (discussed later)

                              pH   log[ H  ]

             pH  pOH   log[ K w ]  14.00 at 25 o C
             A solution is acidic if [H+] > [OH-]
             A solution is basic if [H+] < [OH-]




             An aqueous solution has a neutral pH if [H+]=[OH-]
              - This occurs when [H+] = [OH-] = 10-7M or pH = 7
 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+?
 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
   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
                                 Ka                                 [ H  ][ A ]
                                                               Ka 
Equivalent                       Ka                                    [ HA]

                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
                                                                      [ BH  ][OH  ]
                                                                 Kb 
                                Kb
Equivalent                      Kb                                         [ B]
     Chemical Equilibrium
Some Common Weak Acids (carboxylic acids)

      ACID          FORMULA           Ka      pKa            ACID            FORMULA          Ka       pKa
acetic acid         H(C2H3O2)     1.74 E-5    4.76    hydrocyanic acid      HCN           6.17 E-10   9.21
ascorbic acid (1)   H2(C6H6O6)    7.94 E-5    4.10    hydrofluoric acid     HF            6.31 E-4    3.20
ascorbic acid (2)   (HC6H6O6)-    1.62 E-12   11.79   lactic acid           H(C3H5O3)     8.32 E-4    3.08
boric acid (1)      H3BO3         5.37 E-10   9.27    nitrous acid          HNO2          5.62 E-4    3.25
boric acid (2)      (H2BO3)-      1.8 E-13    12.7    octanoic acid         H(C8H15O2)    1.29 E-4    4.89
boric acid (3)      (HBO3)=       1.6 E-14    13.8    oxalic acid (1)       H2(C204)      5.89 E-2    1.23
butanoic acid       H(C4H7O2)     1.48 E-5    4.83    oxalic acid (2)       (HC2O4)-      6.46 E-5    4.19
carbonic acid (1)   H2CO3         4.47 E-7    6.35    pentanoic acid        H(C5H9O2)     3.31 E-5    4.84
carbonic acid (2)   (HCO3)-       4.68 E-11   10.33   phosphoric acid (1)   H3PO4         6.92 E-3    2.16
chromic acid (1)    H2CrO4        1.82 E-1    0.74    phosphoric acid (2)   (H2PO4)-      6.17 E-8    7.21
chromic acid (2)    (HCrO4)-      3.24 E-7    6.49    phosphoric acid (3)   (HPO4)=       2.09 E-12   12.32
citric acid (1)     H3(C6H5O7)    7.24 E-4    3.14    propanoic acid        H(C3H5O2)     1.38 E-5    4.86
citric acid (2)     (H2C6H5O7)-   1.70 E-5    4.77    sulfuric acid (2)     (HSO4)-       1.05 E-2    1.98
citric acid (3)     (HC6H5O7)=    4.07 E-7    6.39    sulfurous acid (1)    H2SO3         1.41 E-2    1.85
formic acid         H(CHO2)       1.78 E-4    3.75    sulfurous acid (2)    (HSO3)-       6.31 E-8    7.20
heptanoic acid      H(C7H13O2)    1.29 E-5    4.89    uric acid             H(C5H3N4O3)   1.29 E-4    3.89
hexanoic acid       H(C6H11O2)    1.41 E-5    4.84
  Chemical Equilibrium
Some Common Weak Acids (Metals cations)
  Chemical Equilibrium
Some Common Weak Bases (amines)

                  BASE           FORMULA          Kb       pKb
             alanine          C3H5O2NH2        7.41 E-5   4.13
             Ammonia          NH3 (NH4OH)      1.78 E-5   4.75
             dimethylamine    (CH3)2NH         4.79 E-4   3.32
             ethylamine       C2H5NH2          5.01 E-4   3.30
             glycine          C2H3O2NH2        6.03 E-5   4.22
             hydrazine        N2H4             1.26 E-6   5.90
             methylamine      CH3NH2           4.27 E-4   3.37
             trimethylamine   (CH3)3N          6.31 E-5   4.20


           The Ka or Kb of an acid or base may also be written in terms of “pKa” or “pKb”

                   pK a   log( K a )            pK b   log( K b )
           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
 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
 Chemical Equilibrium
Acids and Bases

   8.)   Relationship Between Ka and Kb

                                           [ H  ][ A ]
                                      Ka 
                                              [ HA]

                                            [ HA][OH  ]
                                     Kb 
                                                [ A ]
                                K w  Ka  Kb
                                  [ H  ][ A ] [ HA ][ OH  ]
                                                              [ H  ][OH  ]
                                     [ HA ]         [ A ]

                    K w  Ka  Kb
 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-.

				
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