Chemical Equilibrium Chemical Equilibrium 1 Equilibrium by qingyunliuliu

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									Chemical Equilibrium




                       1
                           Equilibrium




   Initially all liquid
                           Gas only, produced
                                                Balance of gas and liquid
                                                     production
                                                                        2
                 Equilibrium
   When compounds react, they eventually
    form a mixture of products and unreacted
    reactants, in a dynamic equilibrium.
    – A dynamic equilibrium consists of a
      forward reaction, in which substances react to
      give products, and a reverse reaction, in
      which products react to give the original
      reactants.


                                                   3
     Chemical Equilibrium
                   H2 + I2 < -- > 2HI
– Initially only H2 and I2 are present.
   – The rxn. proceeds  only
– As HI concentration increases, some HI is able
  to decompose back into H2 and I2
   – Rxn. proceeds < -- >
– At some point
   – The rate of H2 + I2 -- > 2HI equals
   – The rate of 2HI -- > H2 + I2


  Chemical equilibrium is the state reached by a
   reaction mixture when the rates of the forward
    and reverse reactions have become equal.
                                                    4
See Le Chatelier
          Chemical Equilibrium
   For example, the Haber process for
    producing ammonia from N2 and H2 does
    not go to completion.

        N 2 ( g )  3H 2 ( g )     2NH3 (g)

     – It establishes an equilibrium state where all
       three species are present. (see Figure 15.3)


                                                       6
         Chemical Equilibrium
   Chemical Equilibrium is a fundamentally
    important concept to master because
    most chemical reactions fail to go to
    completion.




                                              7
         A Problem to Consider
   Applying Stoichiometry to an Equilibrium
    Mixture.
     – Suppose we place 1.000 mol N2 and 3.000 mol H2
       in a reaction vessel at 450 oC and 10.0
       atmospheres of pressure. The reaction is

         N 2 ( g )  3H 2 ( g )        2NH3 (g)
     – What is the composition of the equilibrium mixture
       if it contains 0.080 mol NH3?

                                                            8
          A Problem to Consider
    Using the information given, set up an ICE table.


         N 2 ( g )  3H 2 ( g )        2NH3 (g)
  Initial    1.000     3.000                0
 Change        -x       -3x               +2x
Equilibrium 1.000 -   3.000 -        2x = 0.080 mol
               x        3x
        – The equilibrium amount of NH3 was given as
          0.080 mol. Therefore, 2x = 0.080 mol NH3 (x =
          0.040 mol).
                                                          9
           A Problem to Consider
    Using the information given, set up the following
     table.

          N 2 ( g )  3H 2 ( g )                2NH3 (g)
  Initial    1.000          3.000                   0
 Change        -x            -3x                  +2x
Equilibrium 1.000 -        3.000 -           2x = 0.080 mol
                 x           3x
      Equilibrium amount of N2 = 1.000 - 0.040 = 0.960 mol N2
      Equilibrium amount of H2 = 3.000 - (3 x 0.040) = 2.880 mol H2
      Equilibrium amount of NH3 = 2x = 0.080 mol NH3
                                                                      10
       The Equilibrium Constant
   Every reversible system has its own
    “position of equilibrium” under any given
    set of conditions.
     – The ratio of products produced to unreacted
       reactants for any given reversible reaction
       remains constant under constant conditions of
       pressure and temperature.
     – The numerical value of this ratio is called the
       equilibrium constant for the given reaction.


                                                         11
   The Equilibrium Constant

                 H2 + I2 < -- > 2HI
Rate of forward rxn:
                Ratef = kf [H2][I2]
Rate of reverse rxn:
                   Rater = kr[HI]2
At equilibrium:
                kf [H2][I2] =kr[HI]2
Therefore:
                     kf = [HI]2
                     kr [H][I]

                    Kc = [HI]2
                        [H][I]         12
       The Equilibrium Constant
   The equilibrium-constant expression for a reaction
    is obtained by multiplying the concentrations of
    products, dividing by the concentrations of reactants,
    and raising each concentration to a power equal to its
    coefficient in the balanced chemical equation.

            aA  bB                  cC  dD
     – For the general equation above, the               c   d
                                                    [C] [D]
       equilibrium-constant expression would
       be:
                                               Kc       a    b
                                                    [ A ] [B ]

                                                             13
       The Equilibrium Constant
   The equilibrium-constant expression for a reaction
    is obtained by multiplying the concentrations of
    products, dividing by the concentrations of reactants,
    and raising each concentration to a power equal to its
    coefficient in the balanced chemical equation.


            aA  bB                  cC  dD
     – The molar concentration of a substance             c   d
                                                     [C] [D]
       is denoted by writing its formula in
       square brackets.
                                                Kc       a    b
                                                     [ A ] [B ]
                                                              14
       The Equilibrium Constant
   The equilibrium constant, Kc, is the
    value obtained for the equilibrium-
    constant expression when equilibrium
    concentrations are substituted.
     – A large Kc indicates large concentrations of products
       at equilibrium.
     – A small Kc indicates large concentrations of
       unreacted reactants at equilibrium.


                                                           15
       The Equilibrium Constant
   The law of mass action states that the value
    of the equilibrium constant expression Kc is
    constant for a particular reaction at a given
    temperature, whatever equilibrium
    concentrations are substituted.

     – Consider the equilibrium established in the Haber
       process.

          N 2 ( g )  3H 2 ( g )       2NH3 (g)
                                                           16
       The Equilibrium Constant
   The equilibrium-constant expression would
    be                     2
                       [NH 3 ]
                 Kc              3
                      [N 2 ][H 2 ]
     – Note that the stoichiometric coefficients in the
       balanced equation have become the powers to
       which the concentrations are raised.

          N 2 ( g )  3H 2 ( g )        2NH3 (g)
                                                          17
        Calculating Equilibrium
           Concentrations

   Once you have determined the equilibrium
    constant for a reaction, you can use it to
    calculate the concentrations of substances
    in the equilibrium mixture.




                                            18
       Calculating Equilibrium
          Concentrations
– For example, consider the following equilibrium.

       CO(g )  3 H 2 (g)      CH4 (g)  H 2O(g)

– Suppose a gaseous mixture contained 0.30 mol CO,
  0.10 mol H2, 0.020 mol H2O, and an unknown
  amount of CH4 per liter.
– What is the concentration of CH4 in this mixture? The
  equilibrium constant Kc equals 3.92.




                                                          19
        Calculating Equilibrium
           Concentrations
   First, calculate concentrations from moles of substances.

        CO(g )  3 H 2 (g)       CH4 (g)  H 2O(g)
       0.30 mol   0.10 mol                0.020 mol
                                    ??
         1.0 L      1.0 L                   1.0 L




                                                           20
         Calculating Equilibrium
            Concentrations
   First, calculate concentrations from moles of substances.

        CO(g )  3 H 2 (g)       CH4 (g)  H 2O(g)

         0.30 M   0.10 M           ??      0.020 M

    – The equilibrium-constant expression is:

                  [CH 4 ][H 2O]
             Kc             3
                   [CO][H 2 ]

                                                           21
        Calculating Equilibrium
           Concentrations
   First, calculate concentrations from moles of substances.

        CO(g )  3 H 2 (g)       CH4 (g)  H 2O(g)

         0.30 M   0.10 M           ??      0.020 M

– Substituting the known concentrations and the value
  of Kc gives:
                    [CH 4 ](0.020M )
            3.92                   3
                   (0.30M )(0.10M )

                                                           22
        Calculating Equilibrium
           Concentrations
   First, calculate concentrations from moles of substances.

        CO(g )  3 H 2 (g)       CH4 (g)  H 2O(g)

         0.30 M   0.10 M           ??      0.020 M

– You can now solve for [CH4].

                ( 3.92)(0.30M )(0.10M )3
      [CH 4 ]                            0.059
                        (0.020M )
– The concentration of CH4 in the mixture is 0.059 mol/L.
                                                           23
        Calculating Equilibrium
           Concentrations

   Suppose we begin a reaction with known
    amounts of starting materials and want to
    calculate the quantities at equilibrium.




                                            24
      Calculating Equilibrium
         Concentrations
   Consider the following equilibrium.

      CO(g )  H 2O(g )            CO2 (g)  H 2 (g)
      • Suppose you start with 1.000 mol each of carbon
        monoxide and water in a 50.0 L container.
        Calculate the molarity of each substance in the
        equilibrium mixture at 1000 oC.
      • Kc for the reaction is 0.58 at 1000 oC.




                                                      25
          Calculating Equilibrium
             Concentrations
–   First, calculate the initial molarities of CO and H2O.

          CO(g )  H 2O(g )               CO2 (g)  H 2 (g)
          1.000 mol    1.000 mol
            50.0 L       50.0 L




                                                             26
          Calculating Equilibrium
             Concentrations
–   First, calculate the initial molarities of CO and H2O.

          CO(g )  H 2O(g )                   CO2 (g)  H 2 (g)
         0.0200 M      0.0200 M                 0M            0M

          • The starting concentrations of the products are 0.
          • We must now set up a table of concentrations (starting, change,
            and equilibrium expressions in x).




                                                                     27
            Calculating Equilibrium
               Concentrations
   –   Let x be the moles per liter of product formed.

           CO(g )  H 2O(g )          CO2 (g)  H 2 (g)
  Initial    0.0200     0.0200           0         0
 Change        -x         -x             +x        +x
Equilibrium 0.0200-x   0.0200-x          x          x

       – The equilibrium-constant expression is:
                            [CO2 ][H 2 ]
                       Kc 
                            [CO][H 2O]
                                                         28
            Calculating Equilibrium
               Concentrations
  –   Solving for x.

           CO(g )  H 2O(g )            CO2 (g)  H 2 (g)
  Initial    0.0200 0.0200                  0         0
 Change        -x       -x                  +x        +x
Equilibrium 0.0200-x 0.0200-x                x         x
  – Substituting the values for equilibrium concentrations, we
    get:
                               ( x )(x )
               0.58 
                      (0.0200  x )(0.0200  x )
                                                            29
              Calculating Equilibrium
                 Concentrations
  –   Solving for x.

              CO(g )  H 2O(g )           CO2 (g)  H 2 (g)
  Initial    0.0200 0.0200                   0       0
 Change        -x       -x                   +x      +x
Equilibrium 0.0200-x 0.0200-x                 x       x
      – Or:
                                      2
                                   x
                       0.58 
                              (0.0200  x ) 2

                                                          30
             Calculating Equilibrium
                Concentrations
  –   Solving for x.

            CO(g )  H 2O(g )             CO2 (g)  H 2 (g)
  Initial    0.0200 0.0200                    0        0
 Change        -x       -x                    +x       +x
Equilibrium 0.0200-x 0.0200-x                  x        x
      – Taking the square root of both sides we get:
                                   x
                       0.76 
                              (0.0200  x )
                                                            31
             Calculating Equilibrium
                Concentrations
  –   Solving for x.

            CO(g )  H 2O(g )               CO2 (g)  H 2 (g)
  Initial    0.0200 0.0200                     0       0
 Change        -x       -x                     +x      +x
Equilibrium 0.0200-x 0.0200-x                   x       x
      – Rearranging to solve for x gives:
                   0.0200  0.76
                x                0.0086
                       1.76
                                                            32
             Calculating Equilibrium
                Concentrations
  –    Solving for equilibrium concentrations.

             CO(g )  H 2O(g )                CO2 (g)  H 2 (g)
  Initial    0.0200 0.0200                        0          0
 Change        -x       -x                        +x         +x
Equilibrium 0.0200-x 0.0200-x                      x          x
      – If you substitute for x in the last line of the table you
        obtain the following equilibrium concentrations.
           0.0114 M CO            0.0086 M CO2
           0.0114 M H2O           0.0086 M H2
                                                                    33
         Calculating Equilibrium
            Concentrations
   The preceding example illustrates the
    three steps in solving for equilibrium
    concentrations.

     1. Set up a table of concentrations (starting, change, and
        equilibrium expressions in x).
     2. Substitute the expressions in x for the equilibrium
        concentrations into the equilibrium-constant equation.
     3. Solve the equilibrium-constant equation for the values
        of the equilibrium concentrations.

                                                           34
        Calculating Equilibrium
           Concentrations
   In some cases it is necessary to solve a
    quadratic equation to obtain equilibrium
    concentrations.
   The next example illustrates how to solve
    such an equation.




                                                35
        Calculating Equilibrium
           Concentrations
–   Consider the following equilibrium.
            H 2 (g )  I 2 (g )       2HI(g)
        • Suppose 1.00 mol H2 and 2.00 mol I2 are placed in
          a 1.00-L vessel. How many moles per liter of each
          substance are in the gaseous mixture when it
          comes to equilibrium at 458 oC?
        • Kc at this temperature is 49.7.




                                                         36
         Calculating Equilibrium
            Concentrations
    The concentrations of substances are as follows.
              H 2 (g )  I 2 (g )       2HI(g)
  Initial      1.00      2.00               0
 Change         -x        -x               +2x
Equilibrium   1.00-x    2.00-x              2x

 – The equilibrium-constant expression is:
                                    2
                             [HI]
                       Kc 
                            [H 2 ][I 2 ]
                                                   37
         Calculating Equilibrium
            Concentrations
    The concentrations of substances are as follows.
              H 2 (g )  I 2 (g )     2HI(g)
  Initial      1.00      2.00            0
 Change         -x        -x            +2x
Equilibrium   1.00-x    2.00-x           2x

– Substituting our equilibrium concentration expressions
  gives:                           2
                          ( 2x )
              Kc 
                   (1.00  x)(2.00  x)
                                                           38
          Calculating Equilibrium
             Concentrations
–    Solving for x.
              H 2 (g )  I 2 (g )    2HI(g)
  Initial      1.00      2.00           0
 Change         -x        -x           +2x
Equilibrium   1.00-x    2.00-x          2x


    – Because the right side of this equation is not a
      perfect square, you must solve the quadratic
      equation.
                                                         39
          Calculating Equilibrium
             Concentrations
–    Solving for x.
              H 2 (g )  I 2 (g )   2HI(g)
  Initial      1.00      2.00         0
 Change         -x        -x         +2x
Equilibrium   1.00-x    2.00-x        2x


    – The equation rearranges to give:

                0.920x  3.00x  2.00  0
                          2


                                             40
         Calculating Equilibrium
            Concentrations
–   Solving for x.
               H 2 (g )  I 2 (g )      2HI(g)
  Initial       1.00      2.00            0
 Change          -x        -x            +2x
Equilibrium    1.00-x    2.00-x           2x

 – The two possible solutions to the quadratic
   equation are:
              x  2.33            and      x  0.93
                                                      41
           Calculating Equilibrium
              Concentrations
–    Solving for x.
               H 2 (g )  I 2 (g )       2HI(g)
  Initial       1.00      2.00               0
 Change          -x        -x               +2x
Equilibrium    1.00-x    2.00-x              2x

    – However, x = 2.33 gives a negative value to 1.00 - x (the
      equilibrium concentration of H2), which is not possible.

                Only x  0.93 remains.
                                                             42
         Calculating Equilibrium
            Concentrations
   Solving for equilibrium concentrations.
              H 2 (g )  I 2 (g )         2HI(g)
  Initial      1.00       2.00               0
 Change         -x         -x               +2x
Equilibrium   1.00-x     2.00-x              2x

 – If you substitute 0.93 for x in the last line of the table
   you obtain the following equilibrium concentrations.

        0.07 M H2         1.07 M I2        1.86 M HI
                                                                43
         Le Chatelier’s Principle
   Obtaining the maximum amount of
    product from a reaction depends on the
    proper set of reaction conditions.
     – Le Chatelier’s Principle states that when a
       system in a chemical equilibrium is disturbed by a
       change of temperature, pressure, or
       concentration, the equilibrium will shift in a way
       that tends to counteract this change.
     – See LeChatelier’s Principle animation


                                                            44
     Removing Products or Adding
             Reactants
    Let’s refer an illustration of a U-tube.


“reactants”   “products” – It’s a simple concept to see that if
                            we were to remove products
                            (analogous to dipping water out
                            of the right side of the tube) the
                            reaction would shift to the right
                            until equilibrium was
                            reestablished.


                                                             45
     Removing Products or Adding
             Reactants
    Let’s refer back to the illustration of the U-tube
     in the first section of this chapter.

“reactants”   “products” – Likewise, if more reactant is
                           added (analogous to pouring
                           more water in the left side of the
                           tube) the reaction would again
                           shift to the right until equilibrium
                           is reestablished.



                                                            46
     Effects of Pressure Change
   A pressure change caused by changing the
    volume of the reaction vessel can affect the
    yield of products in a gaseous reaction only if
    the reaction involves a change in the total moles
    of gas present (see Figure 15.12).


              CO + 3H2 CH4+H2O
              3 mol   9 mol   3 mol   3 mol




                                                    47
     Effects of Pressure Change
   If the products in a gaseous reaction contain
    fewer moles of gas than the reactants, it is
    logical that they would require less space.
• So, reducing the volume of the reaction vessel
  would favor the products.
• If the reactants require less volume (that is,
  fewer moles of gaseous reactant)
   • decreasing the volume of the reaction vessel
      would shift the equilibrium to the left (toward
      reactants).
                                                    48
       Effects of Pressure Change
    Literally “squeezing” the reaction will cause a shift in
     the equilibrium toward the fewer moles of gas.

    • It’s a simple step to see that reducing the pressure in
      the reaction vessel by increasing its volume would have
      the opposite effect.
    • In the event that the number of moles of gaseous
      product equals the number of moles of gaseous
      reactant, vessel volume will have no effect on the
      position of the equilibrium.



                                                                49
    Effect of Temperature Change
   Temperature has a significant effect on
    most reactions (see Figure 15.13).
     – Reaction rates generally increase with an increase
       in temperature. Consequently, equilibrium is
       established sooner.
     – In addition, the numerical value of the
       equilibrium constant Kc varies with
       temperature.




                                                            50
    Effect of Temperature Change
   Let’s look at “heat” as if it were a product
    in exothermic reactions and a
    reactant in endothermic reactions.
• We see that increasing the temperature is
  analogous to adding more product (in the
  case of exothermic reactions) or adding
  more reactant (in the case of endothermic
  reactions).
• This ultimately has the same effect as if heat
  were a physical entity.
                                                   51
    Effect of Temperature Change
  Exothermic
            A + B → C + D + heat
(-)∆H
How would adding heat effect the equilibrium?
•   Increasing temperature would be analogous to
    adding more product, causing the equilibrium
    to shift left.
•   Since “heat” does not appear in the equilibrium-
    constant expression, this change would result in
    a smaller numerical value for Kc.                52
    Effect of Temperature Change
   Endothermic
                  A + B + heat → C + D
(+)∆H
 How would adding heat effect the equilibrium?

    • Increasing temperature would be analogous to
      adding more reactant, causing the
      equilibrium to shift right.
    • This change results in more product at
      equilibrium, and a larger numerical value for
      Kc.
                                                  53
    Effect of Temperature Change
   In summary:
     – For an endothermic reaction (DH positive) the
       amounts of products are increased at equilibrium by
       an increase in temperature (Kc is larger at higher
       temperatures).
     – For an exothermic reaction (DH is negative) the
       amounts of reactants are increased at equilibrium by
       an increase in temperature (Kc is smaller at higher
       temperatures).


                                                          54
        Obtaining Equilibrium
       Constants for Reactions

   Equilibrium concentrations for a reaction
    must be obtained experimentally and
    then substituted into the equilibrium-
    constant expression in order to calculate
    Kc.



                                                55
        Obtaining Equilibrium
       Constants for Reactions
   Consider the reaction below


    CO(g )  3 H 2 (g)            CH4 (g)  H 2O(g)

     – Suppose we started with initial concentrations of
       CO and H2 of 0.100 M and 0.300 M, respectively.




                                                           56
        Obtaining Equilibrium
       Constants for Reactions
   Consider the reaction below (see Figure 15.5).

    CO(g )  3 H 2 (g)            CH4 (g)  H 2O(g)
     – When the system finally settled into equilibrium we
       determined the equilibrium concentrations to be as
       follows.
                Reactants                 Products
             [CO] = 0.0613 M           [CH4] = 0.0387 M
              [H2] = 0.1893 M          [H2O] = 0.0387 M


                                                             57
        Obtaining Equilibrium
       Constants for Reactions
   Consider the reaction below


    CO(g )  3 H 2 (g)            CH4 (g)  H 2O(g)
     – The equilibrium-constant expression for this
       reaction is:

                     [CH 4 ][H 2O]
                Kc             3
                      [CO][H 2 ]
                                                      58
        Obtaining Equilibrium
       Constants for Reactions
   Consider the reaction below


    CO(g )  3 H 2 (g)             CH4 (g)  H 2O(g)
     – If we substitute the equilibrium concentrations,
       we obtain:

            (0.0387M )(0.0387M )
      Kc                       3
                                   3.93
           (0.0613M )(0.1839M )
                                                          59
        Obtaining Equilibrium
       Constants for Reactions
   Consider the reaction below


    CO(g )  3 H 2 (g)             CH4 (g)  H 2O(g)
     – Regardless of the initial concentrations (whether
       they be reactants or products), the law of mass
       action dictates that the reaction will always settle
       into an equilibrium where the equilibrium-constant
       expression equals Kc.


                                                              60
        Obtaining Equilibrium
       Constants for Reactions
   Consider the reaction below


    CO(g )  3 H 2 (g)              CH4 (g)  H 2O(g)
     – As an example, let’s repeat the previous
       experiment, only this time starting with initial
       concentrations of products:

[CH4]initial = 0.1000 M and [H2O]initial = 0.1000 M


                                                          61
       Obtaining Equilibrium
      Constants for Reactions
   Consider the reaction below


    CO(g )  3 H 2 (g)             CH4 (g)  H 2O(g)
    – We find that these initial concentrations result in
      the following equilibrium concentrations.
          Reactants                    Products
        [CO] = 0.0613 M            [CH4] = 0.0387 M
        [H2] = 0.1893 M             [H2O] = 0.0387 M

                                                            62
        Obtaining Equilibrium
       Constants for Reactions
   Consider the reaction below
    CO(g )  3 H 2 (g)            CH4 (g)  H 2O(g)
    – Substituting these values into the equilibrium-
      constant expression, we obtain the same result.
            (0.0387M )(0.0387M )
      Kc                       3
                                   3.93
           (0.0613M )(0.1839M )
     – Whether we start with reactants or products, the
       system establishes the same ratio.
                      (see Figure 15.5).
                                                          63
    The Equilibrium Constant, Kp
   In discussing gas-phase equilibria, it is often
    more convenient to express concentrations in
    terms of partial pressures rather than molarities
       – It can be seen from the ideal gas equation,
                            PV = nRT
         that the partial pressure of a gas is proportional to
         its molarity.

                      n
                 P  ( )RT  MRT
                      V
                                                                 64
    The Equilibrium Constant, Kp
   If we express a gas-phase equilibria in
    terms of partial pressures, we obtain Kp.
     – Consider the reaction below.

    CO(g )  3 H 2 (g)          CH4 (g)  H 2O(g)
     – The equilibrium-constant expression in terms of
       partial pressures becomes:
                       PCH PH O
                Kp         4     2

                                      3
                        PCO PH   2
                                                         65
    The Equilibrium Constant, Kp
   In general, the numerical value of Kp
    differs from that of Kc.
     – From the relationship n/V=P/RT, we can show that
                                  Dn
               K p  K c ( RT)
       where Dn is the sum of the moles of gaseous
       products in a reaction minus the sum of the moles
       of gaseous reactants.
     Animation: Pressure and Concentration of a Gas).

                                                           66
         A Problem to Consider
   Consider the reaction

        2SO 2 (g )  O 2 (g )           2 SO 3 (g)
     – Kc for the reaction is 2.8 x 102 at 1000 oC.
       Calculate Kp for the reaction at this temperature.




                                                            67
         A Problem to Consider
   Consider the reaction

        2SO 2 (g )  O 2 (g )           2 SO 3 (g)
     – We know that
                                   Dn
                 K p  K c ( RT)
      From the equation we see that Δn = -1. We can
      simply substitute the given reaction temperature and
      the value of R (0.08206 L.atm/mol.K) to obtain Kp.


                                                             68
        A Problem to Consider
   Consider the reaction

        2SO 2 (g )  O 2 (g )           2 SO 3 (g)
     – Since
                                       Dn
                     K p  K c ( RT)

K p  2.8  10   2            Latm
                     (0.08206 mol K    1000 K)  3.4
                                                 -1




                                                      69
     Equilibrium Constant for the
          Sum of Reactions
   Similar to the method of combining reactions
    that we saw using Hess’s law in Chapter 6, we
    can combine equilibrium reactions whose Kc
    values are known to obtain Kc for the overall
    reaction.
    – With Hess’s law, when we reversed reactions or
      multiplied them prior to adding them together, we had
      to manipulate the DH’s values to reflect what we had
      done.
    – The rules are a bit different for manipulating Kc.

                                                          70
     Equilibrium Constant for the
          Sum of Reactions
1.   If you reverse a reaction, invert the value of Kc.
2.   If you multiply each of the coefficients in an equation by
     the same factor (2, 3, …), raise Kc to the same power
     (2, 3, …).
3.   If you divide each coefficient in an equation by the same
     factor (2, 3, …), take the corresponding root of Kc
     (i.e., square root, cube root, …).
4.   When you finally combine (that is, add) the individual
     equations together, take the product of the equilibrium
     constants to obtain the overall Kc.

                                                            71
    Building Equations with K values
   Weak acids and bases are assigned a Ka
    or Kb values based on the degree to which
    they ionize in water.
   Larger K values indicate a greater degree
    of ionization (strength).
   Ka and Kb, along with other K values that
    we will study later (Ksp, KD, Kf) are all
    manipulated in the same manner.

                                            72
    Building Equations with K values
   When equations are added K values are
    multiplied.
    MnS ↔ Mn+2 + S-2        K= 5.1 x 10-15
    S-2 + H2O ↔HS- + OH-    K= 1.0 x 10-19
   2H+ + HS- OH- ↔ H2S +H2O K= 1.0 x 10-7
   MnS + 2H+ ↔ Mn+2+ H2 K= 5.1 x 10-41



                                            73
    Building Equations with K values

   When equations are reversed the K values
    are reciprocated.
    Al(OH)3 ↔ Al+3 + 3OH- K = 1.9 x 10-33
   3OH- + Al+3 ↔ Al(OH)3 K = 1/1.9 x 10-33
                              = 5.2 x 1032



                                           74
    Building Equations with K values
   When equations are multiplied the K values
    are raised to the power.
   NH3 + H2O ↔ NH4++ OH- K = 1.8 x 10-5
   2NH3 + 2H2O ↔ 2NH4++ 2OH-
                              K = (1.8 x 10-5)2
                                = 3.24 x 10-10



                                           75
    Building Equations with K values
   When equations are divided the root of the
    K values are taken.
   2HPO4-2 ↔ 2H+ + 2PO43- K = 1.3 x 10-25
   HPO4-2 ↔ H+ + PO43-       K = √1.3 x 10-25
                                = 3.6 x 10-13




                                           76
       Equilibrium Constant for the
            Sum of Reactions
     For example, nitrogen and oxygen can combine
      to form either NO(g) or N2O (g) according to the
      following equilibria.

(1)    N 2 (g )  O 2 (g )     2 NO(g)       Kc = 4.1 x 10-31
(2)    N 2 (g )  1 O 2 (g )
                  2             N 2O(g)      Kc = 2.4 x 10-18
        – Using these two equations, we can obtain Kc for
          the formation of NO(g) from N2O(g):

(3)   N 2O(g )  1 O 2 (g )
                 2              2 NO(g)           Kc = ?

                                                            77
Animation: Equilibrium Decomposition of
                  N2O4




     (Click here to open QuickTime animation)




                 Return to Slide 15
                                                100
 Figure 15.5: Some equilibrium
compositions for the methanation
           reaction.




          Return to Slide 35
                                   101
 Figure 15.5: Some equilibrium
compositions for the methanation
           reaction.




          Return to Slide 21
                                   102
 Figure 15.5: Some equilibrium
compositions for the methanation
           reaction.




          Return to Slide 22
                                   103
 Figure 15.5: Some equilibrium
compositions for the methanation
           reaction.




          Return to Slide 23
                                   104
 Figure 15.5: Some equilibrium
compositions for the methanation
           reaction.




          Return to Slide 24
                                   105
 Figure 15.5: Some equilibrium
compositions for the methanation
           reaction.




          Return to Slide 25
                                   106
 Figure 15.5: Some equilibrium
compositions for the methanation
           reaction.




          Return to Slide 26
                                   107
 Figure 15.5: Some equilibrium
compositions for the methanation
           reaction.




         Return to Slide 27
                                   108
Figure 15.6: The concentration of a gas at
a given temperature is proportional to the
                pressure.




              Return to Slide 43
                                       109
Animation: Pressure and Concentration
               of a Gas




    (Click here to open QuickTime animation)




               Return to Slide 28
                                               110
Figure 15.12 A-C




     Slide 21
                   111
 Figure
 15.13:
 The effect
 of changing
 the
 temperatur
 e on
 chemical
 equilibrium.
 Photo courtesy
 of American
 Color.

Return to Slide 24
                     112
 Figure
 15.15:
 Oxidation
 of
 ammonia
 using a
 copper
 catalyst.
 Photo
 courtesy of
 James
 Scherer.

Return to Slide 85
                     113

								
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