Chapter 14: Chemical Equilibrium by uExqTD45

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									Chapter 14: Chemical
    Equilibrium
CHE 124: General Chemistry II
 Dr. Jerome Williams, Ph.D.
     Saint Leo University
                   Overview
• The Equilibrium Condition: A Second Example
• Law of Mass Action: Equilibrium Constants
• Homogeneous Equilibrium
                       H2(g) + I2(g)  2 HI(g)
                                                At time 0, there are only reactants in the
                                                mixture, so only the forward reaction
                                                can take place



                                                    [H2] = 8, [I2] = 8, [HI] = 0



                                                At time 16, there are both reactants and
                                                products in the mixture, so both the
                                                forward reaction and reverse reaction
                                                can take place



                                                   [H2] = 6, [I2] = 6, [HI] = 4

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                       H2(g) + I2(g)  2 HI(g)
                                                At time 32, there are now more products
                                                than reactants in the mixture − the
                                                forward reaction has slowed down as
                                                the reactants run out, and the reverse
                                                reaction accelerated




                                                   [H2] = 4, [I2] = 4, [HI] = 8

                                                At time 48, the amounts of products and
                                                reactants in the mixture haven’t
                                                changed – the forward and reverse
                                                reactions are proceeding at the same
                                                rate – it has reached equilibrium




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                       H2(g) + I2(g)  2 HI(g)
      And As the reactant
     Becausethe reaction proceeds, the
           because the product                   Because the [HI] at equilibrium is larger
      concentration[I2] increasing, the
     concentrations is decrease, and the
           [H2] and are decreasing,                At equilibrium the forward reaction
                                                Onceequilibrium,[Iis established,position of
                                                 than the [H2] or 2], we say the the
           [HI] increases                          rate is the same as the reverse
                                                concentrations no longer change
      reverse reaction rate speeds up
     the forward reaction rate slows             equilibrium favors products
                                                   reaction rate
     down




                                                        Equilibrium established
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                            Equilibrium  Equal
         • The rates of the forward and reverse reactions are
           equal at equilibrium
         • But that does not mean the concentrations of
           reactants and products are equal
         • Some reactions reach equilibrium only after almost all
           the reactant molecules are consumed – we say the
           position of equilibrium favors the products
         • Other reactions reach equilibrium when only a small
           percentage of the reactant molecules are consumed –
           we say the position of equilibrium favors the
           reactants




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                                 An Analogy:
                              Population Changes




                 When after a A citizens feel overcrowded,
              However, Countrytime, emigration will occur in some will emigrate to
                 Country B
              both directions at the same rate, leading to
              populations in Country A and Country B that are
              constant, though not necessarily equal




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                         Equilibrium Constant
   • Even though the concentrations of reactants and products
     are not equal at equilibrium, there is a relationship
     between them
   • The relationship between the chemical equation and the
     concentrations of reactants and products is called the Law
     of Mass Action
   • For the general equation aA + bB  cC + dD, the Law of
     Mass Action gives the relationship below
         – the lowercase letters represent the coefficients of the balanced
           chemical equation
         – always products over reactants
   • K is called the equilibrium constant
         – unitless

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                  Writing Equilibrium Constant
                           Expressions
    • For the reaction             aA(aq) +
      bB(aq)  cC(aq) + dD(aq) the
      equilibrium constant expression is


   • So for the reaction
     2 N2O5(g)  4 NO2(g) + O2(g)
     the equilibrium constant
     expression is

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       What Does the Value of Keq Imply?
     • When the value of Keq >> 1, we know that when the
       reaction reaches equilibrium there will be many more
       product molecules present than reactant molecules
           – the position of equilibrium favors products
     • When the value of Keq << 1, we know that when the
       reaction reaches equilibrium there will be many more
       reactant molecules present than product molecules
           – the position of equilibrium favors reactants



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              A Large Equilibrium Constant




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               A Small Equilibrium Constant




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     Practice – Write the equilibrium constant expressions,
        K, and predict the position of equilibrium for the
                            following

  2 SO2(g) + O2(g)  2 SO3(g) K = 8 x 1025
                                                  favors products



  N2(g) + 2 O2(g)  2 NO2(g)                     K = 3 x 10−17
                                                 favors reactants


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                           Relationships between K
                           and Chemical Equations

   • When the reaction is written backward, the
     equilibrium constant is inverted
For the reaction aA + bB  cC + dD               For the reaction cC + dD  aA + bB
the equilibrium constant expression              the equilibrium constant expression
is                                               is




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                           Relationships between K
                           and Chemical Equations
   • When the coefficients of an equation are multiplied by
     a factor, the equilibrium constant is raised to that factor

  For the reaction aA + bB  cC                  For the reaction 2aA + 2bB  2cC
  the equilibrium constant                       the equilibrium constant
  expression is                                  expression is




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                           Relationships between K
                           and Chemical Equations
   • When you add equations to get a new equation, the
     equilibrium constant of the new equation is the product
     of the equilibrium constants of the old equations

  For the reactions (1) aA  bB                  For the reaction aA  cC
  and (2) bB  cC the equilibrium                the equilibrium constant
  constant expressions are                       expression is




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   Example 14.2: Compute the equilibrium constant at 25 °C
       for the reaction NH3(g)  0.5 N2(g) + 1.5 H2(g)
             Given: for N2(g) + 3 H2(g)  2 NH3(g), K = 3.7 x 108 at 25 °C

           Find: K for NH3(g)  0.5N2(g) + 1.5H2(g), at 25 °C
       Conceptual         K               K’
            Plan:
                         Kbackward = 1/Kforward, Knew = Koldn
  Relationships:
      Solution:
       N2(g) + 3 H2(g)  2 NH3(g)                           K1 = 3.7 x 108

       2 NH3(g)  N2(g) + 3 H2(g)


       NH3(g)  0.5 N2(g) + 1.5 H2(g)


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      Practice – When the reaction A(aq)  2 B(aq) reaches
      equilibrium [A] = 1.0 x 10−5 M and [B] = 4.0 x 10−1 M.
    When the reaction 2 B(aq)  Z(aq) reaches equilibrium [B]
              = 4.0 x 10−3 M and [Z] = 2.0 x 10−6 M.
      Calculate the equilibrium constant for each of these
    reactions and the equilibrium constant for the reaction 3
                           Z(aq)  3 A(aq)
                                      Krxn 1 = 1.6 x 104
                                      Krxn 2 = 5.0 x 10−1
            2BA                             K3 = 1/Krxn 1 = 6.25 x 10−5
            Z2B                             K4 = 1/Krxn 2 = 2
            ZA                              K3K4 = 1.25 x 10−4
            3Z3A                           (K3K4)3 = 2.0 x 10−12
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