Chemical Equilibrium by yurtgc548


									Chemical Equilibrium

  Brown, LeMay Ch 15
     AP Chemistry
 Monta Vista High School

 15.1: Chemical Equilibrium
Occurs when opposing reactions are
proceeding at the same rate
   Forward rate = reverse rate of reaction
      Vapor pressure: rate of vaporization = rate
       of condensation
      Saturated solution: rate of dissociation =
       rate of crystallization

Expressing concentrations:
 Gases: partial pressures, PX
 Solutes in liquids: molarity, [X]
Forward reaction:   A→B         Rate = kforward [A]
Reverse reaction:   B→A         Rate = kreverse [B]
                   PV  nRT
                      n      P
                   M 
                      V RT
                  PA                PB
           [ A]             [ B]       R = 0.0821 L•atm
                  RT                RT              mol•K
Forward reaction: Rate  k f

Reverse reaction:   Rate  k r
                               RT                     3
 If equilibrium:    A↔B
          forward rate = reverse rate
                            PA      PB
k f [A]  k r [B]   or   kf     kr
                            RT      RT

    [B] k f              PB k f
                   or          constant  K eq
    [A] k r              PA k r

            Figure 1: Reversible reactions
            [A]0 or PA0 / RT

               [A] or PA / RT

                                Equilibrium is established
PX or [X]

               [B] or PB / RT

                     Time →

           Reversible Reactions and Rate

                 Forward rate
 Reaction Rate

                                Equilibrium is established:

                                Forward rate = Backward rate

     Backward rate
When equilibrium is achieved:
    [A] ≠ [B] and         kf/kr = Keq
    15.2: Law of Mass Action
Derived from rate laws by Guldberg and
Waage (1864)
   For a balanced chemical reaction
    in equilibrium:
           aA+bB↔cC+dD                       Cato Guldberg
                                                             Peter Waage
   Equilibrium constant expression (Keq):
                                                c            d
         [C] [D]c     d
                                     (PC ) (PD )
    Kc     a    b
                           or   Kp       a      b
         [A] [B]                     (PA ) (PB )
Keq is strictly based on stoichiometry of the reaction (is
independent of the mechanism).
Units: Keq is considered dimensionless (no units)
     Relating Kc and Kp
   Convert [A] into PA:
              n  P                          PA  [ A]RT
            M 
              V RT

     (PC )c (PD )d ([C]RT) c ([D]RT) d [C] c [D] d (RT) c  d
Kp       a      b
                          a         b
                                           a    b      ab
     (PA ) (PB ) ([A]RT) ([B]RT)         [A] [B] (RT)

                          (cd) - (a b)
        Kp  Kc (RT)                        K c (RT)   Dn

   where Dn =
   = change in coefficents of products – reactants (gases only!)
   = (c+d) - (a+b)

    Magnitude of Keq
Since Keq a [products]/[reactants], the magnitude of
Keq predicts which reaction direction is favored:

   If Keq > 1   then [products] > [reactants]
                 and equilibrium “lies to the right”

   If Keq < 1   then [products] < [reactants]
                 and equilibrium “lies to the left”

   Relationship Between Q and K
 Reaction Quotient (Q): The particular ratio
 of concentration terms that we write for a
 particular reaction is called reaction
 For a reaction, A B, Q= [B]/[A]
 At equilibrium, Q= K
Reaction Direction: Comparing Q and K
 Q<K, reaction proceeds to right, until
 equilibrium is achieved (or Q=K)
 Q>K, reaction proceeds to left, until Q=K
   Value of K
For the    For the      For the       For the rxn,
reference  reverse rxn, reaction,     A > C
rxn, A>B, B >A,       2A > 2B      C > B

K(ref)=    K= 1/K(ref)   K= K(ref)2   K (overall)=
[B]/[A]                               K1 X K2

 15.3: Types of Equilibria
Homogeneous: all components in same
phase (usually g or aq)
       1 N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
                      c           d
               (PC ) (PD )
          KP       a      b
                                           Fritz Haber
                                          (1868 – 1934)
               (PA ) (PB )
                    (PNH3 )
          KP             1           3
                 (PN 2 ) (PH 2 )
Heterogeneous: different phases
        CaCO3 (s) ↔ CaO (s) + CO2 (g)
Definition:                What we use:
              [CaO] (PCO 2 )
     K eq                                K p  PCO 2
                [CaCO3 ]
   Concentrations of pure solids and pure liquids are not
    included in Keq expression because their concentrations do
    not vary, and are “already included” in Keq (see p. 548).
   Even though the concentrations of the solids or liquids do
    not appear in the equilibrium expression, the substances
    must be present to achieve equilibrium.
   15.4: Calculating Equilibrium Constants
Steps to use “ICE” table:
   1. “I” = Tabulate known initial and equilibrium
      concentrations of all species in equilibrium
   2. “C” = Determine the concentration change for
      the species where initial and equilibrium are
         • Use stoichiometry to calculate
            concentration changes for all other
            species involved in equilibrium
   3. “E” = Calculate the equilibrium concentrations

Ex: Enough ammonia is dissolved in 5.00 L of water
at 25ºC to produce a solution that is 0.0124 M
ammonia. The solution is then allowed to come to
equilibrium. Analysis of the equilibrium mixture
shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at
25ºC for the reaction:
NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

    NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

                          H 2O
              NH3 (aq)             NH41+ (aq)      OH1- (aq)

  Initial     0.0124 M    X           0M              0M
  Change         -x
                          X          +x              +x

Equilibrium   0.0119 M    X       4.64 x 10-4 M   4.64 x 10-4 M

              x = 4.64 x 10-4 M
            [NH ][OH1- ] (4.64 10 -4 ) 2
       Kc       4
                                          1.81x 10 -5
               [NH3 ]        0.0119

Ex: A 5.000-L flask is filled with 5.000 x 10-3 mol of
H2 and 1.000 x 10-2 mol of I2 at 448ºC. The value of
Keq is 1.33. What are the concentrations of each
substance at equilibrium?
           H2 (g) + I2 (g) ↔ 2 HI (g)

     H2 (g) + I2 (g) ↔ 2 HI (g)
                     H2 (g)              I2 (g)         HI (g)
   Initial      1.000x10-3 M       2.000x10-3 M          0M
  Change            -xM                -xM             + 2x M
Equilibrium   (1.000x10-3 – x) M (2.000x10-3 – x) M      2x M
           [HI]2                   (2x) 2
     Kc                                              1.33
          [H2 ][I2 ] (1.000  10 - x)(2.000  10 - x)
                                -3              -3

4x2 = 1.33[x2 + (-3.000x10-3)x + 2.000x10-6]
0 = -2.67x2 – 3.99x10-3x + 2.66x10-6
Using quadratic eq’n: x = 5.00x10-4 or –1.99x10-3; x = 5.00x10-4
Then [H2]=5.00x10-4 M; [I2]=1.50x10-3 M; [HI]=1.00x10-3 M
 15.6: Le Châtelier’s Principle
If a system at equilibrium is
disturbed by a change in:
 Concentration of one of the
 Pressure, or
 Temperature
                                   Henri Le Châtelier
                                     (1850 – 1936)

…the system will shift its equilibrium
position to counteract the effect of
the disturbance.

   4 Changes that do not affect Keq:
1. Concentration
   Upon addition of a reactant or product,
     equilibrium shifts to re-establish equilibrium
     by consuming part of the added substance.
   Upon removal of reactant or product,
     equilibrium shifts to re-establish equilibrium
     by producing more of the removed substance.
  Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l)

     •Add HCl, temporarily inc forward rate
     •Add H2O, temporarily inc reverse rate

       2. Volume, with a gas present (T is constant)
       Upon a decrease in V (thereby increasing P),
        equilibrium shifts to reduce the number of moles of gas.
       Upon an increase in V (thereby decreasing P),
        equilibrium shifts to produce more moles of gas.

   Ex: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
       If V of container is decreased, equilibrium shifts right.
                  XN2 and XH2 dec
                  XNH3 inc
        (PNH3 )    2
                                   (X NH3 PT )   2                       2
                                                                (X NH3 ) PT                     (X NH3 ) 2
KP                    3
                                                      3
                                                                            3       4
                                                                                                          3        2
       PN 2 (PH 2 )            (X N 2 PT )(XH 2 PT )           X N 2 (XH 2 ) PT              X N 2 (XH 2 ) PT
                                                                                         ( NH3 ) 2
 Since PT also inc, KP remains constant. K P 
                                                                                  N 2 ( H 2 )   3   2
 3. Pressure, but not Volume
Usually addition of a noble gas, p. 560
   Avogadro’s law: adding more non-reacting particles “fills
    in” the empty space between particles.
   In the mixture of red and blue gas particles, below,
    adding green particles does not stress the system, so
    there is no Le Châtelier shift.

 4. Catalysts
Lower the activation energy of both forward and
reverse rxns, therefore increases both forward and
reverse rxn rates.
Increase the rate at which equilibrium is achieved,
but does not change the ratio of components of
the equilibrium mixture (does not change the Keq)
                                  Ea, uncatalyzed

    Energy                        Ea, catalyzed

             Rxn coordinate                       23
 1 Change that does affect Keq:
Temperature: consider “heat” as a part of the reaction
 Upon an increase in T, endothermic reaction is favored
  (equilibrium shifts to “consume the extra heat”)
 Upon a decrease in T, equilibrium shifts to produce more
Effect on Keq
1. Exothermic equilibria: Reactants ↔ Products + heat
   • Inc T increases reverse reaction rate which
     decreases Keq
2. Endothermic equilibria: Reactants + heat ↔ Products
   • Inc T increases forward reaction rate increases Keq

Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l); DH=+?
    •Inc T temporarily inc forward rate
    •Dec T temporarily inc reverse rate                          24
 Vant Hoff’s Equation
Vant Hoff’s equation shows mathematically
how the equilibrium constant is affected by
changes in temp.

 ln K2 = -d H0   rxn   (1 – 1)
    K1      R            T2 T1

     Effect of Various Changes on Equilibrium
Disturbance             Net Direction of       Effect of Value
                        Rxn                    of K
  Increase (reactant)   Towards formation of   None

  Decrease(reactant)    Towards formation of   None
   Increase (product)   Towards formation of   None
   Decrease (product)   Towards formation of   None

   Effect of Pressure on Equilb.
Pressure         Towards          None
  Increase P     formation of
  (decrease V)   fewer moles of
  Decrease P     Towards          None
  (Increase V)   formation of
                 more moles of
   Increase P    None,            None
   ( Add inert   concentrations
gas, no change   unchanged
in V)
   Effect of Temperature on Equilb
Temperature      Towards           Increases if
  Increase T     absorption of     endothermic
                 heat              Decreases if

  Decrease T     Towards release   Increases if
                 of heat           exothermic
                                   Decreases if
Catalyst Added   None, forward     None
                 and reverse

To top