Chapter 15 Chemical Equilibrium - TamAPChemistryHart

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Chapter 15 Chemical Equilibrium - TamAPChemistryHart Powered By Docstoc
					 Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr., and
              Bruce E. Bursten




   Chapter 15
Chemical Equilibrium


                                                Equilibrium
Overview of Equilibrium – Chapter 15
 • write Keq expression, in units of pressure or
   concentration

 • assess the Keq with regard to relative concentrations
   of reactants and products

 • manipulate chemical equations and Keq – reciprocal,
   multiplication, application of Hess’s Law

 • heterogeneous equilibria – what is included in Keq?

 • calculate Keq, including use of RICE tables
                                                         Equilibrium
Overview of Equilibrium – Chapter 15
 • calculate Q and predict the direction of a reaction by
   relating Q to Keq

 • calculate equilibrium concentrations of reactants
   and/or products when given Keq

 • apply Le Châtelier’s Principle to predict responses of
   the equilibrium system to changes in reactant or
   product concentrations, changes in pressure or
   volume, changes in temperature

 • describe and explain the effects of catalysts on
   equilibrium
                                                        Equilibrium
  The Concept of Equilibrium




Chemical equilibrium occurs when a
reaction and its reverse reaction proceed at   Equilibrium
the same rate.
The Concept of Equilibrium
             • As a system
               approaches equilibrium,
               both the forward and
               reverse reactions are
               occurring.
             • At equilibrium, the
               forward and reverse
               reactions are
               proceeding at the same
               rate.
                                Equilibrium
     A System at Equilibrium

Once equilibrium is
achieved, the
amount of each
reactant and product
remains constant.




                               Equilibrium
    Depicting Equilibrium

Since, in a system at equilibrium, both
the forward and reverse reactions are
being carried out, we write its equation
with a double arrow.

       N2O4 (g)         2 NO2 (g)


                                       Equilibrium
        Dynamic Equilibrium

• A dynamic equilibrium exists when the rates
  of the forward and reverse reactions are
  equal.

• There is no further change in [reactant] or
  [product].

• How fast you get to equilibrium depends on
  kinetics.
                                                Equilibrium
   The
Equilibrium
 Constant
              Equilibrium
   The Equilibrium Constant

• Forward reaction:
  N2O4 (g) ¾¾® 2 NO2 (g)

• Rate Law:
  Rate = kf [N2O4]



                              Equilibrium
   The Equilibrium Constant

• Reverse reaction:
  2 NO2 (g) ¾¾® N2O4 (g)

• Rate Law:
  Rate = kr [NO2]2



                              Equilibrium
   The Equilibrium Constant

• Therefore, at equilibrium
               Ratef = Rater
           kf [N2O4] = kr [NO2]2
• Rewriting this, it becomes
                             2
             kf   [NO2]
                =
             kr   [N2O4]
                                   Equilibrium
 The Equilibrium Constant

The ratio of the rate constants is a
constant at that temperature, and the
expression becomes
            kf        [NO2]2
      Keq =         =
            kr        [N2O4]

                                        Equilibrium
                Summary

•   At equilibrium, the concentrations of
    reactants and products no longer change
    with time.

2. For equilibrium to occur, neither reactants
   nor products can escape from the system.

3. At equilibrium a particular ratio of
   concentration terms equals a constant.
                                                 Equilibrium
   The Equilibrium Constant

• Consider the generalized reaction

        aA + bB          cC + dD

• The equilibrium expression for this
  reaction would be
                 [C]c[D]d
            Kc =
                 [A]a[B]b               Equilibrium
 The Equilibrium Constant

Since pressure is proportional to
concentration for gases in a closed
system, the equilibrium expression can
also be written
             (PCc) (PDd)
        Kp =
             (PAa) (PBb)

                                     Equilibrium
                  Keq vs Kc vs Kp

•   Keq = the general expression for equilibrium constant
    expressions.

•   Kc = Keq for which molar concentrations were used to evaluate
    the constant (i.e. subscript “c” = concentration).

•   Kc includes
     – Ka (weak acids) and Kb (weak bases) (Chapter 16)
     – Ksp (solubility product) (Chapter 17)
     – Kw (water)

•   Kp = Keq where “p” stands for pressure


                                                                Equilibrium
 Relationship Between Kc and Kp
  • From the Ideal Gas Law we know that
                             PV = nRT
  • Rearranging it, we get

                              n
                           P=   RT
                              V
If we express volume in liters the quantity (n/V) is equivalent to molarity.
                                                                       Equilibrium
 Relationship Between Kc and Kp
     Plugging this into the expression for Kp
     for each substance, the relationship
     between Kc and Kp becomes

                  Kp = Kc (RT)Dn
     where
Dn = (moles of gaseous product) - (moles of gaseous reactant)

                                                       Equilibrium
Sample Exercise 15.1 (p. 632)
 Write the equilibrium expression for Keq
 for these three reactions:

 a) 2 O3(g) D 3 O2(g)

 b) 2 NO(g) + Cl2(g) D 2 NOCl(g)

 c) Ag+(aq) + 2 NH3(g) D Ag(NH3)2+(aq)
                                         Equilibrium
   Practice Exercise 15.1

Write the equilibrium expression for Keq
for these three reactions:

a) H2(g) + I2(g) D 2 HI(g)

b) Cd2+(aq) + 4 Br-(aq) D CdBr42-(aq)

                                        Equilibrium
Equilibrium Can Be Reached from
          Either Direction



As you can see, the ratio of [NO2]2 to [N2O4] remains
constant at this temperature no matter what the initial
concentrations of NO2 and N2O4 are.


                                                 Equilibrium
Equilibrium Can Be Reached from
          Either Direction
                 This is the data from
                 the last two trials from
                 the table on the
                 previous slide.




                                     Equilibrium
Equilibrium Can Be Reached from
          Either Direction




  It doesn’t matter whether we start with N2 and
  H2 or whether we start with NH3: we will have
  the same proportions of all three substances
  at equilibrium.
                                              Equilibrium
Sample Exercise 15.2 (p. 634)
 In the synthesis of ammonia from nitrogen
 and hydrogen,
 N2(g) + 3 H2(g) D 2 NH3(g)

 Kc = 9.60 at 300oC,
 R = 0.0821 Latm/molK

 Calculate Kp for this reaction at this
 temperature.
 (4.34 x 10-3)
                                             Equilibrium
  Practice Exercise 15.2
For the equilibrium
2 SO3(g) D 2 SO2(g) + O2(g),
Kc is 4.08 x 10-3 at 1000 K.

Calculate the value for Kp.

(0.335)
                               Equilibrium
Equilibrium Constants and Units

 • Keq – described using activities rather
   than [ ] or P à no units
 • Activity in an ideal mixture
      = [ ] relative to 1 M ([reference])
                  or
         P relative to 1 atm (reference P)

                                             Equilibrium
Equilibrium Constants and Units

     e.g. 0.010 M à activity = 0.010 M = 0.010
                                  1M
                                          same number, no units

     e.g. Keq = (PNO2/Pref)2
               (PN2O4/Pref)

 •   Results = units cancel
     Molarity and pressure can be used in the same Keq expression

 •   For pure substances (ie. solids or liquids),
     the [reference] = itself, à 1

                                                                  Equilibrium
What Does the Value of K Mean?
               • If K>>1, the reaction is
                 product-favored;
                 product predominates
                 at equilibrium.




                                     Equilibrium
What Does the Value of K Mean?
               • If K>>1, the reaction is
                 product-favored;
                 product predominates
                 at equilibrium.


               • If K<<1, the reaction is
                 reactant-favored;
                 reactant predominates
                 at equilibrium.
                                     Equilibrium
        Sample Exercise 15.3
The following diagrams represent three different systems at
equilibrium, all in the same size containers.




a)   Without doing any calculations, rank the three
     systems in order of increasing equilibrium
     constant, Kc.

b)   If the volume of the containers is 1.0 L and each sphere
     represents 0.10 mol, calculate Kc for each system.         Equilibrium
  Practice Exercise 15.3
The equilibrium constant for the
reaction H2(g) + I2(g) D 2 HI(g) varies
with
temperature as follows:

Kp = 792 at 298 K; Kp = 54 at 700 K.

Is the formation of HI favored more at
the higher or lower temperature?
                                          Equilibrium
Equilibrium
Sample Exercise 15.4 (p. 637)

 The equilibrium constant for the reaction of N2
 with O2 to form NO equals Kc = 1 x 10-30 at
 25oC.
           N2(g) + O2(g) D 2 NO(g)

 Using this information, write the equilibrium
 constant expression and calculate the
 equilibrium constant for the following
 reaction:
           2 NO(g) D N2(g) + O2(g)
                                                 Equilibrium
  Practice Exercise 15.4

For the formation of NH3 from N2 and
H2, N2(g) + H2(g) D 2 NH3(g),
Kp = 4.34 x 10-3 at 300oC.

What is the value of Kp for the reverse
reaction?

(2.30 x 102)
                                       Equilibrium
Overview of Equilibrium – Chapter 15
 Where we have been:

 • write Keq expression, in units of
   pressure or concentration

 • assess the Keq with regard to relative
   concentrations of reactants and
   products
                                            Equilibrium
Overview of Equilibrium – Chapter 15
 Where we are going:

 • manipulate chemical equations and Keq – reciprocal,
   multiplication, application of Hess’s Law

 • heterogeneous equilibria – what is included in Keq?

 • calculate Keq, including use of RICE tables

 • calculate Q and predict the direction of a reaction by
   relating Q to Keq

 • calculate equilibrium concentrations of reactants
   and/or products when given Keq
                                                         Equilibrium
Overview of Equilibrium – Chapter 15

 • apply Le Châtelier’s Principle to predict
   responses of the equilibrium system to
   changes in reactant or product
   concentrations, changes in pressure or
   volume, changes in temperature

 • describe and explain the effects of
   catalysts on equilibrium
                                           Equilibrium
Manipulating Equilibrium Constants
  •    The equilibrium constant of a reaction in the
       reverse reaction is the reciprocal of the equilibrium
       constant of the forward reaction.

  2.   The equilibrium constant of a reaction that has
       been multiplied by a number is the equilibrium
       constant raised to a power that is equal to that
       number.

  3.   The equilibrium constant for a net reaction made up
       of two or more steps is the product of the
       equilibrium constants for the individual steps.
                                                          Equilibrium
Manipulating Equilibrium Constants
    1. The equilibrium constant of a
    reaction in the reverse reaction is the
    reciprocal of the equilibrium constant of
    the forward reaction.
                                In the opposite direction:

     N2O4 (g) D 2 NO2 (g)          2 NO2 (g) D N2O4 (g)

    Keq = [NO2]2 = 0.212            Keq = [N2O4] = 4.72
          [N2O4]                          [NO2]2

  Keq for these reactions are reciprocals of each other.
                                                       Equilibrium
Manipulating Equilibrium Constants
2. The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to a power that is equal to that
number.

N2O4(g) D 2 NO2(g)         Kc = [NO2]2 = 0.212
                                [N2O4]

2 N2O4(g) D 4 NO2(g)       Kc = [NO2]4 = (0.212)2
                                [N2O4]2
                                                 Equilibrium
Manipulating Equilibrium Constants

   3. The equilibrium constant for a net
   reaction made up of two or more steps
   is the product of the equilibrium
   constants for the individual steps.




                                       Equilibrium
Sample Exercise 15.5 (p. 638)
  Given the following information,

   HF(aq) D H+(aq) + F-(aq)              Kc = 6.8
  x 10-4

H2C2O4(aq) D 2 H+(aq) + C2O42-(aq)       Kc = 3.8
  x 10-6

  determine the value of Kc for the following reaction:

  2 HF(aq) + C2O42-(aq) D 2 F-(aq) + H2C2O4(aq)
                                                          Equilibrium
  (0.12)
   Practice Exercise 15.5

Given the following information at 700 K,
   H2(g) + I2(g) D 2HI(g)  Kp = 54.0
N2(g) + 3 H2(g) D 2 NH3(g) Kp = 1.04 x 10-4

determine the value of Kp (at 700 K)
2 NH3(g) + 3 I2(g) D 6 HI(g) + N2(g)

(1.51 x 109)
                                              Equilibrium
Heterogeneous
 Equilibrium
           Equilibrium
   Heterogeneous Equilibria

• Equilibria in which all reactants and
  products are present in the same phase
  are called homogeneous equilibria.

• Equilibria in which one or more
  reactants or products are present in a
  different phase are called
  heterogeneous equilibria.
                                           Equilibrium
The Concentrations of Solids and
Liquids Are Essentially Constant

  Both can be obtained by multiplying the
  density of the substance by its molar
  mass — and both of these are constants
  at constant temperature.




                                      Equilibrium
The Concentrations of Solids and
Liquids Are Essentially Constant

  Therefore, the concentrations of solids
  and liquids do not appear in the
  equilibrium expression.
      PbCl2(s) D Pb2+(aq) + 2 Cl-(aq)

             Kc = [Pb2+] [Cl-]2

                                            Equilibrium
 CaCO3(s) D CaO(s) + CO2(g)
As long as some CaCO3 or CaO remain in
the system, the amount of CO2 above the
solid will remain the same.




                                          Equilibrium
 Heterogeneous Equilibria

Systems where the solvent is involved
as a reactant or product and the solutes
are present at low concentrations.
e.g. dissociation of a weak acid

H2O(l) + CO32–(aq) D OH–(aq) + HCO3–(aq)

Kc = [OH–][HCO3–] / [CO32–]
                                       Equilibrium
Sample Exercise 15.6 (p. 640)

 Write the equilibrium-constant Kc for
 each of the following reactions:

 a) CO2(g) + H2(g) D CO(g) + H2O(l)
    Kc =

 b) SnO2(s) + 2 CO(g) D Sn(s) + 2 CO2(g)
    Kc =
                                           Equilibrium
   Practice Exercise 15.6

Write the equilibrium-constant expressions for
each of the following reactions:

a) Cr(s) + 3 Ag+(aq) D Cr3+(aq) + 3 Ag(s)
   Kc =

b) 3 Fe(s) + 4 H2O(g) D Fe3O4(s) + 4 H2(g)
   Kp =
                                             Equilibrium
Sample Exercise 15.7 (p. 641)
 Each of the following mixtures was placed in a closed
 container and allowed to stand. Which of these
 mixtures is capable of attaining the equilibrium
            CaCO3(s) D CaO(s) + CO2(g)

 a) pure CaCO3

 b) CaO and a pressure of CO2 > Kp

 c) some CaCO3 and a pressure of CO2 > Kp

 d) CaCO3 and CaO
                                                   Equilibrium
   Practice Exercise 15.7

When added to Fe3O4(s) in a closed
container, which one of the following
substances – H2(g), H2O(g), O2(g) - will
allow equilibrium to be established in
the reaction

3 Fe(s) + 4 H2O(g) D Fe3O4(s) + 4 H2(g)
                                           Equilibrium
Equilibrium
Calculations
               Equilibrium
Sample Exercise 15.8 (p. 642)

 A mixture of hydrogen and nitrogen in a
 reaction vessel is allowed to attain equilibrium
 at 472oC. The equilibrium mixture of gases
 was analyzed and found to contain 7.38 atm
 H2, 2.46 atm N2, and 0.166 atm NH3. From
 these data calculate the equilibrium constant,
 Kp, for
           N2(g) + 3 H2(g) D 2 NH3(g)

 (2.79 x 10-5)
                                              Equilibrium
     Practice Exercise 15.8
An aqueous solution of acetic acid is found to
have the following equilibrium concentrations at
25oC: [HC2H3O2] = 1.65 x 10-2 M; [H+] = 5.44 x
10-4 M; and [C2H3O2-] = 5.44 x 10-4 M.

Calculate the equilibrium constant, Kc, for the
ionization of acetic acid at 25oC. The reaction is
    HC2H3O2(aq) D H+(aq) + C2H3O2-(aq)

(1.79 x 10-5)
                                               Equilibrium
           Equilibrium Calculations
1. Tabulate all the known initial and equilibrium concentrations of the
   species that appear in the equilibrium-constant expression.

2. For those species for which both the initial and equilibrium
   concentrations are known, calculate the change in concentration that
   occurs as the system reaches equilibrium.

3. Use the stoichiometry of the reaction (that is, use the coefficients in the
   balanced chemical equation) to calculate the changes in concentration
   of all the other species in the equilibrium.

4. From the initial concentrations and the changes in concentration,
   calculate the equilibrium concentrations. These are then used to
   evaluate the equilibrium constant.




                                                                       Equilibrium
An Equilibrium Problem (Sample
    Exercise 15.9, p. 643)
A closed system initially containing 1.000 x
10-3 M H2 and 2.000 x 10-3 M I2 at 448 °C is
allowed to reach equilibrium.
Analysis of the equilibrium mixture shows
that the concentration of HI is 1.87 x 10-3 M.

Calculate Kc at 448 °C for the reaction taking
place, which is
        H2(g) + I2(g) D 2 HI(g)
                                          Equilibrium
                 What Do We Know?
Reaction:           H2(g)       +       I2(g)      D    2 HI(g)


                  [H2], M             [I2], M          [HI], M
Initially        1.000 x 10-3       2.000 x 10-3          0

Change

At equilibrium                                     1.87 x 10-3




                                                              Equilibrium
            [HI] Increases by 1.87 x 10-3 M
Reaction:            H2(g)       +       I2(g)      D    2 HI(g)


                   [H2], M             [I2], M          [HI], M
Initially         1.000 x 10-3       2.000 x 10-3          0

Change                                              +1.87 x 10-3

At equilibrium                                      1.87 x 10-3




                                                               Equilibrium
Stoichiometry tells us [H2] and [I2] decrease
             by half as much.
 Reaction:           H2(g)       +       I2(g)      D    2 HI(g)

                   [H2], M             [I2], M          [HI], M
 Initially        1.000 x 10-3       2.000 x 10-3          0

 Change           -9.35 x 10-4       -9.35 x 10-4   +1.87 x 10-3

 At equilibrium                                     1.87 x 10-3




                                                               Equilibrium
 We can now calculate the equilibrium
concentrations of all three compounds…
Reaction:           H2(g)       +       I2(g)      D    2 HI(g)

                  [H2], M             [I2], M          [HI], M
Initially        1.000 x 10-3       2.000 x 10-3          0

Change           -9.35 x 10-4       -9.35 x 10-4   +1.87 x 10-3

At equilibrium    6.5 x 10-5        1.065 x 10-3   1.87 x 10-3




                                                              Equilibrium
…and, therefore, the equilibrium constant.

                  2
            [HI]
      Kc =
           [H2] [I2]
                 (1.87 x 10-3)2
         =
           (6.5 x 10-5)(1.065 x 10-3)
         = 51
                                      Equilibrium
   Practice Exercise 15.9

Sulfur trioxide decomposes at high
temperature in a sealed container:
    2 SO3(g) D 2 SO2(g) + O2(g)

Initially the vessel is charged at 1000 K with
SO3(g) at a partial pressure of 0.500 atm. At
equilibrium the SO3 partial pressure is 0.200
atm. Calculate the value of Kp at 1000 K.

(0.338)
                                             Equilibrium
   The Reaction Quotient (Q)

• Q gives the same ratio
  the equilibrium
  expression gives, but
  for a system that is not
  at equilibrium.

• To calculate Q, one
  substitutes the initial
  concentrations of
  reactants and products
  into the equilibrium
  expression.
                               Equilibrium
      The Reaction Quotient

• Note: Q = Keq only at equilibrium.

• If Q < Keq then the forward reaction must
  occur to reach equilibrium.

• If Q > Keq then the reverse reaction must
  occur to reach equilibrium.
  Products are consumed, reactants are
  formed.
  Q decreases until it equals Keq.
                                              Equilibrium
          If Q = K,
the system is at equilibrium.




                                Equilibrium
              If Q > K,
there is too much product, and the
   equilibrium shifts to the left.




                                 Equilibrium
              If Q < K,
there is too much reactant, and the
   equilibrium shifts to the right.




                                  Equilibrium
Sample Exercise 15.10 (p. 645)

  At 448oC the equilibrium constant, Kc, for the
  reaction H2(g) + I2(g) à 2 HI(g) is 51.

  Predict how the reaction will proceed to reach
  equilibrium at 448oC if we start with
  2.0 x 10-2 mol of HI, 1.0 x 10-2 mole of H2,
  and 3.0 x 10-2 mol of I2 in a 2.00-L container.

  (Q = 1.3, so reaction must proceed from left
  to right)
                                                 Equilibrium
  Practice Exercise 15.10

At 1000 K the value of Kp for the reaction
    2 SO3(g) D 2 SO2(g) + O2(g) is 0.338.
Calculate the value for Qp, and predict the
direction in which the reaction will proceed
toward equilibrium if the initial partial
pressures of reactants are PSO3 = 0.16 atm;
PSO2 = 0.41 atm; PO2 = 2.5 atm.

(Qp = 16; Qp > Kp, so reaction will proceed
from right to left)
                                               Equilibrium
Sample Exercise 15.11 (p. 646)

  For the Haber process,
  N2(g) + 3 H2(g) D 2 NH3(g), Kp = 1.45 x 10-5 at
  500oC.
  In an equilibrium mixture of the three gases at
  500oC, the partial pressure of H2 is 0.928 atm
  and that of N2 is 0.432 atm. What is the
  partial pressure of NH3 in this equilibrium
  mixture?

  (2.24 x 10-3 atm)
                                               Equilibrium
  Practice Exercise 15.11
At 500 K the reaction
PCl5(g) D PCl3(g) + Cl2(g) has Kp = 0.497.
In an equilibrium mixture at 500 K, the
partial pressure of PCl5 is 0.860 atm
and that of PCl3 is 0.350 atm. What is
the partial pressure of Cl2 in the
equilibrium mixture?

(1.22 atm)
                                         Equilibrium
       Warmup – Equilibrium
           p.661, 15.31

    A mixture of 0.100 mol of NO, 0.050 mol of
    H2, and 0.100 mol of H2O is placed in a 1.0
    L vessel at 300 K. The following equilibrium
    is established:
2 NO(g) + 2 H2(g) D N2(g) + 2 H2O(g)
At equilibrium, there are 0.062 mol of NO.
a) Calculate the equilibrium concentrations of
    H2, N2, and H2O.
b) Calculate Keq .
                                              Equilibrium
       Warmup – Equilibrium
           p.661, 15.31

    A mixture of 0.100 mol of NO, 0.050 mol of
    H2, and 0.100 mol of H2O is placed in a 1.0
    L vessel at 300 K. The following equilibrium
    is established:
2 NO(g) + 2 H2(g) N2(g) + 2 H2O(g)
At equilibrium, there are 0.062 mol of NO.
a) Calculate the equilibrium concentrations of
    H2, N2, and H2O.
    (0.012 mol, 0.019 mol, 0.138 mol)
b) Calculate Keq . (650)
                                              Equilibrium
Sample Exercise 15.12 (p. 646)
  A 1.000-L flask is filled with 1.000 mol of H2
  and 2.000 mol of I2 at 448oC. The value of
  the equilibrium constant, Kc, for the reaction
             H2(g) + I2(g) D 2 HI(g)
  at 448oC is 50.5.
  What are the partial pressures of H2, I2, and
  HI in the flask at equilibrium?

  ([H2] = 0.065 M, [I2] = 1.065 M, [HI] = 1.87 M)
                                                   Equilibrium
  Practice Exercise 15.12
For the equilibrium, PCl5(g) D PCl3(g) + Cl2(g),
the equilibrium constant, Kp, has the value
0.497 at 500 K.
A gas cylinder at 500 K is charged with PCl5(g)
at an initial pressure of 1.66 atm. What are
the equilibrium pressures of PCl5, PCl3, and
Cl2 at this temperature?

(PPCl5 = 0.967 atm, PPCl3 = PCl2 = 0.693 atm)
                                                Equilibrium
Le Châtelier’s
  Principle
             Equilibrium
   Le Châtelier’s Principle

“If a system at equilibrium is disturbed
by a change in temperature, pressure,
or the concentration of one of the
components, the system will shift its
equilibrium position so as to counteract
the effect of the disturbance.”


                                       Equilibrium
            The Haber Process




As P increases, the amount of ammonia present at equilibrium increases.
As T increases, the amount of ammonia at equilibrium decreases.
                  Can this be predicted?                            Equilibrium
           The Haber Process
The transformation of nitrogen and hydrogen into
ammonia (NH3) is of tremendous significance in
agriculture, where ammonia-based fertilizers are of
utmost importance.




                                                Equilibrium
The Haber Process

          If H2 is added to the
          system, N2 will be
          consumed and the
          two reagents will
          form more NH3.




                             Equilibrium
The Haber Process

          This apparatus
          helps push the
          equilibrium to the
          right by removing
          the ammonia (NH3)
          from the system as
          a liquid.


                           Equilibrium
 Effects of Volume and Pressure
             Changes

Le Châtelier’s principle predicts that if
  pressure is increased, the system will
  shift to counteract the increase.
• That is, the system shifts to remove
  gases and decrease pressure.
• An increase in pressure favors the
  direction that has fewer moles of gas.
                                            Equilibrium
 Effects of Volume and Pressure
             Changes

• In a reaction with the same number of
  moles of gas in the products and
  reactants, changing the pressure has
  no effect on the equilibrium.

• No change will occur if we increase the
  total gas pressure by the addition of a
  gas that is not involved in the reaction.
                                          Equilibrium
         The Effect of Changes in
              Temperature
Co(H2O)62+(aq) + 4Cl–(aq) D CoCl42–(aq) + 6H2O(l) DH > 0




                                                           Equilibrium
Sample Exercise 15.13 (p. 653)
  Consider the following equilibrium:
     N2O4(g) D 2 NO2(g)     DHo = 58.0 kJ

  In what direction will the equilibrium shift
  when each of the following changes is made
  to a system at equilibrium:
  a) add N2O4
  b) remove NO2
  c) increase the total pressure by adding N2(g)
  d) increase the volume
  e) decrease the temperature?
                                              Equilibrium
  Practice Exercise 15.13

For the reaction
PCl5(g) D PCl3(g) + Cl2(g) DHo = 87.9 kJ
in what direction will the equilibrium shift
when
a) Cl2(g) is removed;
b) the temperature is decreased;
c) the volume of the reaction system is
    increased;
d) PCl3(g) is added?
                                               Equilibrium
Sample Exercise 15.14 (p. 653)

a)Using the standard heat of formation data in
  Appendix C, determine the standard enthalpy
  change for the reaction
      N2(g) + 3 H2(g) D 2 NH3(g)
      0           0     2 mol x -46.19 kJ/mol
                             = -92.38 kJ/mol
b)Determine how the equilibrium constant for
  this reaction should change with temperature.
                                             Equilibrium
Equilibrium
  Practice Exercise 15.14

The enthalpy change for the reaction
   2 POCl3(g) D 2 PCl3(g) + O2(g)
  -542.2 kJ/mol    -288.07 kJ/mol
   = (2 mol x -288.07 kJ/mol) – (2 mol x -542.2 kJ/mol
   = + 508.3 kJ
Use this result to determine how the
equilibrium constant for the reaction should
change with temperature.
                                                         Equilibrium
Catalysts

            Equilibrium
Catalysts

      Catalysts increase
      the rate of both the
      forward and reverse
      reactions.




                        Equilibrium
Catalysts

      When one uses a
      catalyst, equilibrium
      is achieved faster,
      but the equilibrium
      composition remains
      unaltered.



                        Equilibrium
Sample Integrative Exercise 15
   At temperatures near 800oC, steam passed over hot
   coke (a form of carbon obtained from coal) reacts to
   form CO and H2:
               C(s) + H2O(g) D CO(g) + H2(g)
   The mixture of gases that results is an important
   industrial fuel called water gas.

 a) At 800oC the equilibrium constant for this reaction is
    Kp = 14.1. What are the equilibrium partial pressures
    of H2O, CO, and H2 in the equilibrium mixture at this
    temperature if we start with solid carbon and 0.100
    mol of H2O in a 1.00-L vessel?

 b) What is the minimum amount of carbon required to
    achieve equilibrium under these conditions?      Equilibrium
Sample Integrative Exercise 15

 c) What is the total pressure in the vessel at
   equilibrium?

 d) At 25oC the value of Keq for this reaction is
   1.7 x 10-21. Is the reaction exothermic or
   endothermic?

 e) To produce the maximum amount of CO and
   H2 at equilibrium, should the pressure of the
   system be increased or decreased?
                                                    Equilibrium

				
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