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					CHAPTER 14: CHEMICAL
     EQUILIBRIUM

     Vanessa N. Prasad-Permaul
     Valencia Community College
              CHM 1046




                                  1
                                           Introduction


1. How far does a reaction proceed toward completion
   before it reaches a state of chemical equilibrium?

2. Chemical equilibrium
    a) The state reached when the concentrations of
       reactants and products remain constant over time
    b) A state in which the concentration of reactants
       and products no longer change (net)

3. Equilibrium mixture
    A mixture of reactants and products in the
    equilibrium state

                                                          2
                                           Introduction



4.   What are we interested in?

     a) What is the relationship between the
       concentration of reactants and products in an
       equilibrium mixture?

     b) How can we determine equilibrium
       concentrations from initial concentrations?

     c) What factors can be exploited to alter the
       composition of an equilibrium mixture?

                                                          3
                                The Equilibrium State


 In previous chapters we have generally assumed
  that chemical reactions result in complete
  conversion of reactants to products
      Many reactions do not go to completion!!
Example1:




                                                        4
The Equilibrium State




                        5
                                    The Equilibrium State



The two experiments demonstrate that the
  interconversion of N2O4 and NO2 is reversible and
  that the same equilibrium state is reached starting
  from either substance.
  1. This is why we use a  instead of 

  2. Since both NO2 and N2O5 are products and
  reactants we will call the chemical on the left
  reactants and on the right products.

  3. All chemical reactions are reversible

                                                            6
                                     The Equilibrium State


4.   We call a reaction irreversible when it proceed
      nearly to completion
       a. Equilibrium mixture contains almost all
             products and almost no reactants
       b. Reverse reaction is too slow to be detected

5.   In an equilibrium state the reaction does not stop
     at particular concentrations of reactants and
     products, the rates of the forward and reverse
     reactions become equal.

          Important: reaction does not stop

                                                             7
                                  The Equilibrium State

6. Chemical equilibrium is a dynamic state in which
  forward and reverse reactions continue at equal rates
  so that there is no net conversion of reactants to
  products




                                                          8
EXAMPLE 14.1:

               CO(g) + 3H2(g)          CH4(g) + H2O(g)

When 1.000 mol CO & 3.000 mol H2 are placed in a 10.00 L vessel
@ 927oC and allowed to come to equilibrium; the mixture is found to
contain 0.387 mol H2O. What is the molar composition of the
equilibrium mixture? (How many moles of each substance are
present?)
              CO(g) + 3H2(g)           CH4(g) + H2O(g)
Starting      1.000       3.000         0         0
Change          -x         -3x         +x        +x
Equilibrium   1.000 - x   3.000 - 3x    x       x = 0.387

CO = 1.000 – x = 1.000 – 0.387 = 0.613 mol
H2 = 3.000 – x = 3.000 – 3(0.387) = 1.839 mol
CH4 = x = 0.387 mol
EXERCISE 14.1:

             CO(g) + H2O(g)           CO2(g) + H2(g)

Suppose there is a mixture containing 1.00 mol CO and 1.00 mol
H2O. When equilibrium is reached at 1000oC, the mixture contains
0.43 mol H2. What is the molar composition of the equilibrium
mixture?
                           The Equilibrium Constant, Kc



a A + bB  cC + dD

 Kc = [C]c [D]d
     [A]a [B]b

 If we write the equation in the reverse direction

 cC + dD  aA + bB

 K’c = [A]a [B]b = 1
      [C]c [D]d kc

                                                          11
                            The Equilibrium Constant, Kc


General equation:     aA + bB  cC + dD

Equilibrium equation: Kc = [C]c [D]d     products
                           [A]a [B]b    reactants

  The substances in the equilibrium equation must
  be gases or molecules and ions in solution:
           NO SOLIDS! NO PURE LIQUIDS!

  Kc units are omitted but you must say at what
  temperature!
                                                           12
EXAMPLE 14.2:

Write the equilibrium-constant expression Kc for catalytic methylation

              CO(g) + 3H2(g)           CH4(g) + H2O(g)

                          Kc = [CH4][H2O]
                                [CO][H2]3

Write the equilibrium-constant expression Kc for the reverse reaction

            CH4(g) + H2O(g)               CO(g) + 3H2(g)

                          Kc =    [CO][H2]3
                                 [CH4][H2O]
EXAMPLE 14.2:

Write the equilibrium-constant expression Kc the synthesis of NH3

                   N2(g) + 3H2(g)             2NH3(g)

                           Kc =      [NH3]2
                                    [N2][H2]3

Write the equilibrium-constant expression Kc for the following rxn.

                 1/2N2(g) + 3/2H2(g)              NH3(g)

                           Kc =        [NH3]
                                    [N2]1/2 [H2]3/2
EXERCISE 14.2:

Write the equilibrium-constant expression Kc for the following
reaction:

           2NO2(g) + 7H2(g)            2NH3(g) + 4H2O(g)


Write the equilibrium-constant expression Kc for the following
reaction:

           NO2(g) + 7/2H2(g)            NH3(g) + 2H2O(g)
                             The Equilibrium Constant Kp


Kp = equilibrium constant with respect to partial
  pressures of reactants and products
               a A + bB  cC + dD

  Kp = (PC)c (PD)d
      (PA)a (PB)b

          Relationship between Kc and Kp
                     Kp = Kc(RT)n


                                                           16
                             Heterogeneous Equilibria


Introduction

  1. So far we have been talking about
     homogeneous equilibria, in which all reactants
     and products are in a single phase (gas or
     solution)

  2. Heterogeneous equilibria are those in which
     reactants and products are present in more
     than one phase

                                                        17
                           Using the Equilibrium Constant


Introduction

  Knowing the value of the equilibrium constant for a
  chemical reaction lets us:

  1. Judge the extent of the reaction

  2. Predict the direction of the reaction

  3. Calculate the equilibrium concentrations from
     any initial concentrations
                                                            18
                                   Using the Equilibrium Constant


The numerical value of the equilibrium constant for a reaction
    indicates the extent to which reactants are converted to
    products

1.   Large value for Kc > 103 reaction proceeds essentially to 100%
     (mostly products)

2.   Small value for Kc < 10-3 reaction proceeds hardly at all before
     equilibrium is reached (mostly reactants)

3.   If a reaction has an intermediate value of Kc = 103 to 10-3
         a. Appreciable concentrations of both reactants and
         products are present in the equilibrium mixture
                                                                        19
EXAMPLE 14.3:

What is the value of Kc for the decomposition of HI at room temp.?

               2HI(g)          H2(g) + I2(g)                   CONC
Starting        0.800 M        0          0              4.00mol/5.00L = 0.800M
Change          -2x            +x        +x            0.442mol/5.00L = 0.0884M
Equilibrium   4.00 mol – 2x     x         x       (0.800 – 2(0.0884))M = 0.623M


                              Kc =    [H2] [I2]
                                        [HI]2

                                 = [0.0884M][0.0884M]
                                         [0.623M]2

                                 = 0.0201
EXERCISE 14.3:

Hydrogen sulfide, a colorless gas with a foul odor, dissociates on
heating:

                    2H2S(g)          2H2(g) + S2(g)

When 0.100 mol H2S was put into a 10.0L vessel and heated to
1132oC, the equilibrium mixture contained 0.0285 mol H2. What is the
value of Kc at this temperature?
EXAMPLE 14.4:

Quicklime (calcium oxide, CaO), is prepared by heating a source of
calcium carbonate CaCO3 such as limestone or seashells.

               CaCO3(s)               CaO(s) + CO2(g)

Write the expression for Kc.
Kc = [CO2]

An equilibrium constant can also be written for a physical equilibrium.

                    H2O(l)           H2O(g)

Write the expression for Kc for the vaporization of water.
Kc = [H2O(g)]
EXERCISE 14.4:

The Mond process for purifying nickel involves the formation of
nickel tetracarbonyl Ni(CO)4, a volatile liquid, from nickel metal and
carbon monoxide. Carbon monoxide is passed over impure nickel to
form nickel carbonyl vapor, which, when heated, decomposes and
deposits pure nickel.

               Ni(s) + 4CO(g)            Ni(CO)4(g)

Write the expression for Kc for this reaction
                          Predicting the direction of Reaction


Reaction Quotient = Qc
  1. Not necessarily equilibrium concentrations, at some time,
     t, snapshot of reaction

  2. As time passes, Qc changes toward the value of Kc

  3. When the equilibrium state is reached Qc = Kc

  4. Qc allows us to predict the direction of reaction by
      comparing the values of Kc and Qc
             a) If Qc< Kc, net reaction goes from left to
               right, (reactant to products)
             b) If Qc > Kc, net reaction goes from right to
                         left, (products to reactants)
             c) If Qc = Kc, no net reaction occurs
                                                                 24
EXAMPLE 14.5:

A 50.0 L reaction vessel contains 1.00 mol N2, 3.00 mol H2, and 0.500
mol NH3. Will more ammonia be formed or will it dissociate when a
mixture goes to equilibrium @ 400oC? Kc is 0.500 @ 400oC.

               N2(g) + 3H2(g)            2NH3(g)

Qc =       [NH3]2
          [N2][H2]3

     =         [0.500mol/50.0L]2                   =        [0.0100]2
         [1.00mol/50.0L][3.00mol/50.0L]3               [0.0200][0.0600]3

Qc = 23.1 which is greater than Kc = 0.500
 the reaction will go to the left or ammonia will dissociate.
EXAMPLE 14.5:

A 10.0 L reaction vessel contains 0.0015 mol CO2 and 0.10 mol CO.
If a small amount of carbon is added to this vessel and the temperature
is raised to 1000oC, will more CO form?

               CO2(g) + C(s)            2CO(g)


The value of Kc is 1.17 at 1000oC
EXAMPLE 14.6:
A gaseous mixture contains 0.30 mol CO, 0.10 mol H2 and 0.020 mol
H2O plus an unknown amount of CH4, in each liter. The mixture is at
equilibrium @ 1200K. What is the concentration of CH4 in this
mixture? Kc = 3.92

              CO(g) + 3H2(g)             CH4(g) + H2O(g)

Kc   =     [CH4][H2O]
            [CO][H2]3

3.92 =        [CH4][0.020mol/1.0L]         = [CH4][0.020M]
         [0.30mol/1.0L][0.10mol/1.0L]3       [0.30M][0.10M]3

3.92[3.00 x 10-4M] = [CH4] = 0.059M
     [0.020M]
EXAMPLE 14.6:
Phosphorus pentachloride gives an equilibrium mixture of PCl5, PCl3,
and Cl2 when heated.

                    PCl5(g)          PCl3(g) + Cl2(s)

A 1.00L vessel contains an unknown amount of PCl5 and 0.020 mol of
each PCl3 and Cl2 at equilibrium at 250oC.How many moles of PCl5
are in the vessel if Kc for this reaction is 0.0415 @ 250oC?
  Altering an Equilibrium Mixture: Changes in Pressure and

                                                    Volume

In general Le Chatelier’s Principle predicts that:

  1. An increase in pressure by reducing the
     volume will bring about net reaction in
     the direction that decreases the number
     of moles of gas

  2. A decrease in pressure by enlarging the
     volume will bring about net reaction in
     the direction that increases the number
     of moles of gas.
                                                             29
      Factors that Alter the Composition of an Equilibrium Mixture

Introduction
   One of the principal goals of chemical synthesis is to maximize
   the conversion of reactants to products while minimizing the
   expenditure of energy.
       1. Can be achieved if the reaction goes nearly to
          completion at mild temperatures and pressures.
       2. If the equilibrium mixture is high in reactants and poor
           in products, the experimental conditions must be
          changed.
       3. Several factors can be exploited to alter the composition
          of an equilibrium mixture.

              A.     The concentration of reactants or products
              B.     The pressure and volume
              C.     The temperature

                                                                 30
                                         Le Chatelier’s Principle


Le Chatelier’s Principle
   If a stress is applied to a reaction mixture at equilibrium, net
   reaction occurs in the direction that relieves the stress

   1. Stress means a change in the concentration, pressure,
       volume, or temperature that disturbs the original
       equilibrium


  2. Reaction then occurs to change the composition of
       the mixture until a new state of equilibrium is reached


   3. The direction that the reaction takes (reactants to
       products or products to reactants) is the one that
       reduces the stress
                                                                      31
 Altering an Equilibrium Mixture:   Changes in Concentration

In general, when an equilibrium is disturbed by the
   addition or removal of any reactant or product, Le
   Chatelier’s principle predicts that:

  1. The concentration stress of an added reactant or
     product is relieved by net reaction in the direction
     that consumes the added substance

  2. The concentration stress of a removed reactant or
     product is relieved by net reaction in the direction
     that replenishes the removed substance

                                                            32
EXAMPLE 14.6:

Predict the direction of reaction when H2 is removed from a mixture
(lowering the concentration) in which the following equilibrium has
been established:

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

When H2 is removed from the reaction mixture, HI will dissociate to
partially restore the H2 that was removed

 The reaction will go to the left
EXAMPLE 14.6:

Consider each of the following equilibria, which are disturbed as
indicated. Predict the direction of the reaction:

The following equilibria is disturbed by increasing pressure of carbon
dioxide:
                CaCO3(s)             CaO(s) + CO2(g)


The following equilibrium is disturbed by increasing the concentration
of hydrogen:
          2Fe(s) + 3H2O(g)            Fe2O3(s) + 3H2(g)
EXAMPLE 14.7:

Will the products increase, decrease or have no effect if the pressure is
increased:
                 CO(g) + Cl2(g)            COCl2(g)
Reaction decreases the number of molecules of gas  an increase in pressure
increases the amount of product and pushes the reaction to the right.

                      2H2S(g)               2H2(g) + S2(g)
Reaction increases the number of molecules of gas  an increase of pressure
decreases the amount of products and the reaction shifts to the left.

                         C(s)   +   S2(g)         CS2(g)
Reaction does not change the number of molecules. Only look at gas volumes when
decide the effect of pressure change on equilibrium composition. Pressure change
has no effect.
EXERCISE 14.7:

Can the amount of product be increased in each of the following
reactions by increasing the pressure? explain:

                 CO2(g) + H2(g)              CO(g) + H2O(g)


                 4CuO(s)               2Cu2O(s) + O2(g)


                 2SO2(g)   +   O2(g)           2SO3(g)
       Altering the Equilibrium Mixture: Changes in Temperature


In general, the temperature dependence of the equilibrium
   constant depends on the sign of H for the reaction

  1.      The equilibrium constant for an exothermic reaction
         (negative H) decreases as the temperature increases

  2.      The equilibrium constant for an endothermic reaction
         (positive H) increases as the temperature increases.

  3.     H = standard enthalpy of reaction, enthalpy change
         measured under standard conditions

  4. Standard conditions = most stable form of a substance at 1
     atm pressure and at a specified temperature, usually 25C;
  1 M concentration for all substances

                                                                  37
     Altering the Equilibrium Mixture: Changes in Temperature



Le Chatelier’s Principle says that if heat is added to an equilibrium
   mixture (increasing the temperature) net reaction occurs in the
   direction that relieves the stress of the added heat.

  1. For an endothermic reaction heat is absorbed by reaction
  in the forward direction. The equilibrium shifts to the right at
     the higher temperatures, Kc increases with increasing
     temperature

  2. For an exothermic heat is absorbed by net reaction in the
      reverse direction, so Kc decreases with temperature, and
  the reaction would flow to the left (reactants)


                                                                        38
EXAMPLE 14.8:

Carbon monoxide is formed when carbon dioxide reacts with solid
carbon:
          CO2(g) + C(s)          2CO(g) ; Ho = 172.5kJ

Is a high or low temperature more favorable to the formation of carbon
monoxide?

Increase products need to shift equilibrium to the right
Endothermic
The temperature needs to be raised
High temperature is more favorable to the formation of carbon
monoxide.
EXAMPLE 14.8:

Carbon monoxide is formed when carbon dioxide reacts with solid
carbon:

           CO2(g) + H2(g)            CO(g) + H2O(g)

Is a high or a low temperature more favorable to the production of
carbon monoxide? Explain.
                        The Effect of a Catalyst on Equilibrium


A catalyst increases the rate of a chemical reaction by making
   available a new, lower-energy pathway for conversion of
   reactants to products.

  1. Since the forward and reverse reaction pass through the
  same       transition state, a catalyst lowers the activation
  energy for both

  2. The rates of the forward and reverse reactions increase by
  the same factor

  3. Catalyst accelerates the rate at which equilibrium is reached

  4. Catalyst does not affect the composition of the equilibrium
     mixture

                                                                   41
The Effect of a Catalyst on Equilibrium




                                          42
  The Link Between Chemical Equilibrium and Chemical Kinetics


                        A+BC+D

Assuming that the forward and reverse reactions occur in a single
  bimolecular step, elementary reactions, we can write the
  following rate laws

               Rate of forward reaction = kf [A] [B]
               Rate of reverse reaction = kr [C] [D]
When t=0 [C] = [D] = 0
As A and B are converted to C and D the rate of the forward
  reaction decreases and the rate of the reverse reaction is
  increasing, until they are equal, chemical equilibrium

                       kf [A] [B] = kr [C] [D]

                                                                    43
The Link Between Chemical Equilibrium and Chemical Kinetics



                         kf = [C] [D]
                         kr    [A] [B]
  The right side of this equation is the equilibrium constant
  expression for the forward reaction, which equals the
  equilibrium constant Kc

                        Kc = [C] [D]
                             [A] [B]

  Therefore the equilibrium constant is simply the ratio of
  the rate constants for the forward and reverse reactions:

                           Kc = kf
                                kr

                                                                44
                                               Example 1:


Which of the following is correct?

1.   Some reactions are truly not reversible

2.   All reactants go to all products in all reactions

3.   All reactions are reversible to some extent

4.   The rates of the forward and reverse reactions
     will never be equal



                                                            45
                                       Example 2:



Write the equilibrium equation for each of the
 following reaction:

 a)   N2(g) + 3 H2(g)  2 NH3(g)


 b) 2 NH3(g)  N2(g) + 3 H2(g)




                                                    46
                                                    Example 3:



 The oxidation of sulfur dioxide to give sulfur trioxide is
  an important step in the industrial process for
  synthesis of sulfuric acid. Write the equilibrium
  equation for each of the following reactions:

  a)   2 SO2(g) + O2(g)  2 SO3(g)
  b)   2 SO3(g)  2 SO2(g) + O2(g)

The following equilibrium concentrations were measured at 800
  K:
     [SO2] = 3.0 x 10-3 M   [O2] = 3.5 x 10-3 M
     [SO3] = 5.0 x 10-2 M

  Calculate the equilibrium constant at 800 K a and b

                                                                 47
                                            Example 4:



In the industrial synthesis of hydrogen, mixtures of
   CO and H2O are enriched in H2 by allowing the
   CO to react with steam. The chemical equation
   for this so-called water-gas shift reaction is:

          CO(g) + H2O(g)  CO2(g) + H2(g)

What is the value of Kp at 700 K if the partial
 pressures in an equilibrium mixture at 700 K are
 1.31 atm of CO, 10.0 atm of H2O, 6.12 atm of
 CO2, and 20.3 atm H2?


                                                         48
                                       Example 5:


When will kc = kp ?

1. 2 SO2(g) + O2(g)  2 SO3(g)

2.   CO(g) + H2O(g)  CO2(g) + H2(g)

3. N2(g) + 3 H2(g)  2 NH3(g)




                                                    49
                                              Example 6:



Nitric oxide reacts with oxygen to give nitrogen
    dioxide, an important reaction in the Ostwald
    process for the industrial synthesis of nitric
    acid:
             2 NO(g) + O2(g)  2 NO2(g)


a)   If Kc = 6.9 x 105 @ 227C, what is the value of
        Kp @ 227C?

b) If Kp = 1.3 x 10-2 @ 1000 K, what is the value of
       Kc @ 1000 K?

                                                           50
                                             Example 7:



For each of the following reactions, write the
  equilibrium constant expression for        Kc

  a) 2 Fe(s) + 3 H2O(g)  Fe2O3(s) + 3 H2(g)

  b) 2 H2O(l)  2 H2(g) + O2(g)

  c) SiCl4(g) + 2 H2(g)  Si(s) + 4 HCl(g)

  d) Hg22+(aq) + 2 Cl-(aq)  Hg2Cl2(s)

                                                          51
                                           Example 8:


Which of the following has a Heterogeneous
   equilibria?

1. 2 SO2(g) + O2(g)  2 SO3(g)

2.   CO(g) + H2O(g)  CO2(g) + H2(g)

3. SiCl4(g) + 2 H2(g)  Si(s) + 4 HCl(g)




                                                        52
                                                      Example 9:


The equilibrium constant for the reaction

                 2 NO(g) + O2(g)  2 NO2(g)

  is 6.9 x 105 @ 500 K. A 5.0 L reaction vessel at this
  temperature was filled with 0.060 mol of NO, 1.0 mol
  O2, and 0.80 mol NO2.

  a) Is the reaction mixture at equilibrium? If not, in which
  direction does the net reaction proceed?

  b) What is the direction of the net reaction if the initial
  amounts are 5.0 x 10-3 mol of NO, 0.20 mol of O2 and 4.0 mol of
  NO2?
                                                                    53
                                                    Example 10:


Consider the equilibrium for the water-gas shift reaction:

                CO(g) + H2O(g)  CO2(g) + H2(g)

     Use Le Chatelier’s principle to predict how the
     concentration of H2 will change and what direction the
     reaction will flow when the equilibrium is disturbed by:

1.    Adding CO
2.    Adding CO2
3.    Removing H2O
4.    Removing CO2



                                                                  54
                                              Example 11:



In the following reaction, if I take away CO, which
     direction will the reaction proceed to equilibrium?

            CO2(g) + H2(g) CO(g) + H2O(g)

1. Products 
2. Reactants 




                                                            55
                                           Example 12:


Which direction will the reaction flow if the
 following equilibria is subjected to an increase in
 pressure by decreasing the volume?

1.   CO(g) + H2O(g)  CO2(g) + H2(g)


2.   2 CO(g)  C(s) + CO2(g)


3.   N2O4(g)  2 NO2(g)


                                                         56
                                             Example 13:


If I increase the pressure by decreasing the volume,
       which direction will the reaction flow to reach
       equilibrium?

              C(s) + CO2(g)  2 CO(g)

1. Products 
2. Reactants 




                                                           57
                                         Example 14:



When air is heated at very high temperatures in an
 automobile engine, the air pollutant nitric oxide
 is produced by the reaction

    N2(g) + O2(g)  2 NO(g)     H = 180.5 kJ

  1. How does the equilibrium amount of NO
     vary with an increase in temperature?

  2. What direction is the net reaction flowing?

                                                       58
                                                     Example 15:


A platinum catalyst is used in automobile catalytic converters to
     hasten the oxidation of carbon monoxide:
           2 CO(g) + O2(g)  2 CO2(g)     H = -566 kJ

     Suppose that you have a reaction vessel containing an
     equilibrium mixture. Will the amount of CO increase,
     decrease, or remain the same when:

1.   A platinum catalyst is added
2.   The temperature is increased
3.   The pressure is increased by decreasing the volume
4.   The pressure is increased by adding argon gas
5.   The pressure is increased by adding O2 gas

                                                                    59
                                                                 Example 16:


Nitric oxide emitted from the engines of supersonic transport planes can
       contribute to the destruction of stratospheric ozone:

                          NO(g) + O3(g)  NO2(g) + O2(g)

     This reaction is highly exothermic (E = -200 kJ), and its equilibrium
     constant Kc is 3.4 x 1034 at 300 K


1.   Which rate constant is larger, kf or kr?


2.   The value of kf at 300 K is 8.5 x 106 M-1 s-1. What is the value of kr at the
     same temperature?


3.   A typical temperature in the stratosphere is 230 K. Do the values of kf,
     kr, and Kc increase or decrease when the temperature is lowered from
     300 K to 230 K?
                                                                                     60
                                           Example 17:



If I increase the temperature of reaction which
      way will the reaction flow to equilibrium?

     NO2(g) + O2(g)  NO(g) + O3(g)   H = 200 kJ



1. Products 
2. Reactants 




                                                         61

				
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