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					                   Chemical Equilibrium

                           Concept of
                        Dynamic Equilibrium


   Heterogeneous                                Homogeneous
                        Equilibrium Constant
    Equilibrium                                  Equilibrium



                            Applications



                           Le Chatelier’s
                             Principle



     Temperature        Reactants/Products     Pressure/Volume
11/25/2011
      The Concept of Equilibrium

• Consider colorless frozen N2O4. At room temperature, it
  decomposes to brown NO2:
                   N2O4(g)  2NO2(g).
• At some time, the color stops changing and we have a
  mixture of N2O4 and NO2.
• Chemical equilibrium is the point at which the
  concentrations of all species are constant.
The Concept of Equilibrium
      The Concept of Equilibrium

• At equilibrium, as much N2O4 reacts to form NO2 as NO2
  reacts to re-form N2O4:
                   N2O4(g)     2NO2(g)
• The double arrow implies the process is dynamic.
• Consider
          Forward reaction: A  B Rate = kf[A]
           Reverse reaction: B  A Rate = kr[B]
• At equilibrium kf[A] = kr[B].
                                           NO2 – equilibrium
The Concept of Equilibrium
        The Equilibrium Constant

• For a general reaction in the gas phase
                 aA + bB           cC + dD
  the equilibrium constant expression is
                                c d
                               PC PD
                       Keq 
                                a b
                               PA PB

  where Keq is the equilibrium constant.
        The Equilibrium Constant

• For a general reaction
                 aA + bB       cC + dD
  the equilibrium constant expression for everything in
  solution is

                      K eq 
                             Cc Dd
                             A a Bb
  where Keq is the equilibrium constant.
The Equilibrium Constant
    N2O4(g)        2NO2(g)




                   2
                 P
        K eq 
                   NO2

                 PN 2O 4
The Equilibrium Constant
 The Magnitude of Equilibrium Constants
        The Equilibrium Constant
    The Direction of the Chemical Equation and Keq

• In the reverse direction:
                  2NO2(g)       N2O4(g)

                      P                 1
                      N 2O 4
               K eq          0.155 
                      P2               6.46
                       NO 2
         The Equilibrium Constant
                 Other Ways to Manipulate Chemical
                          Equations and Keq Values
• The reaction
                 2N2O4(g)          4NO2(g)
   has
                              P4
                             NO 2
                      Keq 
                            P2
                             N 2O 4

   which is the square of the equilibrium constant for
                  N2O4(g)         2NO2(g)
        The Equilibrium Constant
                   Other Ways to Manipulate Chemical
                               Equations and Keq Values
• Equilibrium constant for the reverse direction is the
  inverse of that for the forward direction.
• When a reaction is multiplied by a number, the
  equilibrium constant is raised to that power.
• The equilibrium constant for a reaction which is the sum
  of other reactions is the product of the equilibrium
  constants for the individual reactions.
       The Equilibrium Constant
Given at 700K:
(i) H2(g) + I2(g)  2 HI(g)                    Keq = 54.0
(ii) N2(g) + 3 H2(g)  2 NH3(g)                Keq’ = 1.04x10-4


Determine Keq’’ at 700K for:
2 NH3(g) + 3 I2(g)  6 HI(g) + N2(g)          Answer: 1.51x109



Unit(s) of the Equilibrium Constant
Dimensionless because:
    Partial Pressures divided by Reference Pressure of 1 atm.
    Molarities divided by Reference Concentration of 1 Molar.
Given at 700K:
(i) H2(g) + I2(g)  2 HI(g)            Keq = 54.0
(ii) N2(g) + 3 H2(g)  2 NH3(g)        Keq’ = 1.04x10-4
Determine Keq’’ at 700K for:
2 NH3(g) + 3 I2(g)  6 HI(g) + N2(g)




                                               Answer: 1.51x109
         Heterogeneous Equilibria

• When all reactants and products are in one phase, the
  equilibrium is homogeneous.
• If one or more reactants or products are in a different
  phase, the equilibrium is heterogeneous.
• Consider:
             CaCO3(s)        CaO(s) + CO2(g)
   – experimentally, the amount of CO2 does not seem to depend on
     the amounts of CaO and CaCO3. Why?
         Heterogeneous Equilibria

• The concentration of a solid or pure liquid is its density
  divided by molar mass.
• Neither density nor molar mass is a variable, the
  concentrations of solids and pure liquids are constant.
• For the decomposition of CaCO3:
                [CaO]
        K eq            [CO 2 ]  constant  [CO 2 ]
               [CaCO3 ]
        Heterogeneous Equilibria

• We ignore the concentrations of pure liquids and pure
  solids in equilibrium constant expressions.
• The amount of CO2 formed will not depend greatly on
  the amounts of CaO and CaCO3 present.

• Examples:
              CO 2 ( g )  H 2 ( g )  CO(g)  H 2 O(l)


              Sn(s)  2 H  (aq)  Sn 2 (aq)  H 2 (g)
    Calculating Equilibrium Constants

• Proceed as follows:
   – Tabulate initial and equilibrium concentrations (or partial
     pressures) given.
   – If an initial and equilibrium concentration is given for a species,
     calculate the change in concentration.
   – Use stoichiometry on the change in concentration line only to
     calculate the changes in concentration of all species.
   – Deduce the equilibrium concentrations of all species.
• Usually, the initial concentration of products is zero.
  (This is not always the case.)
  Calculating Equilibrium Constants

1. A mixture of H2(g) and N2(g) is allowed to attain equilibrium at 472oC. At
   equilibrium, the partial pressures of the gases were determined to be 7.38
   atm H2, 2.46 atm N2 and 0.166 atm NH3. Calculate Keq .


2. Enough ammonia is dissolved in 5.00 L of water at 25oC to produce a
   solution that is 0.0124 M in ammonia. The solution is then allowed to come
   to equilibrium. Analysis of the equilibrium mixture shows that the
   concentration of OH- is 4.64x10-4 M. Calculate Keq at 25oC for the reaction.


3. Gaseous sulfur trioxide decomposes to gases of sulfur dioxide and oxygen
   at high temperature in a sealed container. 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 Keq at 1000 K.
A mixture of H2(g) and N2(g) is allowed to attain equilibrium at 472oC. At equilibrium,
   the partial pressures of the gases were determined to be 7.38 atm H2, 2.46 atm N2
   and 0.166 atm NH3. Calculate Keq .




                                                               Answer: 2.79x10-5
Enough ammonia is dissolved in 5.00 L of water at 25oC to produce a solution that is 0.0124 M in
ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture
shows that the concentration of OH- is 4.64x10-4 M. Calculate Keq at 25oC for the reaction.




                                                                       Answer: 1.80x10-5
Gaseous sulfur trioxide decomposes to gases of sulfur dioxide and oxygen at high temperature in a
sealed container. 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 Keq at 1000 K.




                                                                          Answer: 0.338
       Applications of Equilibrium Constants

                 Predicting the Direction of Reaction
• We define Q, the reaction quotient, for a general reaction
                aA + bB          cC + dD
  as
                                c d
                               PC PD
                        Q
                                a b
                               PA PB

• Q = K only at equilibrium.
    Applications of Equilibrium Constants

                 Predicting the Direction of Reaction
• If Q > K then the reverse reaction must occur to reach
  equilibrium (i.e., products are consumed, reactants are
  formed, the numerator in the equilibrium constant
  expression decreases and Q decreases until it equals K).
• If Q < K then the forward reaction must occur to reach
  equilibrium.
     Applications of Equilibrium Constants

                     Calculating Equilibrium Constants
•   The same steps used to calculate equilibrium constants
    are used.
•   Generally, we do not have a number for the change in
    concentration line.
•   Therefore, we need to assume that x mol/L of a species is
    produced (or used).
•   The equilibrium concentrations are given as algebraic
    expressions.
    Applications of Equilibrium Constants

                                            Predicting the Direction of Reaction



1. At 1000 K the value of Keq for the reaction 2SO3(g)  2SO2(g) + O2(g) is 0.338 .
   Calculate the value for Q, and predict the direction in which the reaction will
   proceed toward equilibrium if the initial partial pressures are 0.16 atm for SO3,
   0.41 atm for SO2 , and 2.5 atm for O2 .


2. At 448oC, Keq is 51 for: H2(g) + I2(g)  2HI(g) . Predict how the reaction will
   proceed to reach equilibrium at 448oC if we start with 2.0x10-2 mol HI , 1.0x10-2
   mol H2 , and 3.0x10-2 mol I2 in a 2.00-L container.
   [Hint: Calculate partial pressures using Ideal Gas Law. Calculate Q.]
   [Answer: Q=1.33 Reaction proceeds from left to right (i.e. more products)]
    Applications of Equilibrium Constants

                                        Calculating Equilibrium Concentrations


1. At 500 K the reaction: PCl5(g)  PCl3(g) + Cl2(g) , has Keq = 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?


2. 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 Keq for the reaction: H2(g) + I2(g)  2HI(g) , at 448oC is 50.5.
   What are the partial pressures of H2 , I2 , and HI in the flask at equilibrium?
   [Hint: Solve for initial partial pressures using Ideal Gas law. Set up
   initial/change/equil’m table and define x as change in partial pressures of
   reactants. Solve for x in Keq expression using quadratic formulation.]
           [Answer: x = 55.3 atm; H2 = 3.90 atm, I2 = 63.1 atm, HI = 111 atm.]
2.   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 Keq for the reaction:
     H2(g) + I2(g)  2HI(g) , at 448oC is 50.5. What are the partial pressures of H2 , I2 , and HI in the flask at equilibrium?
     [Hint: Solve for initial partial pressures using Ideal Gas law. Set up initial/change/equil’m table and define x as change in
     partial pressures of reactants. Solve for x in Keq expression using quadratic formulation.]
               [Answer: x = 55.3 atm; H2 = 3.90 atm, I2 = 63.1 atm, HI = 111 atm.]
         Le Châtelier’s Principle
                      Consider the production of ammonia
              N2(g) + 3H2(g)         2NH3(g)
• As the pressure increases, the amount of ammonia
  present at equilibrium increases.
• As the temperature decreases, the amount of ammonia at
  equilibrium increases.
• Can this be predicted?
Le Châtelier’s Principle: if a system at equilibrium is
  disturbed, the system will move in such a way as to
  counteract the disturbance.
          Le Châtelier’s Principle
                      Change in Reactant or Product
                                    Concentrations
• Consider the Haber process
              N2(g) + 3H2(g)          2NH3(g)
• If H2 is added while the system is at equilibrium, the
  system must respond to counteract the added H2 (by Le
  Châtelier).
• The system must consume the H2 and produce products
  until a new equilibrium is established.
• So, [H2] and [N2] will decrease and [NH3] increases.
           Le Châtelier’s Principle
                          Change in Reactant or Product
                                             Concentrations
•   Adding a reactant or product shifts the equilibrium away
    from the increase.
•   Removing a reactant or product shifts the equilibrium
    towards the decrease.
•   To optimize the amount of product at equilibrium, we
    need to flood the reaction vessel with reactant and
    continuously remove product (Le Châtelier).
•   We illustrate the concept with the industrial preparation
    of ammonia.
          Le Châtelier’s Principle
              Effects of Volume and Pressure Changes
•   As volume is decreased pressure increases.
•   Le Châtelier’s Principle: 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.
•   In a reaction with the same number of product and
    reactant moles of gas, pressure has no effect.
         Le Châtelier’s Principle
           Effects of Volume and Pressure Changes
                  N2O4(g)        2NO2(g)
• An increase in pressure (by decreasing the volume)
  favors the formation of colorless N2O4.
• The instant the pressure increases, the system is not at
  equilibrium and the concentration of both gases has
  increased.
• The system moves to reduce the number moles of gas
  (i.e. backward reaction is favored).
         Le Châtelier’s Principle
                      Effect of Temperature Changes

• The equilibrium constant is temperature dependent.
• For an endothermic reaction, H > 0 and heat can be
  considered as a reactant.
• For an exothermic reaction, H < 0 and heat can be
  considered as a product.
            Le Châtelier’s Principle
                          Effect of Temperature Changes

• Adding heat (i.e. heating the vessel) favors away from
  the increase:
   – if H > 0, adding heat favors the forward reaction,
   – if H < 0, adding heat favors the reverse reaction.


• Removing heat (i.e. cooling the vessel), favors towards
  the decrease:
   – if H > 0, cooling favors the reverse reaction,
   – if H < 0, cooling favors the forward reaction.
          Le Châtelier’s Principle
                              Effect of Temperature Changes
• Consider
 Cr(H2O)62+(aq) + 4Cl-(aq)                    CoCl42-(aq) + 6H2O(l)
  for which H > 0.
  – Co(H2O)62+ is pale pink and CoCl42- is blue.
  – If a light purple room temperature equilibrium mixture is placed in a beaker
    of warm water, the mixture turns deep blue.
  – Since H > 0 (endothermic), adding heat favors the forward reaction, i.e. the
    formation of blue CoCl42-.
  – If the room temperature equilibrium mixture is placed in a beaker of ice
    water, the mixture turns bright pink.
  – Since H > 0, removing heat favors the reverse reaction which is the
    formation of pink Co(H2O)62+.
           Le Châtelier’s Principle
                                  The Effect of Catalysis

• A catalyst lowers the activation energy barrier for the
  reaction.
• Therefore, a catalyst will decrease the time taken to reach
  equilibrium.
• A catalyst does not affect the composition of the
  equilibrium mixture.
               Chemical Equilibrium

                       Concept of
                    Dynamic Equilibrium


Heterogeneous                                  Homogeneous
                    Equilibrium Constant
 Equilibrium                                    Equilibrium


         c d
Keq 
        PC PD           Applications
                                           K eq 
                                                  Cc Dd
         a b
        PA PB                                     A a Bb
                       Le Chatelier’s
                         Principle



 Temperature        Reactants/Products        Pressure/Volume

				
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