Chapter 11 equilibrium lecture notes

Chapter 11
              Chemical Equilibrium

11.1 The Equilibrium Condition
11.2 The Equilibrium Constant
11.3 Equilibrium Expressions Involving Pressures
11.4 The Concept of Activity
11.5 Heterogeneous Equilibria
11.6 Applications of the Equilibrium Constant
11.7 Solving Equilibrium Problems
11.8 Le Chatelier's Principle
11.9 Equilibria Involving Real Gases
The Equilibrium Condition (General)

Thermal equilibrium indicates two systems in thermal contact
with each do not exchange energy by heat. If two bricks are in
thermal equilibrium their temperatures are the same.

Chemical equilibrium indicates no unbalanced potentials (or
driving force). A system in equilibrium experiences no change
over time, even infinite time.

The opposite of equilibrium systems are non-equilibrium
systems that are off balance and change with time.
Example 1 atm O2 + 2 atm H2 at 298K
The Equilibrium Condition (Chem Rxn)


aA + bB          cC + dD

The same equilibrium state
is achieved whether
starting with pure reactants
or pure products.
The equilibrium state can
change with temperature.
The Equilibrium State (Chem Rxn)




H2O (g) + CO (g)   H2 (g) + CO2 (g)
                           Change
                           [CO] to PCO
                           [H2O] to PH2O
                           etc
Chemical Reactions and Equilibrium
As the equilibrium state is approached, the
forward and backward rates of reaction
approach equality. At equilibrium the rates
are equal, and no further net change occurs
in the partial pressures of reactants or
products.


Fundamental characteristics of equilibrium states:

1. No macroscopic evidence of change.
2. Reached through spontaneous processes.
3. Show a dynamic balance of forward and backward processes.
4. Same regardless of the direction from which they are approached.
5. No change over time.
Arrows: Chemical Symbolism




           Use this in an
           equilibrium expression.

↔          Use this to indicate
           resonance.
Chemical Reactions and Equilibrium
 The equilibrium condition for every reaction can be
 described in a single equation in which a number, the
 equilibrium constant (K) of the reaction, equals an
 equilibrium expression, a function of properties of the
 reactants and products.

H2O(l)    H2O(g) @ 25oC   Temperature (oC)   Vapor Pressure (atm)
                                 15.0             0.01683
                                 17.0             0.01912
    K = 0.03126                  19.0             0.02168
H2O(l)    H2O(g) @ 30oC          21.0             0.02454
                                 23.0             0.02772
    K = 0.04187                  25.0             0.03126
                                 30.0             0.04187
                                 50.0             0.1217
Law of Mass Action (1)

Partial pressures and concentrations of products appear in the
numerator and those of the reactants in the denominator.
Each is raised to a power equal to its coefficient in the
balanced chemical equation.

         aA + bB                              cC +
if gases dD                     if concentrations
       c        d                     c       d
( PC ) ( PD )                   [C] [D]
                    =K
        a       b                   a    b =K
( PA ) ( PB )                   [ A] [B]
Law of Mass Action (2)
1. Gases enter equilibrium expressions as partial pressures, in
atmospheres. E.g., PCO2
2. Dissolved species enter as concentrations, in molarity (M)
moles per liter. E.g., [Na+]

3. Pure solids and pure liquids are represented in equilibrium
expressions by the number 1 (unity); a solvent taking part in a
chemical reaction is represented by unity, provided that the
solution is dilute. E.g., I2(s) ↔ I2(aq) [I2 (aq) ] = K

     I 2 ( s ) ↔ I 2 ( aq )
        [ I 2 (aq )] [ I 2 ( aq )]
     K=               =            = [ I 2 ( aq )]
         [ I 2 ( s )]      1
Activities

     The concept of Activity (i-th component)

     = ai = Pi / P reference

H2O (l)      H2O (g) Kp = P H2O        @ 25oC Kp = 0.03126 atm

      PH2O
              =K    Pref is numerically equal to 1
      Pref                                        K = 0.03126
    The convention is to express all pressures in atmospheres
    and to omit factors of Pref because their value is unity. An
    equilibrium constant K is a pure number.
The Equilibrium State




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


                                   PH 2 PCO 2
                              Kp =
                                   PH 2 O PCO
The Equilibrium Expressions


              aA + bB                          cC +
              dD
In a chemical reaction in which a moles of species A and b moles of
species B react to form c moles of species C and d moles of species D,


      The partial pressures at equilibrium are related through

                        K = PcCPdD/PaAPbB
Write equilibrium expressions for the
following reactions


    3 H2(g) + SO2(g)      H2S(g) + 2 H2O(g)




 2 C2F5Cl(g) + 4 O2(g) Cl2(g) + 4 CO2(g) + 5 F2(g)
Heterogeneous Equilibrium




Gases and    CaCO3(s)            CaO(s) + CO2(g)
Solids
                              K=PCO2


            K is independent of the amounts
                of CaCO3(s) or CaO(s)
Heterogeneous Equilibrium


Liquids      H2O(l)         H2O(g)
                         K=PH2O

              I2(s)      I2(aq)
Solutions
                          K=[I2]
Relationships Among the K’s of Related Reactions
#1: The equilibrium constant for a reverse reaction is always
the reciprocal of the equilibrium constant for the
corresponding forward reaction.

                        1
             K for   =             or K for K rev = 1
                       K rev
aA + bB                  cC + dD     versus   cC + dD           aA + bB

                                         (PH2O)2
#1   2 H2 (g) + O2 (g)     2 H2O (g)
                                        (PH2)2(PO2) = K1
                                                             K1 = 1/K2
                     2 H2 (g) + O2 (g) (PH2) (PO22)
                                              2
#2   2 H2O (g)                                        = K2
                                         (PH2O)
Relationships Among the K’s of Related Reactions
     # 2: When the coefficients in a balanced chemical
     equation are all multiplied by a constant factor, the
     corresponding equilibrium constant is raised to a power
     equal to that factor.

                                                     (PH2O)2
#1      2 H2 (g) + O2 (g)     2 H2O (g) Rxn 1       (PH2)2(PO2) = K1
#3       H2 (g) + ½ O2 (g)     H2O (g)   Rxn 3 = Rxn 1 times 1/2


           (PH2O)
                             = K3        K3 = K1½
           (PH2)(PO2)½
Relationships Among the K’s of Related Reactions
   # 3: when chemical equations are added to give a new
   equation, their equilibrium constants are multiplied to
   give the equilibrium constant associated with the new
   equation.
   2 BrCl (g) ↔ Br2 (g) + Cl2 (g)      (PBr2)(PCl2)
                                                      = K1 = 0.45 @ 25oC
              wrong arrow
                                         (PBrCl)2
   Br2 (g) + I2 (g) ↔ 2 IBr (g)          (PIBr)2
                     wrong arrow                    = K2 = 0.051 @ 25oC
                                       (PBr2) (PI2)

2 BrCl (g) + I2 (g) ↔ 2 IBr (g) + Cl2(g)
                      wrong arrow
                                                              = K1K2
                                                         = (0.45)(0.051)
     (PBr2)(PCl2)           (PIBr)2
                        X (P ) (P ) = K1K2 = K3
       (PBrCl)2             Br2     I2                   =0.023 @ 25oC
Calculating Equilibrium Constants
Consider the equilibrium
                            4 NO2(g)↔ 2 N2O(g) + 3 O2(g)
The three gases are introduced into a container at partial pressures of 3.6
atm (for NO2), 5.1 atm (for N2O), and 8.0 atm (for O2) and react to reach
equilibrium at a fixed temperature. The equilibrium partial pressure of the
NO2 is measured to be 2.4 atm. Calculate the equilibrium constant of the
reaction at this temperature, assuming that no competing reactions occur.


                                      4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)
 initial partial pressure (atm)
 change in partial pressure (atm)
 equilibrium partial pressure (atm)
Calculate the equilibrium constant of the reaction at this temperature,
  assuming that no competing reactions occur.
                                       4 NO2(g) ↔ 2 N2O(g) + 3 O2(g)
  initial partial pressure (atm)         3.6        5.1       8.0
  change in partial pressure (atm)       – 4x       +2x      +3x
  equilibrium partial pressure (atm)    2.4      5.1 + 2x   8.0 + 3x

                                5.1 + 2(0.3 atm) = 5.7 atm N2O
3.6 – 4x = 2.4 atm NO2;
                    x = 0.3 atm 8.0 + 3(0.3 atm) = 8.9 atm O2

         (PN2O)2(PO2)3
      K=                           =
           (PNO2)4
The compound GeWO4(g) forms at high temperature in the reaction
                  2 GeO (g) + W2O6(g) ↔ 2 GeWO4(g)
Some GeO (g) and W2O6 (g) are mixed. Before they start to react,
their partial pressures both equal 1.000 atm. After their reaction at
constant temperature and volume, the equilibrium partial pressure of
GeWO4(g) is 0.980 atm. Assuming that this is the only reaction that
takes place, (a) determine the equilibrium partial pressures of
GeO and W2O6, and (b) determine the equilibrium constant for
the reaction.


                                    2 GeO (g) + W2O6 (g) ↔ 2 GeWO4(g)
 initial partial pressure (atm)        1.000     1.000        0
 change in partial pressure (atm)       – 2x      –x        +2x
 equilibrium partial pressure (atm)
(a) determine the equilibrium partial pressures of GeO and W2O6, and
(b) determine the equilibrium constant for the reaction.
                                       2 GeO(g) + W2O6(g) ↔ 2 GeWO4(g)
    initial partial pressure (atm)        1.000     1.000       0
    change in partial pressure (atm)       –2x       –x         +2x
    equilibrium partial pressure (atm) 1.000 – 2x   1.000 – x   0.980

   0 + 2x = 0.980 atm GeWO4;             1.000 – 2(0.490) = 0.020 atm GeO
          x = 0.490 atm                  1.000 – 0.490 = 0.510 atm W2O6


              (PGeWO4)2
        K = (P )2(P
              GeO    W2O6 ) =
• Skip Solving quadratic equations
• Will utilize approximation method
  – Systems that have small equilibrium
    constants.
  – Assume “x” (the change in concentration)
    is small (less than 5%) of the initial
    concentration.
A vessel holds pure CO (g) at a pressure of 1.282 atm and a temperature of
354K. A quantity of nickel is added, and the partial pressure of CO (g)
drops to an equilibrium value of 0.709 atm because of the reaction
                           Ni (s) + 4CO (g) ↔ Ni(CO)4 (g)
Compute the equilibrium constant for this reaction at 354K.


        PNi(CO)4           PNi(CO)4
K=        4
                      =
     (PCO ) [Ni(s)]       (PCO ) 4 (1)

                                      Ni (s) + 4CO (g) ↔ Ni(CO)4 (g)
Construct an “ICE” table                   P CO (atm) P Ni(CO)4 (atm)
     initial partial pressure (atm)           1.282         0
     change in partial pressure (atm)         -4x           +1x
     equilibrium partial pressure (atm)       0.709          x

 At equil. Pco=                                       x = PNi(CO)4 =
Equilibrium Calculations
At a particular temperature, K = 2.0 x 10-6 mol/L for the
reaction
                 2CO2 (g)          2CO (g) + O2 (g)
If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate
the equilibrium concentrations of all species.

                                  2CO2 (g)      2CO (g) + O2 (g)

 initial partial pressure (mol/L)     0.4        0          0
 change in partial pressure (mol/L)   – 2x       +2x        +1x
 equilibrium partial pressure (mol/L) 0.4 -2x             2x        1x
At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction
                               2CO2 (g)        2CO (g) + O2 (g)
     If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium
     concentrations of all species.
                                              2CO2 (g)          2CO (g) + O2 (g)

       initial partial pressure (mol/L)     0.4                   0            0
       change in partial pressure (mol/L)   – 2x                  +2x          +1x
       equilibrium partial pressure (mol/L) 0.4 -2x                          2x          1x

     [CO]2 [O 2 ]                (2 x) 2 ( x)
Kp =              = 2.0x10-6 =
      [CO 2 ]2                 (0.40 − 2 x) 2
                                                         check assumption
assume 2x << 0.40
                                                          2x 2(4.3x10-3)
                                                              =              x 100 = 2.2%
                                                         0.40      0.40
           (2x) 2 (x)                                    assumpiton valid, less than 5%
2.0x10-6 =            ⇒ x = 4.3x10-3 M
            (0.40) 2
At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction
                              2CO2 (g)        2CO (g) + O2 (g)
    If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium
    concentrations of all species.
                                             2CO2 (g)         2CO (g) + O2 (g)

      initial partial pressure (mol/L)     0.4                  0            0
      change in partial pressure (mol/L)   – 2x                 +2x          +1x
      equilibrium partial pressure (mol/L) 0.4 -2x                         2x                1x

     [CO]2 [O 2 ]
Kp =              = 2.0x10-6
      [CO 2 ]2                           [CO 2 ] = 0.40 - 2x =
    (2 x) 2 ( x)
=
  (0.40 − 2 x) 2                         = 0.40 - 2(4.3x10-3 ) = 0.39M
x = 4.3x10-3M
                                                                                        -3
                                             [CO] = 2x = 2(4.3 x 10 )
                                  -3
[O 2 ] = x = 4.3 x 10 M                      = 8.6 x 10-3 M
Non-Equilibrium Conditions:
                The Reaction Quotient (1)
                            φορωαρδ
                            →
                    aA + bB ρεϖερσε cC + dD     wrong arrow
                           ← 
                              

(PC ) (PD )
    c      d
                      (PC ) (PD )
                          c      d
                                     =Q
                                               ?
               =K                             Q= K
(PA ) (PB )
     a     b
                      (PA ) (PB )
                           a     b



 K (the Equilibrium Constant) uses equilibrium
 partial pressures
 Q (the reaction quotient) uses prevailing partial
 pressures, not necessarily at equilibrium
The Reaction Quotient (2)
             φορωαρδ
          arrow→
                 
   aA + bB ρεϖερσε cC + dD
           wrong

          ←   
(PC )c (PD )d   =Q
                      ?
                     Q= K
                            (PC )c (PD )d   =K
(PA )a (PB )b               (PA )a (PB )b
  If Q < K, reaction proceeds in a
  forward direction (toward
  products);

 If Q > K, reaction proceeds in a backward direction
 (toward reactants);

 If Q = K, the reaction is in equilibrium.
The equilibrium constant for the reaction P4(g) ↔ 2 P2(g) is
1.39 at 400oC. Suppose that 2.75 mol of P4(g) and 1.08 mol of
P2(g) are mixed in a closed 25.0 L container at 400oC. Compute
Q(init) (the Q at the moment of mixing) and state the direction in
which the reaction proceeds.
                                                  ?
    (PP 2 )2
             = K = 1.39                          Q= K                 PV = nRT
    (PP 4 )1

       K = 1.39 @ 400oC; nP4(init) = 2.75 mol; nP2(init) = 1.08 mol

 PP4(init) = nP4(init)RT/V   = [(2.75mol)(0.08206 atm L mol-1 K-1)(273.15+400oC)]/(25.0L)
                             = 6.08 atm
 PP2(init) = nP2(init)RT/V   = [(1.08mol)(0.08206 atm L mol-1 K-1)(273.15+400oC)]/(25.0L)
                             = 2.39 atm
 Q=
Henri Louis Le Châtelier (1850-1936)
Highlights
   – 1884 Le Chatelier's Principle: A system in
     equilibrium that is subjected to a stress
     reacts in a way that counteracts the stress
   – If a chemical system at equilibrium
     experiences a change in concentration,
     temperature or total pressure the
     equilibrium will shift in order to minimize
     that change.
   – Industrial chemist involved with industrial
     efficiency and labor-management
     relations
Moments in a Life
   – Le Chatelier was named "chevalier"
     (knight) of the Légion d'honneur in 1887,
     decoration established by Napoléon
     Bonaparte in 1802.
Effects of External Stresses on Equilibria:
          Le Châtelier’s Principle
A system in equilibrium that is subjected to a
stress reacts in a way that counteracts the stress.

Le Châtelier’s Principle provides a way to predict the
response of an equilibrium system to an external
perturbation, such as…

1. Effects of Adding or Removing Reactants or Products

2. Effects of Changing the Volume (or Pressure) of the System

3. Effects of Changing the Temperature
Effects of Adding or Removing Reactants or Products
  PCl5(g)          PCl3(g) + Cl2(g)    K = 11.5 @ 300oC = Q

     add extra PCl5(g)
     add extra PCl3(g)
     remove some PCl5(g)
     remove some PCl3(g)




A system in equilibrium that is subjected to a stress reacts in a
way that counteracts the stress. In this case adding or
removing reactants or products
Effects of Changing the Volume of the System
             PCl5(g)           PCl3(g) + Cl2(g)
             1 mole            1+1 = 2 moles
  Let’s decrease the volume of the reaction container
      Less room :: less amount (fewer moles)
          Shifts reaction             to restore equilibrium
  Let’s increase the volume of the reaction container
      More room :: more amount (greater moles)
          Shifts reaction            to restore equilibrium


A system in equilibrium that is subjected to a
stress reacts in a way that counteracts the stress.
In this case a change in volume
Volume Decreased                    Volume Increased
                               (Pressure Increased)               (Pressure Decreased)
                           Equilibrium shift right        Equilibrium Shifts left
V reactants > V products
                                  (toward products)                  (toward reactants)
                           Equilibrium Shifts left        Equilibrium shift right
V reactants < V products
                                  (toward reactants)                 (toward products)
V reactants = V products      Equilibrium not affected            Equilibrium not affected


                      2 P2(g)       P4 (g)
                PCl5(g)       PCl3(g) + Cl2(g)
          CO (g) +H2O (g)                                CO2 (g) + H2 (g)

Boyles Law:
PV = Constant
      A system in equilibrium that is subjected to a stress reacts in a way
      that counteracts the stress. In this case a change in volume (or
      pressure)
Effects of Changing the Temperature

 Endothermic: heat is aborbed by a reaction
         Reactants + heat gives Products


   Exothermic: heat is liberated by a reaction
           Reactants gives Products + heat

A system in equilibrium that is subjected to a stress reacts
in a way that counteracts the stress. In this case a change
in temperature
Effects of Changing the Temperature
Endothermic: absorption of heat by a reaction
      Reactants + heat gives Products
Let’s increase the temperature of the reaction, what
direction does the equilibrium reaction shift


Let’s decrease the temperature of the reaction
Effects of Changing the Temperature
Exothermic: heat liberated by a reaction
        Reactants ↔ Products + heat
   Let’s increase the temperature of the reaction


   Let’s decrease the temperature of the reaction




A system in equilibrium that is subjected to a stress reacts
in a way that counteracts the stress. In this case a change
in temperature
Temperature Raised             Temperature Lowered
Endothermic
                    Equilibrium shift right     Equilibrium Shifts left
  Reaction
                           (toward products)               (toward reactants)
(absorb heat)
 Exothermic
                    Equilibrium Shifts left     Equilibrium shift right
   Reaction
                           (toward reactants)              (toward products)
(liberate heat)


  If a forward reaction is exothermic,
                      forward
     aA + bB → cC + dD + Heat
  Then the reverse reaction must be endothermic
                                       reverse
      cC + dD + Heat  aA + bB
                     →
                                     φορωαρδ
                          →
                  aA + bB ρεϖερσε cC + dD
                         ← 
                            
Driving Reactions to Completion/ Increasing Yield
Industrial Synthesis of Ammonia (Haber)
N2 (g) + 3H2 (g) ↔ 2NH3 (g)
Forward reaction is exothermic
What conditions do we need to increase the yield, i.e.,
produce more ammonia?
                     Volume Decreased           Volume Increased
                    (Pressure Increased)       (Pressure Decreased)

V reactants >         Equilibrium shift    Equilibrium Shifts left
                       right                               (toward
V products           (toward products)               reactants)
                    Temperature Raised       Temperature Lowered
    Exothermic      Equilibrium Shifts left Equilibrium shift right
     Reaction                  (toward                      (toward
  (liberate heat)         reactants)                  products)
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Chapter 11 equilibrium lecture notes

  • 1. Chapter 11 Chemical Equilibrium 11.1 The Equilibrium Condition 11.2 The Equilibrium Constant 11.3 Equilibrium Expressions Involving Pressures 11.4 The Concept of Activity 11.5 Heterogeneous Equilibria 11.6 Applications of the Equilibrium Constant 11.7 Solving Equilibrium Problems 11.8 Le Chatelier's Principle 11.9 Equilibria Involving Real Gases
  • 2. The Equilibrium Condition (General) Thermal equilibrium indicates two systems in thermal contact with each do not exchange energy by heat. If two bricks are in thermal equilibrium their temperatures are the same. Chemical equilibrium indicates no unbalanced potentials (or driving force). A system in equilibrium experiences no change over time, even infinite time. The opposite of equilibrium systems are non-equilibrium systems that are off balance and change with time. Example 1 atm O2 + 2 atm H2 at 298K
  • 3. The Equilibrium Condition (Chem Rxn) aA + bB cC + dD The same equilibrium state is achieved whether starting with pure reactants or pure products. The equilibrium state can change with temperature.
  • 4. The Equilibrium State (Chem Rxn) H2O (g) + CO (g) H2 (g) + CO2 (g) Change [CO] to PCO [H2O] to PH2O etc
  • 5. Chemical Reactions and Equilibrium As the equilibrium state is approached, the forward and backward rates of reaction approach equality. At equilibrium the rates are equal, and no further net change occurs in the partial pressures of reactants or products. Fundamental characteristics of equilibrium states: 1. No macroscopic evidence of change. 2. Reached through spontaneous processes. 3. Show a dynamic balance of forward and backward processes. 4. Same regardless of the direction from which they are approached. 5. No change over time.
  • 6. Arrows: Chemical Symbolism Use this in an equilibrium expression. ↔ Use this to indicate resonance.
  • 7. Chemical Reactions and Equilibrium The equilibrium condition for every reaction can be described in a single equation in which a number, the equilibrium constant (K) of the reaction, equals an equilibrium expression, a function of properties of the reactants and products. H2O(l) H2O(g) @ 25oC Temperature (oC) Vapor Pressure (atm) 15.0 0.01683 17.0 0.01912 K = 0.03126 19.0 0.02168 H2O(l) H2O(g) @ 30oC 21.0 0.02454 23.0 0.02772 K = 0.04187 25.0 0.03126 30.0 0.04187 50.0 0.1217
  • 8. Law of Mass Action (1) Partial pressures and concentrations of products appear in the numerator and those of the reactants in the denominator. Each is raised to a power equal to its coefficient in the balanced chemical equation. aA + bB cC + if gases dD if concentrations c d c d ( PC ) ( PD ) [C] [D] =K a b a b =K ( PA ) ( PB ) [ A] [B]
  • 9. Law of Mass Action (2) 1. Gases enter equilibrium expressions as partial pressures, in atmospheres. E.g., PCO2 2. Dissolved species enter as concentrations, in molarity (M) moles per liter. E.g., [Na+] 3. Pure solids and pure liquids are represented in equilibrium expressions by the number 1 (unity); a solvent taking part in a chemical reaction is represented by unity, provided that the solution is dilute. E.g., I2(s) ↔ I2(aq) [I2 (aq) ] = K I 2 ( s ) ↔ I 2 ( aq ) [ I 2 (aq )] [ I 2 ( aq )] K= = = [ I 2 ( aq )] [ I 2 ( s )] 1
  • 10. Activities The concept of Activity (i-th component) = ai = Pi / P reference H2O (l) H2O (g) Kp = P H2O @ 25oC Kp = 0.03126 atm PH2O =K Pref is numerically equal to 1 Pref K = 0.03126 The convention is to express all pressures in atmospheres and to omit factors of Pref because their value is unity. An equilibrium constant K is a pure number.
  • 11. The Equilibrium State H2O (g) + CO (g) H2 (g) + CO2 (g) PH 2 PCO 2 Kp = PH 2 O PCO
  • 12. The Equilibrium Expressions aA + bB cC + dD In a chemical reaction in which a moles of species A and b moles of species B react to form c moles of species C and d moles of species D, The partial pressures at equilibrium are related through K = PcCPdD/PaAPbB
  • 13. Write equilibrium expressions for the following reactions 3 H2(g) + SO2(g) H2S(g) + 2 H2O(g) 2 C2F5Cl(g) + 4 O2(g) Cl2(g) + 4 CO2(g) + 5 F2(g)
  • 14. Heterogeneous Equilibrium Gases and CaCO3(s) CaO(s) + CO2(g) Solids K=PCO2 K is independent of the amounts of CaCO3(s) or CaO(s)
  • 15. Heterogeneous Equilibrium Liquids H2O(l) H2O(g) K=PH2O I2(s) I2(aq) Solutions K=[I2]
  • 16. Relationships Among the K’s of Related Reactions #1: The equilibrium constant for a reverse reaction is always the reciprocal of the equilibrium constant for the corresponding forward reaction. 1 K for = or K for K rev = 1 K rev aA + bB cC + dD versus cC + dD aA + bB (PH2O)2 #1 2 H2 (g) + O2 (g) 2 H2O (g) (PH2)2(PO2) = K1 K1 = 1/K2 2 H2 (g) + O2 (g) (PH2) (PO22) 2 #2 2 H2O (g) = K2 (PH2O)
  • 17. Relationships Among the K’s of Related Reactions # 2: When the coefficients in a balanced chemical equation are all multiplied by a constant factor, the corresponding equilibrium constant is raised to a power equal to that factor. (PH2O)2 #1 2 H2 (g) + O2 (g) 2 H2O (g) Rxn 1 (PH2)2(PO2) = K1 #3 H2 (g) + ½ O2 (g) H2O (g) Rxn 3 = Rxn 1 times 1/2 (PH2O) = K3 K3 = K1½ (PH2)(PO2)½
  • 18. Relationships Among the K’s of Related Reactions # 3: when chemical equations are added to give a new equation, their equilibrium constants are multiplied to give the equilibrium constant associated with the new equation. 2 BrCl (g) ↔ Br2 (g) + Cl2 (g) (PBr2)(PCl2) = K1 = 0.45 @ 25oC wrong arrow (PBrCl)2 Br2 (g) + I2 (g) ↔ 2 IBr (g) (PIBr)2 wrong arrow = K2 = 0.051 @ 25oC (PBr2) (PI2) 2 BrCl (g) + I2 (g) ↔ 2 IBr (g) + Cl2(g) wrong arrow = K1K2 = (0.45)(0.051) (PBr2)(PCl2) (PIBr)2 X (P ) (P ) = K1K2 = K3 (PBrCl)2 Br2 I2 =0.023 @ 25oC
  • 19. Calculating Equilibrium Constants Consider the equilibrium 4 NO2(g)↔ 2 N2O(g) + 3 O2(g) The three gases are introduced into a container at partial pressures of 3.6 atm (for NO2), 5.1 atm (for N2O), and 8.0 atm (for O2) and react to reach equilibrium at a fixed temperature. The equilibrium partial pressure of the NO2 is measured to be 2.4 atm. Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur. 4 NO2(g) ↔ 2 N2O(g) + 3 O2(g) initial partial pressure (atm) change in partial pressure (atm) equilibrium partial pressure (atm)
  • 20. Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur. 4 NO2(g) ↔ 2 N2O(g) + 3 O2(g) initial partial pressure (atm) 3.6 5.1 8.0 change in partial pressure (atm) – 4x +2x +3x equilibrium partial pressure (atm) 2.4 5.1 + 2x 8.0 + 3x 5.1 + 2(0.3 atm) = 5.7 atm N2O 3.6 – 4x = 2.4 atm NO2; x = 0.3 atm 8.0 + 3(0.3 atm) = 8.9 atm O2 (PN2O)2(PO2)3 K= = (PNO2)4
  • 21. The compound GeWO4(g) forms at high temperature in the reaction 2 GeO (g) + W2O6(g) ↔ 2 GeWO4(g) Some GeO (g) and W2O6 (g) are mixed. Before they start to react, their partial pressures both equal 1.000 atm. After their reaction at constant temperature and volume, the equilibrium partial pressure of GeWO4(g) is 0.980 atm. Assuming that this is the only reaction that takes place, (a) determine the equilibrium partial pressures of GeO and W2O6, and (b) determine the equilibrium constant for the reaction. 2 GeO (g) + W2O6 (g) ↔ 2 GeWO4(g) initial partial pressure (atm) 1.000 1.000 0 change in partial pressure (atm) – 2x –x +2x equilibrium partial pressure (atm)
  • 22. (a) determine the equilibrium partial pressures of GeO and W2O6, and (b) determine the equilibrium constant for the reaction. 2 GeO(g) + W2O6(g) ↔ 2 GeWO4(g) initial partial pressure (atm) 1.000 1.000 0 change in partial pressure (atm) –2x –x +2x equilibrium partial pressure (atm) 1.000 – 2x 1.000 – x 0.980 0 + 2x = 0.980 atm GeWO4; 1.000 – 2(0.490) = 0.020 atm GeO x = 0.490 atm 1.000 – 0.490 = 0.510 atm W2O6 (PGeWO4)2 K = (P )2(P GeO W2O6 ) =
  • 23. • Skip Solving quadratic equations • Will utilize approximation method – Systems that have small equilibrium constants. – Assume “x” (the change in concentration) is small (less than 5%) of the initial concentration.
  • 24. A vessel holds pure CO (g) at a pressure of 1.282 atm and a temperature of 354K. A quantity of nickel is added, and the partial pressure of CO (g) drops to an equilibrium value of 0.709 atm because of the reaction Ni (s) + 4CO (g) ↔ Ni(CO)4 (g) Compute the equilibrium constant for this reaction at 354K. PNi(CO)4 PNi(CO)4 K= 4 = (PCO ) [Ni(s)] (PCO ) 4 (1) Ni (s) + 4CO (g) ↔ Ni(CO)4 (g) Construct an “ICE” table P CO (atm) P Ni(CO)4 (atm) initial partial pressure (atm) 1.282 0 change in partial pressure (atm) -4x +1x equilibrium partial pressure (atm) 0.709 x At equil. Pco= x = PNi(CO)4 =
  • 25. Equilibrium Calculations At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction 2CO2 (g) 2CO (g) + O2 (g) If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. 2CO2 (g) 2CO (g) + O2 (g) initial partial pressure (mol/L) 0.4 0 0 change in partial pressure (mol/L) – 2x +2x +1x equilibrium partial pressure (mol/L) 0.4 -2x 2x 1x
  • 26. At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction 2CO2 (g) 2CO (g) + O2 (g) If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. 2CO2 (g) 2CO (g) + O2 (g) initial partial pressure (mol/L) 0.4 0 0 change in partial pressure (mol/L) – 2x +2x +1x equilibrium partial pressure (mol/L) 0.4 -2x 2x 1x [CO]2 [O 2 ] (2 x) 2 ( x) Kp = = 2.0x10-6 = [CO 2 ]2 (0.40 − 2 x) 2 check assumption assume 2x << 0.40 2x 2(4.3x10-3) = x 100 = 2.2% 0.40 0.40 (2x) 2 (x) assumpiton valid, less than 5% 2.0x10-6 = ⇒ x = 4.3x10-3 M (0.40) 2
  • 27. At a particular temperature, K = 2.0 x 10-6 mol/L for the reaction 2CO2 (g) 2CO (g) + O2 (g) If 2.0 mol CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species. 2CO2 (g) 2CO (g) + O2 (g) initial partial pressure (mol/L) 0.4 0 0 change in partial pressure (mol/L) – 2x +2x +1x equilibrium partial pressure (mol/L) 0.4 -2x 2x 1x [CO]2 [O 2 ] Kp = = 2.0x10-6 [CO 2 ]2 [CO 2 ] = 0.40 - 2x = (2 x) 2 ( x) = (0.40 − 2 x) 2 = 0.40 - 2(4.3x10-3 ) = 0.39M x = 4.3x10-3M -3 [CO] = 2x = 2(4.3 x 10 ) -3 [O 2 ] = x = 4.3 x 10 M = 8.6 x 10-3 M
  • 28. Non-Equilibrium Conditions: The Reaction Quotient (1) φορωαρδ  → aA + bB ρεϖερσε cC + dD wrong arrow ←   (PC ) (PD ) c d (PC ) (PD ) c d =Q ? =K Q= K (PA ) (PB ) a b (PA ) (PB ) a b K (the Equilibrium Constant) uses equilibrium partial pressures Q (the reaction quotient) uses prevailing partial pressures, not necessarily at equilibrium
  • 29. The Reaction Quotient (2) φορωαρδ arrow→  aA + bB ρεϖερσε cC + dD wrong ←   (PC )c (PD )d =Q ? Q= K (PC )c (PD )d =K (PA )a (PB )b (PA )a (PB )b If Q < K, reaction proceeds in a forward direction (toward products); If Q > K, reaction proceeds in a backward direction (toward reactants); If Q = K, the reaction is in equilibrium.
  • 30. The equilibrium constant for the reaction P4(g) ↔ 2 P2(g) is 1.39 at 400oC. Suppose that 2.75 mol of P4(g) and 1.08 mol of P2(g) are mixed in a closed 25.0 L container at 400oC. Compute Q(init) (the Q at the moment of mixing) and state the direction in which the reaction proceeds. ? (PP 2 )2 = K = 1.39 Q= K PV = nRT (PP 4 )1 K = 1.39 @ 400oC; nP4(init) = 2.75 mol; nP2(init) = 1.08 mol PP4(init) = nP4(init)RT/V = [(2.75mol)(0.08206 atm L mol-1 K-1)(273.15+400oC)]/(25.0L) = 6.08 atm PP2(init) = nP2(init)RT/V = [(1.08mol)(0.08206 atm L mol-1 K-1)(273.15+400oC)]/(25.0L) = 2.39 atm Q=
  • 31. Henri Louis Le Châtelier (1850-1936) Highlights – 1884 Le Chatelier's Principle: A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress – If a chemical system at equilibrium experiences a change in concentration, temperature or total pressure the equilibrium will shift in order to minimize that change. – Industrial chemist involved with industrial efficiency and labor-management relations Moments in a Life – Le Chatelier was named "chevalier" (knight) of the Légion d'honneur in 1887, decoration established by Napoléon Bonaparte in 1802.
  • 32. Effects of External Stresses on Equilibria: Le Châtelier’s Principle A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. Le Châtelier’s Principle provides a way to predict the response of an equilibrium system to an external perturbation, such as… 1. Effects of Adding or Removing Reactants or Products 2. Effects of Changing the Volume (or Pressure) of the System 3. Effects of Changing the Temperature
  • 33. Effects of Adding or Removing Reactants or Products PCl5(g) PCl3(g) + Cl2(g) K = 11.5 @ 300oC = Q add extra PCl5(g) add extra PCl3(g) remove some PCl5(g) remove some PCl3(g) A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case adding or removing reactants or products
  • 34. Effects of Changing the Volume of the System PCl5(g) PCl3(g) + Cl2(g) 1 mole 1+1 = 2 moles Let’s decrease the volume of the reaction container Less room :: less amount (fewer moles) Shifts reaction to restore equilibrium Let’s increase the volume of the reaction container More room :: more amount (greater moles) Shifts reaction to restore equilibrium A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in volume
  • 35. Volume Decreased Volume Increased (Pressure Increased) (Pressure Decreased) Equilibrium shift right Equilibrium Shifts left V reactants > V products (toward products) (toward reactants) Equilibrium Shifts left Equilibrium shift right V reactants < V products (toward reactants) (toward products) V reactants = V products Equilibrium not affected Equilibrium not affected 2 P2(g) P4 (g) PCl5(g) PCl3(g) + Cl2(g) CO (g) +H2O (g) CO2 (g) + H2 (g) Boyles Law: PV = Constant A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in volume (or pressure)
  • 36. Effects of Changing the Temperature Endothermic: heat is aborbed by a reaction Reactants + heat gives Products Exothermic: heat is liberated by a reaction Reactants gives Products + heat A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in temperature
  • 37. Effects of Changing the Temperature Endothermic: absorption of heat by a reaction Reactants + heat gives Products Let’s increase the temperature of the reaction, what direction does the equilibrium reaction shift Let’s decrease the temperature of the reaction
  • 38. Effects of Changing the Temperature Exothermic: heat liberated by a reaction Reactants ↔ Products + heat Let’s increase the temperature of the reaction Let’s decrease the temperature of the reaction A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. In this case a change in temperature
  • 39. Temperature Raised Temperature Lowered Endothermic Equilibrium shift right Equilibrium Shifts left Reaction (toward products) (toward reactants) (absorb heat) Exothermic Equilibrium Shifts left Equilibrium shift right Reaction (toward reactants) (toward products) (liberate heat) If a forward reaction is exothermic, forward aA + bB → cC + dD + Heat Then the reverse reaction must be endothermic reverse cC + dD + Heat  aA + bB → φορωαρδ  → aA + bB ρεϖερσε cC + dD ←  
  • 40. Driving Reactions to Completion/ Increasing Yield Industrial Synthesis of Ammonia (Haber) N2 (g) + 3H2 (g) ↔ 2NH3 (g) Forward reaction is exothermic What conditions do we need to increase the yield, i.e., produce more ammonia? Volume Decreased Volume Increased (Pressure Increased) (Pressure Decreased) V reactants > Equilibrium shift Equilibrium Shifts left right (toward V products (toward products) reactants) Temperature Raised Temperature Lowered Exothermic Equilibrium Shifts left Equilibrium shift right Reaction (toward (toward (liberate heat) reactants) products)

Notas del editor

  1. When a reaction is endothermic, an increase in the temperature (as heat is put into the system) causes its equilibrium to shift toward the formation of more products; and a decrease in the temperature causes its equilibrium to shift in the direction of more reactants. When a reaction is exothermic, an increase in the temperature causes its equilibrium to shift in the direction of more reactants; and a decrease in the temperature causes its equilibrium to shift toward the formation of more products.
  2. When a reaction is endothermic, an increase in the temperature (as heat is put into the system) causes its equilibrium to shift toward the formation of more products; and a decrease in the temperature causes its equilibrium to shift in the direction of more reactants. When a reaction is exothermic, an increase in the temperature causes its equilibrium to shift in the direction of more reactants; and a decrease in the temperature causes its equilibrium to shift toward the formation of more products.