Chemistry Concentration Worksheet - PowerPoint

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					                     AP Chem Thurs. 3/9
                    Complete Worksheets:
                         Worksheet 1 & 2
                    AP FR 1995/NChO Problems
                          Practice Test

  Equilibrium/gas laws/equations test on
                MONDAY
    All current worksheets due MONDAY
LOHS AP Chemistry             Schrempp          1
        Kinetics & Equilibrium
                              A --> B

                     Equilibrium Concentrations

                Ao                       [B] - Product Curve
Concentration




                BE
                AE

                                         [A] -Reactant Curve
                Bo
                            time
LOHS AP Chemistry                  Schrempp                    2
The Equilibrium Constant


•No matter the
     starting
composition of
 reactants and
 products, the
 same ratio of
concentrations
 is achieved at
  equilibrium.



LOHS AP Chemistry   Schrempp   3
The Equilibrium Constant
• For a general reaction
                aA + bB(g)              pP + qQ
  the equilibrium constant expression is

                     Kc 
                            P p Qq

                            Aa B b

  where Kc is the equilibrium constant.




 LOHS AP Chemistry                Schrempp        4
The Equilibrium Constant
• Kc is based on the molarities of reactants and products
  at equilibrium.
• We generally omit the units of the equilibrium
  constant.
• Note that the equilibrium constant expression has
  products over reactants.




 LOHS AP Chemistry        Schrempp                   5
     Rate Laws and K
                                   kf
                 aA + bB            kb
                                             cC + dD
        At Equilibrium; the Forward Reaction
         Rate Equals the Back Reaction Rate
                     Rf = Rb
               kf[A]m[B]n = kb[C]x[D]y
                               x         y       c     d
                   kf
                   [C ] [ D]    [C ] [ D]
            K        m    n
                                   a    b
               kb [ A] [ B]     [ A] [ B]
(for multistep mechanisms, K is a product of the ratio of the rate laws for
  all of the elementary steps (whose order of reaction can be determined
from their stoichiometric coefficients). This allows us to determine K in
                                      coefficients of the balanced eq. 6
          terms of the stoichiometricSchrempp
    LOHS AP Chemistry
   Kc & Kp

 Kc is expressed in terms of concentrations
 Kp is expressed in terms of partial pressures
                         [C ]c [ D]d               d
                                              PCc PD
                    Kc      a     b
                                     and K p  a b
                         [ A] [ B]            PA PB
                              n A PA
          Using      [ A]                Relate Kc to KP
                              V    RT


                                 Dn   product stoichiometric coef
KP = KC(RT)Dn
LOHS AP Chemistry
                                        reactant stoichiometric coef
                                    Schrempp                        7
The Equilibrium Constant
The Equilibrium Constant in Terms of Pressure
• If KP is the equilibrium constant for reactions
  involving gases, we can write:
                           PP  Q
                                p P q   
                     KP 
                                a  P b
                           PA  B
• KP is based on partial pressures measured in
  atmospheres.
• We can show that
                      PA = [A](RT)

 LOHS AP Chemistry          Schrempp                8
The Equilibrium Constant
The Equilibrium Constant in Terms of Pressure
                     PA = [A](RT)
• This means that we can relate Kc and KP:

                     K P  K c  RT Dn


  where Dn is the change in number of moles of gas.
• It is important to use:
           Dn = ngas(products) - ngas(reactants)


 LOHS AP Chemistry         Schrempp                   9
The Equilibrium Constant
The Magnitude of Equilibrium Constants
• The equilibrium constant, K, is the ratio of products
  to reactants.
• Therefore, the larger K the more products are present
  at equilibrium.
• Conversely, the smaller K the more reactants are
  present at equilibrium.
• If K >> 1, then products dominate at equilibrium and
  equilibrium lies to the right.
• If K << 1, then reactants dominate at equilibrium and
  the equilibrium lies to the left.
 LOHS AP Chemistry       Schrempp                  10
The Equilibrium Constant
The Magnitude of Equilibrium Constants




 LOHS AP Chemistry   Schrempp            11
The Equilibrium Constant
The Magnitude of Equilibrium Constants
• An equilibrium can be approached from any
  direction.
• Example:
                     N2O4(g)              2NO2(g)
• has
                       Kc 
                            NO2 2  0.212
                            N 2O4 



 LOHS AP Chemistry             Schrempp             12
    Kf & Kr
    Relate K for the forward reaction (Kf) to that
     of the reverse (back) reaction (Kr or K b)

                 aA + bB              cC + dD
                 cC + dD              aA + bB
                     c    d                   a        b
     [C ] [ D ]                    [ A] [ B]
Kf      a     b
                              Kr      c     d
     [ A] [ B ]                    [C ] [ D]
                                                  •The equilibrium constant for a
                              1
                         Kf 
                                                  reaction in one direction is the
                                                  reciprocal of the equilibrium constant
                              Kr                  of the reaction in the opposite
                                                  direction.
 LOHS AP Chemistry                 Schrempp                                      13
The Equilibrium Constant
Heterogeneous Equilibria
• When all reactants and products are in one phase
  (solid, liquid, gas), the equilibrium is homogeneous.
• If one or more reactants or products are in a different
  phase, the equilibrium is heterogeneous.




 LOHS AP Chemistry        Schrempp                   14
The Equilibrium Constant
Heterogeneous Equilibria
• Consider:
            CaCO3(s)         CaO(s) + CO2(g)
• The concentration of a solid or pure liquid is its
  density divided by molar mass.
• Since neither density nor molar mass is a variable, the
  concentrations of solids and pure liquids are constant.

We ignore the concentrations of pure liquids and pure
 solids in equilibrium constant expressions.

 LOHS AP Chemistry        Schrempp                   15
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.)
• When in doubt, assign the change in concentration a
  variable.

 LOHS AP Chemistry           Schrempp                      16
Coupled Reactions
         What is a Coupled Reaction?
                               K1
                    NO2 +NO2             NO + NO3
                               K2
                    NO3 + CO             NO2 + CO2
                               K3
                    NO2 + CO             NO + CO2

   Two reactions are coupled by an intermediate
   (NO3), which is a product of the first step and
           a reactant of the second step.
LOHS AP Chemistry                   Schrempp         17
Coupled Reactions
What is the Enthalpy for the Couple Step?
                    NO2 +NO2
                               K1
                                         NO + NO3    DH1
                    NO3 + CO
                               K2
                                         NO2 + CO2   DH2
                    NO2 + CO
                               K3
                                         NO + CO2    DH3
    Hess’s Law: Energies are Additive

                       DH3 = DH1 + DH2
      How do you express K3 in terms of K1 & K2
LOHS AP Chemistry                   Schrempp               18
                               K1
                    NO2 +NO2              NO + NO3
                               K2
                    NO3 + CO              NO2 + CO2
                               K3
                    NO2 + CO              NO + CO2

        [ NO ][ NO3 ]        [ NO2 ][CO2 ]           [ NO ][ CO2 ]
   K1             2
                      , K2                and K 3 
           [ NO2 ]            [ NO3 ][CO ]           [ NO2 ][CO ]


                               K3=?
                          K3=K1K2
LOHS AP Chemistry                   Schrempp                     19
                               K1
                    NO2 +NO2              NO + NO3
                               K2
                    NO3 + CO              NO2 + CO2
                               K3
                    NO2 + CO              NO + CO2

        [ NO ][ NO3 ]        [ NO2 ][CO2 ]           [ NO ][ CO2 ]
   K1             2
                      , K2                and K 3 
           [ NO2 ]            [ NO3 ][CO ]           [ NO2 ][CO ]


                          K3=K1K2
     Energy is Additive for Coupled Equations,
     Equilibrium Constants are Multiplicative.
LOHS AP Chemistry                   Schrempp                     20
                                  K1
                       NO2 +NO2              NO + NO3
                                  K2
                       NO3 + CO              NO2 + CO2
                                  K3
                       NO2 + CO              NO + CO2
          [ NO ][ NO3 ]        [ NO2 ][CO2 ]           [ NO ][ CO2 ]
     K1             2
                        , K2                and K 3 
             [ NO2 ]            [ NO3 ][CO ]           [ NO2 ][CO ]
  Note, this example uses the elementary steps of a proposed mechanism
for the observed rate law of R=k[NO]2, where the first step was the slow
 step. In this example the coupled reactions are elementary steps and the
orders of reaction can be determined from their coefficients. This allows
K3 to be directly determine from the coefficients of step 3 (intermediates
                                cancel out)

   LOHS AP Chemistry        K3=K1K2    Schrempp                        21
   Reaction Quotient
                                   c    d
                           [C ] [ D ]
                        Q     a     b
                           [ A] [ B ]

                 Where [ ] represents the
               concentration at any time, not
                     just equilibrium


LOHS AP Chemistry            Schrempp           22
   Reaction Quotient
        What is the value of Q when a
      reaction has reached equilibrium?
                    Q=K
Relate Q to K if a system is “Product Loaded”

                    Q>K
Relate Q to K if a system is “Reactant Loaded”

                    Q<K
LOHS AP Chemistry         Schrempp        23
Predicting Reaction Direction From Q

                      Q>K
            -Product Concentration Goes Down
            -Reactant Goes Up Until Q=K

                      Q<K
      -Reactant Concentration Goes Down
      -Product Goes Up Until Q=K
 LOHS AP Chemistry       Schrempp              24
    Reaction Quotient
Consider the following reaction @ 2400oC & K=0.0035
                      N2(g) + O2(g)  2 NO(g)
Describe the direction the reaction will proceed in when:


      a.) [N2]= .3M [O2]=.6M [NO]=.2M
      b.) [N2]= .5M [O2]=.7M [NO]=.035M
      c.) [N2]= .2M [O2]=.5M [NO]=.001M

  LOHS AP Chemistry               Schrempp                  25
    Reaction Quotient
Consider the following reaction @ 2400oC & K=0.0035
                                      NO  @ any time
                                        2

 N2(g) + O2(g)  2 NO(g)        Q
                                    N O 
                                    2   2




 a.) [N2]= .3M [O2]=.6M [NO]=.2M
      QC=.222 TOWARDS REACTANTS
  b.) [N2]= .5M [O2]=.7M [NO]=.035M
       QC=.0035 AT EQUILIBRIUM

  c.) [N2]= .2M [O2]=.5M [NO]=.001M
       QC=0.00001 TOWARDS PRODUCTS
  LOHS AP Chemistry      Schrempp                 26
  Calculating Equilibrium Constants
Consider the Following Reaction:

   2NO(g) + 2H2(g) <==> N2(g) + 2H2O(g)

 Scenario 1: You mix known amounts of the
 above species and monitor the concentration
 of one



LOHS AP Chemistry     Schrempp                 27
Type 1: Calculation of K from initial
concentrations of all species and equilibrium
concentration of 1

Class Problem: Calculate K and equilibrium
concentrations of all species for the reaction:
 2NO(g) + 2H2(g) <==> N2(g) + 2H2O(g)
         if the equilibrium concentration of
         NO = 0.062M after 0.100 mol NO,
        0.050 mol H2 and 0.100 mol H2O are
               mixed in a 1 L container
LOHS AP Chemistry         Schrempp                28
   Type 1: Calculation of K from initial
   concentrations of all species and equilibrium
   concentration of 1
1. Write balanced equation
2. Under eq: Write a table that lists for each substance
   a. I Initial Concentration [ ]I
   b. C Extent of Reaction (change, D[ ] to reach equilibrium) in
              terms of X times the stoichiometric coef.
   c. E Equilibrium Concentration
   3. Determine X from known Equilibrium
        Concentration
   4. Substitute X into equilibrium expression and
        solve for K
  LOHS AP Chemistry                 Schrempp               29
   Class Problem: Calculate K and equilibrium
   concentrations of all species for the reaction:

   2NO(g) + 2H2(g) <==> N2(g) + 2H2O(g)

           if the equilibrium concentration of
           NO = 0.062M after 0.100 mol NO,
          0.050 mol H2 and 0.100 mol H2O are
                 mixed in a 1 L container



LOHS AP Chemistry          Schrempp                  30
                     2NO + 2H2 --> N2 + 2H2O
            I        0.1   0.05          0.0   0.10
            C        -2x    -2x           +x    +2x

           E         .1-2x .05-2x         x    .1+2x
Solve for X from known equilibrium concentration of [NO]
                [NO]E = 0.062M = 0.1-2x
                       X = 0.019
 LOHS AP Chemistry            Schrempp                 31
                    2NO + 2H2 --> N2 + 2H2O
          I         0.01       0.05             0.0       0.10
        C           -2x        -2x              +x        +2x

         E          0.062 .012                  .019      .138

                           2                          2
         [ N 2 ][H 2O]    [.019][.138]
      K         2    2
                              2       2
                                          653
         [ NO] [ H 2 ] [.062] [.012]

LOHS AP Chemistry                    Schrempp                    32
   Type 2: Calculate Eq. Conc Knowing K &
   Initial Concentrations

  For the Reaction
CO(g) + Cl2(g)                       COCl2(g), Kc =26.7
Determine the equilibrium concentrations
if 1.0 mol CoCl2 is placed in a 10 L vessel
        Since we start with CoCl2, lets
        rewrite the eq so it is a reactant

  COCl2
LOHS AP Chemistry
                    CO + Cl2
                          Schrempp
                                          K = 1/26.7   33
    Type 2: Calculate Eq. Conc Knowing K &
    Initial Concentrations
1. Write balanced equation
2. Under eq. Write a table that lists for each substance
        a. I Initial Concentration [ ]i
        b. C Extent of Reaction (change, D[ ] to reach equilibrium)
              in terms of x times the stoichiometric coef.
        c. E Equilibrium Concentration
   3. Substitute Equilibrium Concentrations into
         Equilibrium Equation and solve for x.

   4. Calculate Equilibrium Concentration by Substituting
         in X-values
  LOHS AP Chemistry            Schrempp                      34
                    COCl2          CO + Cl2

         I           .10           0         0

         C           -x           +x         +x

        E           .10-x         x              x

                                                     2
                          1     [CO][Cl2 ]     x
                      K                  
                         26.7     COCl2      .1  x
LOHS AP Chemistry                 Schrempp               35
                    How do we solve for x?
                                                     2
                        1     [CO][Cl2 ]     x
                    K                  
                       26.7     COCl2      .1  x

                        ax2 + bx + c = 0
                        27x2 + x - 0.1 = 0
                             b  b 2  4ac
                         x
                                  2a

                           1  (1) 2  4(27)(.1)
                       x
                                   2(27)

       X=0.045 or -0.082, which makes sense?
LOHS AP Chemistry                  Schrempp              36
CO(g) + Cl2(g)                            COCl2(g), Kc =26.7

        X=0.045 or -0.082, which makes sense?

                    COCl2 = .1-x = .055M
                    CO = x = .045M
                    Cl2 = x = .045M




LOHS AP Chemistry              Schrempp                 37
  Completing the Square/Cube
       -Special Case Where Reactants Are in
     Stoichiometric Ratios and # of Species Are
          Conserved As Reaction Proceeds

      Cl2(g) + 2HBr(g)  Br2(g) + 2HCl(g)

              If K = 15, How Much Bromine Is
             Produced If 0.200 Moles of Chlorine
             React With 0.400 Moles of Hydrogen
                Bromide in a 1 liter container?
LOHS AP Chemistry           Schrempp               38
                   Cl2 +        2HBr --> Br2 + 2HCl

           [ ]i    .2             .4        0.0         0
           Dx          -x        -2x        +x         +2x

     [ ]E          .2-x         .4-2x         x         2x

   [ Br2 ][ HCl ]           2
                       [ x ][2 x ]            2                  [ x ]  4[ x ]2 
K                                    
    [ HBr ] [Cl ] [.4  2 x ] [.2  x ]  2[.2  x ] 2 [.2  x ]
            2                2


                                        3
     4[ x ]      [ x] 
                   3
                                        [ x]
                             K   [.2  x ]
                                   1
K                               3

   4[.2  x ]  [.2  x ] 
             3



K        [.2  x ]  [ x ]  [ x ] 1  K 1/ 3   .2 K 1/ 3
       1
       3




         .2 K 1/ 3     .2 3 15
[ x ]  1/ 3          3         0.14
  LOHS APK       
         Chemistry 1
                         15  1      Schrempp                                        39
                    Cl2 +   2HBr --> Br2 + 2HCl

      [ ]i          .2        .4     0.0       0
      Dx             -x      -2x     +x        +2x

    [ ]E            .2-x    .4-2x     x         2x



         [Cl2]= .200-(.14)=0.060M
         [HBr] = .400-2(.14) = .12M
         [Br2] = .14M
         [HCl]=2.8M
LOHS AP Chemistry                   Schrempp         40
   Le Chatelier’s Principle
          A System at Equilibrium Will React
             to Alleviate an Applied Stress
              3 Types of Stress We will Consider

                    1. D[ ]
                    2. DP(total)
                    3. DT



LOHS AP Chemistry             Schrempp             41
   D Concentration
       1. Adding a Substance shifts equilibria
            to consume substance added
      2. Removing a Substance shifts equilibria
           to create substance removed

             Relate D [ ] to the Reaction
                     Quotient Q

LOHS AP Chemistry         Schrempp                42
   Relating LeChatlier’s Principle to Q

 Consider a System at Equilibrium: Q=K
             •Q>K when you add products or
                 remove reactants
             •Q<K when you remove products or
                 add reactants



LOHS AP Chemistry          Schrempp             43
LeChatlier’s Principle

                      Add A
  Ao



Conc


    Bo

                         Remove B        t
  LOHS AP Chemistry           Schrempp       44
D Pressure
         aA(g) +bB(g) < == > cC(g) + dD(g)

How does a change in the total pressure effect
    the concentration for an isothermal system?
                    PTV=nTRT

Look at the coefficients of the balance eq. If
the sum is not equal, you can reduce or
decrease n by shifting the equilibria

LOHS AP Chemistry        Schrempp                45
D Pressure
          2NH3(g) ---> N2(g) + 3H2(g)
                    PTV=nTRT
1. Increasing P increases product concentration
      (n=1+3) more than reactant (n=2), so
      equilibria shifts to left

2. Decreasing P decreases product concentration
     (n=1+3) more than reactant (n=2), so
     equilibria shifts to right
LOHS AP Chemistry        Schrempp             46
D Pressure
Consider the following reaction
           N2O4(g) ---> 2N2O(g)
  How does a change in the total pressure effect
      the concentration for an isothermal system?

                     PV=Constant
                    P V and [ ]

                    P V and [ ]
LOHS AP Chemistry         Schrempp            47
D Pressure
Look at the Haber Process for Ammonia Production
           2NH3(g) ---> N2(g) + 3H2(g)
How will doubling P effect the above reaction

            Since There Are More Gas Phase
        Products Than Reactants, It Will Increase
         the Product Concentration More, Thus
           More Ammonia Will Be Produced
          According to Le Chatelier’s Principle

LOHS AP Chemistry         Schrempp                  48
            Consider the reaction of Nitric Oxide
                        with Ozone


            NO(g) +O3(g) --> NO2(g) + O2(g)


               How Does the Pressure Effect
                    This Reaction?



LOHS AP Chemistry           Schrempp                49
 K & D Temperature

           How Does a Change in Temperature
           Effect the Equilibrium Distribution?

               It Depends If the Reaction Is
                Exothermic or Endothermic


LOHS AP Chemistry          Schrempp               50
   K, DT & Exothermic Reactions
 1. Increasing Temperature
        T           P -> R              R
       (chemical reaction will
                                                       P
          absorb the heat)              <- endothermicity

   2. Decreasing Temperature

          T         R -> P               R
         (chemical reaction will
                                                        P
             release heat)                  exothemicity->
LOHS AP Chemistry            Schrempp                        51
   K, DT & Endothermic Reactions
 1. Increasing Temperature
        T           R -> P                            P
                                         R
       (chemical reaction will
          absorb the heat)              endothermicity ->

   2. Decreasing Temperature
                                                      P
          T         P -> R
                                         R
         (chemical reaction will
             release heat)                <- exothermicity
LOHS AP Chemistry            Schrempp                        52
Summary of Temperature Effect


        1. Increasing Temperature, Reaction Shifts
             in Direction of Endothermicity to
             Absorb the Heat Added

       2. Decreasing Temperature, Reaction Shifts
            in Direction of Exothermicity to
            Replace the Heat Lost

 LOHS AP Chemistry        Schrempp                   53
Cobalt Complexes


     CoCl4-2 + 6H2O  [Co(H2O)6]+2 +4Cl-
         blue             pink




 LOHS AP Chemistry     Schrempp            54
    Effects of a Catalyst

                How does a Catalyst Effect
                 the Equilibrium Ratio?

                      It Does Not
            It Effects the Rate With Which the
               Reaction Reaches Equilibrium


LOHS AP Chemistry           Schrempp             55
Le Châtelier’s Principle
Effects of Volume and Pressure
• 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.
• Consider
                     N2O4(g)              2NO2(g)



 LOHS AP Chemistry             Schrempp              56
Le Châtelier’s Principle
Effects of Volume and Pressure
• 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. the forward reaction is favored).
• A new equilibrium is established in which the mixture
  is lighter because colorless N2O4 is favored.


 LOHS AP Chemistry         Schrempp                   57
Le Châtelier’s Principle
Effect of Temperature Changes
• The equilibrium constant is temperature dependent.
• For an endothermic reaction, DH > 0 and heat can be
  considered as a reactant.
• For an exothermic reaction, DH < 0 and heat can be
  considered as a product.
• Adding heat (i.e. heating the vessel) favors away from
  the increase:
   – if DH > 0, adding heat favors the forward reaction,
   – if DH < 0, adding heat favors the reverse reaction.


 LOHS AP Chemistry            Schrempp                     58
Le Châtelier’s Principle
Effect of Temperature Changes
• Removing heat (i.e. cooling the vessel), favors towards
  the decrease:
   – if DH > 0, cooling favors the reverse reaction,
   – if DH < 0, cooling favors the forward reaction.
• Consider
Cr(H2O)6(aq) + 4Cl-(aq)                  CoCl42-(aq) + 6H2O(l)
  for which DH > 0.
   – Co(H2O)62+ is pale pink and CoCl42- is blue.




 LOHS AP Chemistry            Schrempp                    59
Le Châtelier’s Principle
Effect of Temperature Changes




 LOHS AP Chemistry   Schrempp   60
Le Châtelier’s Principle
Effect of Temperature Changes
Cr(H2O)6(aq) + 4Cl-(aq)    CoCl42-(aq) + 6H2O(l)
   – If a light purple room temperature equilibrium mixture is
     placed in a beaker of warm water, the mixture turns deep
     blue.
   – Since DH > 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 DH > 0, removing heat favors the reverse reaction
     which is the formation of pink Co(H2O)62+.



 LOHS AP Chemistry           Schrempp                     61
Le Châtelier’s Principle
The Effect of Catalysts
• 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 effect the composition of the
  equilibrium mixture.




 LOHS AP Chemistry        Schrempp                   62

				
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