Chapter 6: Chemical Equilibrium - 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The...

download Chapter 6: Chemical Equilibrium - 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium

of 87

  • date post

    16-Oct-2019
  • Category

    Documents

  • view

    50
  • download

    6

Embed Size (px)

Transcript of Chapter 6: Chemical Equilibrium - 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The...

  • Chapter 6: Chemical Equilibrium

    6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions Involving Pressures 6.4 The Concept of Activity 6.5 Heterogeneous Equilibria 6.6 Applications of the Equilibrium Constant 6.7 Solving Equilibrium Problems 6.8 LeChatelier’s Principle 6.9 OMIT: Equilibria Involving Real Gases

    2 HWs due Satd at 6 PM: HW #7: Chapter 5, part 2 HW #8: Chapter 6, part 1

  • A new HW assignment (HW#8) over 1st part of Ch. 6 is also due this Satd..

    The 2nd part of Ch. 6 (HW#9) will be due 7 AM on Friday, Nov. 19. Exam on Nov. 22.

    ------------------- All Discussion Sessions after 12:00 on Nov. 18 have been

    rescheduled to: Wed. Nov. 17th from 2:45 to 3:45

    in Bagley Hall room 154 (run by Emily Sprafka, a TA)

    --------------------------- She will also hold a Review for Exam 2 on Fri. Nov. 19th from 3:30 to 5:00 PM

    in BAG 131.

  • Molecular Picture of Establishment of Equilibrium

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

    INITIAL

    7 CO(g) + 7 H2O(g) + 0 CO2(g) + 0 H2(g) AFTER IT STOPS CHANGING

    2 CO(g) + 2 H2O(g) + 5 CO2(g) + 5 H2(g)

  • Chemical Equilibrium • Previously, we assumed that a chemical reaction goes to completion as written.

    H2O(g) + CO (g) → H2(g) + CO2(g) • In general this is not correct. Instead, a stable state of the system in reached, which includes both reactants and products. It is called the equilibrium state, or simply “equilibrium”.

    H2O(g) + CO (g) ⇌ H2(g) + CO2(g)

    How far toward completion it goes depends on the specific reaction, and on temperature.

  • Concentration vs. Time CO(g) + H2O(g) → CO2(g) + H2(g)

    Ratio depends on temperature

  • Characteristics of Chemical Equilibrium States

    • Reaching equilibrium requires reactions to occur.

    • Once reached, they show no macroscopic evidence of further change.

    • Reached through dynamic balance of forward and reverse reaction rates.

  • Figure 5.21: The collision rate of gas particles defines

    the maximum blue-pink reaction rate!

    Z = Collision rate (of one pink with blues)= # collisions with blues per s = 4 [Nblue/V] d2 (πRT/M)1/2 = 4 [blues] d2 (πRT/M)1/2

  • Equilibrium arises through dynamic balance between forward and back reactions

    Forward rate = k1[NO2][NO2] = k1[NO2]2

    Back rate = k-1[NO3][NO]

    The rate constants k1 and k-1 reflect probabilities that one collision leads to a successful reaction.

    Reverse: NO3(g) + NO(g) → 2NO2(g) k -1

  • Kinetics of Approach to Equilibrium

    [ ] [ ][ ] [ ][ ] [ ]

    2 1 2 1 3

    31 2

    1 2

    NO NO NO

    NO NO

    NO

    k k

    k k

    =

    = = K

    At eqbm.: Forward rate = back rate

    ⇑ “Equilibrium constant”

  • The Equilibrium Constant - Definition Consider the generalized chemical reaction:

    a A + b B ⇌ c C + d D A, B, C and D represent chemical species and a, b, c, and d are their stoichiometric coefficients in the balanced chemical equation. At equilibrium,

    The square brackets indicate the concentrations of the species in equilibrium.

    K is a constant called the equilibrium constant.

    K depends only on T , and not on concentrations.

    [ ] [ ] [ ] [ ]

    c d

    a b

    C D K

    A B = Note: The “units” for K are

    concentration units raised to some power = c+d–(a+b)

  • Reaching Equilibrium on the Macroscopic and Molecular Level

    N2O4 (g) 2 NO2 (g) Colorless Brown

    K = [NO2]2 / [N2O4] Units are mol/L.

  • Equilibrium from Different Starting Points

    CO(g) + 2 H2(g) ⇌ CH3OH(g)

    [ ] [ ][ ]

    3 2

    2

    has same value at equil. for all 3 starting points

    CH OH K

    CO H =

  • N2(g) + 3 H2(g) 2 NH3(g) Fe

    K = [NH3] 2

    [N2][H2]3

  • Key Stages in the Haber Synthesis of Ammonia

    Gerhard Ertl, 2007 Nobel Prize in Chemistry

  • Small K N2 (g) + O2 (g) 2 NO(g) K = 1 x 10 –30

    Essentially only reactants at eqbm. (1015 x products)

    Intermediate K 2 BrCl(g) Br2 (g) + Cl2 (g) K = 5

    Comparable amounts of products and reactants at eqbm.

    Large K 2 CO(g) + O2 (g) 2 CO2 (g) K = 2.2 x 1022

    Essentially only products at eqbm.

    Equilibrium Constants can have a wide range of values

  • CS2(g) + 3 O2(g) ⇌ CO2(g) + 2 SO2(g) • The equilibrium expression for a reaction written in

    reverse is the reciprocal of that for the original expression.

    CO2(g) + 2 SO2(g) ⇌ CS2(g) + 3 O2(g)

    • If the original reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power n.

    2 CS2(g) + 6 O2(g) ⇌ 2 CO2(g) + 4 SO2(g)

    [ ][ ] [ ][ ]322

    2 22

    1 OCS SOCO

    =K

    [ ][ ] [ ][ ] 1222

    3 22

    2 1

    SOCO OCS

    K K ==

    [ ] [ ] [ ] [ ]

    ( ) 2 4

    22 2 3 12 6

    2 2

    CO SO CS O

    K K= =

  • Example: Calculation of the Equilibrium Constant from equilibrium amounts

    At 454 K, the following reaction takes place:

    3 Al2Cl6(g) ⇌ 2 Al3Cl9(g)

    At this temperature, the eqbm concentration of Al2Cl6(g) is 1.00 M and the equilibrium concentration of Al3Cl9(g) is 1.02 x 10-2 M. Compute the equilibrium constant at 454 K.

    Strategy: Substitute values into K

    [ ] [ ]

    ( ) ( )

    14 3

    22

    3 62

    2 93 1004.1

    00.1 1002.1

    ClAl ClAl −−− ×=×== M

    M MK

  • Like Example 6.1 (P 201) - I The following equilibrium concentrations were observed for the reaction between CO and H2 to form CH4 and H2O at 927oC:

    CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g) [CO] = 0.613 mol/L [CH4] = 0.387 mol/L [H2] = 1.839 mol/L [H2O] = 0.387 mol/L

    a) Calculate the value of K at 927oC for this reaction. b) Calculate the value of the equilibrium constant at 927oC for:

    H2O(g) + CH4 (g) ⇌ CO(g) + 3 H2 (g) c) Calculate the value of the equilibrium constant at 927oC for:

    1/3 CO(g) + H2 (g) ⇌ 1/3 CH4 (g) + 1/3 H2O(g) Solution: a) For the first balanced reaction above:

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

  • Like Example 6.1 (P 201) - I The following equilibrium concentrations were observed for the Reaction between CO and H2 to form CH4 and H2O at 927oC.

    CO(g) + 3 H2 (g) ⇌ CH4 (g) + H2O(g) [CO] = 0.613 mol/L [CH4] = 0.387 mol/L [H2] = 1.839 mol/L [H2O] = 0.387 mol/L

    a) Calculate the value of K at 927oC for this reaction. b) Calculate the value of the equilibrium constant at 927oC for:

    H2O(g) + CH4 (g) ⇌ CO(g) + 3 H2 (g) c) Calculate the value of the equilibrium constant at 927oC for:

    1/3 CO(g) + H2 (g) ⇌ 1/3 CH4 (g) + 1/3 H2O(g) Solution: a) For the first balanced reaction above:

    K = = = 0.0393 L2/mol2[CO] [H2]3 [CH4] [H2O] (0.387 mol/L) (0.387 mol/L)

    (0.613 mol/L) (1.839 mol/L)3

  • Like Example 6.1 (P 201) - II b) Calculate the value of the equilibrium constant at 927oC for:

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

    K = = [CO] [H2]3

    [CH4][H2O]

    Easier way: This K is just the reciprocal of K from part a: 1 Ka

    K = 1

    c) Calculate the value of the equilibrium constant at 927oC for:

    1/3 CO(g) + H2 (g) ⇌ 1/3 CH4 (g) + 1/3 H2O(g)

    K = = [CO]1/3 [H2]

    [H2O]1/3[CH4]1/3

    Easier way: K = (K from part a)1/3

  • Like Example 6.1 (P 201) - II b) Calculate the value of the equilibrium constant at 927oC for:

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

    K = = = 25.45 mol2/L2 [CO] [H2]3

    [CH4][H2O] (0.613 mol/L) (1.839 mol/L)3

    (0.387 mol/L) (0.387 mol/L)

    1 Ka

    K = = = 25.45 mol2/L2 1

    0.0393 L2/mol2 c) Calculate the value of the equilibrium constant at 927oC for:

    1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g)

    K = = [CO]1/3 [H2]

    [H2O]1/3[CH4]1/3 (0.387mol/L)1/3 (0.387 mol/L)1/3

    (0.613 mol/L)1/3 (1.839 mol/L)

    Easier way: K = (Ka)1/3 = (0.0393L2/mol2)1/3

    Easier way: This K is just the reciprocal of K from part a:

  • Determining Equilibrium Concentrations from K Example: Methane can be made by reacting carbon disulfide with hydrogen gas. K for this reaction is 27.8 (L/mol)2 at 900°C.

    CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2S (g)

    At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 mol CS2, 1.10 mol of H2, and 0.45 mol of H2S. How much methane was formed?

    Strategy: (1) Calculate the equilibrium concentrations from the moles given and the volume of the container. (2)Use the value of K to solve for the concentration of methane. (3) Calculate the number of moles of methane from M and V.

  • [ ]2 0.250 molCS 0.0532 M

    4.70 L = =