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CHAPTER 16:

CHAPTER 16:. PRINCIPLES OF REACTIVITY: Chemical Equilibria. 16.0 OBJECTIVES. State the characteristics of a system in dynamic equilibrium as applied to reversible reactions .

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CHAPTER 16:

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  1. CHAPTER 16: PRINCIPLES OF REACTIVITY: Chemical Equilibria

  2. 16.0 OBJECTIVES • State the characteristics of a system in dynamic equilibrium as applied to reversible reactions. • Calculate the value of and interpret the meaning of the reaction quotient based on either partial pressures or concentrations. • Understand the factors that will change the values of the reaction quotient. • Use LeChatelier’s Principle to describe changes in reactions based on stresses to the equilibrium system.

  3. Homework • #1 - 9, 11, 13, 15, 37, 39, 45 • Basic Equilibrium Calculations • #2 - 17, 19, 21, 23, 41 • ICE Problems • #3 - 27, 29, 31, 43, 51, 53, 55, 57 • More ICE and Conceptual Equilibrium • #4 - 67, 33, 35, 47, 49 • Le’Chatelier’s Principle

  4. Static Equilibrium No movement observed. In a reversible chemical change, a static equilibrium is never established.

  5. Dynamic Equilibrium Movement IS observed Rate of water in-flow = Rate of water out-flow The level of water is constant within the bucket. In a reversible chemical change: a static equilibrium is never established, but Dynamic Equilibrium is!

  6. Examples of Equilibria Phase changes such as H2O(g)↔H2O(liq)

  7. Examples of Chemical Equilibria Formation of stalactites and stalagmites CaCO3(s) +H2O(liq) + CO2(g)↔Ca2+ (aq) + 2 HCO3(aq)

  8. 16.1 NATURE OF THE EQUILIBRIUM STATE • A. Requirements • Reaction that forms products to a certain point (equilibrium!), then reforms reactants (NOT ALL RXNS will go to completion and only form products) • Equ. will not occur as soon as the reversible reaction begins • Like all good things, it takes time!

  9. 16.1 NATURE OF THE EQUILIBRIUM STATE • B. Characteristics of a Specific System • ChemEqn has Double Headed Arrow ↔ • Beginning of Rxn Form lots of products (rate laws) • Moments Later  Forming both products and reactants (equilibrium) • **Not Necessarily at Equal Rates yet! • Dynamic Equilibrium- “two opposing processes occur at exactly the same rate” • Reactants  Products = Products  Reactants

  10. 16.1 NATURE OF THE EQUILIBRIUM STATE • C. Graphs

  11. 16.1 NATURE OF THE EQUILIBRIUM STATE • N2O4 2NO2 Equilibrium Demo • Decomposition Reaction • NO2 = Smog (dark brown color) • N2O4 = Propellant (light brown/clear) • **Observe and Write down some observations!

  12. Nitrogen Dioxide EquilibriumN2O4(g)  2 NO2(g)

  13. 16.2 THE REACTION QUOTIENT AND EQUILIBRIUM CONSTANT • A. For aA(g) + bB(g)  cC(g) + dD(g) Kc = [C]c [D]d … [A]a [B]b ... REACTION QUOTIENT (“Q”) at equilibrium EQUILIBRIUM CONSTANT For a given reaction at equilibrium, the ratio of Products over Reactants raised to their coefficients is CONSTANT. That RATIO will always equal K no matter the conc.

  14. 16.2 THE REACTION QUOTIENT AND EQUILIBRIUM CONSTANT • B. Example 16.1 Write the Kc for • a. H2(g) + I2(g)  2 HI (g) • b. 2 SO2(g) + O2(g)  2SO3(g)

  15. 16.2 THE REACTION QUOTIENT AND EQUILIBRIUM CONSTANT • C. Kc is • independent of • Initial Concentrations • dependent on • Temperature • Reaction/Stoichiometry aA(g) + bB(g)  cC(g) + dD(g) Kc = [C]c [D]d … [A]a [B]b ...

  16. 16.2 THE REACTION QUOTIENT AND EQUILIBRIUM CONSTANT • D. No term for pure liquids, solvents, or solids • Their concentrations do not change during the reaction aA+ bB(s) + nH2O(l) cC + dD Kc = [C]c [D]d _ [A]a [B]b[H2O]n

  17. Meaning of K? Usefulness of K? • Dictates weather a reaction if product or reactant favored aA(g) + bB(g)  cC(g) + dD(g) Kc = [C]c [D]d … [A]a [B]b ... A+B A+B C+D C+D

  18. 16.2 THE REACTION QUOTIENT AND EQUILIBRIUM CONSTANT In pairs

  19. 16.3 DETERMINING AN EQUILIBRIUM CONSTANT • A. Determine the value of Kc when all equilibrium concentrations are known: • Products over Reactants raised to their coefficients

  20. 16.3 DETERMINING AN EQUILIBRIUM CONSTANT • 1. Example 16.3 For NH4Cl(s)  NH3(g) + HCl(g) at 500oC, at equilibrium there are 2.0 moles of ammonia, 2.0 moles of hydrochloric acid and 1.0 mole of ammonium chloride present in a 5.0L container. Calculate the equilibrium constant of the system at this temperature.

  21. 16.3 DETERMINING AN EQUILIBRIUM CONSTANT • 1. Example 16.3 For NH4Cl(s)  NH3(g) + HCl(g) at 500oC, at equilibrium there are 2.0 moles of ammonia, 2.0 moles of hydrochloric acid and 1.0 mole of ammonium chloride present in a 5.0L container. Calculate the equilibrium constant of the system at this temperature. Kc = [NH3]1 [HCl]1 Kc = [2.0/5.0] [2.0/5.0] Kc = 0.16 *note: units never given with K!

  22. 16.3 DETERMINING AN EQUILIBRIUM CONSTANT In pairs • 2. Example 16.4 For the system 2SO3(g)  2SO2(g) + O2(g) at a given temperature the equilibrium concentrations are [SO2] = [O2] = 0.10M and [SO3] = 0.20M. Calculate the Kc at this temperature.

  23. 16.3 DETERMINING AN EQUILIBRIUM CONSTANT • 2. Example 16.4 For the system 2SO3(g)  2SO2(g) + O2(g) at a given temperature the equilibrium concentrations are [SO2] = [O2] = 0.10M and [SO3] = 0.20M. Calculate the Kc at this temperature. Kc = [SO2]2 [O2]1 [SO3]2 Kc = [0.10] [0.10] [0.20]2 Kc = 0.25

  24. 16.3 DETERMINING AN EQUILIBRIUM CONSTANT • B. Determine the value of Kc when some equilibrium concentrations are missing: Kc = Equation Initial(M) Change(M) Equilibrium(M) 2HI(g)  I2(g) + H2(g) 0.100 0 0 - + + 0.010

  25. 16.3 DETERMINING AN EQUILIBRIUM CONSTANT Ex. 16.5 Equation Initial(M) 0.100 0 0 Change(M) - 0.020 +0.010 + 0.010 Equilibrium(M) 0.100-0.020.010 0.010 =(0.080) 2HI(g)  I2(g) + H2(g) Kc = [I2]1 [H2]1 = [0.010]1 [0.010]1 = [HI]2 [0.080]2 0.016

  26. 16.3 DETERMINING AN EQUILIBRIUM CONSTANT In pairs 2HI(g)  I2(g) + H2(g) Equation Initial(M) Change(M) Equilibrium(M)

  27. 16.3 DETERMINING AN EQUILIBRIUM CONSTANT 2HI(g)  I2(g) + H2(g) 0 0.100 0 Equation Initial(M) Change(M) Equilibrium(M) + X + X -2X 0.074 X=0.013 X=0.013 0.100 – 2X = 0.074 X = 0.013 Kc = [I2]1 [H2]1 = [HI]2 0.031 =(0.013)(0.013)/(0.074)2 QUIZ!!!!!!!

  28. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS • A. Predicting the direction of shift of a reaction to reach equilibrium • 1. For aA(g) + bB(g)  cC(g) + dD(g) Kc = [C]c [D]d …_ [A]a [B]b ... Q = [C]c [D]d …_ [A]a [B]b ... REACTION QUOTIENT (Q) EQUILIBRIUM CONSTANT (Kc) is essentially REACTION QUOTIENT (Q) at equilibrium In Text Book, Q is introduced in 16.2, pg 663

  29. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS • 2. Relationship between Kc and Q • Q is the ratio of Products/Reactants at ANY given point in time, not necessarily at equilibrium • Q values can change as reaction proceeds • Kc is the equilibrium constant which is established for EQUILIBRIUM CONDITIONS ONLY! Pg. 665

  30. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS aA(g) + bB(g) cC(g) + dD(g) Kc = [C]c [D]d …_ [A]a [B]b ... Q = [C]c [D]d …_ [A]a [B]b ... Pg. 665

  31. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS • 3. Example 16.7 N2(g) + 3H2(g)  2NH3(g) Kc= 5 x 108 at 25oC • If the original concentrations are [N2] = [H2] = 2.0M, which way will the reaction run? Q = Compare Q to Kc:

  32. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS • 3. Example 16.7 N2(g) + 3H2(g)  2NH3(g) Kc= 5 x 108 at 25oC • If the original concentrations are [N2] = [H2] = 2.0M, which way will the reaction run? • Q = [NH3]2 _ [N2]1 [H2]3 • Q = [0]2 _ [2.0]1 [2.0]3 • Q< Kc, products favored At equilibrium, Q = Kc = 5 x 108

  33. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS In pairs • 4. Example 16.8 For the system N2O4(g)  2NO2(g) Kc = 0.36 at 100oC: • Predict the direction the reaction will shift to reach equilibrium if the original concentrations are: • a. 0.20 moles of N2O4 in a 4.0L container • b. 0.20 moles of N2O4 and 0.20 moles of NO2 in a 4.0L container • c. 1.00M NO2 and N2O4 c. b. a.

  34. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS • 4. Example 16.8 For the system N2O4(g)  2NO2(g) Kc = 0.36 at 100oC: • Predict the direction the reaction will shift to reach equilibrium if the original concentrations are: • a. 0.20 moles of N2O4 in a 4.0L container • b. 0.20 moles of N2O4 and 0.20 moles of NO2 in a 4.0L container • c. 1.00M NO2 and N2O4 Q = [NO2]2 / [N2O4], Kc = 0.36 c. Q = [0]2 / [.2/4] Q=0 Q = [1.0]2 / 1.0] Q=1 b. a. Q = [.2/4]2 / [.2/4] Q=0.05 Q<K, products favored Q>K, reactants favored Q<K, products favored

  35. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS • B. Determine the equilibrium concentration of one reactant or product: • Example 16.9 For N2(g) + O2(g)  2NO(g) Kc = 1.0 x 10-30at 25oC • Calculate the equilibrium concentration of NO(g) if at equilibrium the concentration of N2(g) is 0.04M and that of O2 is 0.01M.

  36. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS • B. Determine the equilibrium concentration of one reactant or product: • Example 16.9 For N2(g) + O2(g)  2NO(g) Kc = 1.0 x 10-30at 25oC • Calculate the equilibrium concentration of NO(g) if at equilibrium the concentration of N2(g) is 0.04M and that of O2 is 0.01M. Kc =1x10-30= [NO]2 / [N2][O2] 1x10-30= [NO]2 / [0.04][0.01] [NO] = 2x10-17

  37. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS • C. Determine the equilibrium concentrations of all reactants and products from original concentrations and Kc. • Example 16.10For CO2(g) + H2(g)  CO(g) + H2O(g) K c = 0.64 at 900K. Starting with both reactants at a concentration of 0.100M, what are the equilibrium concentrations of all species?

  38. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS • C. Determine the equilibrium concentrations of all reactants and products from original concentrations and Kc. • Example 16.10For CO2(g) + H2(g)  CO(g) + H2O(g) K c = 0.64 at 900K. Starting with both reactants at a concentration of 0.100M, what are the equilibrium concentrations of all species? CO2(g) + H2(g)  CO(g) + H2O(g) I 0.1 0.1 0 0 C -x -x +x +x E 0.1-x 0.1-x x x K c = 0.64 = [x][x]/[0.1-x][0.1-x] Solve for x: x =0.044M X = 0.044 M [CO] = [H2O]

  39. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS CO2(g) + H2(g)  CO(g) + H2O(g) K c = 0.64 at 900K • Example 16.11 For the same system as above, calculate the equilibrium concentrations of all species if the reaction system is started with [CO2] = 0.100M and [H2] = 0.200M. Quad eqn example!

  40. 2) Solve for x 1) Set up Kc 3) Use x to find equil concs

  41. 16.4 USING EQUILIBRIUM CONSTANTS IN CALCULATIONS In pairs • Example 16.12 For N2O4(g)  2NO2(g) Kc = 0.36 at 100oC. Starting with a concentration of 0.100 moles/L for N2O4, what are the equilibrium concentrations of both species?

  42. 1) Set up Kc 2) Solve for x 3) Use x to find equil concs

  43. 16.5 MORE ABOUT BALANCED EQUATIONS AND EQUILIBRIUM CONSTANTS • A. KP • Ratio of Partial Pressures of reactants/products at Equilibrium KP = PCc PDd …_ PAa PBb ...

  44. 16.5 MORE ABOUT BALANCED EQUATIONS AND EQUILIBRIUM CONSTANTS • B. Equation Relationship betweenKp and Kc (from PV=nRT) (pg. 661, “A Closer Look”) • Kp = Kc (RT) n gas • n = change in number of moles going from reactants to products • Sometimes, Kp = Kc

  45. 16.5 MORE ABOUT BALANCED EQUATIONS AND EQUILIBRIUM CONSTANTS • C. Example 16.13 For N2(g) + 3H2(g)  2NH3(g) Kc = 9.5 at 300oC. Calculate the KP at this same temperature. Kp = Kc (RT) n gas

  46. 16.5 MORE ABOUT BALANCED EQUATIONS AND EQUILIBRIUM CONSTANTS • D. Equilibrium constants and different forms of the equation For N2O4(g)  2NO2(g) Kc = 0.36 at 100oC • 1. Kc of 2NO2(g)  N2O4(g) • Reverse Equation: 1/K Kc forward rxn = Kc reverse rxn =

  47. 16.5 MORE ABOUT BALANCED EQUATIONS AND EQUILIBRIUM CONSTANTS • 2. Kc of 4NO2(g)  2N2O4(g) • Multiply by coefficent: Kcoefficient • 3. Kc of 1/2N2O4(g)  NO2(g) • See above • 4. Summation of Equations Rule • Multiply K values for all equations

  48. 16.5 MORE ABOUT BALANCED EQUATIONS AND EQUILIBRIUM CONSTANTS • 5. Example 16.14 Given: 3/2O2(g)  O3(g) K = 2.5 x 10-29 State the K for: a. 3O2(g)  2O3(g) b. 2O3(g)  3O2(g)

  49. 16.5 MORE ABOUT BALANCED EQUATIONS AND EQUILIBRIUM CONSTANTS • 6. Example 16.15 Given at 500K: • H2(g) + Br2(g)  2HBr(g) Kp = 7.9 x 1011 • H2(g)  2H(g) Kp = 4.8 x 10-41 • Br2(g)  2Br(g) Kp = 2.2 x 10-15 Calculate the Kp for H(g) + Br(g)  HBr(g) at the same temperature QUIZ!!!! HW # 1, 2 and 3 Due!!!

  50. “When a system is stressed (changing conc., temp, pressure, or volume), the system will respond by attaining new equilibrium conditions that counteract the change” 16.6 DISTURBING A CHEMICAL EQUILIBRIUM: Le Chatelier's Principle • A. Statement

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