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Chemistry: The Molecular Science Moore, Stanitski and Jurs

Chemistry: The Molecular Science Moore, Stanitski and Jurs. Chapter 14: Chemical Equilibrium. Characteristics of Chemical Equilibrium. At equilibrium, equilibrium [Reactants] stop decreasing and [Products] stop increasing. If there are more: products than reactants = product favored

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Chemistry: The Molecular Science Moore, Stanitski and Jurs

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  1. Chemistry: The Molecular Science Moore, Stanitski and Jurs Chapter 14: Chemical Equilibrium

  2. Characteristics of Chemical Equilibrium At equilibrium,equilibrium [Reactants] stop decreasing and [Products] stop increasing. If there are more: products than reactants = product favored reactants than products = reactant favored C A + B (aqeous or gas phase)

  3. Equilibrium is Dynamic a A + b B c C + d D Reactants produce products BUT products can also produce reactants. • Equilibrium occurs when [A], [B], [C] and [D] no longer change. • A dynamic equilibrium: • Rate of formation = rate of removal CH3COOH + H2O CH3COO- + H3O+ 14 14

  4. Equilibrium is Independent of Direction of Approach N2(g) + 3 H2(g) 2 NH3(g) start with eitheror 2 mol NH3 1 mol N2 + 3 mol H2 and get the same equilibrium mixture. (@ given T)

  5. Equilibrium and Catalysts Catalystsdo not affect equilibrium concentrations. A catalyst • speeds up a forward reaction… • …but also speeds up the reverse reaction. • decreases the time needed to reach equilibrium but does not improve yield.

  6. H H3C CH3 H3C C=C C=C CH3 H H H The Equilibrium Constant cis-2-butene trans-2-butene At equilibrium: rate forward = rate in reverse kforward[cis] = kreverse[trans] Keq = kforward = [trans] = [products] kreverse [cis] [reactants]

  7. The Equilibrium Constant At equilibrium the concentrations become constant.

  8. kforward kreverse [trans] [cis] [trans] [cis] or = kforward kreverse Kc = = = 1.65 (at 500 K) The Equilibrium Constant We had: kforward[cis] = kreverse[trans] This ratio is named the equilibrium constant, Kc: note…these are [ ] at equilibrium

  9. Equilibrium Constant Equilibrium constant Kc is a quotient of equilibrium concentrations of reactant and product • constant value for a given reaction at a given temperature • varies with temperature since values of rate constant vary with temperature

  10. H H3C CH3 H3C C=C C=C CH3 H H H Practice: 14.3 cis trans After a mixture of cis-2-butene and trans-2-butene has reached equilibrium at 600K, with Keq=1.47, half of the cis-2-butene is suddenly removed. Answer these question: Is the new mixture at equilibrium? In the new mixture, which rate is faster, cistrans or trans  cis? In an equilibrium mixture, which concentration is larger, cis or trans? If the concentration of cis at equilibrium is 0.10 M, what will be the concentration of trans?

  11. kforward kreverse [C]c [D]d [A]a [B]b Kc = = The Equilibrium Constant For a general reaction: a A + b B c C + d D Since kf or kr can change with T, Kc is T dependant.

  12. [NO]2 [N2] [O2] N2(g) + O2(g) 2 NO(g) S8(s) + O2(g) SO2(g) Kc = [SO2] [O2] Kc = 1 8 The Equilibrium Constant Examples S8 is ignored. All pure solids (or liquids) are omitted from Kc.

  13. density mol. wt g / L g / mol Equilibria Involving Pure Liquids and Solids [Solid] is constant throughout a reaction. • pure solid concentration(mol/L) = • [S8] = dS8 / mol. wt S8 • d and mol. wt. are constants, so [S8] is constant. • This constant factor is “absorbed” into Kc. Pure liquids are omitted for the same reason.

  14. [NH4+][OH-] [NH3] NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Kc = Equilibria in Dilute Solutions Water is omitted from Kc in dilute solution reactions: • It is in large excess (pure water = 55.5 M). • Even if it appears in the reaction equation, [H2O] ≈ constant. • The constant factor is incorporated into Kc. = 1.8 x 10-5 (at 25°C) (Units are customarily omitted from Kc)

  15. [H2O]2 [SiH4][O2]2 Write down Kc for: 2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O(l) [Na2SO4] [H2SO4][NaOH]2 The Equilibrium Constant What is the equilibrium constant for: SiH4(g) + 2 O2(g) SiO2(s) + 2 H2O(g) ? Kc = Kc =

  16. Calculating Equilibrium Constant An equilibrium mixture contains 1.50 mol of nitrogen, 4.45 mol of hydrogen and 7.05 mol of ammonia in a 5.00 L reaction vessel. Calculate Kc for the Haber-Bosh process. N2(g) + 3 H2(g) 2 NH3(g)

  17. Calculating Equilibrium Constant Consider the following data for reaction: 2HI(g) H2(g) + I2(g) [HI]:1.000 reactants only 0.0185 [H2]:0.000 [I2]: 0.000 [HI]:0.000 products only [H2]:1.000 0.0183 [I2]: 1.000 [HI]:1.000 [H2]:1.000 mixture 0.0183 [I2]: 1.000 Reactant or product favored? Kc><1?

  18. Calculating Equilibrium Constant Experiment 1. Only reactants are present: Experiment 2. Only products are present: Experiment 3. Mixture is present: Equilibrium constant expression has a constant value regardless of the initial concentrations of reactants and products.

  19. N2(g) + 3 H2(g) 2 NH3(g) [NH3] [N2]½[H2] [NH3]2 [N2][H2]3 Kc1 = Kc2 = ½ N2(g) + H2(g) NH3(g) 3 2 3 2 Kc for Related Reactions Change the stoichiometry… …change Kc reaction divided by 2 • Kc2 is the square root of Kc1. • Multiply an equation by a factor… • Raise Kc to the power of that factor.

  20. [NH3]2 [N2][H2]3 Kc1 = 2 NH3(g) N2(g) + 3 H2(g) [N2][H2]3 [NH3]2 Kc = Kc for Related Reactions N2(g) + 3 H2(g) 2 NH3(g) Reverse a reaction… … Invert Kc1:

  21. step 1 N2(g) + O2(g) 2NO(g) [NO2]2 [NO]2[O2] [NO2]2 [NO]2[O2] [NO2]2 [N2][O2]2 [NO2]2 [N2][O2]2 [NO]2 [N2][O2] [NO]2 [N2][O2] Kc = Kc = Kc = 2 NO(g) + O2(g) 2 NO2(g) N2(g) + 2 O2(g) 2 NO2(g) Kc(step1) x Kc(step2) = = Kc for a Reaction that Combines Reactions If a reaction can be written as a series of steps: step 2 overall The overall Kc is the product of the steps:

  22. Practice For the following equation, Kc1 = 7.9x1011 @500K. H2(g) + Br2(g) 2 HBr (g) • ½H2(g) + ½ Br2(g) 1 HBr (g) Kc2=? • 2HBr (g) H2(g) + Br2(g) Kc3=? • 4HBr (g) 2H2(g) + 2Br2(g) Kc4=? Kc2=8.9x105, Kc3=1.3x10-12,Kc4=1.7x10-24

  23. Practice 14.2 When carbon dioxide dissolves in water it reacts to produce carbonic acid, H2CO3(aq), which can ionize in two steps H2CO3(aq) + H2O(aq) HCO-3(aq) + H3O+(aq) Kc1=4.2x10-7 HCO-3(aq) + H2O(aq) CO3-2(aq) + H3O+(aq) Kc2=4.8x10-11 Calculate the equilibrium constant for the reaction H2CO3(aq) + 2H2O(aq) CO3-2(aq) + 2H3O+(aq) Kc3=? Kc3 = Kc1*Kc2 = 2.0x10-17

  24. Questions • What is the equilibrium constant for a reaction? • At equilibrium, do products or reactants dominate? • Given initial concentrations, which direction will a reaction go? • What concentrations of reactants and products are present at equilibrium?

  25. kforward kreverse [C]c [D]d [A]a [B]b Kc = = a A + b B c C + d D Example Consider: 2 A (aq) B (aq) At equilibrium [A] = 2.0 M & [B] = 4.0 M. What is Kc? Determining Equilibrium Constants If all the equilibrium concentrations are known it’s easy to calculate Kc for a reaction…. Kc = [B]/[A]2 = 4.0/(2.0)2 = 1.0

  26. Determining Equilibrium Constants Example H2(g) and I2(g) were added to a heated container. [H2]initial = 0.0100 mol/L and [I2]initial = 0.00800 mol/L. At equilibrium [I2] = 0.00560 mol/L. Determine Kc. H2(g) + I2(g) 2 HI(g) … In other cases stoichiometry is used to find some concentrations.

  27. H2(g) + I2(g) 2 HI(g) [ ]initial 0.01000 0.00800 0 Determining Equilibrium Constants Write a balanced equation; fill in [ ]initial: [ ]change 2x -x -x [ ]equilib 0.01000- x 0.00800 - x 2x Since [I2] = 0.00560 M at equilibrium 0.00560 = 0.00800 - x -0.00240 M = - x x = 0.00240 M

  28. (0.00480)2 (0.00760)(0.00560) [HI]2 [H2][I2] Kc = = Determining Equilibrium Constants Calculate [ ]equilib and then Kc [H2]equilib =0.01000- (0.00240) = 0.00760 M [I2]equilib =0.00800- (0.00240) = 0.00560 M [HI]equilib =2(0.00240) = 0.00480 M Kc = 0.541

  29. Lab Data Fe+3 +SCN- FeSCN+2 Abs. Kc = 3.29x10-5 = 204 (9.67x10-4)(1.67x10-4)

  30. Calculating Equilibrium Constant • 1.00 mol N2 and 3.00 mol H2 are placed in a 5.00 L reaction vessel at 450 oC and 10.0 atmospheres of pressure. The reaction is N2(g) + 3 H2(g) 2 NH3(g) • What is the value of equilibrium constant if the equilibrium mixture contains 0.0160 mol NH3? • Is this reaction product or reactant favored?

  31. Calculating Equilibrium Constant 1.00 mol N2 and 3.00 mol H2 are placed in a 5.00 L reaction vessel at 450 oC and 10.0 atmospheres of pressure. What is the value of equilibrium constant if the equilibrium mixture contains 0.0160 mol NH3? N2 (g) + 3 H2 (g) 2 NH3 (g) First, calculate the concentrations given…

  32. Calculating Equilibrium Constant Then put them into an “ICE” table … N2 (g) + 3 H2 (g) 2 NH3 (g) 0.200 0.600 0 0.200 -x 0.600-3x 2x = 0.00320 2x = 0.00320 x = 0.00160

  33. Calculating Equilibrium Constant N2(g) + 3 H2(g) 2 NH3(g) 0.200 0.600 0.200 -x 0.600-3x (2x =) 0.00320 Equilibrium concentration of NH3 = 0.00320 M Equilibrium concentration of H2 = 0.600 – 3*(0.00160) = 0.5952 M Equilibrium concentration of N2 = 0.200 – 1*(0.00160) = 0.1984 M Kc << 1, reaction is strongly reactant favored.

  34. Calculating Equilibrium Concentrations If Kc is known, [ ]equilib can be calculated… Problem H2(g) + I2(g) 2 HI(g) 0.0500 mol of HI is placed in an empty 1.00 L flask at 600K. What will be [HI], [H2]and [I2] at equilibrium? Kc = 76

  35. Kc = = (0.0500 – 2x)2 (x)(x) [HI]2 [H2][I2] [ ]equilib +x +x 0.0500 – 2x H2(g) + I2(g) 2 HI(g) [ ]initial 0 0 0.0500 Calculating Equilibrium Concentrations [ ]change +x +x -2x

  36. Kc = 76 = [HI]2 [H2][I2] (0.0500 – 2x)2 x2 Take the square root of both sides: 8.718 x = ±(0.0500 – 2x) Calculating Equilibrium Concentrations 76 x2 = (0.0500 – 2x)2 10.718 x = 0.0500 or 6.718 x = -0.0500 x = 4.67 x 10-3or x = -7.44 x 10-3

  37. [HI]2 [H2][I2] (0.0407 M)2 (0.00467 M)(0.00467 M) Kc = = = 76. Calculating Equilibrium Concentrations So: [H2]equilib =x = 0.00467 M [I2]equilib =x = 0.00467 M [HI]equilib =0.0500 – 2x = 0.0407 M It’s a good idea to check your work: Matches Kc given in the problem

  38. Calculating Equilibrium Concentrations Consider: A(g) B(g) + C(g) Kc = 2.500 A is added to an empty container. [A]initial = 0.1000 M. What are [A], [B] and [C] at equilibrium?

  39. x2 (0.1000 – x) 2.500 = (x)(x) (0.1000 – x) [B][C] [A] Kc = = Calculating Equilibrium Concentrations A B + C [ ]Initial 0.100 0 0 [ ]change -xxx [ ]equilib (0.100 – x) x x A quadratic equation x2 + 2.500 x – 0.2500 = 0

  40. has two roots (solutions): x = -b ± b2 - 4ac 2a −2.500± (2.500)2− (4)(1)(−0.250) 2(1) x = Calculating Equilibrium Concentrations x2 + 2.500 x – 0.2500 = 0 ax2 + bx + c = 0 Here: 1x2+ 2.500 x– 0.2500= 0 abc x = +0.0963 M or -2.596 M

  41. Calculating Equilibrium Concentrations [ A ] = 0.1000 - x = 0.0037 M [ B ] = x = 0.0963 M [ C ] = x = 0.0963 M Most of the A is converted to products. A(g) B(g) + C(g) Kc = 2.500 A is added to an empty container. [A]initial = 0.1000 M. What are [A], [B] and [C] at equilibrium? Check: (0.0963)(0.0963)/0.0037 = 2.5

  42. Calculating Equilibrium Concentrations A gaseous mixture containing 0.050 mol each of CO and H2O in a 1.00 L closed container is heated to 420ºC The reaction is: CO(g) + H2O(g) CO2(g) + H2(g) Kc= 0.100 (at 420ºC) What is the total pressure when the equilibrium is achieved? What is the partial pressure of CO2 in the mixture? [CO]eq=0.038, [CO2]eq=0.012, PCO2=0.68atm, Ptotal = 5.7atm

  43. Example Consider the following equilibrium: CO (g) + 3 H2(g) CH4(g) + H2O (g) A 1.00 L of a gaseous mixture contained 0.300 mol CO, 0.100 mol H2, 0.0200 mol H2O, and an unknown amount of CH4 at equilibrium. What is the concentration of CH4 in this mixture? Kc = 3.92.

  44. Example Consider the following equilibrium: • Suppose you start with 1.000 mol each of carbon monoxide and water in a 50.0 L container. Calculate the molarity of each substance in the equilibrium mixture at 1000 oC. • Kc for the reaction is 0.580 at 1000 oC.

  45. Meaning of the Equilibrium Constant When: • Kc >> 1Reaction is strongly product favored. • very little reactant remains • often written as a forward reaction only. • assume reaction goes to completion. • Kc << 1Reaction is strongly reactant favored. • very little product forms • usually written as “no reaction” or NR • Kc≈ 1Reactants & products present at equilibrium. • use equilibrium methods discussed here.

  46. For the reaction: aA + bB cC + dD [C]c [D]d [A]a [B]b Q = Predicting the Direction of a Reaction The reaction quotient, Qcan be used to predict the direction of a reaction: Q looks identical to Kc, BUT… the concentrations are not equilibrium values. (Kc= Q whenever a system is at equilibrium)

  47. [Products] [Reactants] [Products]equilib [Reactants]equilib Q = Kc= Predicting the Direction of a Reaction Reactants Products Kc is a constant (only changes if T changes). • If Q < Kc, Q must increase to reach equilibrium. • make more product (and less reactant). • move forward. • If Q > Kc, Q must decrease to reach equilibrium. • make less product (and more reactant). • move back.

  48. (0.250)2 (0.085)2(0.100) [ SO3]2 [SO2]2[O2] Q = = Predicting the Direction of a Reaction 2 SO2(g) + O2(g) 2 SO3(g) Kc = 245 at 1000 K Predict the direction of the reaction if SO2 (0.085 M), O2 (0.100 M) and SO3 (0.250 M) are mixed in a reactor at 1000 K. = 86.5 • Q < Kc, Q must increase to reach equilibrium. • Need more product (less reactant). • Forward direction

  49. Le Chatelier’s Principle “If a system is at equilibrium and the conditions are changed so that it is no longer at equilibrium, the system will react to reach a new equilibrium in a way that partially counteracts the change.” • A system at equilibrium resists change. • If “pushed”, it “pushes back”

  50. “reactants” “products” Removing Products or Adding Reactants If we remove products(analogous to removing water out of the right side of the tube) the reaction would shift to the rightuntil equilibrium was reestablished.

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