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Equilibrium student 2014 2

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  • 1. Chemical Equilibrium When a system is at equilibrium, the forward and reverse reaction are proceeding at the same rate The concentrations of all species remains constant over time, but both the forward and reverse reaction never cease That means this is a dynamic equilibrium, not like a beach ball on a seal’s nose. Equilibrium is signified with double arrows or the equal sign Symbol found in “Arrows” section of Insert equations in Microsoft Office.
  • 2. Dynamic Equilibrium
  • 3. For given overall system composition Always reach same equilibrium concentrations Whether equilibrium is approached from forward or reverse direction
  • 4. Mass Action Expression
  • 5. Mass Action Expression Uses stoichiometric coefficients as exponent for each reactant For reaction: aA + bB cC + dD c d [C ] [D ] Q= b a [ A] [B ] Reaction quotient Numerical value of mass action expression Equals “Q ” at any time, and Equals “K ” only when reaction is known to be at equilibrium 6
  • 6. Calculation Results Q = __[HI]2_ = same for all = K if all [H2][I2] experiments at equilibrium Expt I II III IV [H2] .0222 .0350 .0150 .0442 [I2] .0222 .0450 .0135 .0442 [HI] .156 .280 .100 .311 Kc 49.4 49.8 49.4 49.5
  • 7. Equilibrium established
  • 8. Equilibrium Law At 440oC equilibrium law: Kc = _[HI]2_ = 49.5 [H2][I2] Equilibrium constant = Kc at given T Use Kc as normally working with molarity At equilibrium Q = Kc
  • 9. Predicting Equilibrium Law
  • 10. Predicting Equilibrium Law Where only concentrations that satisfy this equation are equilibrium concentrations Numerator Multiply [products] raised to their stoichiometric coefficients Denominator Multiply [reactants] raised to their stoichiometric coefficients Kc = [products]f [reactants]d
  • 11. Check 3H2(g) + N2(g) 2NH3(g) Kc = 4.26 × 108 at 25 °C What is equilibrium law?
  • 12. Check 3H2(g) + N2(g) 2NH3(g) Kc = 4.26 × 108 at 25 °C What is equilibrium law? 2 Kc = [NH3 ] 3 [H2 ] [N2 ] = 4.26 ´ 10 8
  • 13. Example In the reaction: N2O4 Q = [NO2]2 [N2O4] 2 NO2
  • 14. Check Which of the following is the correct mass action expression for the reaction: Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)? A. B. C. Q = Q= [Cu(NH3 )2+ ] 4 [Cu2+ ][NH3 ]4 [Cu(NH3 )2+ ] 4 [Cu2+ ][NH3 ] Q = [Cu2+ ][NH3 ]4 [Cu(NH3 )2+ ] 4 D. none of these 15
  • 15. Manipulating Equations
  • 16. Reverse Equilibrium
  • 17. Kc Related to Equations II
  • 18. Kc Related to Equations III
  • 19. Example
  • 20. Adding Equations
  • 21. Adding Equations
  • 22. Adding Equations II
  • 23. Adding Equations III Adding equations gives the overall equation and multiplying K values gives the overall Kc:Kc = [CO2] x [H2][O2]½ = 1.9 x 105 [CO][O2]½ [H2O] Kc = (5.7 x 1045) x (3.3 x 10-41) = 1.9 x 105
  • 24. Practice 1
  • 25. Practice 2
  • 26. Equilibrium Constant, Kc Constant value equal to ratio of product concentrations to reactant concentrations raised to their respective exponents Kc = [products]f [reactants]d Changes with temperature (van’t Hoff Equation) Depends on solution concentrations Assumes reactants and products are in solution
  • 27. Equilibrium Constant, Kp Based on reactions in which substances are gaseous Assumes gas quantities are expressed in atmospheres in mass action expression Use partial pressures for each gas in place of concentrations Ex. N2 (g) + 3 H2 (g)  2 NH3 (g) Kp = _P2NH3 PN2.P3
  • 28. How are Kp and Kc Related? Start with ideal gas law PV = nRT Rearranging gives æn ö P = ç ÷ RT = MRT èV ø Substituting P/RT for molar concentration into Kc results in pressure-based formula ∆n = moles of gas in product – moles of gas in reactant Kp = Kc(RT ) Dn Kp = Kc when Δn = 0 29
  • 29. Check Consider the reaction A(g) + 2B(g) 4C(g) If the Kc for the reaction is 0.99 at 25ºC, what would be the Kp? A. 0.99 B. 2.0 C. 24. D. 2400 E. None of these
  • 30. Solution
  • 31. Check 2
  • 32. Solution 2
  • 33. Practice 3
  • 34. Heterogeneous Equilibrium Homogeneous reaction/equilibrium All reactants and products in same phase Can mix freely Heterogeneous reaction/equilibrium Reactants and products in different phases Can’t mix freely Solutions are expressed in M Gases are expressed in M Governed by Kc
  • 35. Heterogeneous Equilibrium II
  • 36. Heterogeneous Equilibrium III 2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g) Kc = [Na2CO3(s)] [H2Og][CO2(g)] = [H2Og][CO2(g)] [NaHCO3(s)]2 1 mol NaHCO3 , V = 38.9 mL M = 1 mol/.0389 L = 25.7 M 2 mol NaHCO3 , V = 77.8 mL M = 2 mol/.0778 L = 25.7 M
  • 37. Check Write equilibrium laws for the following: Ag+(aq) + Cl– (aq)  AgCl(s) Kc = ____1___ [Ag+][Cl-] H3PO4(aq) + H2O(ℓ)  H3O+(aq) + H2PO4(aq) Kc = [H3O+][H2PO4-] [H3PO4]
  • 38. Check Given the reaction: 3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s) What is the mass action expression? A. Q = B. Q = C. Q = D. Q = [Ca2+ ]3 [PO3- ]2 4 [Ca3 (PO 4 )2 ] [Ca2+ ]3 [PO3- ]2 4 [1] [Ca3 (PO 4 )2 ] [Ca2+ ]3 [PO3- ]2 4 [1] [Ca2+ ]3 [PO3- ]2 4 39
  • 39. Large Kc Kc >> 1 Rich in product Goes toward complete 2SO2(g) + O2(g)  2SO3(g) Kc = 7.0  1025 at 25 ° C
  • 40. Small Kc Kc << 1 Rich in reactant Few products H2(g) + Br2(g)  2HBr(g) Kc = 1.4  10–21 at 25 °C
  • 41. K near 1
  • 42. Check 43
  • 43. Equilibrium and Changes Equilibrium positions Combination of concentrations that allow Q = K Infinite number of possible equilibrium positions Le Châtelier’s principle System at equilibrium (Q = K) when upset by disturbance (Q ≠ K) will shift to offset stress
  • 44. Reaction Direction Can us value of Q to predict reaction direction 1. Qc > Kc reaction goes to left; reverse 2. Qc < Kc reaction goes to right; forward 3. Qc = Kc reaction at equilibrium. 4. If only reactants Q = 0; forward 5. If only reactants Q = ∞; reverse
  • 45. Relationship Between Q and K Q=K Q<K reaction at equilibrium reactants go to products Too many reactants Must convert some reactant to product to move reaction toward equilibrium: shift to right Q>K products go to reactants Too many products Must convert some product to reactant to move reaction toward equilibrium: shift to left 46
  • 46. Le Châtelier’s Principle 1. 2. 3. 4. 5. Concentration Pressure and volume Temperature Catalysts Adding inert gas to system at constant volume
  • 47. 1. Effect of Change in Concentration Cu(H2O)42+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O blue yellow Equilibrium mixture is blue-green Kc  [CuCl2  (aq )][H2 O]4 4 [Cu(H2 O)2  (aq )][Cl (aq )]4 4  Kc  Kc [H2 O] 4  [CuCl2  (aq )] 4 [Cu(H2 O)2  (aq )][Cl (aq )]4 4 Add excess Cl– (conc HCl) Equilibrium shifts to products Makes more yellow CuCl42– Solution becomes green 48
  • 48. 1. Effect of Change in Concentration Cu(H2O)42+(aq) + 4Cl–(aq) blue CuCl42–(aq) + 4H2O yellow Kc  2 [CuCl4 (aq )] [Cu(H2 O)2  (aq )][Cl (aq )]4 4 Add Ag+ Removes Cl–: Ag+(aq) + Cl–(aq)  AgCl(s) Equilibrium shifts to reactants Makes more blue Cu(H2O)42+ Solution becomes bluer Add H2O? 49
  • 49. Addition of Product Add NO2(g) to N2O4(g) 2NO2(g) System is disturbed so Q now > K; system responds by converting more NO2(g) into N2O4(g) until again Q = K but with larger values for all concentrations.
  • 50. Check 2SO2(g) + O2(g) → 2SO3(g) Kc = 2.4 x 10-3 at 700 oC Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture ? A. Towards the products B. Towards the reactants C. No change will occur
  • 51. Concentration Effect Summary When changing concentrations of reactants or products Equilibrium shifts to remove reactants or products that have been added Equilibrium shifts to replace reactants or products that have been removed
  • 52. Effect of Change in P & V
  • 53. Effect of Change in P & V If Vexpect P To reduce P, look at each side of reaction Which has less moles of gas Reactants = 3 + 1 = 4 moles gas Products = 2 moles gas Reaction favors products (shifts to right)
  • 54. Effect of Change in P & V II 2NaHSO3(s)  NaSO3(s) + H2O(g) + SO2(g) Kp = PH2O.PSO2 If you V of reaction, Reactants: no moles gas = all solids Products: 2 moles gas V, causes P Reaction shifts to left (reactants), as this has fewer moles of gas
  • 55. Effect of Change in Temperature Cu(H2O)42+(aq) + 4Cl–(aq) + Heat  blue CuCl42–(aq) + 4H2O yellow Reaction endothermic Adding heat shifts equilibrium toward products Cooling shifts equilibrium toward reactants 56
  • 56. Temperature Change H2O(s)  H2O(ℓ) H° =+6 kJ (at 0 °C) Energy + H2O(s)  H2O(ℓ) Energy is reactant Add heat, shift reaction right 3H2(g) + N2(g)  2NH3(g) Hf°= –47.19 kJ 3 H2(g) + N2(g)  2 NH3(g) + energy Energy is product Add heat, shift reaction left
  • 57. Effect of T Change  T shifts reaction in direction that produces endothermic (heat absorbing) change  T shifts reaction in direction that produces exothermic (heat releasing) change Changes in T change value of mass action expression at equilibrium, so K changed
  • 58. Effect of T Change II K depends on T T of exothermic reaction makes K smaller More heat (product) forces equilibrium to reactants T of endothermic reaction makes K larger More heat (reactant) forces equilibrium to products
  • 59. Catalysts Catalysts lower Ea Change in Ea has equal effect on kf & kr Result is no effect on K
  • 60. Addition of Inert Gas Will not affect concentrations or partial pressures of any components Will not affect reaction
  • 61. Learning Check: Consider: H3PO4(aq) + 3OH–(aq) Q 3H2O(ℓ) + PO43–(aq) 3 [PO4 ]  3 [OH ] [H3PO4 ] What will happen if PO43– is removed?  Q is proportional to [PO43– ]   [PO43– ],  Q  Q < K equilibrium shifts to right 62
  • 62. Learning Check: The reaction H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq) is exothermic. What will happen if system is cooled? Q     heat 3 [PO4 ] [OH ]3 [H3PO4 ] Since reaction is exothermic, heat is product Heat is directly proportional to Q T,  Q Q < K equilibrium shifts to right 63
  • 63. Check The equilibrium between aqueous cobalt ion and the chlorine ion is shown: [Co(H2O)6]2+(aq) + 4Cl–(aq)  [Co(Cl)4]2–(aq) + 6H2O(ℓ) blue It is noted that heating a pink sample causes it to turn violet. The reaction is: A. endothermic B. exothermic C. cannot tell from the given information pink
  • 64. Calculating Kc from [Equilibrium] Most direct way. Measure all reactant & product concentrations. Kc always be the same as long as T constant 65 Tro: Chemistry: A Molecular Approach, 2/e
  • 65. Equilibrium Calculations For solution reactions, must use KC For gaseous reactions, use either KP or KC Calculate K from equilibrium [X] or PX Find one or more equilibrium [X] or PX from KP or KC
  • 66. Kc From [X}s at Equilibrium Use mass action expression to relate [X] values Can use: Equilibrium [reactant]s & [product]s From initial and one final [product] find remaining {X}s then find Kc
  • 67. Kc from Equilibrium [X] 1 N2O4(g)  2NO2(g) If you place 0.0350 mol N2O4 in 1 L flask at equilibrium, what is KC? [N2O4]eq = 0.0292 M [NO2]eq = 0.0116 M Kc = _[NO2}2 = (0.0292)2 = 4.61 x 10-3 [N2O4] (0.0292)
  • 68. Practice 4 For the reaction: 2A(aq) + B(aq)  3C(aq) the equilibrium concentrations are: A = 2.0 M, B = 1.0 M and C = 3.0 M. What is the expected value of Kc at this temperature? A. 14 B. 0.15 C. 1.5 D. 6.75
  • 69. From [X]i and [Y]f Set up ICE table Changes in same ratio as coefficients [X]eq = [X]initial - [X]change 1. Assign x as coefficient, + for materials on side it is going to or – for opposite 2. Solve for x, if 2nd order, take square root, use quadratic equation or simplify if K very large or very small.
  • 70. Generic Use concentration tables set up as follows: A + B  2C [A] [B] [C] I P Q R C -x -x +2x E P–x Q–x R + 2x Kc = _ (R + 2x)2__ (P - x)(Q - x)
  • 71. Example 2SO2(g) + O2(g)  2SO3(g) 1.000 mol SO2 and 1.000 mol O2 are placed in a 1.000 L flask at 1000 K. At equilibrium 0.925 mol SO3 has formed. Calculate KC for this reaction. [SO2]i = [O2]i = 1.00 mol = 1.00 M 1.00 L [SO3]eq = 0.925 mol = 0.925 M 1.00 L
  • 72. ICE Table 2SO2(g) + O2(g)  2SO3(g) I 1.00 1.00 0.000 C -0.925 -0.462 +0.925 E 0.075 0.538 0.925 Kc = __[SO3]2__ = _(0.925)2__ 2.8 x 102 [SO2]2[O2] (.075)2(.538)
  • 73. Practice 5 CH4(g) + H2O(g)  CO(g) + 3H2(g) Kc = 5.67 [CH4] = 0.400 M; [H2] = 0.800M; [CO] =0.300M What is [H2O] at equilibrium?
  • 74. Practice H2(g) + I2(g)  2HI(g) at 425 °C KC = 55.64 If one mole each of H2 and I2 are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI? Kc = __[HI]2_= 55.64 [H2][I2] [H2]i = [I2]i = 1.00 mol = 2.00 M 0.500 L
  • 75. Solution H2(g) + I2(g)  2HI(g) I 2.00 2.00 0.00 C -x -x +2x E 2.00 – x2.00 – x 2x 55.64 = ____(2x)2______ = ____(2x)2___ (2.00 – x)(2.00 – x) (2.00 – x)2 Take square root of both sides
  • 76. Solution II  55.64 = 7.459 = __(2x)__ (2.00 – x) Rearrange: (7.459)(2.00 – x) = 2x 14.918 – 7.459x = 2x 14.918 = 9.459x x = 14.918 = 1.58 9.459 [H2]eq = [I2]eq = 2.00 – 1.58 = 0.42 M [HI]eq = 2x = 2(1.58) = 3.16 M
  • 77. Example 2 H2(g) + I2(g)  2HI(g) at 425 °C KC = 55.64 If one mole each of H2, I2 and HI are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI? Now have product as well as reactants initially Kc = __[HI]2_= 55.64 [H2][I2] H2]i = [I2]i = [HI] 1.00 mol = 2.00 M 0.500 L
  • 78. Solution H2(g) + I2(g)  2HI(g) I 2.00 2.00 2.00 C -x -x +2x E 2.00 – x2.00 – x 2.00 + 2x 55.64 = __(2.00 + 2x)2__ = (2.00 + 2x)2 (2.00 – x)(2.00 – x) (2.00 – x)2 Take square root of both sides
  • 79. Solution  55.64 = 7.459 = (2.00 + 2x)_ (2.00 – x) Rearrange: (7.459)(2.00 – x) = 2.00 + 2x 14.918 – 7.459x = 2.00 + 2x 12.918 = 9.459x x = 12.918 = 1.37 9.459 [H2]eq = [I2]eq = 2.00 – 1.37 = 0.63 M [HI]eq = 2.00 + 2(1.37) = 2.00 + 2.74 = 4.74 M
  • 80. Practice 6 N2(g) + O2(g) → 2NO(g) Kc = 0.0123 at 3900 oC If 0.25 moles of N2 and O2 are placed in a 250 mL container, what are the equilibrium concentrations of all species ? A. 0.0526 M, 0.947 M, 0.105 M B. 0.947 M, 0.947 M, 0.105 M C. 0.947 M 0.105 M, 0.0526 M D. 0.105 M, 0.105 M, 0.947 M
  • 81. Practice CH3CO2H(aq) + C2H5OH(aq)  CH3CO2C2H5(aq) +H2O(l) acetic acid ethanol ethyl acetate KC = 0.11 An aqueous solution of ethanol and acetic acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?
  • 82. Solution Kc = __[CH3CO2C2H5]__ [CH3CO2H][C2H5OH] CH3CO2H C2H5OH CH3CO2C2H5 I 0.810 0.810 0.000 C -x -x +x E .810 – x .810 – x x 0.11 = ________x________ (.810 – x)(.810 – x)
  • 83. Solution II Rearranging: 0.11 x(0.6561 – 1.62x + x2) = x Expand then write as quadratic equation ax2 + bx + c = 0 0.07217 – 0.1782x + 0.11x2 - x = 0 0.11x2 – 1.1782x + 0.07217 = 0 x = -b  2 – 4ac) (b 2a
  • 84. Solution III 2  (1.1782)  (1.1782)  4(0.11)(0.07217) x  2(0.11) 1.1782  (1.388)  (0.032) 1.1782  1.164 x   0.22 0.22 This gives two roots: x = 10.6 and x = 0.064 Only x = 0.064 is possible x = 10.6 is >> 0.810 initial concentrations 0.810 – 10.6 = negative concentration, which is impossible 85
  • 85. Solution IV [CH3CO2C2H5]eq = x = 0.064 M [CH3CO2H] = [C2H5OH] = 0.810 – x = 0.810 – 0.064 = 0.746 M
  • 86. Simplifications If get trinomial use successive approximations If K is very small x is also small can drop x in change part only Valid if [X]I = 400x > K
  • 87. More Problems 1. Finding Kc from change in reactant concentrations. PCl 3(g) + Cl2(g)  PCl5(g) 0.200 mol PCl3 and 0.100 mol Cl2 put in 1.00 L flask, at equilibrium found 0.120 mol PCl3. Means change in [PCl3] is 0.200 0.120 = 0.080 mol, change must be same for other components. As volume is 1.0 L, molarity has same value.
  • 88. Example Using Kc to calculate equilibrium concentrations: A flask contains a mixture of Br2 and Cl2 ,both with an initial concentration of 0.0250 M. Find the equilibrium concentration of all molecules from the following equilibrium information.
  • 89. Example Continued [BrCl] I 0 C +2x E 2x Kc = [Br2][Cl2] [BrCl]2 [Br2] [Cl2] .0250 .0250 -x -x .0250 – x .0250 - x = (.0250 - x)2 = .145 (2x)2
  • 90. Continued Kc = [Br2][Cl2] = (.0250 - x)2 = .145 [BrCl]2 (2x)2 Take square root: (.0250 - x) = .381 2x .0250 - x = .762 or .0250 = 1.762 x x = .0250/1.762 = .0142 for Br2 & Cl2 BrCl = 2x = .0284 M
  • 91. Practice Find equilibrium concentrations with very small Kc. N2O + NO2  3NO Kc = 1.4 x 10-10 Start with .200 mol N2O & .400 mol NO2 in 4.00 L
  • 92. Solution [N2O] = .200 mol/4.00 L = 0.0500 M [N2O] = .400 mol/4.00 L = 0.100 M I C E [N2O] .0500 -x .0500 – x [NO2] .100 -x .100 – x [NO] 0 +3x 3x
  • 93. Problem 3 (II) Kc = [NO]3 = [N2O][NO2] (3x)3 =1.4 x10-10 (.0500 - x)(.100 - x) Simplify reactant concentrations as x very small: 27x3 = 1.4 x 10-10 (.0500)(.100) x3 = 2.6 x 10-14 or x = 3.0 x 10 -5 [NO] = 3x = 9.0 x 10-5