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Chapter 15

Chapter 15. Chemical Equilibrium. Overview. Equilibrium Reactions Equilibrium Constants K c & K p Equilibrium Expression product/reactant Reaction Quotient Q Calculations Le Chatelier’s Principle disturbing the equilibrium. Overview, cont’d. All reactions are reversible

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Chapter 15

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  1. Chapter 15 Chemical Equilibrium

  2. Overview • Equilibrium Reactions • Equilibrium Constants Kc & Kp • Equilibrium Expression product/reactant • Reaction Quotient Q • Calculations • Le Chatelier’s Principle • disturbing the equilibrium

  3. Overview, cont’d • All reactions are reversible • Dynamic Equilibrium • When the rates of the forward and • reverse reactions are equal • Reactions do not “go to completion” • Cannot use stoichiometric methods to calculate amount of products formed

  4. Equilibrium and Rates A B kf [A] = kr [B] forward rate = reverse rate [B ] = kf = constant [A] kr

  5. Equilibrium Constant Expression • Ratio of Products to Reactant • raised to each coefficient • for example, • N2 + 3H2 2NH3 • K = [NH3]2 [N2][H2]3

  6. cont’d N2 + 3H2 2NH3 Large K, more product Product Favored Small K, more reactant Reactant Favored

  7. In General: aA + bB cC + dD Kc = [C]c[D]d [A]a[B]b products reactants concentration

  8. Forms of Eq. Constant Expression CO(g) + H2O(g) CO2(g) + H2(g) Kc = [CO2][H2] [CO][H2O] 4HCl(g) + O2(g) 2H2O(g) + 2Cl2(g) Kc = [Cl2]2[H2O]2 [O2][HCl]4

  9. Cont’d 2HI H2 + I2 Kf = [H2][I2] [HI]2 H2 + I2 2HI Kr = [HI]2 = 1 [H2][I2] Kf

  10. Cont’d 2{ 2HI H2 + I2 } Kf = [H2][I2] [HI]2 4HI 2H2 + 2I2 K = Kf2 = [H2]2[I2]2 [HI]4

  11. Example N2O4 2NO2 Initial Initial Equilibrium Equilibrium KcN2O4 NO2 N2O4 NO2 0.0 0.02 0.0014 0.0172 0.211 0.0 0.03 0.0028 0.0243 0.211 0.0 0.04 0.0045 0.0310 0.213 0.02 0.0 0.0045 0.0310 0.213 Kc = [NO2]2 [N2O4] generally unitless

  12. Review values of K: N2 + 3H22NH3 Large K, more product K > > 1 Product Favored N2 + 3H22NH3 Small K, more reactant Reactant Favored K < < 1

  13. Example: N2 + 3H2 2NH3 Kc = 4.34 x 10-3 at 300°C = [NH3]2 [H2]2 [N2] • What is K for reverse reaction? • What is K for 2N2 + 6H2 4NH3 ? • What is K for 4NH3 6H2 + 2N2 ? Kc reverse = 230 Kc = 1.88 x 10-5 Kc = 5.31 x 104

  14. Heterogeneous Equilibria: • When pure solid or liquid is involved • Pure solids & liquids do not appear in the equilibrium constant expression • When H2O is a reactant or product and is the solvent • H2O does not appear in the equilibrium constant expression

  15. both are constant Examples: • CaCO3(s) CaO(s) + CO2(g) • conc. = mol = g/cm3 = density cm3 g/mol MM • Kc = [CaO] [CO2] = (constant 1) [CO2] [CaCO3] (constant 2) • Kc’ = Kc (constant 2) = [CO2] (constant 1)

  16. Multi-Step Equilibria AgCl(s) Ag+(aq) + Cl-(aq)K1 = [Ag+][Cl-] Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq)K2 = [Ag(NH3)2+] [Ag+][NH3]2 AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl-(aq) Ktot = K1 K2 = [Ag(NH3)2+][Cl-] [NH3]2

  17. Problem: 1.00 mole of H2 & 1.00 mole of I2 are placed in a 1.0 L container at 520 K and allowed to come to equilibrium. Analysis reveals 0.12 mol of HI present at equilibrium.Calculate Kc. H2(g) + I2(g) 2HI(g) initial 1.00 1.00 0 change -0.06 -0.06 +0.12 equil. 0.94 mol 0.94 mol +0.12 Kc = (0.12)2 = 1.6 x 10-2 (0.94)(0.94)

  18. Conversion between Kp and Kc • Kc • Equilibrium constant using concentrations • Kp • Equilibrium constant using partial pressures Kp = Kc (RT)DnP = n RT V R = 0.0821 L atm/mol K T = Temperature in KDn = tot. mol product - tot. mol reactant

  19. Problem: For 2SO3(g) 2SO2(g) + O2(g) Kc = 4.08 x 10-3 at 1000 K. Calculate Kp. Kp = Kc(RT)Dn = 4.08 x 10-3 (0.0821 x 1000)1 Kp = 0.0335

  20. Problem: For 3H2(g) + N2(g) 2NH3(g) Kc = 0.105 at 472°C. Calculate Kp. Kp = Kc(RT)Dn = 0.105 (0.0821 x 745)-2 Kp = 2.81 x 10-5

  21. Applications of Eq. Constants • Reaction Quotient • Non-equilibrium concentrations used in the equilibrium constant expression Q = K Reaction is at equilibrium Q > K Reaction will shift left to equilibrium Q < K Reaction will shift right to equilibrium

  22. Problem: Kc = 5.6 x 10-12 at 500 K for I2(g) 2I(g) [I2] = 0.020 M & [I] = 2.0 x 10-8 M. Is the reaction at equilibrium? Which direction will it shift to reach equilibrium? Q = [I ]2 = (2.0 x 10-8)2 = 2.0 x 10-14 [I2] (0.020) Q < K (2.0 x 10-14)<(5.6 x 10-12) Reaction Shifts Right to get to equilibrium not at equilibriumbecause Q  K

  23. Calculation of Eq. Concentrations • use the stoichiometry of reaction • initial concentration of all species • change that occurs to all species • equilibrium concentration of all species • reaction will occur to reach the equilibrium point no matter the direction of reaction

  24. Problem: Cyclohexane, C6H12(g), can isomerize to form methylcyclopentane, C5H9CH3(g). The equilibrium constant at 25°C is 0.12. If the original amount was 0.045 mol cyclohexane in a 2.8 L flask, what are the concentrations at equilibrium? C6H12 C5H9CH3 initial 0.045 mol 0 change -x +x equil. 0.045 - x x

  25. Cont’d 0.12 = (x) (0.045 - x) Solve for x which is the equilibrium concentration of methylcyclopentane or the product x = 4.8 x 10-3 mol C5H9CH3 in 2.8 L flask [C5H9CH3] = 1.7 x 10-3 M [C6H12] = 1.4 x 10-2 M

  26. Problem: For the reaction H2 + I2 2HI the Kc = 55.64. You start with 1.00 mol H2 and 1.00 mol I2 in a 0.500 L flask. Calculate the equilibrium concentrations of all species? H2 + I2 2HI initial 2.00 M 2.00 M 0 change -x -x +2x equil. 2.00 - x 2.00 - x 2x

  27. Cont’d perfect square 55.64 = (2x)2 (2.00 -x)2 (55.64)½ = (2x)2 (2.00 -x)2 x = 1.58 M [HI] = 3.16 M [H2] = 0.42 M [I2] = 0.42 M = 2x = 2.00 - x = 2.00 - x ½

  28. Problem: For the reaction H2 + I2 2HI the Kc = 55.64. You start with 1.00 mol H2 and 0.50 mol I2 in a 0.500 L flask. Calculate the equilibrium concentrations of all species? H2 + I2 2HI initial 2.00 M 1.00 M 0 change -x -x +2x equil. 2.00 - x 1.00 - x 2x

  29. Cont’d Kc = [HI]2 [H2][I2] 55.64 = (2x)2 (2.00 -x)(1.00 -x) reduces to a quadratic equation: x2 - 3.232 x + 2.155 = 0 not a perfect square

  30. Quadratic Equation x = - b ± b2 - 4ac 2a for ax2 + bx + c = 0 ½

  31. Cont’d ½ x = +3.232 ± 10.446 - 8.62 2 x = 1.616 ± 0.6755 [HI] = 1.88 M [H2] = 1.06 M [I2] = 0.060 M = 2x = 2.00 - x = 1.00 - x

  32. Le Chatelier’s Principle • When a stress is applied to an equilibrium reaction, the equilibrium shifts to reduce the stress • Types of Stress • Addition or removal of reactant • Addition or removal or product • Increase or decrease of temperature • Change in pressure or volume

  33. 2NOCl(g) Cl2(g) + 2NO(g)DH° = +77 kJ reaction shifts temp.NOCl Cl2NO temp.NOCl Cl2NO reaction shifts pressure reactionshifts volume

  34. Addition of Reactant or Product [C6H12] [C5H9CH3] initial 1.4 x 10-2 + 1.0 x 10-2 M 1.7 x 10-3 M change -x +x equil. 2.4 x 10-2 - x 1.7 x 10-3 + x 0.12 = (1.7 x 10-3 + x) x = 1.05 x 10-3 M (2.4 x 10-2 - x) [C6H12] = 0.023 M [C5H9CH3] = 0.0028 M

  35. Changes in Temperature • will change K • for an endothermic reaction • increasing T increases K • for an exothermic reaction • increasing T decreases K

  36. withcatalyst withoutcatalyst Effect of a Catalyst rr rf Ea (f) Ea (r) Energy A B Reactions Path

  37. Reaction Mechanisms & Equilibria 2O3(g) 3O2(g) overall rxn O3(g) O2(g) + O(g) fast equil. rate1 = k1[O3] rate2 = k2[O2][O] O(g) + O3(g) 2O2(g) slow rate3 = k3[O][O3] rate 3 includes the conc. of an intermediate and the exptl. rate law will include only species that are present in measurable quantities k1 k2 k3

  38. Substitution Method at equilibrium k1[O3] = k2[O2][O] rate3 =k3[O][O3] [O] = k1 [O3] k2 [O2] rate3 = k3k1 [O3]2 or k2 [O2] overall rate = k’ [O3]2 [O2] substitute

  39. Problem: Derive the rate law for the following reaction given the mechanism step below: OCl -(aq) + I -(aq) OI -(aq) + Cl -(aq) OCl - + H2O HOCl + OH - fast I - + HOCl HOI + Cl - slow HOI + OH - H2O + OI - fast k1 k2 k3 k4

  40. Cont’d rate1 = k1 [OCl -][H2O] = rate 2 = k2 [HOCl][OH -] [HOCl] = k1[OCl -][H2O] k2[OH -] rate 3 = k3 [HOCl][I -] rate 3 = k3k1[OCl -][H2O][I -] k2 [OH -] overall rate = k’ [OCl -][I -][OH -] solvent

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