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CHEMICAL EQUILIBRIUM Chapter 16

CHEMICAL EQUILIBRIUM Chapter 16. PLAY MOVIE. Pb 2+ ( aq ) + 2 Cl – ( aq )  PbCl 2 (s). Properties of an Equilibrium. Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction. PLAY MOVIE. Pink to blue

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CHEMICAL EQUILIBRIUM Chapter 16

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  1. CHEMICAL EQUILIBRIUMChapter 16 PLAY MOVIE Pb2+(aq) + 2 Cl–(aq)  PbCl2(s)

  2. Properties of an Equilibrium Equilibrium systems are • DYNAMIC (in constant motion) • REVERSIBLE • can be approached from either direction PLAY MOVIE Pink to blue Co(H2O)6Cl2 Co(H2O)4Cl2 + 2 H2O Blue to pink Co(H2O)4Cl2 + 2 H2O  Co(H2O)6Cl2

  3. +  Fe(H2O)63+ + SCN- Fe(SCN)(H2O)52+ + H2O Chemical EquilibriumFe3+ + SCN-e FeSCN2+

  4. Chemical EquilibriumFe3+ + SCN-e FeSCN2+ • After a period of time, the concentrations of reactants and products are constant. • The forward and reverse reactions continue after equilibrium is attained. PLAY MOVIE PLAY MOVIE

  5. Examples of Chemical Equilibria Phase changes such as H2O(s) H2O(liq) PLAY MOVIE

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

  7. Chemical Equilibria CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq) At a given T and P of CO2, [Ca2+] and [HCO3-] can be found from the EQUILIBRIUM CONSTANT.

  8. Equilibrium achieved See Active Figure 16.2 Reaction Quotient & Equilibrium Constant Product conc. increases and then becomes constant at equilibrium Reactant conc. declines and then becomes constant at equilibrium

  9. Reaction Quotient & Equilibrium Constant At any point in the reaction H2 + I2 2 HI

  10. Equilibrium achieved Reaction Quotient & Equilibrium Constant In the equilibrium region

  11. The Reaction Quotient, Q In general, ALL reacting chemical systems are characterized by their REACTION QUOTIENT, Q. a A + b B c C + d D If Q = K, then system is at equilibrium.

  12. THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B c C + d D the following is a CONSTANT (at a given T) If K is known, then we can predict concs. of products or reactants.

  13. Determining K 2 NOCl(g)  2 NO(g) + Cl2(g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of an “ICE” table of concentrations [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change Equilibrium 0.66

  14. Determining K 2 NOCl(g)  2 NO(g) + Cl2(g) Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. Solution Set of an “ICE” table of concentrations [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33

  15. Determining K 2 NOCl(g)  2 NO(g) + Cl2(g) [NOCl] [NO] [Cl2] Initial 2.00 0 0 Change -0.66 +0.66 +0.33 Equilibrium 1.34 0.66 0.33

  16. Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. S(s) + O2(g)  SO2(g)

  17. Writing and Manipulating K Expressions Solids and liquids NEVER appear in equilibrium expressions. NH3(aq) + H2O(liq)  NH4+(aq) + OH-(aq)

  18. The Meaning of K 1. Can tell if a reaction is product-favored or reactant-favored. For N2(g) + 3 H2(g)  2 NH3(g) Conc. of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored.

  19. The Meaning of K For AgCl(s)  Ag+(aq) + Cl-(aq) Kc = [Ag+] [Cl-] = 1.8 x 10-5 Conc. of products is much less than that of reactants at equilibrium. The reaction with small K is strongly reactant-favored. Ag+(aq) + Cl-(aq) e AgCl(s) is product-favored.

  20. Product- or Reactant Favored Product-favored K > 1 Reactant-favored K < 1

  21. The Meaning of K K comes from thermodynamics. (See Chapter 19) ∆G˚ < 0: reaction is product favored ∆G˚ > 0: reaction is reactant-favored If K > 1, then ∆G˚ is negative If K < 1, then ∆G˚ is positive

  22. The Meaning of K 2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium. PLAY MOVIE PLAY MOVIE

  23. The Meaning of K If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? If not, which way does the reaction “shift” to approach equilibrium? See Chemistry Now

  24. The Meaning of K All reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium. Q (2.33) < K (2.5) Reaction is NOT at equilibrium, so [iso] must become ________ and [n] must ____________.

  25. Æ Typical Calculations PROBLEM: Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calc. equilibrium concentrations. H2(g) + I2(g)  2 HI(g)

  26. H2(g) + I2(g)  2 HI(g)Kc = 55.3 Step 1. Set up ICE table to define EQUILIBRIUM concentrations. [H2] [I2] [HI] Initial 1.00 1.00 0 Change Equilib

  27. H2(g) + I2(g)  2 HI(g)Kc = 55.3 Step 1. Set up ICE table to define EQUILIBRIUM concentrations. [H2] [I2] [HI] Initial 1.00 1.00 0 Change -x -x +2x Equilib 1.00-x 1.00-x 2x where x is defined as am’t of H2 and I2 consumed on approaching equilibrium.

  28. H2(g) + I2(g)  2 HI(g)Kc = 55.3 Step 2. Put equilibrium concentrations into Kc expression.

  29. H2(g) + I2(g)  2 HI(g)Kc = 55.3 Step 3. Solve Kc expression - take square root of both sides. x = 0.79 Therefore, at equilibrium [H2] = [I2] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M

  30. Nitrogen Dioxide EquilibriumN2O4(g) e 2 NO2(g) e

  31. Nitrogen Dioxide EquilibriumN2O4(g)  2 NO2(g) If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N2O4] [NO2] Initial 0.50 0 Change Equilib

  32. Nitrogen Dioxide EquilibriumN2O4(g)  2 NO2(g) If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an ICE table [N2O4] [NO2] Initial 0.50 0 Change -x +2x Equilib 0.50 - x 2x

  33. Nitrogen Dioxide EquilibriumN2O4(g)  2 NO2(g) Step 2. Substitute into Kc expression and solve. Rearrange: 0.0059 (0.50 - x) = 4x2 0.0029 - 0.0059x = 4x2 4x2 + 0.0059x - 0.0029 = 0 This is a QUADRATIC EQUATION ax2 + bx + c = 0 a = 4 b = 0.0059c = -0.0029

  34. Nitrogen Dioxide EquilibriumN2O4(g)  2 NO2(g) Solve the quadratic equation for x. ax2 + bx + c = 0 a = 4 b = 0.0059c = -0.0029 x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027

  35. Nitrogen Dioxide EquilibriumN2O4(g)  2 NO2(g) x = 0.026 or -0.028 But a negative value is not reasonable. Conclusion: x = 0.026 M [N2O4] = 0.050 - x = 0.47 M [NO2] = 2x = 0.052 M x = -0.00074 ± 1/8(0.046)1/2 = -0.00074 ± 0.027

  36. Solving Quadratic Equations • Recommend you solve the equation exactly on a calculator or use the “method of successive approximations” • See Appendix A.

  37. Writing and Manipulating K Expressions Adding equations for reactions S(s) + O2(g)  SO2(g) SO2(g) + 1/2 O2(g)  SO3(g) Net equation S(s) + 3/2 O2(g)  SO3(g)

  38. Writing and Manipulating K Expressions Changing coefficients S(s) + 3/2 O2(g)  SO3(g) 2 S(s) + 3 O2(g)  2 SO3(g)

  39. Writing and Manipulating K Expressions Changing direction S(s) + O2(g)  SO2(g) SO2(g)  S(s) + O2(g)

  40. Writing and Manipulating K Expressions Concentration Units We have been writing K in terms of mol/L. These are designated by Kc But with gases, P = (n/V)·RT = conc · RT P is proportional to concentration, so we can write K in terms of P. These are designated by Kp. Kc and Kp may or may not be the same.

  41. Writing and Manipulating K Expressions K using concentration and pressure units Kp = Kc(RT)∆n For S(s) + O2(g)  SO2(g) ∆n = 0 and Kp = Kc For SO2(g) + 1/2 O2(g)  SO3(g) ∆n = –1/2 and Kp = Kc(RT)–1/2

  42. EQUILIBRIUM AND EXTERNAL EFFECTS • Temperature, catalysts, and changes in concentration affect equilibria. • The outcome is governed by LE CHATELIER’S PRINCIPLE • “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”

  43. EQUILIBRIUM AND EXTERNAL EFFECTS Henri Le Chatelier 1850-1936 Studied mining engineering. Interested in glass and ceramics.

  44. EQUILIBRIUM AND EXTERNAL EFFECTS • Temperature change  change in K • Consider the fizz in a soft drink CO2(aq) + HEATCO2(g) + H2O(liq) • K = P (CO2) / [CO2] • Increase T. What happens to equilibrium position? To value of K? • K increases as T goes up because P(CO2) increases and [CO2] decreases. • Decrease T. Now what? • Equilibrium shifts left and K decreases.

  45. Temperature Effects on Equilibrium N2O4 (colorless) + heat 2 NO2 (brown) ∆Ho = + 57.2 kJ Kc (273 K) = 0.00077 Kc (298 K) = 0.0059 PLAY MOVIE

  46. Temperature Effects on Equilibrium See Figure 16.8

  47. EQUILIBRIUM AND EXTERNAL EFFECTS • Add catalyst  no change in K • A catalyst only affects the RATE of approach to equilibrium. Catalytic exhaust system

  48. Haber-Bosch Process for NH3 • N2(g) + 3 H2(g)  2 NH3(g) + heat • K = 3.5 x 108 at 298 K

  49. Haber-Bosch Ammonia Synthesis Fritz Haber 1868-1934 Nobel Prize, 1918 Carl Bosch 1874-1940 Nobel Prize, 1931

  50. EQUILIBRIUM AND EXTERNAL EFFECTS • Concentration changes • no change in K • only the equilibrium composition changes.

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