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CHEMICAL EQUILIBRIUM Chapter 16

CHEMICAL EQUILIBRIUM Chapter 16. Pb 2+ (aq) + 2 Cl – (aq) Æ PbCl 2 (s). Properties of an Equilibrium. Equilibrium systems are DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction. Pink to blue Co(H 2 O) 6 Cl 2 ---> Co(H 2 O) 4 Cl 2 + 2 H 2 O.

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CHEMICAL EQUILIBRIUM Chapter 16

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  1. CHEMICAL EQUILIBRIUMChapter 16 Pb2+(aq) + 2 Cl–(aq) Æ PbCl2(s)

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

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

  4. Examples of Chemical Equilibria Phase changes such as H2O(s) ÆH2O(liq)

  5. Examples of Chemical Equilibria Formation of stalactites and stalagmites CaCO3(s) + H2O(liq) + CO2(g)ÆCa2+(aq) + 2 HCO3-(aq)

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

  7. Equilibrium achieved Screen 16.4 & Active Figure 16.3 Reaction Quotient & Equilibrium Constant Product conc. increases and then becomes constant at equilibrium Reactant conc. declines and then becomes constant at equilibrium

  8. Reaction Quotient & Equilibrium Constant At any point in the reaction H2 + I2Æ 2 HI

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

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

  11. THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B Æ 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.

  12. Determining K 2 NOCl(g) Æ 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

  13. Determining K 2 NOCl(g) Æ 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 a 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

  14. Determining K 2 NOCl(g) Æ 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

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

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

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

  18. The Meaning of K For AgCl(s) Æ 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) Æ AgCl(s) is product-favored.

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

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

  21. 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 Screen 16.9

  22. 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 ____________.

  23. 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) Æ 2 HI(g)

  24. H2(g) + I2(g) Æ 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

  25. H2(g) + I2(g) Æ 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.

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

  27. H2(g) + I2(g) Æ 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

  28. Nitrogen Dioxide EquilibriumN2O4(g) Æ 2 NO2(g) Æ

  29. Nitrogen Dioxide EquilibriumN2O4(g) Æ 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

  30. Nitrogen Dioxide EquilibriumN2O4(g) Æ 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

  31. Nitrogen Dioxide EquilibriumN2O4(g) Æ 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

  32. Nitrogen Dioxide EquilibriumN2O4(g) Æ 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

  33. Nitrogen Dioxide EquilibriumN2O4(g) Æ 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

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

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

  36. 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.

  37. 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.”

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

  39. EQUILIBRIUM AND EXTERNAL EFFECTS • Temperature change ---> change in K • Consider the fizz in a soft drink CO2(aq) + HEATÆCO2(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.

  40. Temperature Effects on Equilibrium N2O4 (colorless) + heatÆ 2 NO2 (brown) ∆Ho = + 57.2 kJ Kc (273 K) = 0.00077 Kc (298 K) = 0.0059

  41. Temperature Effects on Equilibrium Figure 16.8

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

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

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

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

  46. 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.”

  47. Le Chatelier’s Principle • Change T • change in K • therefore change in P or concentrations at equilibrium • Use a catalyst: reaction comes more quickly to equilibrium. K not changed. • Add or take away reactant or product: • K does not change • Reaction adjusts to new equilibrium “position”

  48. Æ Nitrogen Dioxide EquilibriumN2O4(g) Æ 2 NO2(g) Increase P in the system by reducing the volume (at constant T).

  49. Nitrogen Dioxide EquilibriumN2O4(g) Æ 2 NO2(g) Increase P in the system by reducing the volume. In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). Therefore, reaction shifts LEFTand P of NO2 decreases and P of N2O4 increases.

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