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Chapter 16 Principles of Reactivity: Chemical Equilibria

Chapter 16 Principles of Reactivity: Chemical Equilibria. Important – Read Before Using Slides in Class

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Chapter 16 Principles of Reactivity: Chemical Equilibria

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  1. Chapter 16Principles of Reactivity: Chemical Equilibria

  2. Important – Read Before Using Slides in Class Instructor: This PowerPoint presentation contains photos and figures from the text, as well as selected animations and videos. For animations and videos to run properly, we recommend that you run this PowerPoint presentation from the PowerLecture disc inserted in your computer. Also, for the mathematical symbols to display properly, you must install the supplied font called “Symb_chm,” supplied as a cross-platform TrueType font in the “Font_for_Lectures” folder in the "Media" folder on this disc. If you prefer to customize the presentation or run it without the PowerLecture disc inserted, the animations and videos will only run properly if you also copy the associated animation and video files for each chapter onto your computer. Follow these steps: 1. Go to the disc drive directory containing the PowerLecture disc, and then to the “Media” folder, and then to the “PowerPoint_Lectures” folder. 2. In the “PowerPoint_Lectures” folder, copy the entire chapter folder to your computer. Chapter folders are named “chapter1”, “chapter2”, etc. Each chapter folder contains the PowerPoint Lecture file as well as the animation and video files. For assistance with installing the fonts or copying the animations and video files, please visit our Technical Support at http://academic.cengage.com/support or call (800) 423-0563. Thank you.

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

  4. 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)6Cl2f Co(H2O)4Cl2 + 2 H2O Blue to pink Co(H2O)4Cl2 + 2 H2O f Co(H2O)6Cl2

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

  6. 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

  7. Examples of Chemical Equilibria Phase changes such as H2O(s) eH2O(liq) PLAY MOVIE

  8. Examples of Chemical Equilibria Formation of stalactites and stalagmites CaCO3(s) + H2O(liq) + CO2(g)eCa2+(aq) + 2 HCO3-(aq)

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

  10. 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

  11. Reaction Quotient & Equilibrium Constant At any point in the reaction H2 + I2e 2 HI

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

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

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

  15. Determining K 2 NOCl(g) e 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

  16. Determining K 2 NOCl(g) e 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

  17. Determining K 2 NOCl(g) e 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

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

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

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

  21. The Meaning of K For AgCl(s) e 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.

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

  23. 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

  24. 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

  25. 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

  26. 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 ____________.

  27. 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) e 2 HI(g)

  28. H2(g) + I2(g) e 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

  29. H2(g) + I2(g) e 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.

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

  31. H2(g) + I2(g) e 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

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

  33. Nitrogen Dioxide EquilibriumN2O4(g) e 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

  34. Nitrogen Dioxide EquilibriumN2O4(g) e 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

  35. Nitrogen Dioxide EquilibriumN2O4(g) e 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

  36. Nitrogen Dioxide EquilibriumN2O4(g) e 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

  37. Nitrogen Dioxide EquilibriumN2O4(g) e 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

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

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

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

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

  42. 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.

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

  44. 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.”

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

  46. EQUILIBRIUM AND EXTERNAL EFFECTS • Temperature change f change in K • Consider the fizz in a soft drink CO2(aq) + HEATeCO2(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.

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

  48. Temperature Effects on Equilibrium See Figure 16.8

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

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

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