1. Chapter 15 Chemical Equilibrium

2. Review Section of Chapter 15 Test Calculating an Empirical Formula Stoichiometry (mass – mass) Empirical vs. Molecular Formulas Bonds –Comparison of single/double/triple bonds –Resonance VSEPR theory –Molecular Shape –Bond Angles

3. Reversible Reactions- most chemical reactions are reversible under the correct conditions

4. Reversible Reactions In a reversible reaction, there is both a forward and a reverse reaction. Suppose SO 2 and O 2 are present initially. As they collide, the forward reaction begins. 2SO 2 (g) + O 2 (g) → 2SO 3 (g) As SO 3 molecules form, they also collide in the reverse reaction that forms reactants. A reversible reaction is written with a double arrow. forward 2SO 2 (g) + O 2 (g) 2SO 3 (g) reverse

5. Chemical Equilibrium At equilibrium The rate of the forward reaction becomes equal to the rate of the reverse reaction. The forward and reverse reactions continue at equal rates in both directions. Figure 15.5 Page 454

6. Chemical Equilibrium When equilibrium is Reached: There is no further change in the amounts (concentrations) of reactant and product.

7. N 2 (g) + O 2 (g) ↔ 2NO(g) At equilibrium the concentrations remain constant however the forward and reverse reactions are still occuring.

8. Writing Equilibrium Constant-Expressions

9. The Equilibrium Constant, K eq Two methods for describing the equilibrium constant of a reaction. 1.K c - describes the equilibrium of a reaction where the concentrations of the materials is known. 2.K p - describes the equilibrium of a gaseous reaction using partial pressures instead of concentrations.

10. K c c = concentration aA + bB  cC + dD Coefficients in the chemical equation become exponents in the equilibrium constant expression. Include only substances in the gas or aqueous phase. Solid’s and liquid’s concentrations do not change during a chemical reaction.

11. Write the equilibrium expression (K c ) for the reaction below. Ni (s) + 4CO (g)  Ni(CO) 4 (g)

12. K c = [Ni(CO) 4 ] [CO] 4

13. Determine the equilibrium constant. A ↔ B

14. K p p= pressure Equilibrium is described in terms of the partial pressures of the reactants and products. 2CO 2 (g)  2CO (g) + O 2 (g)

15. Write the equilibrium expression (K p ) for the following mixture of gases at equilibrium. 2NO (g) + O 2 (g)  2NO 2 (g) K p = (P NO 2 ) 2 (P NO ) 2 (P O 2 )

16. Properties of the Equilibrium Constant K eq is constant for a particular reaction at a specific temperature. We generally omit the units of the equilibrium constant. Note that the equilibrium constant expression has products over reactants. K>>1 implies products are favored, and K eq lies to the right (towards the products). K<<1 implies reactants are favored, and K eq lies to the left. (towards the reactants).

17. The Magnitude of Equilibrium Constants

18. More Facts About K eq The equilibrium constant of a reaction in the reverse direction is the inverse of the equilibrium constant of the reaction in the forward direction… K eq(forward) = 1/K eq(reverse) For example: At 100 ºC, K eq(forward) = 6.49 At 100 ºC, K eq(reverse) = 1/6.49 = 0.154

19. More Facts About K eq The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps. For example: A + B ↔ X + CK eq(1) = 2.0 X + B ↔ DK eq(2) = 5.0 A + 2B ↔ C + DK eq = K 1 x K 2 = 10.0 When this is written in terms of concentrations at equilibrium… K eq =10.0 = [C][D]/[A][B] 2

20. More Facts About K eq The equilibrium expression only contains the concentrations of gases or aqueous substances and NEVER solids or pure liquids. Why? - Consider the decomposition of calcium carbonate: CaCO 3(s) ↔ CaO (s) + CO 2(g) - The concentrations of solids and pure liquids are constant. Therefore. K eq = K p = P CO 2 Or K eq = K c = [CO 2 ] This is an example of a heterogeneous equilibrium.

21. Note: Although the concentrations of these species are not included in the equilibrium expression, they do participate in the reaction and must be present for an equilibrium to be established!

22. What is the equilibrium-constant expression for this reaction? K c = [Pb 2+ ] [Cl − ] 2

23. The equilibrium constant (K) is related to the rate constant (k)

24. We can show for a reversible reaction having equal rates in both directions that….

25. The Reaction Quotient- Q There are times when a chemist may be given information about a chemical reaction that may or may not be at equilibrium. A “Q” calculation can be used to determine the direction the reaction is heading to reach equilibrium.

26. We define Q, the reaction quotient, for a general reaction Predicting Direction of Reaction

27. The Reaction Quotient- Q If Q > K, then the reaction has too many products and is heading towards the reactant side. If Q < K, then the reaction has too many reactants and is heading towards the product side. If Q = K, then the reaction is at equilibrium.

28. Reaction Shifts At 448 o C the equilibrium constant, K eq, for the reaction: H 2 (g) + I 2 (g) 2HI(g) is 50.5. Predict how the reaction will proceed to reach equilibrium at 448 o C if we start with 2.0 x 10 -2 mol of HI, 1.0 x 10 -2 mol of H 2, and 3.0 x 10 -2 mol of I 2 in a 2.0-L container.

29. Example 15.6 Page 459 At 2000°C, the equilibrium constant for the reaction N 2 (g) + O 2 (g) ↔ 2NO(g) is 4.1 x 10 -4. Find the concentration of NO(g) in a mixture of NO(g), N 2 (g), and O 2 (g) in which, at equilibrium, [N 2 ] = 0.036 mol L -1 and [O 2 ] = 0.0089 mol L -1.

30. Example 15.7 Page 459 The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 x 10 -2. Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26M and 2.09M, respectively.

31. Example 15.5 Page 457 The “ICE” Table Iodine molecules react reversibly with iodide ions to produce triiodide ions. I 2 (aq) + I - (aq) ↔ I 3 - (aq) If a solution is prepared with the concentrations of both I 2 and I - equal to 1.000 x 10 -3 M before reaction and if the concentration of I 2 changes to 6.61 x 10 -4 M at equilibrium, what is the equilibrium constant for the reaction?

32. Example 15.5 Page 457 Iodine molecules react reversibly with iodide ions to produce triiodide ions.I 2 (aq) + I - (aq) ↔ I 3 - (aq) If a solution is prepared with the concentrations of both I 2 and I - equal to 1.000 x 10 -3 M before reaction and if the concentration of I 2 changes to 6.61 x 10 -4 M at equilibrium, what is the equilibrium constant for the reaction?

33. Example 15.10 Page 466 Find the concentration of NO(g) at equilibrium when a mixture O 2 (g) with a concentration of 0.50M and N 2 (g) with a concentration of 0.75M is heated at 700°C. THE 5% RULE N 2 (g) + O 2 (g) ↔ 2NO(g) K = 4.1 x 10 -9.

36. Example 15.8 Page 462THE 5% RULE Under certain conditions the equilibrium constant for the decomposition of PCl 5 (g) into PCl 3 (g) and Cl 2 (g) is 0.0211. What are the equilibrium concentrations of all of three of these gases when the initial concentration of PCl5 was 1.00M?

38. Example 15.8. The Quadratic Equation(See Appendix A.4)

40. See the graphing calculator solutions method in notebook.

41. Example 15.9 Page 464 Acetic acid, CH 3 CO 2 H, reacts with ethanol to form water and ethyl acetate. CH 3 CO 2 H + C 2 H 5 OH ↔ CH 3 CO 2 C 2 H 5 + H 2 O The equilibrium constant for this reaction (when done with dioxane as a solvent) is 4.0. What are the equilibrium concentrations of all chemical species when 0.15 mol of CH 3 CO 2 H, 0.15 mol of C 2 H 5 OH, 0.40 mol of CH 3 CO 2 C 2 H 5, and 0.40 mol of H 2 O are mixed in enough dioxane to make 1.0L of solution.

45. Effect of a Catalyst on Equilibrium A catalyst speeds up the forward and reverse reactions equally and therefore equilibrium is reached faster. However a catalyst has no effect on the value of the equilibrium constant or on the equilibrium concentrations.

46. Le Chatelier’s Principle Le Chatelier's principle states that if a stress is applied to a system at equilibrium, the system will shift to minimize the stress. Stresses include changes in: 1.Concentration 2.Temperature 3.Pressure Reactions can shift: 1.Right (toward products) 2.Left (towards reactants)

47. Stress: Change in concentration

49. SCN - is thiocyanate

50. Stress: Change in concentration Would Adding Water Create a Stress?

51. What can you tell me about the following reaction?

52. It is at equilibrium.

53. Calculate the equilibrium constant.

55. Adjust the graph to illustrate what would happen if some N 2 was added.

56. N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)

57. Stress: Change in Temperature An increase in temperature will always shift a reaction in the endothermic (heat/energy absorbing direction). A decrease in temperature will always shift a reaction in the exothermic (heat/energy releasing direction).

58. How can we tell if a reaction is exothermic or endothermic? Energy term written as a product (exothermic) or reactant (endothermic) within the equation. -∆H (exothermic) or +∆H (endothermic). Warms up (exothermic) or cools down (endothermic).

59. Stress: Change in Pressure An increase in pressure will shift an equilibrium towards fewer moles of gas. An decrease in pressure will shift an equilibrium towards more moles of gas.

60. Example: Describe how the reaction below adjusts its equilibrium in each of the scenarios. CO (g) + 2H 2 (g)  CH 3 OH (g)  H ° = -21.7 Kcal a.Some of the methanol vapor is condensed and removed from the reaction vessel. b.The pressure is increased by decreasing the volume of the reaction vessel. c.The temperature is increased.

61. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l) Which way will the reaction shift? 1.CO 2 is removed. 2.Pressure is increased. 3.How would your answer to #2 change if H 2 O were a gas?

62. CaCO 3 (s) + heat  CaO(g) + CO 2 (g) Which way will the reaction shift? 1.CO 2 is added. 2.CaCO 3 is added. 3.Temperature is increased. 4.Pressure is decreased.

63. browncolorless 2NO 2 (g) ↔N 2 O 4 (g) brown colorless Is this reaction endothermic or exothermic? Exothermic

64. Teacher Note Only do next slide if there is time.

65. shift to a new equilibrium: Then go inside… shift to a new equilibrium: Photochromatic Sunglasses AgCl + energy Ag o + Cl o “energy” Go outside… Sunlight more intense than inside light; GLASSES DARKEN (clear) (dark) “energy” GLASSES LIGHTEN