1. Copyright © Cengage Learning. All rights reserved1 Chemical Equilibrium

2. Copyright © Cengage Learning. All rights reserved2 Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

3. Copyright © Houghton Mifflin Company. All rights reserved.13–3 Equilibrium Is: Macroscopically static. Microscopically dynamic

4. Writing equilibrium equations Equilibrium reactions are written using a double arrow –Each of the arrows only has a single-sided head REACTANTSPRODUCTS

5. The Equilibrium Condition

6. Copyright © Houghton Mifflin Company. All rights reserved.13–6 Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

7. We Are Here

8. Equilibrium Vaporization - change from liquid to gas at boiling point. Evaporation - change from liquid to gas below boiling point Heat (or Enthalpy) of Vaporization (  H vap )- the energy required to vaporize 1 mol at 1 atm.

9. Vaporization is an endothermic process - it requires heat. Energy is required to overcome intermolecular forces. Responsible for cool beaches. Why we sweat.

10. Condensation Change from gas to liquid. Achieves a dynamic equilibrium with vaporization in a closed system. What is a closed system? A closed system means matter can’t go in or out. Put a cork in it. What the heck is a “dynamic equilibrium?”

11. /When first sealed the molecules gradually escape the surface of the liquid Dynamic equilibrium

12. /When first sealed the molecules gradually escape the surface of the liquid /As the molecules build up above the liquid some condense back to a liquid. Dynamic equilibrium

13. /As time goes by the rate of vaporization remains constant / but the rate of condensation increases because there are more molecules to condense. /Equilibrium is reached when Dynamic equilibrium

14. Behavior of a Liquid in a Closed Container

15. The Rates of Condensation and Evaporation

16. Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the forward reaction is possible As C and D build up, the reverse reaction speeds up while the forward reaction slows down. Eventually the rates are equal

17. The Changes with Time in the Rates of Forward and Reverse Reactions

18. Reaction Rate Time Forward Reaction Reverse reaction Equilibrium

19. What is equal at Equilibrium? Rates are equal Concentrations are not. Rates are determined by concentrations and activation energy. The concentrations do not change at equilibrium. or if the reaction is verrrry slooooow.

20. Equilibrium Position The proportion of reactants and products in the equilibrium mixture Reactions that lie to the right produce mainly products Reaction that lie to the left produce mainly reactants Requires experimentation and mathematical calculations to capture this information Copyright © Houghton Mifflin Company. All rights reserved.13–20

21. Quick Review Equilibrium is dynamic Equilibrium is only achieved in a closed system The concentration of the reactants and products remain the same at equilibrium Equilibrium can be reached from any direction Copyright © Houghton Mifflin Company. All rights reserved.13–21

22. Equilibrium Constant and Expression Copyright © Houghton Mifflin Company. All rights reserved.13–22

23. Law of Mass Action For any reaction the equation must be balanced first jA + kB lC + mD K = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power K is called the equilibrium constant. is how we indicate a reversible reaction

24. Copyright © Houghton Mifflin Company. All rights reserved.13–24 The Equilibrium Constant jA + kB lC + mD kj ml [B] [A] [D] [C] K =

25. Playing with K If we write the reaction in reverse. lC + mD jA + kB Then the new equilibrium constant is K ’ = [A] j [B] k = 1/K [C] l [D] m

26. Effects on Equilibrium Expressions Doubling the coefficients: squares the expression K c 2 Tripling the coefficients: Cubes the expression K c 3 Halving the coefficients: Square roots the expression √K c Adding together two reactions: multiplies the two expressions K c i X K c ii Copyright © Houghton Mifflin Company. All rights reserved.13–26

27. Playing with K If we double the coefficients this is the expression njA + nkB nlC + nmD Then the equilibrium constant is K’= [C] nl [D] nm = ( [C] l [D] m ) n = K c 2 [A] nj [B] nk ([A] j [B] k ) n

28. Write the Equilibrium Expression for the following reactions 2K 2 (g) + O 2 (g) 2H 2 O (g) 4NH 3 (g) + 7O 2 (g) 4NO 2(g) + 6H 2 O (g) 2SO 2(g) + O 2(g) 2SO 3(g) Copyright © Houghton Mifflin Company. All rights reserved.13–28

29. Express in K ’ Express the equations on the previous page in K ’, and √K c Copyright © Houghton Mifflin Company. All rights reserved.13–29

30. Magnitude of K c If K c is very small it lies to the left and favors the reactants If K c is very large it lies to the right and favors the products The magnitude of K c does not give us any information on the rate of the reaction and not how quickly equilibrium will be reached Copyright © Houghton Mifflin Company. All rights reserved.13–30

31. The units for K Are determined by the various powers and units of concentrations. They depend on the reaction.

32. K is CONSTANT At any temperature. Temperature affects rate. The equilibrium concentrations don’t have to be the same, only K. Equilibrium position is a set of concentrations at equilibrium. There are an unlimited number.

33. Copyright © Houghton Mifflin Company. All rights reserved.13–33 Equilibrium Constant

34. Calculate K N 2 + 3H 2 2NH 3 Initial At Equilibrium [N 2 ] 0 =1.000 M [N 2 ] = 0.921M [H 2 ] 0 =1.000 M [H 2 ] = 0.763M [NH 3 ] 0 =0 M [NH 3 ] = 0.157M

35. Calculate K N 2 + 3H 2 2NH 3 Initial At Equilibrium [N 2 ] 0 = 0 M [N 2 ] = 0.399 M [H 2 ] 0 = 0 M [H 2 ] = 1.197 M [NH 3 ] 0 = 1.000 M [NH 3 ] = 0.203M K is the same no matter what the amount of starting materials

36. Equilibrium and Pressure Some reactions are gaseous PV = nRT P = (n/V)RT P = CRT C is a concentration in moles/Liter C = P/RT

37. Write the equilibrium constant for the heterogeneous reaction

38. The Reaction Quotient Tells you the direction the reaction will go to reach equilibrium Calculated the same as the equilibrium constant, but for a system not at equilibrium Q = [Products] coefficient [Reactants] coefficient Compare value to equilibrium constant

39. What Q tells us If Q<K : Not enough products : Shift to right If Q>K : Too many products : Shift to left If Q=K system is at equilibrium

40. Example for the reaction 2NOCl(g) 2NO(g) + Cl 2 (g) K = 1.55 x 10 -5 M at 35ºC In an experiment 0.10 mol NOCl, 0.0010 mol NO(g) and 0.00010 mol Cl 2 are mixed in 2.0 L flask. Which direction will the reaction proceed to reach equilibrium?

41. Copyright © Houghton Mifflin Company. All rights reserved.13–41 The Equilibrium Constant and Applications

42. We Are Here

43. Copyright © Houghton Mifflin Company. All rights reserved.13–43 LeChâtelier’s Principle

44. Copyright © Houghton Mifflin Company. All rights reserved.13–44 Le Châtelier’s Principle If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce or minimize the effect of the change. What ever we do to a system at equilibrium the system will respond in the opposite way

45. Le Châtelier’s Principle Add something to the system and it will react to try to remove it Remove something from the system and it will try to add it After a while a new equilibrium will be established with a different composition from the earlier mixture

46. Le Châtelier’s Principle If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress. 3 Types of stress –Concentration –Pressure –Temperature

47. Effects of Changes on the System 1.Concentration: The system will shift away from the added component.  An increase in one of the reactants speeds up the reaction but there on the reactant side while there is no change in the reverse reaction  If the equilibrium is disrupted by a decrease in the concentration of the products a shift in the reaction occurs towards the reactants a new equilibrium position will be achieved

48. Effects of Changes on the System 2.Temperature: K will change depending upon the temperature (treat the energy change as a reactant).  Enthalpy changes of the forward reactions and the reverse reactions are equal and opposite of each other  -∆H = exothermic forward reaction, releases heat  +∆H = endothermic backward reaction, absorbs heat

49. Effects of Changes on the System  Increase in temp favors the reverse reaction so K shifts to the left  Decrease in temp favors the reactants so K shifts to the right

50. Effects of Changes on the System 3.Pressure: a)Addition of inert gas does not affect the equilibrium position. b)Decreasing the volume shifts the equilibrium toward the side with fewer moles. c)Temp must remain the same

51. Effects of Changes on the System 4. Catalysts:  Speeds up the rate of a reaction by providing an alternate reaction pathway that has a transition state with a lower activation energy.  Speeds up the attainment of the equilibrium state  Used in industry to increase the rate of product formation  Used in every biochemical reaction

52. Quick Review EffectChange in positionChange in K ConcentrationChangesNo Change PressureChanges if a change in gas molecules No Change TemperatureChangesChange CatalystsNo change

53. Copyright © Houghton Mifflin Company. All rights reserved.13–53 LeChâtelier’s Principle

54. Equilibrium in Industry The contact process: 3 steps Combustion Oxidation Reaction with water Sulfur –by mass highest produced of any chemical Production of Sulfur used as a economical indicator of a countries industrial strength Used directly or indirectly in nearly all industrial processes Approx 250 million tons are produced annually across all continents Copyright © Houghton Mifflin Company. All rights reserved.13–54

55. Equilibrium in Industry Haber process – The production of ammonia NH 3 Used as fertilizers By plants Without Ammonia fertilizers it is estimated that two billion people would starve Copyright © Houghton Mifflin Company. All rights reserved.13–55

56. Equilibrium in Industry Production of Methanol Converted into plastics Paints Explosives Plywood Bio-diesel Fuel antifreeze Copyright © Houghton Mifflin Company. All rights reserved.13–56

57. Copyright © Houghton Mifflin Company. All rights reserved.13–57 Equilibrium Decomposition of N 2 O 4

58. HL ONLY From this point Copyright © Houghton Mifflin Company. All rights reserved.13–58

59. The Equilibrium Law There are both heterogeneous and homogeneous equilibria We will concentrate on Homogeneous equilibrium There are 2 processes involved  Calculation of the equilibrium concentration  Calculation of the concentration of reactants and products present at equilibrium Copyright © Houghton Mifflin Company. All rights reserved.13–59

60. Solving Equilibrium Problems Given the starting concentrations and one equilibrium concentration. Use stoichiometry to figure out other concentrations and K. Learn to create a table of initial and final conditions.

61. Calculating equilibrium from initial concentrations 3 Steps 1. Write the balanced equation 2. Under the equation make 3 rows label initial, change and equilibrium 3. Write the expression of K Copyright © Houghton Mifflin Company. All rights reserved.13–61

62. Solving Equilibrium Problems

63. Example A student adds 0.20 mols of PCl 3 and 0.10 mols of Cl 2 allowed the reaction to reach equilibrium. At equilibrium there were 0.12 mols of PCl 3 What is the concentration of the reactants and products at equilibrium Copyright © Houghton Mifflin Company. All rights reserved.13–63

64. Example Step 1 PCl 3 + Cl 2 = PCl 5 Copyright © Houghton Mifflin Company. All rights reserved.13–64

65. Example Step 2 PCl 3 + Cl 2 = PCl 5 Initial Change Equilibrium Copyright © Houghton Mifflin Company. All rights reserved.13–65

66. Example Step 2 PCl 3 + Cl 2 = PCl 5 Initial 0.20 0.10 0 Change -0.08 -0.08 +0.08 Equilibrium 0.12 0.02 0.08 Copyright © Houghton Mifflin Company. All rights reserved.13–66

67. Example Step 3 K c = [PCl 5 ] / [PCl 3 ] X [Cl 2 ] K c =.08 /.12 X.02 = 33 Copyright © Houghton Mifflin Company. All rights reserved.13–67

68. Example 2 Consider the reaction CO + 2H 2 ↔ CH 3 OH At equalibrium K = 0.500 CO = 0.200 H 2 = 0.155 Calculate the equilibrium concentration of CH 3 OH Copyright © Houghton Mifflin Company. All rights reserved.13–68

69. Example 2 Write the equilibrium expression K c = [CH 3 OH] / [CO] X [H 2 ] 2 Copyright © Houghton Mifflin Company. All rights reserved.13–69

70. Example 2 Substitute the data Copyright © Houghton Mifflin Company. All rights reserved.13–70 0.500 = [CH 3 OH] / [0.200] X [0.155] 2 [CH 3 OH] = 0.00240 mol

71. Free Energy ∆G - = negative →reaction proceeds in the forward direction ∆G + = positive → reaction proceeds in the reverse direction ∆G - = 0 →reaction is at equilibrium ∆G - = G - products - G - reactants AT Equilibrium G - products = G - reactants Copyright © Houghton Mifflin Company. All rights reserved.13–71

72. Exothermic DH<0 Releases heat Think of heat as a product Raising temperature push toward reactants. Shifts to left.

73. Endothermic DH>0 Produces heat Think of heat as a reactant Raising temperature push toward products. Shifts to right.