1. Chapter Fourteen 1 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Today… Turn in: –Get out POGIL (Day 1) Our Plan: –Test Results/Calendar –POGIL discussion –Intro to Equilibrium Notes Homework (Write in Planner): –Begin Homework Problems –Gizmos Due next class

2. Chapter Fourteen 2 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Test Results A7 B8 C3 D0 F2 Average83.2 High Score98.5

3. Chapter Fourteen 3 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chemical Equilibrium Chapter Fourteen

4. 4 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 What is Equilibrium? Reactions don’t always have a beginning and an end. They find a balance and have to work through stress to find it. Many reaction never EVER finish. –They are reversible –Can go forward or backward without extra energy used

5. Chapter Fourteen 5 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 What is Equilibrium? Equilibrium state is reached when a reaction’s forward progress is perfectly balanced with the reverse process.

6. Chapter Fourteen 6 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Dynamic Nature of Equilibrium When a system reaches equilibrium, the forward and reverse reactions continue to occur … but at equal rates. We are usually concerned with the situation after equilibrium is reached. After equilibrium the concentrations of reactants and products remain constant.

7. Chapter Fourteen 7 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Concentration vs. Time If we begin with only 1 M HI, the [HI] decreases and both [H 2 ] and [I 2 ] increase. Beginning with 1 M H 2 and 1 M I 2, the [HI] increases and both [H 2 ] and [I 2 ] decrease. Beginning with 1 M each of H 2, I 2, and HI, the [HI] increases and both [H 2 ] and [I 2 ] decrease.

8. Chapter Fourteen 8 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Regardless of the starting concentrations; once equilibrium is reached … … the expression with products in numerator, reactants in denominator, where each concentration is raised to the power of its coefficient, appears to give a constant.

9. Chapter Fourteen 9 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Equilibrium Constant Expression For the general reaction: aA + bB  gG + hH The equilibrium expression is: Each concentration is simply raised to the power of its coefficient Products in numerator. Reactants in denominator. [G] g [H] h K c = [A] a [B] b

10. Chapter Fourteen 10 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Equilibrium Constant The equilibrium constant is constant regardless of the initial concentrations of reactants and products. This constant is denoted by the symbol K c and is called the concentration equilibrium constant. The constant is only good at a given temperature. If temperature is changed, so is the constant. Concentrations of the products appear in the numerator and concentrations of the reactants appear in the denominator. The exponents of the concentrations are identical to the stoichiometric coefficients in the chemical equation.

11. Chapter Fourteen 11 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Quick Check Write the equilibrium constant expression for:

12. Chapter Fourteen 12 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.1 If the equilibrium concentrations of COCl 2 and Cl 2 are the same at 395 °C, find the equilibrium concentration of CO in the reaction: CO(g) + Cl 2 (g) COCl 2 (g) K c = 1.2 x 10 3 at 395 °C

13. Chapter Fourteen 13 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Modifying the Chemical Equation What will be the equilibrium constant K' c for the new reaction? [NO 2 ] 2 K c = ––––––––– = 4.67 x 10 13 (at 298 K) [NO] 2 [O 2 ] Now consider the reaction: 2 NO 2 (g) 2 NO(g) + O 2 (g) [NO] 2 [O 2 ] 1 K' c = ––––––––– = ––––––––––– = [NO 2 ] 2 [NO 2 ] 2 ––––––––– [NO] 2 [O 2 ] 1 1 –– = ––––––––– = 2.14 x 10 –14 K c 4.67 x 10 13 Consider the reaction:2 NO(g) + O 2 (g) 2 NO 2 (g)

14. Chapter Fourteen 14 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Modifying the Chemical Equation (cont’d) What will be the equilibrium constant K" c for the new reaction? Consider the reaction:2 NO(g) + O 2 (g) 2 NO 2 (g) [NO 2 ] 2 K c = ––––––––– = 4.67 x 10 13 (at 298 K) [NO] 2 [O 2 ] Now consider the reaction: NO 2 (g) NO(g) + ½ O 2 (g) [NO] [O 2 ] 1/2 1 1/2 K" c = ––––––––– = ––– [NO 2 ] 2 K c = 2.14 x 10 –14 = 1.46 x 10 –7

15. Chapter Fourteen 15 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Modifying the Chemical Equation (cont’d) For the reverse reaction, K is the reciprocal of K for the forward reaction. When an equation is divided by two, K for the new reaction is the square root of K for the original reaction. General rule: –When the coefficients of an equation are multiplied by a common factor n to produce a new equation, we raise the original K c value to the power n to obtain the new equilibrium constant. It should be clear that we must write a balanced chemical equation when citing a value for K c.

16. Chapter Fourteen 16 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Quick Check So… –If we reverse a reaction, to find the new K c we would… –If we double a reaction, to find the new K c we would… –If we third a reaction, to find the new K c we would…

17. Chapter Fourteen 17 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.2 The equilibrium constant for the reaction: ½ H 2 (g) + ½ I 2 (g) HI(g) at 718 K is 7.07. (a)What is the value of K c at 718 K for the reaction HI(g) ½ H 2 (g) + ½ I 2 (g) (b) What is the value of K c at 718 K for the reaction H 2 (g) + I 2 (g) 2 HI(g)

18. Chapter Fourteen 18 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Try It Out N 2(g) + 3H 2(g) ↔2NH 3(g) For the equation above the K c = 3.6 x 10 8 What is the value of K c at the same temp for: NH 3 ↔ ½ N 2 + 3/2 H 2 5.3 x 10 -5

19. Chapter Fourteen 19 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Try It Out! N 2(g) + 3H 2(g) ↔ 2NH 3(g) For the equation above the K c = 3.6 x 10 8 Determine the value for K c for the reaction: 1/3 N 2 + H 2 ↔2/3 NH 3 7.1 x 10 2

20. Chapter Fourteen 20 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Equilibria Involving Gases In reactions involving gases, it is often convenient to measure partial pressures rather than molarities. In these cases, a partial pressure equilibrium constant, K p, is used. K c and K p are related by: K p = K c (RT) Δn(gas) where  n(gas) is the change in number of moles of gas as the reaction occurs in the forward direction. (P G ) g (P H ) h K p = (P A ) a (P B ) b  n(gas) = mol gaseous products – mol gaseous reactants

21. Chapter Fourteen 21 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Remember “R”.0821 atm · L mol · K Units in problem have to be Kelvin and atm!

22. Chapter Fourteen 22 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 An Example 2H 2 S (g) ↔ 2H 2 (g)+ S 2 (g) K p = 1.2 x 10 -2 K c = ? (@ 1065 deg C) 1.1 x 10 -4

23. Chapter Fourteen 23 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.3 Consider the equilibrium between dinitrogen tetroxide and nitrogen dioxide: N 2 O 4 (g) 2 NO 2 (g) K p = 0.660 at 319 K (a) What is the value of K c for this reaction? (b) What is the value of K p for the reaction 2 NO 2 (g) N 2 O 4 (g)? (c) If the equilibrium partial pressure of NO 2 (g) is 0.332 atm, what is the equilibrium partial pressure of N 2 O 4 (g)?

24. Chapter Fourteen 24 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Equilibrium Constant for an Overall Reaction Adding the given equations gives the desired equation. Multiplying the given values of K gives the equilibrium constant for the overall reaction. (To see why this is so, write the equilibrium constant expressions for the two given equations, and multiply them together. Examine the result …) and we’re given: Suppose we need: N 2 O(g) + 3/2 O 2 (g) 2 NO 2 (g) K c (1) = ?? 2 NO(g) + O 2 (g) 2 NO 2 (g) K c (3) = 4.67 x 10 13 N 2 O(g) + ½ O 2 (g) 2 NO(g) K c (2) = 1.7 x 10 –13

25. Chapter Fourteen 25 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Let’s Try One… (#25) What is the value of K p at 298 K for the reaction ½ CH 4 (g) + H 2 O (g) ↔ ½ CO 2 (g) + 2H 2 (g), given the following data at 298 K? CO 2 (g) + H 2 (g) ↔ CO (g) + H 2 O (g) K p = 9.80 x 10 -6 CO (g) + 3 H 2 (g) ↔ CH 4 (g) + H 2 O (g) K p = 8.00 x 10 24

26. Chapter Fourteen 26 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Try It Out! (#27) Determine K c at 298 K for the reaction 2CH 4 (g) ↔ C 2 H 2 (g) + 3 H 2 (g), given the following data at 298 K. CH 4 (g) + H 2 O (g) ↔ CO (g) + 3 H 2 (g) K p = 1.2 x 10 -25 2C 2 H 2 (g) + 3O 2 (g) ↔ 4CO (g) + 2H 2 O (g) K p = 1.1 x 10 2 H 2 (g) + ½ O 2 (g) ↔ H 2 O (g) K p = 1.1 x 10 40

27. Chapter Fourteen 27 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Equilibria Involving Pure Solids and Liquids The equilibrium constant expression does not include terms for pure solid and liquid phases because their concentrations do not change in a reaction. Although the amounts of pure solid and liquid phases change during a reaction, these phases remain pure and their concentrations do not change. [CaO] [CO 2 ] K c = –––––––––– [CaCO 3 ] K c = [CO 2 ] Example: CaCO 3 (s) CaO(s) + CO 2 (g)

28. Chapter Fourteen 28 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Write the K c expression for: C (s) + H 2 O (g) ↔ CO (g) + H 2(g) MgCO 3(s) ↔ MgO (s) + CO 2(g) 2H 2(g) + O 2(g) ↔ 2H 2 O (l) H 2 S (g) + I 2(s) ↔ 2HI (g) + S (s)

29. Chapter Fourteen 29 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.4 The reaction of steam and coke (a form of carbon) produces a mixture of carbon monoxide and hydrogen, called water-gas. This reaction has long been used to make combustible gases from coal: C(s) + H 2 O(g) CO(g) + H 2 (g) Write the equilibrium constant expression for K c for this reaction.

30. Chapter Fourteen 30 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Equilibrium Constants: When Do We Need Them and When Do We Not? A very large numerical value of K c or K p signifies that a reaction goes (essentially) to completion. A very small numerical value of K c or K p signifies that the forward reaction, as written, occurs only to a slight extent. An equilibrium constant expression applies only to a reversible reaction at equilibrium. Although a reaction may be thermodynamically favored, it may be kinetically controlled … Thermodynamics tells us “it’s possible (or not)” Kinetics tells us “it’s practical (or not)”

31. Chapter Fourteen 31 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 2H 2 (g) + O 2(g) ↔ 2 H 2 O (g) K p = 1.4 x 10 83 @ 298 K A term in the denominator of the equilibrium approaches zero and makes the value of the equilibrium constant very large. Very large K c or K p favors forward reaction

32. Chapter Fourteen 32 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 C (s) + H 2 O (g) ↔ CO (g) + H 2(g) K p = 1.6 x 10 -21 @ 298 K To account for a small numerical value of an equilibrium constant, the numerator must be very small. Very small K c or K p favors reverse reaction

33. Chapter Fourteen 33 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 SO….. A reaction is most likely to reach a state of equilibrium in which significant quantities of both reactants and products are present if the numerical value of K c and K p is neither very large nor very small, roughly in the range from 10 -10 to 10 10

34. Chapter Fourteen 34 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Reaction Quotient, Q For nonequilibrium conditions, the expression having the same form as K c or K p is called the reaction quotient, Q c or Q p. The reaction quotient is not constant for a reaction, but is useful for predicting the direction in which a net change must occur to establish equilibrium. To determine the direction of net change, we compare the magnitude of Q c to that of K c.

35. Chapter Fourteen 35 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Reaction Quotient, Q When Q is larger than K, the numerator of Q is too big; we have “too much products.” When Q = K, equilibrium has been reached. When Q is smaller than K, the denominator of Q is too big; we have “too much reactants.”

36. Chapter Fourteen 36 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 A better graphic

37. Chapter Fourteen 37 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 CO (g) + 2 H 2(g) ↔ CH 3 OH (g) If the K c for this reaction = 14.5, then what direction will the reaction proceed if the initial concentrations are CO = 0.100 M H 2 = 0.100 M CH 3 OH = 0 M Left to right

38. Chapter Fourteen 38 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Try It Out! (#49) If a gaseous mixture at 588 K contains 20.0 g each of CO and H 2 O and 25.0 g each of CO 2 and H 2, in what direction will a net reaction occur to reach equilibrium? Explain CO (g) + H 2 O (g) ↔ CO 2 (g) + H 2 (g) K c = 31.4 at 588 K

39. Chapter Fourteen 39 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Stop – Homework Time You should be able to complete problems 20, 22, 24, 26, 28, 30, 32, 50, & 52

40. Chapter Fourteen 40 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Today… Turn in: –Gizmos - basket Our Plan: –Review/Quiz –LeChatelier’s POGIL –Notes –Demonstration Homework (Write in Planner): –Investigation 13 Pre-Lab

41. Chapter Fourteen 41 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Quick Check For the reaction 2NO(g) ↔ N 2 (g) + O 2 (g), K c = 9.92 at 2273 K. 1.Write the equilibrium expression (K c ) for this reaction. 2.What is K c for this reaction: 2N 2 (g) + 2O 2 (g) ↔ 4NO(g) (0.0102) 3.At equilibrium, the concentration of nitrogen and oxygen are both 1.5 x 10 -3 M. What is the concentration of NO? (4.76 x 10 -4 M) 4.What is K p for the reaction? (9.92)

42. Chapter Fourteen 42 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Stresses on Equilibrium Chemists make a living preventing equilibrium. They want to “knock it out of whack” to maximize a certain product. Stresses can be applied to shift equilibrium and maximize the amount of product, as you saw in the POGIL.

43. Chapter Fourteen 43 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Le Châtelier’s Principle When any change in concentration, temperature, pressure, or volume is imposed on a system at equilibrium, the system responds by attaining a new equilibrium condition that minimizes the impact of the imposed change. Analogy: Begin with 100 men and 100 women at a dance. Assume that there are 70 couples dancing, though not always the same couples (dynamic equilibrium). If 30 more men arrive, what happens? The equilibrium will shift, and shortly, more couples will be dancing … but probably not 30 more couples. Real Example – Deer populations

44. Chapter Fourteen 44 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Changing the Amounts of Reacting Species (concentration) At equilibrium, Q = K c. If the concentration of one of the reactants is increased, the denominator of the reaction quotient increases. Q is now less than K c. This condition is only temporary, however, because the concentrations of all species must change in such a way so as to make Q = K c again. In order to do this, the concentrations of the products increase; the equilibrium is shifted to the right.

45. Chapter Fourteen 45 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Concentration An increase in concentration will cause the reaction to shift to the opposite side until a new equilibrium is achieved.

46. Chapter Fourteen 46 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Important Note Remember that this is a change in concentration so changing the amount of a pure solid or liquid which is already in equilibrium should not make the reaction consume the additional solid or liquid.

47. Chapter Fourteen 47 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example Example: 2HgO (g) ↔2Hg (l) + O 2 (g) ∆ H = +43.4 kcal Add HgO, what happens to equilibrium? Add O 2, what happens to equilibrium? Add Hg, what happens to equilibrium?

48. Chapter Fourteen 48 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.7 Water can be removed from an equilibrium mixture in the reaction of 1-octanol and acetic acid, for example, by using a solid drying agent that is insoluble in the reaction mixture. Describe how the removal of a small quantity of water affects the equilibrium. CH 3 (CH 2 ) 6 CH 2 OH(aq) + CH 3 COOH(aq) CH 3 (CH 2 ) 6 CH 2 OCOCH 3 (aq) + H 2 O(soln) H+H+ p. 591 – As water is removed, the reverse reaction is slowed and the forward, water- forming reaction is stimulated. Not all of the water that is removed can be replaced, however, because a new equilibrium could not tolerate a constant concentration of H 2 O at the same time that the other three concentrations change. Thus, in the new equilibrium the amount of water is somewhat less than in the original equilibrium, the amount of octyl acetate is greater, and the amounts of both 1-octanol and acetic acid are less.

49. Chapter Fourteen 49 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Changing External Pressure or Volume in Gaseous Equilibria When the external pressure is increased (or system volume is reduced), an equilibrium shifts in the direction producing the smaller number of moles of gas. When the external pressure is decreased (or the system volume is increased), an equilibrium shifts in the direction producing the larger number of moles of gas. If there is no change in the number of moles of gas in a reaction, changes in external pressure (or system volume) have no effect on an equilibrium. Example: H 2 (g) + I 2 (g) 2 HI (g) What would happen if pressure is increased?

50. Chapter Fourteen 50 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example Example: 2HgO (g) ↔2Hg (l) + O 2 (g) ∆ H = +43.4 kcal Add pressure, what happens to equilibrium? Increase volume, what happens to equilibrium?

51. Chapter Fourteen 51 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Try It Out! Example 14.8 An equilibrium mixture of O 2 (g), SO 2 (g), and SO 3 (g) is transferred from a 1.00-L flask to a 2.00-L flask. In which direction does a net reaction proceed to restore equilibrium? The balanced equation for the reaction is 2 SO 3 (g) ↔ 2 SO 2 (g) + O 2 (g)

52. Chapter Fourteen 52 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Temperature Changes Raising the temperature of an equilibrium mixture shifts equilibrium in the direction of the endothermic reaction; lowering the temperature shifts equilibrium in the direction of the exothermic reaction. –Consider heat as though it is a product of an exothermic reaction or as a reactant of an endothermic reaction, and apply Le Châtelier’s principle.

53. Chapter Fourteen 53 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.9 Is the amount of NO(g) formed from given amounts of N 2 (g) and O 2 (g), N 2 (g) + O 2 (g) ↔ 2 NO(g) ΔH° = +180.5 kJ greater at high or low temperatures?

54. Chapter Fourteen 54 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Catalysts & Equilibrium A catalyst lowers the activation energy … of both the forward and the reverse reaction. Adding a catalyst does not affect an equilibrium state. A catalyst merely causes equilibrium to be achieved faster.

55. Chapter Fourteen 55 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Demo Time! This is going to be similar to Investigation 13! Pay attention!

56. Chapter Fourteen 56 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Today… Turn in: –Nothing Our Plan: –Investigation 13 Homework (Write in Planner): –Lab Report Due Monday

57. Chapter Fourteen 57 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Today… Turn in: –Lab Report (basket) – rubric on top Our Plan: –Review Quiz (use your notes) –Notes – Equilibrium Calculations Homework (Write in Planner): –Homework Problems (due next Monday)

58. Chapter Fourteen 58 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Determining Values of Equilibrium Constants Experimentally When initial amounts of one or more species, and equilibrium amounts of one or more species, are given, the amounts of the remaining species in the equilibrium state and, therefore, the equilibrium concentrations often can be established. A useful general approach is to tabulate under the chemical Reaction: –the concentrations of substances present Initially –Changes in these concentrations that occur in reaching equilibrium –the Equilibrium concentrations. This sort of table is sometimes called a “RICE” table: Reaction/Initial/Change/Equilibrium.

59. Chapter Fourteen 59 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Important Note The “RICE” table may be used with moles, molarity, or Liters (for gaseous equilibria)… never grams.

60. Chapter Fourteen 60 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example Equilibrium involving SO 2, O 2, and SO 3 is important in sulfuric acid production. When a 0.0200 mol sample of SO 3 is introduced into an evacuated 1.52 L flask @ 900 K, 0.0142 mol SO 3 is found to be present at equilibrium. What is the value of K p ?

61. Chapter Fourteen 61 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005

62. Chapter Fourteen 62 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Given in RedReaction 2SO 3 2SO 2 O2O2O2O2 Initial 0.0200 mol 0 mol Change Equil 0.0142 mol Equil Conc. -0.0058 mol+ 0.0058 mol + 0.0029 mol + 0.0058 mol + 0.0029 mol 0.0142/ 1.52 L 9.34 x 10 -3 M 0.0058/1.52L 3.82 x 10 -3 0.0029/ 1.52L 1.91 x 10 -3 M

63. Chapter Fourteen 63 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 K c = (3.82 x 10 -3 )(1.91 x 10 -3 ) 9.34 x 10 -3 K p = K c (RT) ∆ngas K p = 3.1 x 10 -4 (.0821 x 900K) 1 2.3 x 10 -2

64. Chapter Fourteen 64 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.11 In a 10.0-L vessel at 1000 K, 0.250 mol SO 2 and 0.200 mol O 2 react to form 0.162 mol SO 3 at equilibrium. What is K c, at 1000 K, for the reaction that is shown here? 2 SO 2 (g) + O 2 (g) 2 SO 3 (g)

65. Chapter Fourteen 65 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005

66. Chapter Fourteen 66 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Try It Out! 14.11 A A reaction starts with 1.00 mol each of PCl 3 and Cl 2 in a 1.00 L flask. When equilibrium is established at 250 degrees C in the reaction PCl 3 (g) + Cl 2 (g) ↔ PCl 5 (g), the amount of PCl 5 present is 0.82 mol. What is K c for this reaction?

67. Chapter Fourteen 67 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Calculating Equilibrium Quantities from K c and K p Values When starting with initial reactants and no products and with the known value of the equilibrium constant, these data are used to calculate the amount of substances present at equilibrium. Typically, a RICE table is constructed, and the symbol x is used to identify one of the changes in concentration that occurs in establishing equilibrium. Then, all the other concentration changes are related to x, the appropriate terms are substituted into the equilibrium constant expression, and the equation solved for x.

68. Chapter Fourteen 68 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.12 Consider the reaction H 2 (g) + I 2 (g) ↔ 2 HI(g) K c = 54.3 at 698 K If we start with 0.500 mol I 2 (g) and 0.500 mol H 2 (g) in a 5.25-L vessel at 698 K, how many moles of each gas will be present at equilibrium?

69. Chapter Fourteen 69 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005

70. Chapter Fourteen 70 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.14 Carbon monoxide and chlorine react to form phosgene, COCl 2, which is used in the manufacture of pesticides, herbicides, and plastics: CO(g) + Cl 2 (g) COCl 2 (g) K c = 1.2 x 10 3 at 668 K How much of each substance, in moles, will there be at equilibrium in a reaction mixture that initially has 0.0100 mol CO, 0.0100 mol Cl 2, and 0.100 mol COCl 2 in a 10.0-L flask?

71. Chapter Fourteen 71 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005

72. Chapter Fourteen 72 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Example 14.15 A sample of phosgene, COCl 2 (g), is introduced into a constant-volume vessel at 395 °C and observed to exert an initial pressure of 0.351 atm. When equilibrium is established for the reaction CO(g) + Cl 2 (g) COCl 2 (g) K p = 22.5 what will be the partial pressure of each gas and the total gas pressure?

73. Chapter Fourteen 73 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005

74. Chapter Fourteen 74 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Try It Out! 14.13 A: Starting with 0.100 mol CO and 0.200 mol Cl 2 in a 25.0 L flask, how many moles of COCl 2 will be present at equilibrium in the following reaction? CO (g) + Cl 2 (g) ↔ COCl 2 K c = 1.2 x 10 2 at 668 K

75. Chapter Fourteen 75 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005

76. Chapter Fourteen 76 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 The Effect of T on K p For an exothermic reaction; as T increases, K p decreases For an endothermic reaction; as T decreases, K p increases

77. Chapter Fourteen 77 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Today… Turn in: –Nothing Our Plan: –Crash Course Review - Equilibrium –Practice, Practice, Practice RICE Tables (guided worksheet) –Homework Homework (Write in Planner): –Homework due Monday

78. Chapter Fourteen 78 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Today… Turn in: –Nothing Our Plan: –Crash Course Review – Calculations –Practice, Practice, Practice RICE Tables –Quiz –Work on Homework Homework (Write in Planner): –Homework Due Monday

79. Chapter Fourteen 79 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Today… Turn in: –Mark Homework Questions on the Board Our Plan: –Homework Discussion –Station Review –Test Review Homework (Write in Planner): –Test Next Class –Breakfast Club Wednesday at 6 am

80. Chapter Fourteen 80 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Today… Turn in: –Nothing Our Plan: –Questions on Test Review –Unit 10 Test Homework (Write in Planner): –Enjoy your long weekend!