1. Chapter 14 Chemical Equilibrium 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2. Tro, Chemistry: A Molecular Approach2 Hemoglobin Equilibrium System Hb + O 2  HbO 2 the concentrations of Hb, O 2, and HbO 2 are all interdependent the relative amounts of Hb, O 2, and HbO 2 at equilibrium are related to a constant called the equilibrium constant, K the larger the value of K, the more product is found at equilibrium changing the concentration of any one of these necessitates changing the other concentrations to reestablish equilibrium

3. Tro, Chemistry: A Molecular Approach3 Hemoglobin protein (Hb) found in red blood cells that reacts with O 2 enhances the amount of O 2 that can be carried through the blood stream Hb + O 2  HbO 2 the Hb represents the entire protein – it is not a chemical formula the  represents that the reaction is in dynamic equilibrium

4. Tro, Chemistry: A Molecular Approach4 O 2 Transport Hb + O2O2  in the lungs, with high concentration of O 2, the equilibrium shifts to combine the Hb and O 2 together to make more HbO 2 in the cells, with low concentration of O 2, the equilibrium shifts to break down the HbO 2 and increase the amount of free O 2 HbO 2 O 2 in lungs Hb HbO 2 O 2 in cells

5. Tro, Chemistry: A Molecular Approach5 HbF Hb Fetal Hemoglobin, HbF HbF + O 2  HbFO 2 fetal hemoglobin’s equilibrium constant is larger than adult hemoglobin because fetal hemoglobin is more efficient at binding O 2, O 2 is transferred to the fetal hemoglobin from the mother’s hemoglobin in the placenta Hb + O2O2  HbO 2 O2O2 O2O2 Hb + O2O2  HbFO 2 O2O2

6. Tro, Chemistry: A Molecular Approach6 Oxygen Exchange between Mother and Fetus

7. Tro, Chemistry: A Molecular Approach7 Reaction Dynamics when a reaction starts, the reactants are consumed and products are made forward reaction = reactants  products therefore the reactant concentrations decrease and the product concentrations increase as reactant concentration decreases, the forward reaction rate decreases eventually, the products can react to reform some of the reactants reverse reaction = products  reactants assuming the products are not allowed to escape as product concentration increases, the reverse reaction rate increases processes that proceed in both the forward and reverse direction are said to be reversible reactants  products

8. 8 Hypothetical Reaction 2 Red  Blue Time[Red][Blue] 00.4000.000 100.2080.096 200.1900.105 300.1800.110 400.1740.113 500.1700.115 600.1680.116 700.1670.117 800.1660.117 900.1650.118 1000.1650.118 1100.1640.118 1200.1640.118 1300.1640.118 1400.1640.118 1500.1640.118 The reaction slows over time, But the Red molecules never run out! At some time between 100 and 110 sec, the concentrations of both the Red and the Blue molecules no longer change – equilibrium has been established. Notice that equilibrium does not mean that the concentrations are equal! Once equilibrium is established, the rate of Red molecules turning into Blue is the same as the rate of Blue molecules turning into Red

9. Tro, Chemistry: A Molecular Approach9 Hypothetical Reaction 2 Red  Blue

10. Tro, Chemistry: A Molecular Approach10 Reaction Dynamics Time Rate Rate Forward Rate Reverse Initially, only the forward reaction takes place. As the forward reaction proceeds it makes products and uses reactants. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant.

11. Tro, Chemistry: A Molecular Approach11 Dynamic Equilibrium as the forward reaction slows and the reverse reaction accelerates, eventually they reach the same rate dynamic equilibrium is the condition where the rates of the forward and reverse reactions are equal once the reaction reaches equilibrium, the concentrations of all the chemicals remain constant because the chemicals are being consumed and made at the same rate

12. Tro, Chemistry: A Molecular Approach12 H 2 (g) + I 2 (g)  2 HI(g) at time 0, there are only reactants in the mixture, so only the forward reaction can take place [H 2 ] = 8, [I 2 ] = 8, [HI] = 0 at time 16, there are both reactants and products in the mixture, so both the forward reaction and reverse reaction can take place [H 2 ] = 6, [I 2 ] = 6, [HI] = 4

13. Tro, Chemistry: A Molecular Approach13 H 2 (g) + I 2 (g)  2 HI(g) at time 32, there are now more products than reactants in the mixture − the forward reaction has slowed down as the reactants run out, and the reverse reaction accelerated [H 2 ] = 4, [I 2 ] = 4, [HI] = 8 at time 48, the amounts of products and reactants in the mixture haven’t changed – the forward and reverse reactions are proceeding at the same rate – it has reached equilibrium

14. Tro, Chemistry: A Molecular Approach14 H 2 (g) + I 2 (g)  2 HI(g) Concentration  Time  Equilibrium Established Since the [HI] at equilibrium is larger than the [H 2 ] or [I 2 ], we say the position of equilibrium favors products As the reaction proceeds, the [H 2 ] and [I 2 ] decrease and the [HI] increases Since the reactant concentrations are decreasing, the forward reaction rate slows down And since the product concentration is increasing, the reverse reaction rate speeds up Once equilibrium is established, the concentrations no longer change At equilibrium, the forward reaction rate is the same as the reverse reaction rate

15. Tro, Chemistry: A Molecular Approach15 Equilibrium  Equal the rates of the forward and reverse reactions are equal at equilibrium but that does not mean the concentrations of reactants and products are equal some reactions reach equilibrium only after almost all the reactant molecules are consumed – we say the position of equilibrium favors the products other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – we say the position of equilibrium favors the reactants

16. Tro, Chemistry: A Molecular Approach16 An Analogy: Population Changes When Country A citizens feel overcrowded, some will emigrate to Country B. However, as time passes, emigration will occur in both directions at the same rate, leading to populations in Country A and Country B that are constant, though not necessarily equal

17. Tro, Chemistry: A Molecular Approach17 Equilibrium Constant even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them the relationship between the chemical equation and the concentrations of reactants and products is called the Law of Mass Action for the general equation aA + bB  cC + dD, the Law of Mass Action gives the relationship below the lowercase letters represent the coefficients of the balanced chemical equation always products over reactants K is called the equilibrium constant unitless

18. Tro, Chemistry: A Molecular Approach18 Writing Equilibrium Constant Expressions for the reaction aA (aq) + bB (aq)  cC (aq) + dD (aq) the equilibrium constant expression is so for the reaction 2 N 2 O 5  4 NO 2 + O 2 the equilibrium constant expression is:

19. Tro, Chemistry: A Molecular Approach19 What Does the Value of K eq Imply? when the value of K eq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules the position of equilibrium favors products when the value of K eq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules the position of equilibrium favors reactants

20. Tro, Chemistry: A Molecular Approach20 A Large Equilibrium Constant

21. Tro, Chemistry: A Molecular Approach21 A Small Equilibrium Constant

22. Tro, Chemistry: A Molecular Approach22 Relationships between K and Chemical Equations when the reaction is written backwards, the equilibrium constant is inverted for the reaction aA + bB  cC + dD the equilibrium constant expression is: for the reaction cC + dD  aA + bB the equilibrium constant expression is:

23. Tro, Chemistry: A Molecular Approach23 Relationships between K and Chemical Equations when the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor for the reaction aA + bB  cC the equilibrium constant expression is: for the reaction 2aA + 2bB  2cC the equilibrium constant expression is:

24. Tro, Chemistry: A Molecular Approach24 Relationships between K and Chemical Equations when you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations for the reactions (1) aA  bB and (2) bB  cC the equilibrium constant expressions are: for the reaction aA  cC the equilibrium constant expression is:

25. K backward = 1/K forward, K new = K old n Ex 14.2 – Compute the equilibrium constant at 25°C for the reaction NH 3 (g)  0.5 N 2 (g) + 1.5 H 2 (g) for N 2 (g) + 3 H 2 (g)  2 NH 3 (g), K = 3.7 x 10 8 at 25°C K for NH 3 (g)  0.5N 2 (g) + 1.5H 2 (g), at 25°C Solution: Concept Plan: Relationships: Given: Find: N 2 (g) + 3 H 2 (g)  2 NH 3 (g)K 1 = 3.7 x 10 8 K’K 2 NH 3 (g)  N 2 (g) + 3 H 2 (g) NH 3 (g)  0.5 N 2 (g) + 1.5 H 2 (g)

26. 26 Equilibrium Constants for Reactions Involving Gases the concentration of a gas in a mixture is proportional to its partial pressure therefore, the equilibrium constant can be expressed as the ratio of the partial pressures of the gases for aA(g) + bB(g)  cC(g) + dD(g) the equilibrium constant expressions are or

27. Tro, Chemistry: A Molecular Approach27 K c and K p in calculating K p, the partial pressures are always in atm the values of K p and K c are not necessarily the same because of the difference in units K p = K c when  n = 0 the relationship between them is:  n is the difference between the number of moles of reactants and moles of products

28. Tro, Chemistry: A Molecular Approach28 Deriving the Relationship between K p and K c

29. 29 Deriving the Relationship Between K p and K c for aA(g) + bB(g)  cC(g) + dD(g) substituting

30. Ex 14.3 – Find K c for the reaction 2 NO(g) + O 2 (g)  2 NO 2 (g), given K p = 2.2 x 10 12 @ 25°C K is a unitless number since there are more moles of reactant than product, K c should be larger than K p, and it is K p = 2.2 x 10 12 KcKc Check: Solution: Concept Plan: Relationships: Given: Find: KpKp KcKc 2 NO(g) + O 2 (g)  2 NO 2 (g)  n = 2  3 = -1

31. Tro, Chemistry: A Molecular Approach31 Heterogeneous Equilibria pure solids and pure liquids are materials whose concentration doesn’t change during the course of a reaction its amount can change, but the amount of it in solution doesn’t  because it isn’t in solution because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant expression for the reaction aA(s) + bB(aq)  cC(l) + dD(aq) the equilibrium constant expression is:

32. Tro, Chemistry: A Molecular Approach32 Heterogeneous Equilibria The amount of C is different, but the amounts of CO and CO 2 remains the same. Therefore the amount of C has no effect on the position of equilibrium.

33. Tro, Chemistry: A Molecular Approach33 Calculating Equilibrium Constants from Measured Equilibrium Concentrations the most direct way of finding the equilibrium constant is to measure the amounts of reactants and products in a mixture at equilibrium actually, you only need to measure one amount – then use stoichiometry to calculate the other amounts the equilibrium mixture may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same as long as the temperature is kept constant the value of the equilibrium constant is independent of the initial amounts of reactants and products

34. 34 Initial and Equilibrium Concentrations for H 2 (g) + I 2 (g)  2HI(g) @ 445°C Initial Equilibrium Equilibrium Constant [H 2 ][I 2 ][HI][H 2 ][I 2 ][HI] 0.50 0.00.11 0.78 0.0 0.500.055 0.39 0.50 0.165 1.17 1.00.50.00.530.0330.934

35. 35 Calculating Equilibrium Concentrations Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration suppose you have a reaction 2 A (aq) + B (aq)  4 C (aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M. [A][B][C] initial molarity 1.00 0 change in concentration equilibrium molarity 0.50 +0.50-¼(0.50)-½(0.50) 0.75 0.88

36. Tro, Chemistry: A Molecular Approach36 Ex 14.6  Find the value of K c for the reaction 2 CH 4 (g)  C 2 H 2 (g) + 3 H 2 (g) at 1700°C if the initial [CH 4 ] = 0.115 M and the equilibrium [C 2 H 2 ] = 0.035 M Construct an ICE table for the reaction for the substance whose equilibrium concentration is known, calculate the change in concentration +0.035 [CH 4 ][C 2 H 2 ][H 2 ] initial 0.1150.000 change equilibrium 0.035

37. 37 Ex 14.6  Find the value of K c for the reaction 2 CH 4 (g)  C 2 H 2 (g) + 3 H 2 (g) at 1700°C if the initial [CH 4 ] = 0.115 M and the equilibrium [C 2 H 2 ] = 0.035 M use the known change to determine the change in the other materials add the change to the initial concentration to get the equilibrium concentration in each column use the equilibrium concentrations to calculate K c [CH 4 ][C 2 H 2 ][H 2 ] initial 0.1150.000 change equilibrium 0.035 +0.035 -2(0.035)+3(0.035) 0.045 0.105

38. Tro, Chemistry: A Molecular Approach38 The following data were collected for the reaction 2 NO 2 (g)  N 2 O 4 (g) at 100°C. Complete the table and determine values of K p and K c for each experiment.

39. Tro, Chemistry: A Molecular Approach39 The following data were collected for the reaction 2 NO 2 (g)  N 2 O 4 (g) at 100°C. Complete the table and determine values of K p and K c for each experiment.

40. Tro, Chemistry: A Molecular Approach40 The Reaction Quotient if a reaction mixture, containing both reactants and products, is not at equilibrium; how can we determine which direction it will proceed? the answer is to compare the current concentration ratios to the equilibrium constant the concentration ratio of the products (raised to the power of their coefficients) to the reactants (raised to the power of their coefficients) is called the reaction quotient, Q for the gas phase reaction aA + bB  cC + dD the reaction quotient is:

41. Tro, Chemistry: A Molecular Approach41 The Reaction Quotient: Predicting the Direction of Change if Q > K, the reaction will proceed fastest in the reverse direction the [products] will decrease and [reactants] will increase if Q < K, the reaction will proceed fastest in the forward direction the [products] will increase and [reactants] will decrease if Q = K, the reaction is at equilibrium the [products] and [reactants] will not change if a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction if a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction

42. 42 Q, K, and the Direction of Reaction

43. If Q = K, equilibrium; If Q K, reverse Ex 14.7 – For the reaction below, which direction will it proceed if P I2 = 0.114 atm, P Cl2 = 0.102 atm & P ICl = 0.355 atm for I 2 (g) + Cl 2 (g)  2 ICl(g), K p = 81.9 direction reaction will proceed Solution: Concept Plan: Relationships: Given: Find: I 2 (g) + Cl 2 (g)  2 ICl(g)K p = 81.9 QP I2, P Cl2, P ICl since Q (10.8) < K (81.9), the reaction will proceed to the right

44. Ex 14.8 If [COF 2 ] eq = 0.255 M and [CF 4 ] eq = 0.118 M, and K c = 2.00 @ 1000°C, find the [CO 2 ] eq for the reaction given. Units & Magnitude OK Check: Check: Round to 1 sig fig and substitute back in Solution: Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts Concept Plan: Relationships: Strategize: You can calculate the missing concentration by using the equilibrium constant expression 2 COF 2  CO 2 + CF 4 [COF 2 ] eq = 0.255 M, [CF 4 ] eq = 0.118 M [CO 2 ] eq Given: Find: Sort: You’re given the reaction and K c. You’re also given the [X] eq of all but one of the chemicals K, [COF 2 ], [CF 4 ][CO 2 ]

45. Tro, Chemistry: A Molecular Approach45 A sample of PCl 5 (g) is placed in a 0.500 L container and heated to 160°C. The PCl 5 is decomposed into PCl 3 (g) and Cl 2 (g). At equilibrium, 0.203 moles of PCl 3 and Cl 2 are formed. Determine the equilibrium concentration of PCl 5 if K c = 0.0635

46. Tro, Chemistry: A Molecular Approach46 A sample of PCl 5 (g) is placed in a 0.500 L container and heated to 160°C. The PCl 5 is decomposed into PCl 3 (g) and Cl 2 (g). At equilibrium, 0.203 moles of PCl 3 and Cl 2 are formed. Determine the equilibrium concentration of PCl 5 if K c = 0.0635 PCl 5  PCl 3 + Cl 2 equilibrium concentration, M ?

47. Tro, Chemistry: A Molecular Approach47 Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures first decide which direction the reaction will proceed compare Q to K define the changes of all materials in terms of x use the coefficient from the chemical equation for the coefficient of x the x change is + for materials on the side the reaction is proceeding toward the x change is  for materials on the side the reaction is proceeding away from solve for x for 2 nd order equations, take square roots of both sides or use the quadratic formula may be able to simplify and approximate answer for very large or small equilibrium constants

48. Tro, Chemistry: A Molecular Approach48 [I 2 ][Cl 2 ][ICl] initial 0.100 change equilibrium Ex 14.11  For the reaction I 2 (g) + Cl 2 (g)  2 ICl(g) @ 25°C, K p = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations Construct an ICE table for the reaction determine the direction the reaction is proceeding since Q p (1) < K p (81.9), the reaction is proceeding forward

49. Tro, Chemistry: A Molecular Approach49 [I 2 ][Cl 2 ][ICl] initial 0.100 change equilibrium Ex 14.11  For the reaction I 2 (g) + Cl 2 (g)  2 ICl(g) @ 25°C, K p = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations represent the change in the partial pressures in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression and solve for x +2x xx xx 0.100  x 0.100+2x

50. Tro, Chemistry: A Molecular Approach50 [I 2 ][Cl 2 ][ICl] initial 0.100 change equilibrium Ex 14.11  For the reaction I 2 (g) + Cl 2 (g)  2 ICl(g) @ 25°C, K p = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations substitute into the equilibrium constant expression and solve for x +2x xx xx 0.100  x 0.100+2x

51. Tro, Chemistry: A Molecular Approach51 [I 2 ][Cl 2 ][ICl] initial 0.100 change equilibrium Ex 14.11  For the reaction I 2 (g) + Cl 2 (g)  2 ICl(g) @ 25°C, K p = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations substitute x into the equilibrium concentration definition and solve +2x xx xx 0.100  x 0.100+2x 0.027 0.246 -0.0729 2(-0.0729)

52. Tro, Chemistry: A Molecular Approach52 [I 2 ][Cl 2 ][ICl] initial 0.100 change equilibrium -0.0729 Ex 14.11  For the reaction I 2 (g) + Cl 2 (g)  2 ICl(g) @ 25°C, K p = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K 0.027 0.246 -0.0729 2(0.0729) K p (calculated) = K p (given) within significant figures

53. Tro, Chemistry: A Molecular Approach53 For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]?

54. Tro, Chemistry: A Molecular Approach54 For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.500- x2x2x since [I] initial = 0, Q = 0 and the reaction must proceed forward

55. 55 For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.500- x2x2x

56. Tro, Chemistry: A Molecular Approach56 For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.4980.00432 x = 0.00216 0.500  0.00216 = 0.498 [I 2 ] = 0.498 M 2(0.00216) = 0.00432 [I] = 0.00432 M  

57. Tro, Chemistry: A Molecular Approach57 Approximations to Simplify the Math when the equilibrium constant is very small, the position of equilibrium favors the reactants for relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium the [X] equilibrium = ([X] initial  ax)  [X] initial  we are approximating the equilibrium concentration of reactant to be the same as the initial concentration assuming the reaction is proceeding forward

58. Tro, Chemistry: A Molecular Approach58 Checking the Approximation and Refining as Necessary we can check our approximation afterwards by comparing the approximate value of x to the initial concentration if the approximate value of x is less than 5% of the initial concentration, the approximation is valid

59. Tro, Chemistry: A Molecular Approach59 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. Construct an ICE table for the reaction determine the direction the reaction is proceeding since no products initially, Q c = 0, and the reaction is proceeding forward

60. Tro, Chemistry: A Molecular Approach60 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. represent the change in the partial pressures in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +2x 2x2x 2.50E-4  2x 2x2xx

61. 61 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2.50E-4  2x 2x2xx since K c is very small, approximate the [H 2 S] eq = [H 2 S] init and solve for x 2.50E-4

62. 62 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2x2xx check if the approximation is valid by seeing if x < 5% of [H 2 S] init 2.50E-4 the approximation is not valid!!

63. 63 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2.50E-4  2x 2x2xx if approximation is invalid, substitute x current into K c where it is subtracted and re-solve for x new x current = 1.38 x 10 -5

64. [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2.50E-4  2x 2x2xx Substitute x current into K c where it is subtracted and re-solve for x new. If x new is the same number, you have arrived at the best approximation. x current = 1.27 x 10 -5 since x current = x new, approx. OK

65. Tro, Chemistry: A Molecular Approach65 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2.50E-4  2x 2x2xx x current = 1.28 x 10 -5 substitute x current into the equilibrium concentration definitions and solve 2.24E-42.56E-51.28E-5

66. 66 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2.24E-4 2.56E-51.28E-5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K K c (calculated) = K c (given) within significant figures

67. Tro, Chemistry: A Molecular Approach67 For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? (use the simplifying assumption to solve for x)

68. Tro, Chemistry: A Molecular Approach68 For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.500- x2x2x since [I] initial = 0, Q = 0 and the reaction must proceed forward

69. Tro, Chemistry: A Molecular Approach69 For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.500- x2x2x the approximation is valid!!

70. Tro, Chemistry: A Molecular Approach70 For the reaction I 2 (g)  2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.500- x2x2x x = 0.00217 0.500  0.00217 = 0.498 [I 2 ] = 0.498 M 2(0.00217) = 0.00434 [I] = 0.00434 M  

71. Tro, Chemistry: A Molecular Approach71 Disturbing and Re-establishing Equilibrium once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same however if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established the new concentrations will be different, but the equilibrium constant will be the same unless you change the temperature

72. Tro, Chemistry: A Molecular Approach72 Le Ch âtelier’s Principle Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium it says that if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance disturbances all involve making the system open

73. Tro, Chemistry: A Molecular Approach73 An Analogy: Population Changes When the populations of Country A and Country B are in equilibrium, the emigration rates between the two states are equal so the populations stay constant. When an influx of population enters Country B from somewhere outside Country A, it disturbs the equilibrium established between Country A and Country B. The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is established, however the new populations will have different numbers of people than the old ones.

74. Tro, Chemistry: A Molecular Approach74 The Effect of Concentration Changes on Equilibrium Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found that has the same K Removing a product will increase the amounts of the other products and decrease the amounts of the reactants. you can use this to drive a reaction to completion! Equilibrium shifts away from side with added chemicals or toward side with removed chemicals Remember, adding more of a solid or liquid does not change its concentration – and therefore has no effect on the equilibrium

75. Tro, Chemistry: A Molecular Approach75 Disturbing Equilibrium: Adding or Removing Reactants after equilibrium is established, how will adding a reactant affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? What will this cause? How will it affect the value of K? as long as the added reactant is included in the equilibrium constant expression  i.e., not a solid or liquid after equilibrium is established, how will removing a reactant affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? What will this cause? How will it affect the value of K?

76. Tro, Chemistry: A Molecular Approach76 Disturbing Equilibrium: Adding Reactants adding a reactant initially increases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction. the reaction proceeds to the right until equilibrium is re-established. at the new equilibrium position, you will have more of the products than before, less of the non-added reactants than before, and less of the added reactant but not as little of the added reactant as you had before the addition at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

77. Tro, Chemistry: A Molecular Approach77 Disturbing Equilibrium: Removing Reactants removing a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction. so the reaction is going faster in reverse the reaction proceeds to the left until equilibrium is re- established. at the new equilibrium position, you will have less of the products than before, more of the non-removed reactants than before, and more of the removed reactant but not as much of the removed reactant as you had before the removal at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

78. Tro, Chemistry: A Molecular Approach78 The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium adding a gaseous reactant increases its partial pressure, causing the equilibrium to the right increasing its partial pressure increases its concentration does not increase the partial pressure of the other gases in the mixture adding an inert gas to the mixture has no effect on the position of equilibrium does not effect the partial pressures of the gases in the reaction

79. Tro, Chemistry: A Molecular Approach79 The Effect of Concentration Changes on Equilibrium When NO 2 is added, some of it combines to make more N 2 O 4

80. Tro, Chemistry: A Molecular Approach80 The Effect of Concentration Changes on Equilibrium When N 2 O 4 is added, some of it decomposes to make more NO 2

81. Tro, Chemistry: A Molecular Approach81 Effect of Volume Change on Equilibrium decreasing the size of the container increases the concentration of all the gases in the container increases their partial pressures if their partial pressures increase, then the total pressure in the container will increase according to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure the way the system reduces the pressure is to reduce the number of gas molecules in the container when the volume decreases, the equilibrium shifts to the side with fewer gas molecules

82. Tro, Chemistry: A Molecular Approach82 Disturbing Equilibrium: Changing the Volume after equilibrium is established, how will decreasing the container volume affect the total pressure of solids, liquid, and gases? How will it affect the concentration of solids, liquid, solutions, and gases? What will this cause? How will it affect the value of K?

83. Tro, Chemistry: A Molecular Approach83 Disturbing Equilibrium: Reducing the VolumeReducing the Volume for solids, liquids, or solutions, changing the size of the container has no effect on the concentration, therefore no effect on the position of equilibrium decreasing the container volume will increase the total pressure Boyle’s Law if the total pressure increases, the partial pressures of all the gases will increase  Dalton’s Law of Partial Pressures decreasing the container volume increases the concentration of all gases same number of moles, but different number of liters, resulting in a different molarity since the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules shift toward the side with fewer gas molecules at the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same

84. 84 Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products. When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side. The Effect of Volume Changes on Equilibrium

85. Tro, Chemistry: A Molecular Approach85 The Effect of Temperature Changes on Equilibrium Position exothermic reactions release energy and endothermic reactions absorb energy if we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes even though heat is not matter and not written in a proper equation

86. Tro, Chemistry: A Molecular Approach86 The Effect of Temperature Changes on Equilibrium for Exothermic Reactions for an exothermic reaction, heat is a product increasing the temperature is like adding heat according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat adding heat to an exothermic reaction will decrease the concentrations of products and increase the concentrations of reactants adding heat to an exothermic reaction will decrease the value of K how will decreasing the temperature affect the system? aA + bB  cC + dD + Heat

87. Tro, Chemistry: A Molecular Approach87 The Effect of Temperature Changes on Equilibrium for Endothermic Reactions for an endothermic reaction, heat is a reactant increasing the temperature is like adding heat according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat adding heat to an endothermic reaction will decrease the concentrations of reactants and increase the concentrations of products adding heat to an endothermic reaction will increase the value of K how will decreasing the temperature affect the system? Heat + aA + bB  cC + dD

88. Tro, Chemistry: A Molecular Approach88 The Effect of Temperature Changes on Equilibrium

89. Tro, Chemistry: A Molecular Approach89 Not Changing the Position of Equilibrium - Catalysts catalysts provide an alternative, more efficient mechanism works for both forward and reverse reactions affects the rate of the forward and reverse reactions by the same factor therefore catalysts do not affect the position of equilibrium

90. Tro, Chemistry: A Molecular Approach90 Practice - Le Ch âtelier’s Principle The reaction 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) with  H° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established? adding more O 2 to the container condensing and removing SO 3 compressing the gases cooling the container doubling the volume of the container warming the mixture adding the inert gas helium to the container adding a catalyst to the mixture

91. Tro, Chemistry: A Molecular Approach91 Practice - Le Ch âtelier’s Principle The reaction 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) with  H° = -198 kJ is at equilibrium. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established? adding more O 2 to the containershift to SO 3 condensing and removing SO 3 shift to SO 3 compressing the gasesshift to SO 3 cooling the containershift to SO 3 doubling the volume of the containershift to SO 2 warming the mixtureshift to SO 2 adding helium to the containerno effect adding a catalyst to the mixtureno effect