1. 14 Chemical Equilibrium Contents 14-1 Introduction to Chemical Equilibria 14-2 Equilibrium Constants and their Expression 14-3 Determination of Values of Equilibrium Constants 14-4 Calculations Involving Equilibrium Constants 14-5 Using Le Chatelier’s Principle to Predict Shifts in Chemical Equilibria 14-6 Some Industrially Important Chemical Equilibria

2. 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. The reversible reaction is written with a double arrow. forward 2SO 2 (g) + O 2 (g) 2SO 3 (g) reverse 14-1 Introduction to Chemical Equilibria

3. 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. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

4. Chemical Equilibrium When equilibrium is reached  There is no further change in the amounts of reactant and product. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

5. Equilibrium At equilibrium  The forward reaction of N 2 and O 2 forms NO.  The reverse reaction of 2NO forms N 2 and O 2.  The amounts of N 2, O 2, and NO remain constant. N 2 (g) + O 2 (g) 2NO(g) Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

6. Learning Check Write the forward and reverse reactions for the following: CH 4 (g) + 2H 2 S(g) CS 2 (g) + 4H 2 (g) The two single-headed arrows, one pointing to the right and the other to the left, indicate an equilibrium.

7. Solution Write the forward and reverse reactions for the following: Forward: CH 4 (g) + 2H 2 S(g) CS 2 (g) + 4H 2 (g) Reverse: CS 2 (g) + 4H 2 (g) CH 4 (g) + 2H 2 S(g) Or CH 4 (g) + 2H 2 S(g) CS 2 (g) + 4H 2 (g)

8. Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Dynamic equilibria 15.1

9. HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 - (aq) Acetic acid: weak acid; weak eletrolyte When HC 2 H 3 O 2 is dissolved in water, some ions are formed, although most of the acetic acid remains in the form of molecules. (0.1M HAc solution at 25 ℃ : 1.3%H +, 1.3%Ac -, 98.7%HAc) HCl (aq) H + + Cl - Hydrochloric acid: strong acid; strong eletrolyte HCl is assumed to be ionized completely in aqueous solution.

10. Reaching equilibrium on the macroscopic and molecular levels. N 2 O 4 ( g ) 2NO 2 ( g )

11. N 2 O 4 (g; colorless) 2NO 2 (g; reddish brown) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium 15.1

12. constant The NO 2 -N 2 O 4 System at 25ºC

13. N 2 O 4 (g) 2NO 2 (g) = 4.63 x 10 -3 K = [NO 2 ] 2 [N 2 O 4 ] aA + bB cC + dD K = [C] c [D] d [A] a [B] b K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Equilibrium Will 14.1

14. 14-2 Equilibrium Constants and Equilibrium Homogeneous Equilibria: Equilibria that involve only a single phase. H 2 (g) +I 2 (g) 2HI(g) Equilibrium Constants expression: [ ] : Stand for concentration in molarity eq : indicate an equilibrium concentration, usually omitted. Kc is equal to the product of the equilibrium molarities of the products of reaction, each raised to a power given by the coefficient in the equation, divided by the product of the equilibrium molarities of the reactants of reaction, each raised to a power given by the coefficient in the equation.

15. The Equilibrium Constant No matter the starting composition of reactants and products, the same ratio of concentrations is achieved at equilibrium. For a general reaction the equilibrium constant expression is where K c is the equilibrium constant. 14-2 Equilibrium Constants and Equilibrium

16. The Equilibrium Constant K c is based on the molarities of reactants and products at equilibrium. We generally omit the units of the equilibrium constant. Note that the equilibrium constant expression has products over reactants.

17. The Equilibrium Constant The Equilibrium Constant in Terms of Pressure If K P is the equilibrium constant for reactions involving gases, we can write: K P is based on partial pressures measured in atmospheres. We can show that P A = [A](RT)

18. The Equilibrium Constant The Equilibrium Constant in Terms of Pressure P A = [A](RT) This means that we can relate K c and K P : where  n is the change in number of moles of gas. It is important to use:  n = n gas (products) - n gas (reactants)

19. The Equilibrium Constant The Magnitude of Equilibrium Constants The equilibrium constant, K, is the ratio of products to reactants. Therefore, the larger K the more products are present at equilibrium. Conversely, the smaller K the more reactants are present at equilibrium. If K >> 1, then products dominate at equilibrium and equilibrium lies to the right. If K << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.

20. The Equilibrium Constant The Magnitude of Equilibrium Constants

21. The range of equilibrium constants. small K large K intermediate K N 2 (g) + O 2 (g) 2NO(g) K=1×10 -30 Barely proceed; no reaction 2CO (g) + O 2 (g) 2CO 2 (g) K=2.2×10 22 Go to completion 2BrCl(g) Br 2 (g) + Cl 2 (g) K=5 Significant amounts of both reactant and product are present at equilibrium.

22. The Equilibrium Constant The Magnitude of Equilibrium Constants An equilibrium can be approached from any direction. Example: has

23. The Equilibrium Constant The Magnitude of Equilibrium Constants However, has The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction.

24. The Equilibrium Constant Heterogeneous Equilibria When all reactants and products are in one phase, the equilibrium is homogeneous. If one or more reactants or products are in a different phase, the equilibrium is heterogeneous. Equilibria that involve more than one phase. Consider: –experimentally, the amount of CO 2 does not seem to depend on the amounts of CaO and CaCO 3. Why? The concentration of a solid or pure liquid is its density divided by molar mass.

25. The Equilibrium Constant Heterogeneous Equilibria

26. The Equilibrium Constant Heterogeneous Equilibria Neither density nor molar mass is a variable, the concentrations of solids and pure liquids are constant. For the decomposition of CaCO 3 : We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions. The amount of CO 2 formed will not depend greatly on the amounts of CaO and CaCO 3 present.

27. Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) K c = ‘ [CaO][CO 2 ] [CaCO 3 ] [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K c x ‘ [CaCO 3 ] [CaO] K p = P CO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. 14.2

28. P CO 2 = K p CaCO 3 (s) CaO (s) + CO 2 (g) P CO 2 does not depend on the amount of CaCO 3 or CaO 14.2

29. Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. N 2 O 4 (g) 2NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] K p = NO 2 P2P2 N2O4N2O4 P In most cases K c  K p aA (g) + bB (g) cC (g) + dD (g) 14.2 K p = K c (RT)  n  n = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

30. Ways Different States of Matter Can Appear in the Equilibrium Constant, K MolarityPartial Pressure gas, (g) YES aqueous, (aq)YES--- liquid, (l)--- solid, (s)---

31. Writing Equilibrium Constant Expressions 1. Write an equation for the equilibrium. 2. Put the product of the equilibrium concentrations of the products in the numerator. Omit pure solids, pure liquids and solvents in dilute solutions (concentration around 0.1M or less). 3. Write the product of the equilibrium concentrations of the reactants in the denominator. Omit pure solids, pure liquids and solvents in dilute solutions. 4 The exponent of each concentration should be the same as the coefficient of the species in the equation. 14.2

32. Notice: The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents in dilute solutions do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature.

33. 33 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K 1. Can tell if a reaction is product-favored or reactant-favored. For N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

34. 34 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K 1.Can tell if a reaction is product-favored or reactant-favored. For N 2 (g) + 3 H 2 (g) 2 NH 3 (g)

35. 35 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K 1.Can tell if a reaction is product-favored or reactant-favored. For N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Conc. of products is much greater than that of reactants at equilibrium.

36. 36 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K 1.Can tell if a reaction is product-favored or reactant-favored. For N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Conc. of products is much greater than that of reactants at equilibrium. The reaction is strongly product-favored.

37. 37 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K For AgCl(s) Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = 1.8 x 10 -5 K c = [Ag + ] [Cl - ] = 1.8 x 10 -5 Conc. of products is much less than that of reactants at equilibrium. The reaction is strongly reactant- favored. Ag + (aq) + Cl - (aq) AgCl(s) AgCl(s) is product-favored. Ag + (aq) + Cl - (aq) AgCl(s) AgCl(s) is product-favored.

38. 38 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K 2. Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium. H H H H H H H H H HH C H H H H H H—C—C—C—C—HH—C—C—C—H K = n-butane iso-butane [iso] [n] = 2.5

39. 39 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? Which way does the reaction “shift” to approach equilibrium? H H H H H H H H H HH C H H H H H H—C—C—C—C—HH—C—C—C—H K = n-butane iso-butane [iso] [n] = 2.5

40. 40 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Predicting the Direction of Reaction We define Q, the reaction quotient, for a general reaction as where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium. 14-4 Calculations Involving Equilibrium Constants

41. 41 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Applications of Equilibrium Constants Predicting the Direction of Reaction If Q > K then the reverse reaction must occur to reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals K). If Q < K then the forward reaction must occur to reach equilibrium.

42. 42 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Reaction Quotient 1. Q=K, equilibrium 2. Q>K, the system shifts to the left 3. Q<K, the system shifts to the right

43. 43 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The reaction quotient (Q c ) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (K c ) expression. IF Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium 14.4

44. 44 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved Applications of Equilibrium Constants Calculation of Equilibrium Concentrations The same steps used to calculate equilibrium constants are used. Generally, we do not have a number for the change in concentration line. Therefore, we need to assume that x mol/L of a species is produced (or used). The equilibrium concentrations are given as algebraic expressions.

45. 45 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium.

46. 46 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. If Q = K, then system is at equilibrium. If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? Q (2.3) < K (2.5).

47. 47 Copyright (c) 1999 by Harcourt Brace & Company All rights reserved The Meaning of K In general, 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 ____________.

48. 14-3 Determination of Values of Equilibrium Constants The numerical values of equilibrium constants can be found by experiment. See Sample problem 14.4. Initial concentration: the concentrations at the beginning of the experiment, [ ] 0 Equilibrium concentration, [ ] equ Stoichiometry problem Changes in concentration due to reaction

49. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = 0.012 M, [Cl 2 ] = 0.054 M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = 0.14 0.012 x 0.054 = 220 K p = K c (RT)  n  n = 1 – 2 = -1 R = 0.0821T = 273 + 74 = 347 K K p = 220 x (0.0821 x 347) -1 = 7.7 14.2

50. The equilibrium constant K p for the reaction is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = 0.400 atm and P NO = 0.270 atm? 2 2NO 2 (g) 2NO (g) + O 2 (g) 14.2 K p = 2 P NO P O 2 P NO 2 2 POPO 2 = K p P NO 2 2 2 POPO 2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm

51. CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) K c = ‘ [CH 3 COO - ][H 3 O + ] [CH 3 COOH][H 2 O] [H 2 O] = constant K c = [CH 3 COO - ][H 3 O + ] [CH 3 COOH] =K c [H 2 O] ‘ General practice not to include units for the equilibrium constant. 14.2

52. Calculating Equilibrium Constants Proceed as follows: –Tabulate initial and equilibrium concentrations (or partial pressures) given. –If an initial and equilibrium concentration is given for a species, calculate the change in concentration. –Use stoichiometry on the change in concentration line only to calculate the changes in concentration of all species. –Deduce the equilibrium concentrations of all species. Usually, the initial concentration of products is zero. (This is not always the case.) When in doubt, assign the change in concentration a variable.

53. PROBLEM: Place 1.00 mol each of H 2 and I 2 in a 1.00 L flask. Calc. equilibrium concentrations. H 2 (g) + I 2 (g) 2 HI(g) Typical Calculations

54. H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3 Step 1. Set up table to define EQUILIBRIUM concentrations. [H 2 ][I 2 ][HI] [H 2 ][I 2 ][HI] Initial 1.001.000 ChangeEquilib

55. Step 1. Set up table to define EQUILIBRIUM concentrations. [H 2 ][I 2 ][HI] [H 2 ][I 2 ][HI] Initial 1.001.000 Change-x-x+2x Equilib1.00-x1.00-x2x where x is defined as amount of H 2 and I 2 consumed on approaching equilibrium. H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3

56. Step 2. Put equilibrium concentrations into K c expression. H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3

57. Step 3. Solve K c expression - take square root of both sides. x = 0.79 H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3

58. Step 3. Solve K c expression - take square root of both sides. x = 0.79 Therefore, at equilibrium [H 2 ] = [I 2 ] = 1.00 - x = 0.21 M H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3

59. Step 3. Solve K c expression - take square root of both sides. x = 0.79 Therefore, at equilibrium [H 2 ] = [I 2 ] = 1.00 - x = 0.21 M [HI] = 2x = 1.58 M H 2 (g) + I 2 (g) 2 HI(g), K c = 55.3

60. Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g)

61. If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an equilibrium table [N 2 O 4 ][NO 2 ] [N 2 O 4 ][NO 2 ] Initial0.500 ChangeEquilib

62. Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) If initial concentration of N 2 O 4 is 0.50 M, what are the equilibrium concentrations? Step 1. Set up an equilibrium table [N 2 O 4 ][NO 2 ] [N 2 O 4 ][NO 2 ] Initial0.500 Change-x+2x Equilib0.50 - x2x

63. Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) Step 2. Substitute into K c expression and solve. Rearrange: 0.0059 (0.50 - x) = 4x 2 0.0029 - 0.0059x = 4x 2 0.0029 - 0.0059x = 4x 2 4x 2 + 0.0059x - 0.0029 = 0 4x 2 + 0.0059x - 0.0029 = 0 This is a QUADRATIC EQUATION ax 2 + bx + c = 0 ax 2 + bx + c = 0 a = 4b = 0.0059 c = -0.0029

64. Solve the quadratic equation for x. ax 2 + bx + c = 0 a = 4b = 0.0059 c = -0.0029 Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g)

65. Solve the quadratic equation for x. ax 2 + bx + c = 0 a = 4b = 0.0059 c = -0.0029

66. Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) Solve the quadratic equation for x. ax 2 + bx + c = 0 a = 4b = 0.0059 c = -0.0029 x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027

67. Nitrogen Dioxide Equilibrium N 2 O 4 (g) 2 NO 2 (g) x = -0.00074 ± 1/8(0.046) 1/2 = -0.00074 ± 0.027 x = 0.026 or -0.028 But a negative value is not reasonable. Conclusion Conclusion [N 2 O 4 ] = 0.050 - x = 0.47 M [N 2 O 4 ] = 0.050 - x = 0.47 M [NO 2 ] = 2x = 0.052 M [NO 2 ] = 2x = 0.052 M

68. Le Châtelier’s Principle 1.System starts at equilibrium. 2.A change/stress is then made to system at equilibrium. Change in concentration Change in volume Change in pressure Change in Temperature Add Catalyst 3.System responds by shifting to reactant or product side to restore equilibrium.

69. Consider the production of ammonia As the pressure increases, the amount of ammonia present at equilibrium increases. As the temperature decreases, the amount of ammonia at equilibrium increases. Can this be predicted? Le Châtelier’s Principle: if a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance. 14-5 Using Le Chatelier’s Principle to Predict Shifts in Chemical Equilibria

70. Le Châtelier’s Principle

71. Change in Reactant or Product Concentrations Consider the Haber process If H 2 is added while the system is at equilibrium, the system must respond to counteract the added H 2 (by Le Châtelier). That is, the system must consume the H 2 and produce products until a new equilibrium is established. Therefore, [H 2 ] and [N 2 ] will decrease and [NH 3 ] increases.

72. Le Châtelier’s Principle Change in Reactant or Product Concentrations

73. Le Châtelier’s Principle Changes in Concentration continued ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left 14.5 aA + bB cC + dD Add Remove

74. Le Châtelier’s Principle Change in Reactant or Product Concentrations Adding a reactant or product shifts the equilibrium away from the increase. Removing a reactant or product shifts the equilibrium towards the decrease. To optimize the amount of product at equilibrium, we need to flood the reaction vessel with reactant and continuously remove product (Le Châtelier). We illustrate the concept with the industrial preparation of ammonia

75. Le Châtelier’s Principle Change in Reactant or Product Concentrations

76. Le Châtelier’s Principle Change in Reactant or Product Concentrations N 2 and H 2 are pumped into a chamber. The pre-heated gases are passed through a heating coil to the catalyst bed. The catalyst bed is kept at 460 - 550  C under high pressure. The product gas stream (containing N 2, H 2 and NH 3 ) is passed over a cooler to a refrigeration unit. In the refrigeration unit, ammonia liquefies but not N 2 or H 2.

77. Le Châtelier’s Principle Change in Reactant or Product Concentrations The unreacted nitrogen and hydrogen are recycled with the new N 2 and H 2 feed gas. The equilibrium amount of ammonia is optimized because the product (NH 3 ) is continually removed and the reactants (N 2 and H 2 ) are continually being added. Effects of Volume and Pressure As volume is decreased pressure increases. Le Châtelier’s Principle: if pressure is increased the system will shift to counteract the increase.

78. Le Châtelier’s Principle Effects of Volume and Pressure That is, the system shifts to remove gases and decrease pressure. An increase in pressure favors the direction that has fewer moles of gas. In a reaction with the same number of product and reactant moles of gas, pressure has no effect. Consider

79. Le Châtelier’s Principle Effects of Volume and Pressure An increase in pressure (by decreasing the volume) favors the formation of colorless N 2 O 4. The instant the pressure increases, the system is not at equilibrium and the concentration of both gases has increased. The system moves to reduce the number moles of gas (i.e. the forward reaction is favored). A new equilibrium is established in which the mixture is lighter because colorless N 2 O 4 is favored.

80. Le Châtelier’s Principle Effect of Temperature Changes The equilibrium constant is temperature dependent. For an endothermic reaction,  H > 0 and heat can be considered as a reactant. For an exothermic reaction,  H < 0 and heat can be considered as a product. Adding heat (i.e. heating the vessel) favors away from the increase: –if  H > 0, adding heat favors the forward reaction, –if  H < 0, adding heat favors the reverse reaction.

81. Le Châtelier’s Principle Effect of Temperature Changes Removing heat (i.e. cooling the vessel), favors towards the decrease: –if  H > 0, cooling favors the reverse reaction, –if  H < 0, cooling favors the forward reaction. Consider for which  H > 0. –Co(H 2 O) 6 2+ is pale pink and CoCl 4 2- is blue.

82. Le Châtelier’s Principle Effect of Temperature Changes

83. Le Châtelier’s Principle Effect of Temperature Changes –If a light purple room temperature equilibrium mixture is placed in a beaker of warm water, the mixture turns deep blue. –Since  H > 0 (endothermic), adding heat favors the forward reaction, i.e. the formation of blue CoCl 4 2-. –If the room temperature equilibrium mixture is placed in a beaker of ice water, the mixture turns bright pink. –Since  H > 0, removing heat favors the reverse reaction which is the formation of pink Co(H 2 O) 6 2+.

84. Le Châtelier’s Principle The Effect of Catalysts A catalyst lowers the activation energy barrier for the reaction. Therefore, a catalyst will decrease the time taken to reach equilibrium. A catalyst does not effect the composition of the equilibrium mixture.

85. Calculating Equilibrium Concentrations from K and Initial Concentrations 1.Balance chemical equation. 2.Determine Q to see which way reaction will proceed to get to equilibrium 3.Define changes needed to get to equilibrium. Use ‘x’ to represent the change in concentration to get to equilibrium. Determine expression for equilibrium concentration. 4.Write the expression for equilibrium constant, K. 5.Substitute expressions defined in step 3 into the equilibrium constant (step 4). 6.Solve for x and plug value back to table made in step 3. 14.4

86. At 1280 0 C the equilibrium constant (K c ) for the reaction Is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x [Br] 2 [Br 2 ] K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 Solve for x 14.4

87. K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x x = -0.00178x = -0.0105 At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 M At equilibrium, [Br 2 ] = 0.062 – x = 0.0648 M 14.4

88. 14-6 Some Industrially Important Chemical Equilibria A brief summary of the Contact Process for Sulfuric Acid: The Contact Process: makes sulphur dioxide; converts the sulphur dioxide into sulphur trioxide (the reversible reaction at the heart of the process); converts the sulphur trioxide into concentrated sulphuric acid.

89. Making the sulphur dioxide This can either be made by burning sulphur in an excess of air:... or by heating sulphide ores like pyrite in an excess of air: In either case, an excess of air is used so that the sulphur dioxide produced is already mixed with oxygen for the next stage.

90. Converting the sulphur dioxide into sulphur trioxide This is a reversible reaction, and the formation of the sulphur trioxide is exothermic. A flow scheme for this part of the process looks like this:

91. Converting the sulphur trioxide into sulphuric acid This can't be done by simply adding water to the sulphur trioxide - the reaction is so uncontrollable that it creates a fog of sulphuric acid. Instead, the sulphur trioxide is first dissolved in concentrated sulphuric acid: The product is known as fuming sulphuric acid or oleum. This can then be reacted safely with water to produce concentrated sulphuric acid - twice as much as you originally used to make the fuming sulphuric acid.