1. Equilibrium 2007-20081 Chemical Equilibrium

2. Equilibrium 2007-20082 Chemical Equilibrium Few chemical reactions proceed in only one direction. Most are reversible, at least to some extent. At the start of a reversible process, the reaction proceeds toward the formation of products. As soon as some product molecules are formed, the reverse process begins to take place and reactant molecules are formed from product molecules. 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.

3. Equilibrium 2007-20083 Equilibrium is a dynamic process. Physical equilibrium H 2 O(s) H 2 O(l) At equilibrium, liquid water is reforming ice as fast as ice is melting.

4. Equilibrium 2007-20084 Chemical equilibrium N 2 O 4 (g) 2NO 2 (g) colorless brown Equilibrium is a state in which there are no observable changes as time goes by. When a chemical reaction has reached the equilibrium state, the concentrations of reactants and products remain constant over time, and there are no visible changes in the system.

5. Equilibrium 2007-20085 The Law of Mass Action For a reversible reaction at equilibrium and at a constant temperature, the ratio of reactant and product concentrations has a constant value, k (the equilibrium constant). Note that although the concentrations may vary, as long as a given reaction is at equilibrium and the temperature does not change, according to the law of mass action, the value of k remains constant.

6. 6 We can generalize this phenomenon with the following reversible reaction: aA + bB cC + dD where a, b, c and d are the stoichiometric coefficients for the reacting species A, B, C, and D. k = [C] c [D] d [A] a [B] b k c indicates that the concentrations of the reacting species are expressed in molarity or moles per liter. k p is used when equilibrium concentrations are expressed in terms of pressure. Law of Mass Action

7. 7 k = For the reaction, N 2 O 4 (g) 2NO 2 (g) at 25 o C k = [NO 2 ] 2 [N 2 O 4 ] Note that the exponent 2 for [NO 2 ] in this expression is the same as the stoichiometric coefficient for NO 2 in the balanced chemical equation. k is defined as having no units.

8. Equilibrium 2007-20088 Changes in the concentrations of NO 2 and N 2 O 4 with time, in three situations. (a)Initially only NO 2 is present (b)Initially only N 2 O 4 is present (c)Initially a mixture of NO 2 and N 2 O 4 is present Note that even though equilibrium is reached in all cases, the equilibrium concentrations of NO 2 and N 2 O 4 are not the same as in the previous trials.

9. Equilibrium 2007-20089 Solving for k, the equilibrium constant The following table shows some experimental data for the NO 2 -N 2 O 4 system at 25 o C. The gas concentrations are expressed in molarity, which can be calculated from the number of moles of the gases present initially and at equilibrium and the volume of the flask in liters.

10. 10 The NO 2 -N 2 O 4 System at 25 o C Initial Concentrations (M) Equilibrium Concentrations (M) Ratio of Concentrations at Equilibrium [NO 2 ][N 2 O 4 ][NO 2 ][N 2 O 4 ] [NO 2 ] [N 2 O 4 ] [NO 2 ] 2 [N 2 O 4 ] 0.0000.6700.05470.643 0.05000.4460.04570.448 0.03000.5000.04750.491 0.04000.6000.05230.594 0.2000.0000.02040.0898 0.0851 0.102 0.0967 0.0880 0.227 Analysis of the equilibrium data shows that the ratio [NO 2 ]/[N 2 O 4 ] gives scattered values.

11. 11 The NO 2 -N 2 O 4 System at 25 o C Initial Concentrations (M) Equilibrium Concentrations (M) Ratio of Concentrations at Equilibrium [NO 2 ][N 2 O 4 ][NO 2 ][N 2 O 4 ] [NO 2 ] [N 2 O 4 ] [NO 2 ] 2 [N 2 O 4 ] 0.0000.6700.05470.643 0.05000.4460.04570.448 0.03000.5000.04750.491 0.04000.6000.05230.594 0.2000.0000.02040.0898 0.0851 0.102 0.0967 0.0880 0.227 Analysis of the equilibrium data shows that the ratio [NO 2 ] 2 /[N 2 O 4 ] gives a nearly constant value that averages 4.63 x 10 -3

12. Equilibrium 2007-200812 The magnitude of the equilibrium constant tells us whether an equilibrium reaction favors the products or reactants. If k is much greater than 1 (in this context, any number greater than 10 is considered to be much greater than 1) the equilibrium will lie to the right and favors the products. Conversely, if the equilibrium constant is much smaller than 1 (any number less than 0.1 is much less than 1), the equilibrium will lie to the left and favor the reactants.

13. 13 The concentrations of reactants and products in gaseous reactions can also be expressed in terms of their partial pressures. For the equilibrium process N 2 O 4 (g) 2NO 2 (g) we can write k p = (P NO 2 ) 2 (P N 2 O 4 ) Where P NO 2 and P N 2 O 4 are the equilibrium partial pressures (in atm) of NO 2 and N 2 O 4, respectively.

14. 14 Example – Use the Law of Mass Action to write the equilibrium constant expression k c and k p if applicable, for the following system: (NH 4 ) 2 Se(s) 2NH 3 (g) + H 2 Se(g) The “concentration” of a solid or a liquid is a constant, (an intensive property that does not depend on how much of the substance is present) and can be omitted when writing equilibrium expressions. k c = [NH 3 ] 2 [H 2 Se] k = [NH 3 ] 2 [H 2 Se] [(NH 4 ) 2 Se] k p = (P NH 3 ) 2 (P H 2 Se )

15. Equilibrium 2007-200815 Use the Law of Mass Action to write the equilibrium expressions for the following reactions. CH 3 COOH (aq) H + (aq) + CH 3 COOH + (aq) N 2 O 4(g) 2NO 2(g) Fe(OH) 3(s) Fe +3 (aq) + 3OH - (aq) kc = kp = kc =

16. Equilibrium 2007-200816 In general, k c is not equal to k p, since the partial pressures of reactants and products are not equal to their concentrations expressed in moles per liter. A simple relationship exists between k p and k c k p = k c (RT)  n where  n = moles of gaseous product – moles of gaseous reactants R = the gas constant 0.0821 L. atm/mol. K and T = temperature in Kelvins

17. Equilibrium 2007-200817 Consider the following equilibrium at 395 K NH 4 HS (s) NH 3(g) + H 2 S (g) The partial pressure of each gas is.265 atm. Calculate k p and k c for the reaction

18. Equilibrium 2007-200818 Multiple Equilibria The reactions we have considered so far are all relatively simple. A more complicated situation is one in which the product molecules in one equilibrium system are involved in a second equilibrium process: aA + bB cC + dDk’ c cC + dD eE + fFk” c aA + bB eE + fFk c k c = k’ c. k” c

19. Equilibrium 2007-200819 A typical operating temperature of an auto engine (~100 o C), N 2 and O 2 form nitric oxide, which then combines with more O 2 to form nitrogen dioxide, a toxic pollutant: (1) N 2(g) + O 2(g) 2NO (g) kc 1 = 4.3 x 10 -25 (2) 2NO (g) + O 2(g) 2NO 2(g) kc 2 = 6.4 x 10 9 Calculate k c for the overall reaction.

20. Equilibrium 2007-200820 If a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant. The value of k also depends on how the equilibrium equation is balanced. (If you double the chemical equation throughout, the corresponding equilibrium constant will be the square of the original value.) The balanced chemical equation must be written when quoting the numerical value of an equilibrium constant.

21. Equilibrium 2007-200821 Q, the Reaction Quotient The equilibrium constant k c for the formation of hydrogen iodide from molecular hydrogen and molecular iodine in the gas phase H 2 (g) + I 2 (g) 2HI(g) is 54.3 at 430 o C. Suppose that in a certain experiment we place 0.243 mole of H 2, 0.146 mole of I 2, and 1.98 moles of HI all in a 1.00 L container at 430 o C. Will there be a net reaction to form more H 2 and I 2 or more HI? [HI] 0 2 [H 2 ] 0 [I 2 ] 0 [1.98] 2 [0.243][0.146] == 111 Q=

22. Equilibrium 2007-200822 Because the quotient [HI] 0 2 /[H 2 ] 0 [I 2 ] 0 is greater than k c, this system is not at equilibrium. Consequently, some of the HI will react to form more H 2 and I 2 (decreasing the value of the quotient). Thus the net reaction proceeds from right to left to reach equilibrium. For reactions that have not reached equilibrium, such as the formation of HI considered above, we obtain the reaction quotient (Q c ), instead of the equilibrium constant by substituting the initial concentrations into the equilibrium constant expression.

23. Equilibrium 2007-200823 To determine the direction in which the net reaction will proceed to achieve equilibrium, we compare the values of Q c and k c. If Q c > k c, the reaction proceeds from right to left (consuming products, forming reactants) to reach equilibrium If Q c < k c, the reaction proceeds from left to right (consuming reactants, forming products) to reach equilibrium. If Q c = k c, the initial concentrations are equilibrium concentrations. The system is at equilibrium.

24. Equilibrium 2007-200824 Calculating concentrations in an equilibrium system If we know the equilibrium constant for a particular reaction, we can calculate the concentrations in the equilibrium mixture from the initial concentrations. Let’s consider the following system involving two organic compounds, cis-stilbene and trans-stilbene cis-stilbene trans-stilbene The equilibrium constant for this system is 24.0 at 200 o C.

25. Equilibrium 2007-200825 Suppose that initially only cis-stilbene is present at a concentration of 0.850 mol/L. How do we calculate the concentrations of cis and trans-stilbene at equilibrium? Initial0.8500.000 Change-x+x Equilbrium(0.850 – x)x cis-stilbene trans-stilbene

26. 26 Next we set up the equilibrium constant expression k c = [trans-stilbene] [cis-stilbene] 24.0 = x 0.850 - x x = 0.816 M [cis-stilbene] = (0.850- 0.816) M = 0.034 M [trans-stilbene] = 0.816 M

27. Equilibrium 2007-200827 Factors that Affect Equilibrium Chemical equilibrium represents a balance between forward and reverse reactions. In most cases, this balance is quite delicate. Changes in experimental conditions (concentration, pressure, volume and temperature) may disturb the balance and shift the equilibrium position so that more or less of the desired product is formed. When we say that an equilibrium position shifts to the right, for example, we mean that the net reaction proceeds from left to right.

28. Equilibrium 2007-200828 Le Chatelier’s Principle There is a general rule that helps us to predict the direction in which an equilibrium reaction will move when a change in concentration, pressure, volume, or temperature occurs. The rule, known as Le Chatelier’s principle, states that if an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset. The word “stress” here means a change in concentration, pressure, volume, or temperature that removes a system from the equilibrium state.

29. Equilibrium 2007-200829 Changes in concentration FeSCN +2 (aq) Fe +3 (aq) + SCN - (aq) red pale yellow colorless An aqueous solution of Fe(SCN) 3

30. Equilibrium 2007-200830 In this case, the stress applied to the equilibrium system is an increase in the concentration of SCN - (from the dissociation of NaSCN). To offset this stress, some Fe +3 ions react with the added SCN - ions, and the equilibrium shifts from right to left, and the red color of the solution deepens. What happens if we add some sodium thiocyanate, NaSCN, to the solution? FeSCN +2 (aq) Fe +3 (aq) + SCN - (aq) red pale yellow colorless

31. Equilibrium 2007-200831 Similarly, if we added iron (III) nitrate to the original solution, the red color would also deepen because the additional Fe +3 ions (from the Fe(NO 3 ) 3 solution) would shift the equilibrium from right to left. FeSCN +2 (aq) Fe +3 (aq) + SCN - (aq) red pale yellow colorless

32. Equilibrium 2007-200832 Now suppose we add some oxalic acid, H 2 C 2 O 4, to the original solution. Oxalic acid ionizes in water to form the oxalate ion, C 2 O 4 -2, which binds strongly to the Fe +3 ions. This formation of Fe(C 2 O 4 ) 3 -3 removes free Fe +3 ions in solution. Consequently, more FeSCN +2 units dissociate and the equilibrium shifts from left to right, and the solution will turn yellow. FeSCN +2 (aq) Fe +3 (aq) + SCN - (aq) red pale yellow colorless

33. Equilibrium 2007-200833 Changes in pressure do not affect the concentrations of solids, liquids or aqueous solutions because they are virtually incompressible. On the other hand, concentrations of gases are greatly affected by changes in pressure. Changes in Volume and Pressure

34. Equilibrium 2007-200834 Suppose that the equilibrium system N 2 O 4 (g) 2NO 2 (g) is in a cylinder fitted with a movable piston. What happens if we increase the pressure on the gases by pushing down on the piston at constant temperature?

35. Equilibrium 2007-200835 Since the volume decreases, the concentration of both NO 2 and N 2 O 4 increases. Because the concentration of NO 2 is squared in the equilibrium constant expression, the increase in pressure increases the numerator more than the denominator. The system is no longer at equilibrium, so we write [NO 2 ] 0 2 [N 2 O 4 ] 0 Q c =

36. 36 Q c > K c, and the net reaction will shift to the left until Q c = k c. (a)Brown NO 2 (g) and colorless N 2 O 4 (g) in equilibrium. (b)When the volume in the syringe is suddenly decreased, the concentration of NO 2 and N 2 O 4 are both increased, (indicated by the darker brown color). (c)A few seconds after the sudden volume decrease, the color is much lighter brown as the equilibrium shifts the brown NO 2 (g) to colorless N 2 O 4 (g). (a) (b) (c)

37. Equilibrium 2007-200837 In general, an increase in pressure (decrease in volume) favors the net reaction that decreases the total number of moles of gases (the reverse reaction, in this case), and a decrease in pressure (increase in volume) favors the reaction that increases the total number of moles of gases. For the reactions in which there is no change in the number of moles of gases, a pressure (or volume) change has no effect on the position of equilibrium.

38. Equilibrium 2007-200838 We can also increase the pressure of a system without changing its volume by adding an inert gas (helium, for example) to the equilibrium system. Adding helium to the equilibrium mixture at constant volume increases the total gas pressure and decreases the mole fractions of both NO 2 and N 2 O 4 ; but the partial pressure of each gas, given by the product of its mole fraction and total pressure, does not change. Thus the presence of an inert gas does not affect the equilibrium.

39. Equilibrium 2007-200839 Changes in Temperature A change in concentration, pressure, or volume may alter the equilibrium position, but it DOES NOT change the value of the equilibrium constant (k). Only a change in temperature can alter the equilibrium constant.

40. Equilibrium 2007-200840 N 2 O 4 (g) 2NO 2 (g)  H = 58.0 kJ The formation of NO 2 from N 2 O 4 is an endothermic process. Another way to write this reaction would be The formation of N 2 O 4 from NO 2 is an exothermic process. This process could be written 2NO 2 (g) N 2 O 4 (g)  H = -58.0 kJ N 2 O 4 (g) + heat 2NO 2 (g) 2NO 2 (g) N 2 O 4 (g) + heat or

41. Equilibrium 2007-200841 What will happen if the following system N 2 O 4 (g) 2NO 2 (g) is heated at constant volume?

42. Equilibrium 2007-200842 Since endothermic processes absorb heat from the surroundings, heating favors the dissociation of colorless N 2 O 4 into brown NO 2 molecules, (an endothermic process). N 2 O 4 (g) 2NO 2 (g)

43. Equilibrium 2007-200843 Cooling favors the exothermic process (the formation of colorless N 2 O 4 ). N 2 O 4 (g) 2NO 2 (g)

44. Equilibrium 2007-200844 The Effect of a Catalyst A catalyst enhances the rate of a reaction by lowering the reaction’s activation energy of the forward and reverse reaction to the same extent. We can therefore conclude that the presence of a catalyst does not alter the equilibrium constant, nor does it shift the position of an equilibrium system.

45. Equilibrium 2007-200845 Adding a catalyst to a reaction mixture that is not at equilibrium will simply cause the mixture to reach equilibrium sooner. The same equilibrium mixture could be obtained without the catalyst, but we might have to wait much longer for it to happen.

46. Equilibrium 2007-200846 k sp, Solubility Equilibria Precipitation reactions are important in industry, medicine, and everyday life. Although useful, the solubility rules do not allow us to make quantitative predictions about how much of a given ionic compound will dissolve in water.

47. Equilibrium 2007-200847 Consider a saturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s) Ag + (aq) + Cl - (aq) Because salts are considered strong electrolytes, all the AgCl that dissolves in water is assumed to dissociate completely into Ag+ and Cl- ions.

48. Equilibrium 2007-200848 Because the concentration of a solid is constant, the equilibrium constant for the dissolution of AgCl can be written as k sp = [Ag + ][Cl - ] where k sp is called the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium expression.

49. Equilibrium 2007-200849 Write the equilibrium expressions for the dissolution of magnesium fluoride MgF 2 (s) Mg +2 (aq) + 2F - (aq)k sp = [Mg +2 ][F - ] 2 silver carbonate calcium phosphate Ag 2 CO 3 (s) 2Ag + (aq) + CO 3 -2 (aq) k sp = [Ag + ] 2 [CO 3 -2 ] Ca 3 (PO 4 ) 2 (s) 3Ca +2 (aq) + 2PO 4 -3 (aq) k sp = [Ca +2 ] 3 [PO 4 -3 ] 2

50. Equilibrium 2007-200850 The value of k sp indicates the solubility of an ionic compound-the smaller the value, the less soluble the compound in water. However, in using k sp values to compare solubilities, you should choose compounds that have similar formulas, such as AgCl and ZnS, or CaF 2 and Fe(OH) 2.

51. Equilibrium 2007-200851 For equilibrium reactions involving an ionic solid in aqueous solution, any one of the following conditions may exist: the solution is unsaturated, (a solution that contains less solute than it has the capacity to dissolve) the solution is saturated, (at a given temperature, the solution that results when the maximum amount of a substance has dissolved in a solvent), or the solution is supersaturated, (a solution that contains more of the solute than is present in a saturated solution).

52. Equilibrium 2007-200852 Q, the ion product For concentrations that do not correspond to equilibrium conditions, we use the reaction quotient which in this case is called the ion product (Q), to predict whether a precipitate will form.

53. Equilibrium 2007-200853 For example, if we mix a solution containing Ag + ions with one containing Cl - ions, then the ion product is given by Q = [Ag + ] 0 [Cl - ] 0 The subscript 0 reminds us that these are initial concentrations, and do not necessarily correspond to those at equilibrium.

54. Equilibrium 2007-200854 The possible relationships between Q and k sp are Q< k sp [Ag + ] 0 [Cl - ] 0 < 1.6 x 10 -10 Unsaturated solution Q = k sp [Ag + ] 0 [Cl - ] 0 = 1.6 x 10 -10 Saturated solution Q > k sp [Ag + ] 0 [Cl - ] 0 > 1.6 x 10 -10 Supersaturated solution; AgCl will precipitate out until the product of the ionic concentrations is equal to 1.6 x 10 -10

55. Equilibrium 2007-200855 There are two other ways to express a substance’s solubility: molar solubility, which is the number of moles of solute in one liter of a saturated solution (mol/L), and solubility, which is the number of grams in one liter of a saturated solution (g/L). Both solubility and molar solubility can be used to determine k sp.

56. Equilibrium 2007-200856 The solubility of calcium sulfate is found experimentally to be 0.67 g/L. Calculate the value of k sp for calcium sulfate.

57. Equilibrium 2007-200857 Sometimes we are given the value of k sp for a compound and are asked to calculate the compound’s molar solubility. For example, the k sp of silver bromide is 7.7 x 10 -13. Calculate the molar solubility of silver bromide.

58. Equilibrium 2007-200858 Common Ion Effect The solubility product is an equilibrium constant; precipitation of an ionic compound from solution occurs whenever the ion product exceeds k sp for that substance. In a saturated solution of AgCl, for example, the ion product [Ag + ][Cl - ] is, of course, equal to k sp. Furthermore, simple stoichiometry tells us that [Ag + ]=[Cl - ]. But this equality does not hold true in all situations.

59. Equilibrium 2007-200859 Suppose we study a solution containing two dissolved substances that share a common ion, say, AgCl and AgNO 3. In addition to the dissociation of AgCl, the following process also contributes to the total concentration of the common silver ions in solution: AgNO 3 (s) Ag + (aq) + NO 3 - (aq)

60. Equilibrium 2007-200860 If AgNO 3 is added to a saturated AgCl solution, the increase in [Ag + ] will make the ion product greater than the solubility product: Q = [Ag + ] 0 [Cl-] 0 > k sp To reestablish equilibrium, some AgCl will precipitate out of the solution, as LeChatelier’s principle would predict, until the ion product is one again equal to k sp. The effect of adding a common ion, then, is a decrease in the solubility of the salt (AgCl) in solution. Note that in this case [Ag + ] is no longer equal to [Cl - ] at equilibrium; rather, [Ag + ]>[Cl - ].

61. Equilibrium 2007-200861 pH and Solubility The solubilities of many substances also depend on the pH of the solution. Consider the solubility equilibrium of magnesium hydroxide: Mg(OH) 2 (s) Mg +2 (aq) + 2OH - (aq)

62. Equilibrium 2007-200862 Adding OH - ions (increasing the pH) shifts the equilibrium from right to left, thereby decreasing the solubility of Mg(OH) 2. On the other hand, adding H + ions (decreasing the pH) shifts the equilibrium from left to right, (because H + and OH - will form water, thus removing the OH- from solution) and the solubility of Mg(OH) 2 increases. Thus, insoluble bases tend to dissolve in acidic solutions.

63. Equilibrium 2007-200863 Lewis acids and base reactions in which a metal cation (electron pair acceptor) combines with a Lewis base (electron pair donor) result in the formation of complex ions: Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2 + Lewis acidLewis base Transition metals have a particular tendency to form complex ions. For example, a solution of cobalt (II) chloride is pink because of the presence of the Co(H 2 O) 6 +2 ions. When HCl is added, the solution turns blue as a result of the formation of the complex ion CoCl 4 -2 Co +2 (aq) + 4Cl - (aq) CoCl 4 -2

64. Equilibrium 2007-200864 Separation of Ions by Fractional Precipitation In chemical analysis, it is sometimes desirable to remove one type of ion from solution by precipitation while leaving other ions in solution. For instance, the addition of sulfate ions to a solution containing both potassium and barium ions causes BaSO 4 to precipitate out, thereby removing most of the Ba +2 ions from the solution. The other “product” K 2 SO 4, is soluble and will remain in solution. The BaSO 4 precipitate can be separated from the solution by filtration or decanting.

65. Equilibrium 2007-200865 Application of the Solubility Product Principle to Qualitative Analysis Qualitative analysis is the determination of the types of ions present in a solution. (We will focus on the cations.) There are some twenty cations that can be analyzed readily in aqueous solution. These cations can be divided into groups according to the solubility products of their insoluble salts.

66. Equilibrium 2007-200866 The separation of cations into groups is made as selective as possible; that is, the anions that are added as reagents must be such that they will precipitate the fewest types of cations. The separation of cations at each step must be carried out as completely as possible. Since an unknown solution may contain from one to all twenty ions, any analysis must be carried out systematically from Group I through group III.

67. Equilibrium 2007-200867 A flow chart for the separation of cations in qualitative analysis Solution containing ions of all cation groups Solution containing ions of remaining groups Solution contains Na +, K + and NH 4 + ions Group I precipitates AgCl, Hg 2 Cl 2, and PbCl 2 Group II precipitates CoS, FeS, MnS, NiS, ZnS, Al(OH) 3 and Cr(OH) 3 Group III precipitates BaCO 3, CaCO 3 MgCO 3 and SrCO 3 HCl NH 4 OH and (NH 4 ) 2 S decanting (NH 4 ) 2 CO 3

68. Equilibrium 2007-200868 The End