1. 1 Acids and Bases Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 2 Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases

3. 3 A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acidbaseacidbase acid conjugate base base conjugate acid

4. 4 O H H+ O H H O H HH O H - + [] + Acid-Base Properties of Water H 2 O (l) H + (aq) + OH - (aq) H 2 O + H 2 O H 3 O + + OH - acid conjugate base base conjugate acid autoionization of water

5. 5 H 2 O (l) H + (aq) + OH - (aq) The Ion Product of Water K c = [H + ][OH - ] [H 2 O] [H 2 O] = constant K c [H 2 O] = K w = [H + ][OH - ] The ion-product constant (K w ) is the product of the molar concentrations of H + and OH - ions at a particular temperature. At 25 0 C K w = [H + ][OH - ] = 1.0 x 10 - 14 [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic

6. 6 What is the concentration of OH - ions in a HCl solution whose hydrogen ion concentration is 1.3 M? K w = [H + ][OH - ] = 1.0 x 10 - 14 [H + ] = 1.3 M [OH - ] = KwKw [H + ] 1 x 10 -14 1.3 = = 7.7 x 10 -15 M

7. 7 pH – A Measure of Acidity pH = - log [H + ] [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic [H + ] = 1 x 10 -7 [H + ] > 1 x 10 -7 [H + ] < 1 x 10 -7 pH = 7 pH < 7 pH > 7 At 25 0 C pH[H + ]

8. 8 pOH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00 Other important relationships pH Meter

9. 9 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H + ion concentration of the rainwater? pH = - log [H + ] [H + ] = 10 -pH = 10 -4.82 = 1.5 x 10 -5 M The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood? pH + pOH = 14.00 pOH = -log [OH - ]= -log (2.5 x 10 -7 )= 6.60 pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

10. 10 Strong Electrolyte – 100% dissociation NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) Strong Acids are strong electrolytes HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) HClO 4 (aq) + H 2 O (l) H 3 O + (aq) + ClO 4 - (aq) H 2 SO 4 (aq) + H 2 O (l) H 3 O + (aq) + HSO 4 - (aq)

11. 11 HF (aq) + H 2 O (l) H 3 O + (aq) + F - (aq) Weak Acids are weak electrolytes HNO 2 (aq) + H 2 O (l) H 3 O + (aq) + NO 2 - (aq) HSO 4 - (aq) + H 2 O (l) H 3 O + (aq) + SO 4 2- (aq) H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) Strong Bases are strong electrolytes NaOH (s) Na + (aq) + OH - (aq) H2OH2O KOH (s) K + (aq) + OH - (aq) H2OH2O Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) H2OH2O

12. 12 F - (aq) + H 2 O (l) OH - (aq) + HF (aq) Weak Bases are weak electrolytes NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Conjugate acid-base pairs: H 3 O + is the strongest acid that can exist in aqueous solution. The OH - ion is the strongest base that can exist in aqeous solution.

13. 13 Strong Acid (HCl)Weak Acid (HF)

14. 14 What is the pH of a 2 x 10 -3 M HNO 3 solution? HNO 3 is a strong acid – 100% dissociation. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) pH = -log [H + ] = -log [H 3 O + ] = -log(0.002) = 2.7 Start End 0.002 M 0.0 M What is the pH of a 1.8 x 10 -2 M Ba(OH) 2 solution? Ba(OH) 2 is a strong base – 100% dissociation. Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) Start End 0.018 M 0.036 M0.0 M pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6

15. 15 HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) Weak Acids (HA) and Acid Ionization Constants HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] K a is the acid ionization constant KaKa weak acid strength

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17. 17 What is the pH of a 0.5 M HF solution (at 25 0 C)? HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] = 7.1 x 10 -4 HF (aq) H + (aq) + F - (aq) Initial (M) Change (M) Equilibrium (M) 0.500.00 -x-x+x+x 0.50 - x 0.00 +x+x xx K a = x2x2 0.50 - x = 7.1 x 10 -4 Ka  Ka  x2x2 0.50 = 7.1 x 10 -4 0.50 – x  0.50 K a << 1 x 2 = 3.55 x 10 -4 x = 0.019 M [H + ] = [F - ] = 0.019 M pH = -log [H + ] = 1.72 [HF] = 0.50 – x = 0.48 M

18. 18 percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% For a monoprotic acid HA Percent ionization = [H + ] [HA] 0 x 100% [HA] 0 = initial concentration

19. 19 Molecular Structure and Acid Strength H X H + + X - The stronger the bond The weaker the acid HF << HCl < HBr < HI acidity increases

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21. 21 Arrhenius acid is a substance that produces H + (H 3 O + ) in water A Brønsted acid is a proton donor A Lewis acid is a substance that can accept a pair of electrons A Lewis base is a substance that can donate a pair of electrons Definition of An Acid H+H+ H O H + OH - acidbase N H H H H+H+ + acidbase N H H H H +

22. 22 Lewis Acids and Bases N H H H acidbase F B F F + F F N H H H No protons donated or accepted!

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24. 24 Chemical Kinetics Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

25. 25 Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = -  [A] tt rate =  [B] tt  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative.

26. 26 A B rate = -  [A] tt rate = [B][B] tt

27. 27 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector  [Br 2 ]   Absorption red-brown t 1 < t 2 < t 3

28. 28 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) average rate = -  [Br 2 ] tt = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent instantaneous rate = rate for specific instance in time

29. 29 2H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) PV = nRT P = RT = [O 2 ]RT n V [O 2 ] = P RT 1 rate =  [O 2 ] tt RT 1 PP tt = measure  P over time

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31. 31 Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed. rate =  [B] tt rate = -  [A] tt 1 2 aA + bB cC + dD rate = -  [A] tt 1 a = -  [B] tt 1 b =  [C] tt 1 c =  [D] tt 1 d

32. 32 Write the rate expression for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate = -  [CH 4 ] tt = -  [O 2 ] tt 1 2 =  [H 2 O] tt 1 2 =  [CO 2 ] tt

33. 33 The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A] x [B] y Reaction is xth order in A Reaction is yth order in B Reaction is (x + y)th order overall

34. 34 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y Double [F 2 ] with [ClO 2 ] constant Rate doubles x = 1 Quadruple [ClO 2 ] with [F 2 ] constant Rate quadruples y = 1 rate = k [F 2 ][ClO 2 ]

35. 35 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. 1

36. 36 Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq) Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) 10.080.0342.2 x 10 -4 20.080.0171.1 x 10 -4 30.160.0172.2 x 10 -4 rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 k = rate [S 2 O 8 2- ][I - ] = 2.2 x 10 -4 M/s (0.08 M)(0.034 M) = 0.08/M s rate = k [S 2 O 8 2- ][I - ]

37. 37 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life 0 1 2 rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln 2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0

38. 38 Exothermic ReactionEndothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. A + B AB C + D + +

39. 39 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O 2 (g) 2NO 2 (g) N 2 O 2 is detected during the reaction! Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 +

40. 40 2NO (g) + O 2 (g) 2NO 2 (g) Mechanism:

41. 41 Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 + Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. The molecularity of a reaction is the number of molecules reacting in an elementary step. Unimolecular reaction – elementary step with 1 molecule Bimolecular reaction – elementary step with 2 molecules Termolecular reaction – elementary step with 3 molecules

42. 42 Unimolecular reactionA productsrate = k [A] Bimolecular reactionA + B productsrate = k [A][B] Bimolecular reactionA + A productsrate = k [A] 2 Rate Laws and Elementary Steps Writing plausible reaction mechanisms: The sum of the elementary steps must give the overall balanced equation for the reaction. The rate-determining step should predict the same rate law that is determined experimentally. The rate-determining step is the slowest step in the sequence of steps leading to product formation.

43. 43 Sequence of Steps in Studying a Reaction Mechanism

44. 44 The experimental rate law for the reaction between NO 2 and CO to produce NO and CO 2 is rate = k[NO 2 ] 2. The reaction is believed to occur via two steps: Step 1:NO 2 + NO 2 NO + NO 3 Step 2:NO 3 + CO NO 2 + CO 2 What is the equation for the overall reaction? NO 2 + CO NO + CO 2 What is the intermediate? NO 3 What can you say about the relative rates of steps 1 and 2? rate = k[NO 2 ] 2 is the rate law for step 1 so step 1 must be slower than step 2

45. 45 A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. EaEa k rate catalyzed > rate uncatalyzed E a < E a ′ UncatalyzedCatalyzed

46. 46 In heterogeneous catalysis, the reactants and the catalysts are in different phases. In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. Haber synthesis of ammonia Ostwald process for the production of nitric acid Catalytic converters Acid catalysis Base catalysis

47. 47 Catalytic Converters CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter

48. 48 Enzyme Catalysis

49. 49 Binding of Glucose to Hexokinase

50. 50 rate =  [P] tt rate = k [ES] Enzyme Kinetics

51. 51 All of the following may be true concerning catalysts and the reaction which they catalyze EXCEPT a. catalysts are not used up by the reaction b. catalysts lower the activation energy c. catalysts increase the rate of the reverse reaction d. catalysts shift the reaction equlibrium to the right

52. 52 As the temperature is increased in an exothermic gaseous reaction, all of the following increase EXCEPT a. reaction rate b. rate constant c. activation energy d. all of the above

53. 53 Which of the following changes to a reaction will always increase rate constant for that reaction? a. decreaseing the temperature b. increasing the temperature c. increasing the concentration of the reactants d. increasing the concentration of the catalysts

54. 54 When a radioactive isotope undergoes nuclear decay, the concentration of the isotope decreases exponentially with constant half live. It can be determined from this that radioactive decay is a a. zeroth order reaction b. first order reaction c. second order reaction d. third order reactin

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56. 56 Chemical Equilibrium Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

57. 57 Equilibrium is a state in which there are no observable changes as time goes by. 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 Physical equilibrium H 2 O (l) Chemical equilibrium N2O4 (g)N2O4 (g) H 2 O (g) 2NO 2 (g) NO 2

58. 58 N 2 O 4 (g) 2NO 2 (g) Start with NO 2 Start with N 2 O 4 Start with NO 2 & N 2 O 4 equilibrium

59. 59 constant

60. 60 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 Law of Mass Action

61. 61 K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Equilibrium Will K = [C] c [D] d [A] a [B] b aA + bB cC + dD

62. 62 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 P 2 N2O4N2O4 P aA (g) + bB (g) cC (g) + dD (g) K p = K c (RT)  n  n = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b) In most cases K c  K p

63. 63 Homogeneous Equilibrium 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.

64. 64 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) 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

65. 65 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K p = P CO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. [CaO][CO 2 ] [CaCO 3 ] K c = ′ [CaCO 3 ] [CaO] K c x ′

66. 66 If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. Le Châtelier’s Principle Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Add NH 3 Equilibrium shifts left to offset stress

67. 67 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 aA + bB cC + dD Add Remove

68. 68 Le Châtelier’s Principle Changes in Temperature ChangeExothermic Rx Increase temperatureK decreases Decrease temperatureK increases Endothermic Rx K increases K decreases colder hotter N 2 O 4 (g) 2NO 2 (g)

69. 69 Catalyst lowers E a for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium. Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle

70. 70 Le Châtelier’s Principle - Summary ChangeShift Equilibrium Change Equilibrium Constant Concentrationyesno Pressureyes*no Volumeyes*no Temperatureyes Catalystno *Dependent on relative moles of gaseous reactants and products

71. 71 As the temperature is increased, the equilibrium of gaseous reaction will always: a. shift to the right b. shift to the left c. remain constant d. the answer cannot be determined from the information given

72. 72 All of the following are true concerning a reaction at equilibrium EXCEPT: a. the rate of the forward reaction equals the rate of the reverse reaction b. There is no change in the concentrations of both the products and the reactants c. The activation energy has reached zero d. All reactants will start to move forward at equilibrium

73. 73 What is the equilibrium expression for the following reaction CaCO 3(s)  CaO (s) + CO 2(g) a. K=[CO 2 ] b. K=[CaO] [CO 2 ] c. K=[CaO] [CO 2 ]/ [CaO] d. K= [CO 2 ]/ [CaO]