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AP Chemistry Super Saturday Review I tried to include as much review material as possible in this session. Work on practice tests and review the Bozeman.

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Presentation on theme: "AP Chemistry Super Saturday Review I tried to include as much review material as possible in this session. Work on practice tests and review the Bozeman."— Presentation transcript:

1 AP Chemistry Super Saturday Review I tried to include as much review material as possible in this session. Work on practice tests and review the Bozeman videos for other material.

2 AP Chemistry Super Saturday Review 5 Essentials 1.Know the basics – writing formulas, writing and balancing equations, dimensional analysis, atomic theory, acid-base theories (Arrhenius and Bronsted Lowry), VSEPR, kinetic molecular theory, and collision theory.

3 5 Essentials 2. Atomic and Molecular Structures Atomic structures (like electron configurations) will help explain relationships on the periodic table which explains many physical and chemical properties. Molecular structures involves Lewis structures and VSEPR to determine shapes which describes polarities thus describing intermolecular forces which describes many physical properties.

4 5 Essentials 3. Stoichiometric Calculations Basic stoichiometry, limiting reactants, titration calculations, and empirical formulas. 4. Principles of chemical kinetics, equilibrium, And thermodynamics Kinetics- describes the speed in which substances react Equilibrium – used to determine the extent of a reaction or the composition at equilibrium Thermodynamics – explains why chemical reactions happen in terms of kinetic and potential energies

5 5 Essentials 5.Representation and Interpretation Be able to draw what is happening at the molecular level and read and interpret graphs and data tables

6 Net Ionic Equations Graphic: Wikimedia Commons User Tubifex

7 Solubility Rules – AP Chemistry All sodium, potassium, ammonium, and nitrate salts are soluble in water. Memorization of other solubility rules is beyond the scope of this course and the AP Exam. What dissociates (breaks apart) – Aqueous solutions of the following: All strong acids (HCl, HBr, HI, HNO 3, H 2 SO 4, and HClO 4 ) Strong bases (group I and II hydroxides) Soluble salts

8 Write the net ionic for the following: 1. Solutions of lead nitrate and potassium chloride are mixed 2. Solutions of sulfuric acid and potassium hydroxide are mixed. 3. Solid sodium hydroxide is mixed with acetic acid

9 Big Idea #6:Chemical Equilibrium 2NO 2 (g) 2NO(g) + O 2 (g) Sketch a graph of change in concentration vs. Time for the reaction above vs. Time for the reaction above

10 2NO 2 (g) 2NO(g) + O 2 (g) Be able to explain the variance in slope

11 Law of Mass Action For the reaction For the reaction: Where K is the equilibrium constant, and is unitless jA + kB lC + mD

12 Product Favored Equilibrium Large values for K signify the reaction is product favored When equilibrium is achieved, most reactant has been converted to product

13 Reactant Favored Equilibrium Small values for K signify the reaction is reactant favored When equilibrium is achieved, very little reactant has been converted to product

14 Writing an Equilibrium Expression 2NO 2 (g) 2NO(g) + O 2 (g) K = ??? Write the equilibrium expression for the reaction:

15 Conclusions about Equilibrium Expressions The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse 2NO 2 (g) 2NO(g) + O 2 (g ) 2NO(g) + O 2 (g) 2NO 2 (g)

16 Conclusions about Equilibrium Expressions When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power. When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power. 2NO 2 (g) 2NO(g) + O 2 (g ) NO 2 (g) NO(g) + ½O 2 (g )

17 A)0.584 B)4.81 C)0.416 D)23.1 E)0.208 If the equilibrium constant for A + B C is then the equilibrium constant for 2C 2A + 2B is Answer: D

18 Equilibrium Expressions Involving Pressure For the gas phase reaction: 3H 2 (g) + N 2 (g) 2NH 3 (g)

19 Heterogeneous Equilibria The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present Write the equilibrium expression for the reaction: PCl 5 (s) PCl 3 (l) + Cl 2 (g) Pure solid Pure liquid

20 The Reaction Quotient Q K For some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action. jA + kB lC + mD

21 Significance of the Reaction Quotient If Q = K, the system is at equilibrium If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

22 A)To the left. B)To the right. C)The above mixture is the equilibrium mixture. D)Cannot tell from the information given. If you mixed 5.0 mol B, 0.10 mol C, and mol A in a one-liter container, which direction would the reaction initially proceed?

23 LeChateliers Principle When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress and restore a state of equilibrium. Henry Le Chatelier

24 When you take something away from a system at equilibrium, the system shifts in such a way as to replace some of what youve taken away. Le Chatelier Translated: When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what youve added.

25 Do FRQ #1Do FRQ #1

26 Acid Equilibrium and pH Søren Sørensen

27 Acid/Base Definitions Arrhenius Model Acids produce hydrogen ions in aqueous solutions Bases produce hydroxide ions in aqueous solutions Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors

28 Acid Dissociation HA H + + A - Acid Proton Conjugate base Alternately, H + may be written in its hydrated form, H 3 O + (hydronium ion)

29 Dissociation Constants: Strong Acids

30 Dissociation Constants: Weak Acids

31 Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH 3 NH 2 + H 2 O CH 3 NH OH - K b = 4.38 x 10 -4

32 K b for Some Common Weak Bases Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

33 Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H 2 O BH + + OH - (Yes, all weak bases do this – DO NOT try to make this complicated!) Ex. Write the reaction of ammonia with water

34 Self-Ionization of Water H 2 O + H 2 O H 3 O + + OH - At 25, [H 3 O + ] = [OH - ] = 1 x K w is a constant at 25 C: K w = [H 3 O + ][OH - ] K w = (1 x )(1 x ) = 1 x

35 Calculating pH, pOH pH = -log 10 (H 3 O + ) pOH = -log 10 (OH - ) Relationship between pH and pOH pH + pOH = 14 Finding [H 3 O + ], [OH - ] from pH, pOH [H 3 O + ] = 10 -pH [OH - ] = 10 -pOH

36 Calculate the pH of a 0.1M HCl solution? (answer =1)

37 A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC 2 H 3 O 2, K a = 1.8 x ? (answer=4.52)

38 A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH 3, K b = 1.8 x ? (answer=9.48)

39 Acid-Base Properties of Salts To determine if a salt is acidic or basic, determine the stronger parent. Examples: KCl NH 4 Cl NaC 2 H 3 O 2 NaCl KNO 3

40 Acid-Base Properties of Salts If both parents are weak: IF K a for the acidic ion is greater than K b for the basic ion, the solution is acidic IF K b for the basic ion is greater than K a for the acidic ion, the solution is basic IF K b for the basic ion is equal to K a for the acidic ion, the solution is neutral

41 Buffered Solutions A solution that resists a change in pH when either hydroxide ions or protons are added. A solution that resists a change in pH when either hydroxide ions or protons are added. Buffered solutions contain either: Buffered solutions contain either: A weak acid and its salt A weak acid and its salt A weak base and its salt A weak base and its salt

42 Acid/Salt Buffering Pairs The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

43 Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO 3 )

44 Titration of an Unbuffered Solution A solution that is 0.10 M CH 3 COOH is titrated with 0.10 M NaOH

45 Titration of a Buffered Solution A solution that is 0.10 M CH 3 COOH and 0.10 M NaCH 3 COO is titrated with 0.10 M NaOH

46 Comparing Results Buffered Unbuffered

47 Henderson-Hasselbalch Equation This is an exceptionally powerful tool that can be used in your problem solving.

48 Title: A solution that is 0.10 M CH 3 COOH is titrated with 0.10 M NaOH Endpoint is above pH 7

49 Title: A solution that is 0.10 M HCl is titrated with 0.10 M NaOH Endpoint is at pH 7

50 Title: A solution that is 0.10 M NaOH is titrated with 0.10 M HCl Endpoint is at pH 7 It is important to recognize that titration curves are not always increasing from left to right.

51 Title: A solution that is 0.10 M HCl is titrated with 0.10 M NH 3 Endpoint is below pH 7

52 Solubility Equilibria Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate. Graphic: Wikimedia Commons user PRHaney

53 Solving Solubility Problems For the salt AgI at 25 C, K sp = 1.5 x Answer = solubility of AgI in mol/L = 1.2 x M

54 Solving Solubility Problems For the salt PbCl 2 at 25 C, K sp = 1.6 x Answer = solubility of PbCl 2 in mol/L = 1.6 x M

55 Solving Solubility with a Common Ion For the salt AgI at 25 C, K sp = 1.5 x What is its solubility in 0.05 M NaI? Answer = solubility of AgI in mol/L = 3.0 x M

56 Big Idea #5: Spontaneity, Entropy and Free Energy

57 Spontaneity, Entropy and Free Energy G = H - TS

58 Spontaneous Processes and Entropy First Law Energy can neither be created nor destroyed" The energy of the universe is constant Spontaneous Processes Processes that occur without outside intervention Spontaneous processes may be fast or slow –Many forms of combustion are fast –Conversion of diamond to graphite is slow

59 Which of the following reactions is spontaneous? H 2 (g) + I 2 (g) 2HI Kc=49 Br 2 + Cl 2 2BrCl Kc=6.9 HF + H 2 O F- + H Kc=6.8x10 -4

60 Second Law of Thermodynamics "In any spontaneous process there is always an increase in the entropy of the universe" S solid < S liquid << S gas

61 For reactions at constant temperature: G 0 = H 0 - T S 0 Calculating Free Energy Method #1

62 Calculating Free Energy: Method #2 An adaptation of Hess's Law: C diamond (s) + O 2 (g) CO 2 (g) G 0 = -397 kJ C graphite (s) + O 2 (g) CO 2 (g) G 0 = -394 kJ CO 2 (g) C graphite (s) + O 2 (g) G 0 = +394 kJ C diamond (s) C graphite (s) G 0 = C diamond (s) + O 2 (g) CO 2 (g) G 0 = -397 kJ -3 kJ

63 Calculating Free Energy Method #3 Using standard free energy of formation ( G f 0 ): G f 0 of an element in its standard state is zero

64 Free Energy and Equilibrium Equilibrium point occurs at the lowest value of free energy available to the reaction system At equilibrium, G = 0 and Q = K G 0 K G 0 = 0K = 1 G 0 < 0K > 1 G 0 > 0K < 1

65 Bond Energy Breaking bonds require energy (+) Forming bonds releases energy (-) If the reaction A + B C is exothermic, which is larger – the energy needed to break the bonds or the energy released when forming the bonds?

66 TRY FRQ #2

67 Big Idea #4: Kinetics, Rates, and Rate Laws

68 Reaction Rate The change in concentration of a reactant or product per unit of time

69 2NO 2 (g) 2NO(g) + O 2 (g) Reaction Rates: 2. Can measure appearance of products 1. Can measure disappearance of reactants 3. Are proportional stoichiometrically

70 2NO 2 (g) 2NO(g) + O 2 (g) Reaction Rates: 4. Are equal to the slope tangent to that point [NO 2 ] t 5. Change as the reaction proceeds, if the rate is dependent upon concentration

71 Rate Laws Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. Integrated rate laws express (reveal) the relationship between concentration of reactants and time The differential rate law is usually just called the rate law.

72 Writing a (differential) Rate Law 2 NO(g) + Cl 2 (g) 2 NOCl(g) Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s x x x x 10 -6

73 Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s x x x x In experiment 1 and 2, [Cl 2 ] is constant while [NO] doubles. R = k[NO] x [Cl 2 ] y The rate quadruples, so the reaction is second order with respect to [NO] R = k[NO] 2 [Cl 2 ] y

74 Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s x R = k[NO] 2 [Cl 2 ]

75 Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO] 2 [Cl 2 ] Overall order is the sum of the exponents, or orders, of the reactants 2+1 = 3 The reaction is 3 rd order

76 Determining Order with Concentration vs. Time data (the Integrated Rate Law) Zero Order: First Order: Second Order:

77 Solving an Integrated Rate Law Problem: Find the integrated rate law and the value for the rate constant, k A graphing calculator with linear regression analysis greatly simplifies this process!! (Click here to download my Rate Laws program for theTi-83 and Ti-84) (Click here to download my Rate Laws program for theTi-83 and Ti-84)

78 Time vs. [H 2 O 2 ] Time (s) [H 2 O 2 ] y = ax + b a = x b = r 2 = r = Regression results:

79 Time vs. ln[H 2 O 2 ] Time (s) ln[H 2 O 2 ] Regression results: y = ax + b a = x b = r 2 = r =

80 Time vs. 1/[H 2 O 2 ] Time (s) 1/[H 2 O 2 ] y = ax + b a = b = r 2 = r = Regression results:

81 And the winner is… Time vs. ln[H 2 O 2 ] 1. As a result, the reaction is 1 st order 2. The (differential) rate law is: 3. The integrated rate law is: 4. But…what is the rate constant, k ?

82 Finding the Rate Constant, k Method #2: Method #2: Obtain k from the linear regresssion analysis. Now remember: k = -slope k = 8.35 x s -1 Regression results: y = ax + b a = x b = r 2 = r =

83 Rate Laws Summary Zero Order First Order Second Order Rate Law Rate = kRate = k[A]Rate = k[A] 2 Integrated Rate Law [A] = -kt + [A] 0 ln[A] = -kt + ln[A] 0 Plot that produces a straight line [A] versus tln[A] versus t Relationship of rate constant to slope of straight line Slope = -k Slope = k Half-Life

84 Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs. The sum of the elementary steps must give the overall balanced equation for the reaction The mechanism must agree with the experimentally determined rate law

85 Rate-Determining Step In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction. The experimental rate law must agree with the rate-determining step

86 Identifying the Rate-Determining Step For the reaction: 2H 2 (g) + 2NO(g) N 2 (g) + 2H 2 O(g) The experimental rate law is: R = k[NO] 2 [H 2 ] Which step in the reaction mechanism is the rate-determining (slowest) step? Step #1 H 2 (g) + 2NO(g) N 2 O(g) + H 2 O(g) Step #2 N 2 O(g) + H 2 (g) N 2 (g) + H 2 O(g) Step #1 agrees with the experimental rate law

87 Identifying Intermediates For the reaction: 2H 2 (g) + 2NO(g) N 2 (g) + 2H 2 O(g) Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?) Step #1 H 2 (g) + 2NO(g) N 2 O(g) + H 2 O(g) Step #2 N 2 O(g) + H 2 (g) N 2 (g) + H 2 O(g) 2H 2 (g) + 2NO(g) N 2 (g) + 2H 2 O(g) N 2 O(g) is an intermediate

88 Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

89 Collision Model Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy). Colliding particles must be correctly oriented to one another in order to produce a reaction.

90 Factors Affecting Rate Increasing temperature always increases the rate of a reaction. Particles collide more frequently Particles collide more energetically Increasing surface area increases the rate of a reaction Increasing Concentration USUALLY increases the rate of a reaction Presence of Catalysts, which lower the activation energy by providing alternate pathways

91 TRY FRQ #3TRY FRQ #3

92 Big Idea #2 Intermolecular Forces

93 Relative Magnitudes of Forces The types of bonding forces vary in their strength as measured by average bond energy. Covalent bonds (400 kcal/mol) Hydrogen bonding (12-16 kcal/mol ) Dipole-dipole interactions (2-0.5 kcal/mol) London forces (less than 1 kcal/mol) Strongest Weakest

94 London Dispersion Forces The temporary separations of charge that lead to the London force attractions are what attract one nonpolar molecule to its neighbors. Fritz London London forces increase with the size of the molecules. Synonyms: Induced dipoles, dispersion forces, and dispersion-interaction forces

95 London Dispersion Forces

96 Dipole-Dipole Forces of attraction between two polar moleculesForces of attraction between two polar molecules Permanent dipolesPermanent dipoles

97 Hydrogen Bonding Special type of dipole-dipole in which hydrogen bonds with N, O, F. Hydrogen bonding between ammonia and water

98 Hydrogen Bonding in DNA TA Thymine hydrogen bonds to Adenine

99 Boiling point as a measure of intermolecular attractive forces

100 TRY FRQ #4

101 Definition: Half of the distance between nuclei in covalently bonded diatomic molecule Radius decreases across a period Increased effective nuclear charge due to decreased shielding Radius increases down a group Each row on the periodic table adds a shell or energy level to the atom Radius decreases across a period Increased effective nuclear charge due to decreased shielding Radius increases down a group Each row on the periodic table adds a shell or energy level to the atom Atomic Radius Big Idea #1: Periodic Trends

102 Table of Atomic Radii

103 Period Trend: Atomic Radius

104 Increases for successive electrons taken from the same atom Tends to increase across a period Electrons in the same quantum level do not shield as effectively as electrons in inner levels Irregularities at half filled and filled sublevels due to extra repulsion of electrons paired in orbitals, making them easier to remove Tends to decrease down a group Outer electrons are farther from the nucleus and easier to remove Increases for successive electrons taken from the same atom Tends to increase across a period Electrons in the same quantum level do not shield as effectively as electrons in inner levels Irregularities at half filled and filled sublevels due to extra repulsion of electrons paired in orbitals, making them easier to remove Tends to decrease down a group Outer electrons are farther from the nucleus and easier to remove Ionization Energy Definition: the energy required to remove an electron from an atom

105 Ionization Energy: the energy required to remove an electron from an atom Increases for successive electrons taken from the same atom Tends to increase across a period Electrons in the same quantum level do not shield as effectively as electrons in inner levels Irregularities at half filled and filled sublevels due to extra repulsion of electrons paired in orbitals, making them easier to remove Tends to decrease down a group Outer electrons are farther from the nucleus

106 Table of 1st Ionization Energies

107 Periodic Trend: Ionization Energy

108 Electronegativity Definition: A measure of the ability of an atom in a chemical compound to attract electrons o Electronegativity tends to increase across a period o As radius decreases, electrons get closer to the bonding atoms nucleus o Electronegativity tends to decrease down a group or remain the same o As radius increases, electrons are farther from the bonding atoms nucleus o Electronegativity tends to increase across a period o As radius decreases, electrons get closer to the bonding atoms nucleus o Electronegativity tends to decrease down a group or remain the same o As radius increases, electrons are farther from the bonding atoms nucleus

109 Periodic Table of Electronegativities

110 Periodic Trend: Electronegativity

111 Summary of Periodic Trends

112 Ionic Radii Cations Positively charged ions formed when an atom of a metal loses one or more electrons Smaller than the corresponding atom Anions Negatively charged ions formed when nonmetallic atoms gain one or more electrons Larger than the corresponding atom

113 Table of Ion Sizes

114 Determine the element Answer: Hg

115 Determine the element Answer: Na and Mg

116 TRY FRQ #5TRY FRQ #5

117 Electrochemistry

118 Electrochemistry Terminology #1 Oxidation Oxidation – A process in which an element attains a more positive oxidation state Na(s) Na + + e - Reduction Reduction – A process in which an element attains a more negative oxidation state Cl 2 + 2e - 2Cl -

119 Electrochemistry Terminology #2 G ain E lectrons = R eduction An old memory device for oxidation and reduction goes like this… LEO says GER L ose E lectrons = O xidation

120 Electrochemistry Terminology #4 Anode The electrode where oxidation occurs Cathode The electrode where reduction occurs Memory device: Reduction at the Cathode

121 Galvanic (Electrochemical) Cells Spontaneous redox processes have: A positive cell potential, E 0 A negative free energy change, (- G) G=-nFE

122 Zn - Cu Galvanic Cell Zn e - Zn E = -0.76V Cu e - Cu E = +0.34V From a table of reduction potentials:

123 Zn - Cu Galvanic Cell Cu e - Cu E = +0.34V The less positive, or more negative reduction potential becomes the oxidation… Zn Zn e - E = +0.76V Zn + Cu 2+ Zn 2+ + Cu E 0 = V

124 Line Notation Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) An abbreviated representation of an electrochemical cell Anodesolution Anodematerial Cathodesolution Cathodematerial ||||

125 Calculating G 0 for a Cell G 0 = -nFE 0 G 0 = -nFE 0 n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e - E 0 Zn + Cu 2+ Zn 2+ + Cu E 0 = V

126 Concentration Cell Step 1: Determine which side undergoes oxidation, and which side undergoes reduction. Both sides have the same components but at different concentrations. ???

127 Concentration Cell Both sides have the same components but at different concentrations. The 1.0 M Zn 2+ must decrease in concentration, and the 0.10 M Zn 2+ must increase in concentration Zn 2+ (1.0M) + 2e - Zn (reduction) Zn Zn 2+ (0.10M) + 2e - (oxidation) ??? Cathode Anode Zn 2+ (1.0M) Zn 2+ (0.10M)

128 Electrolytic Processes A negative -E 0 cell potential, ( -E 0 ) + G A positive free energy change, ( + G ) NOT Electrolytic processes are NOT spontaneous. They have:

129 Solving an Electroplating Problem Q: What mass of copper is plated out when a current of 10 amps is passed for 30 minutes through a solution of Cu 2+ ? (Amp=C/sec) 10 C Cu e - Cu 1800sec1 mol e C 1 mole Cu 2 mol e g Cu 1 mole Cu = 5.94g Cu sec

130 Good Luck on the Exam!! Try your best. You have worked hard and will do great! I am very proud of each one of you.Good Luck on the Exam!! Try your best. You have worked hard and will do great! I am very proud of each one of you. Mrs. L Mrs. L


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