Presentation on theme: "AP Chemistry Super Saturday Review"— Presentation transcript:
1AP Chemistry Super Saturday Review I tried to include as much review materialas possible in this session. Work on practicetests and review the Bozeman videos forother material.
2AP Chemistry Super Saturday Review 5 EssentialsKnow the basics – writing formulas,writing and balancing equations,dimensional analysis, atomic theory,acid-base theories (Arrhenius and BronstedLowry), VSEPR, kinetic molecular theory,and collision theory.
35 Essentials2. Atomic and Molecular StructuresAtomic structures (like electron configurations)will help explain relationships on the periodictable which explains many physical and chemicalproperties. Molecular structures involves Lewisstructures and VSEPR to determine shapeswhich describes polarities thus describingintermolecular forces which describes manyphysical properties.
45 Essentials3. Stoichiometric CalculationsBasic stoichiometry, limiting reactants,titration calculations, and empirical formulas.4. Principles of chemical kinetics, equilibrium,And thermodynamicsKinetics- describes the speed in whichsubstances reactEquilibrium – used to determine the extent ofa reaction or the composition at equilibriumThermodynamics – explains why chemical reactionshappen in terms of kinetic and potential energies
55 EssentialsRepresentation and InterpretationBe able to draw what is happening at themolecular level and read and interpret graphsand data tables
6Net Ionic EquationsGraphic: Wikimedia Commons User Tubifex
7Solubility 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, HNO3, H2SO4, and HClO4)Strong bases (group I and II hydroxides)Soluble salts
8Write the net ionic for the following: 1 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
9Big Idea #6:Chemical Equilibrium 2NO2(g) 2NO(g) + O2(g)Sketch a graph of change in concentrationvs. Time for the reaction above
102NO2(g) 2NO(g) + O2(g)Be able to explain the variance in slope
11jA + kB lC + mD Law of Mass Action For the reaction: Where K is the equilibrium constant, and is unitless
12Product Favored Equilibrium Large values for K signify the reaction is “product favored”When equilibrium is achieved, most reactant has been converted to product
13Reactant Favored Equilibrium Small values for K signify the reaction is “reactant favored”When equilibrium is achieved, very little reactant has been converted to product
14Writing an Equilibrium Expression Write the equilibrium expression for the reaction:2NO2(g) 2NO(g) + O2(g)K = ???
15Conclusions about Equilibrium Expressions The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse2NO2(g) 2NO(g) + O2(g)2NO(g) + O2(g) 2NO2(g)
16Conclusions 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.2NO2(g) 2NO(g) + O2(g)NO2(g) NO(g) + ½O2(g)
17If the equilibrium constant for A + B C is 0 If the equilibrium constant for A + B C is then the equilibrium constant for 2C A + 2B isA)0.584B)4.81C)0.416D)23.1E)0.208Answer: D
18Equilibrium Expressions Involving Pressure For the gas phase reaction:3H2(g) + N2(g) 2NH3(g)
19Heterogeneous Equilibria The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids presentWrite the equilibrium expression for the reaction:PCl5(s) PCl3(l) + Cl2(g)PuresolidPureliquid
20jA + kB lC + mD The Reaction Quotient 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
21Significance of the Reaction Quotient If Q = K, the system is at equilibriumIf Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achievedIf Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved
22If you mixed 5.0 mol B, 0.10 mol C, and 0.0010 mol A in a one-liter container, which direction would the reactioninitially proceed?A)To the left.B)To the right.C)The above mixture is the equilibrium mixture.D)Cannot tell from the information given.
23LeChatelier’s Principle When a system atequilibrium is placed understress, the system willundergo a change in sucha way as to relieve thatstress and restore astate of equilibrium.Henry Le Chatelier
24Le Chatelier Translated: When you take something away from a system at equilibrium, the system shifts in such a way as to replace some of what you’ve taken away.When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you’ve added.
27Acid/Base Definitions Arrhenius ModelAcids produce hydrogen ions in aqueous solutionsBases produce hydroxide ions in aqueous solutionsBronsted-Lowry ModelAcids are proton donorsBases are proton acceptors
28Acid Dissociation Acid Proton Conjugate base HA H+ + A- Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)
29Dissociation Constants: Strong Acids FormulaConjugate BaseKa Perchloric HClO4 ClO4- Very large Hydriodic HI I- Hydrobromic HBr Br- Hydrochloric HCl Cl- Nitric HNO3 NO3- Sulfuric H2SO4 HSO4- Hydronium ion H3O+ H2O 1.0
30Dissociation Constants: Weak Acids FormulaConjugate BaseKa Iodic HIO3 IO3- 1.7 x 10-1 Oxalic H2C2O4 HC2O4- 5.9 x 10-2 Sulfurous H2SO3 HSO3- 1.5 x 10-2 Phosphoric H3PO4 H2PO4- 7.5 x 10-3 Citric H3C6H5O7 H2C6H5O7- 7.1 x 10-4 Nitrous HNO2 NO2- 4.6 x 10-4 Hydrofluoric HF F- 3.5 x 10-4 Formic HCOOH HCOO- 1.8 x 10-4 Benzoic C6H5COOH C6H5COO- 6.5 x 10-5 Acetic CH3COOH CH3COO- 1.8 x 10-5 Carbonic H2CO3 HCO3- 4.3 x 10-7 Hypochlorous HClO ClO- 3.0 x 10-8 Hydrocyanic HCN CN- 4.9 x 10-10
31Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion:CH3NH2 + H2O CH3NH3+ + OH- Kb = 4.38 x 10-4
32Kb 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?BaseFormulaConjugate AcidKbAmmonia NH3 NH4+ 1.8 x 10-5 Methylamine CH3NH2 CH3NH3+ 4.38 x 10-4 Ethylamine C2H5NH2 C2H5NH3+ 5.6 x 10-4 Diethylamine (C2H5)2NH (C2H5)2NH2+ 1.3 x 10-3 Triethylamine (C2H5)3N (C2H5)3NH+ 4.0 x 10-4 Hydroxylamine HONH2 HONH3+ 1.1 x 10-8 HydrazineH2NNH2 H2NNH3+ 3.0 x 10-6 Aniline C6H5NH2 C6H5NH3+ 3.8 x 10-10 Pyridine C5H5N C5H5NH+ 1.7 x 10-9
33Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion:B + H2O BH+ + OH-(Yes, all weak bases do this – DO NOTtry to make this complicated!)Ex. Write the reaction of ammonia with water
34Self-Ionization of Water H2O + H2O H3O+ + OH-At 25, [H3O+] = [OH-] = 1 x 10-7Kw is a constant at 25 C:Kw = [H3O+][OH-]Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14
35Relationship between pH and pOH Calculating pH, pOHpH = -log10(H3O+)pOH = -log10(OH-)Relationship between pH and pOHpH + pOH = 14Finding [H3O+], [OH-] from pH, pOH[H3O+] = 10-pH[OH-] = 10-pOH
36Calculate the pH of a 0.1M HCl solution? (answer =1)
37A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ?(answer=4.52)
38A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3,Kb = 1.8 x 10-5 ? (answer=9.48)
39Acid-Base Properties of Salts To determine if a salt is acidic or basic, determine the stronger parent.Examples:KClNH4ClNaC2H3O2NaClKNO3
40Acid-Base Properties of Salts If both parents are weak: Type of SaltExamplesCommentpH of solutionCation is the conjugate acid of a weak base, anion is conjugate base of a weak acidNH4C2H3O2NH4CNCation is acidic,Anion is basicSee belowIF Ka for the acidic ion is greater than Kb for the basic ion, the solution is acidicIF Kb for the basic ion is greater than Ka for the acidic ion, the solution is basicIF Kb for the basic ion is equal to Ka for the acidic ion, the solution is neutral
41Buffered SolutionsA solution that resists a change in pH when either hydroxide ions or protons are added.Buffered solutions contain either:A weak acid and its saltA weak base and its salt
42Acid/Salt Buffering Pairs The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)Weak AcidFormulaof the acidExample of a salt of the weak acid Hydrofluoric HF KF – Potassium fluoride Formic HCOOH KHCOO – Potassium formate Benzoic C6H5COOH NaC6H5COO – Sodium benzoate Acetic CH3COOH NaH3COO – Sodium acetate Carbonic H2CO3 NaHCO3 - Sodium bicarbonate Propanoic HC3H5O2 NaC3H5O2 - Sodium propanoate Hydrocyanic HCN KCN - potassium cyanide
43Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)BaseFormula of the baseExample of a salt of the weak acidAmmonia NH3 NH4Cl - ammonium chloride Methylamine CH3NH2 CH3NH2Cl – methylammonium chloride Ethylamine C2H5NH2 C2H5NH3NO3 - ethylammonium nitrate Aniline C6H5NH2 C6H5NH3Cl – aniline hydrochloride Pyridine C5H5N C5H5NHCl – pyridine hydrochloride
44Titration of an Unbuffered Solution A solution that is0.10 M CH3COOHis titrated with0.10 M NaOH
45Titration of a Buffered Solution A solution that is0.10 M CH3COOH and0.10 M NaCH3COO is titrated with0.10 M NaOH
47Henderson-Hasselbalch Equation This is an exceptionally powerful tool that can be used in your problem solving.
48Title: Endpoint is above pH 7 A solution that is 0.10 M CH3COOH is titrated with0.10 M NaOH
49Title: Endpoint is at pH 7 A solution that is 0.10 M HCl is titrated with0.10 M NaOH
50Title: A solution that is 0.10 M NaOH is titrated with 0.10 M HCl Endpoint is atpH 7It is important to recognize that titration curves are not always increasing from left to right.
51Title: A solution that is Endpoint is below 0.10 M HCl is titrated with0.10 M NH3Endpoint is belowpH 7
52Solubility Equilibria Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate.Graphic: Wikimedia Commons user PRHaney
53Solving Solubility Problems For the salt AgI at 25C, Ksp = 1.5 x 10-16Answer = solubility of AgI in mol/L = 1.2 x 10-8 M
54Solving Solubility Problems For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5Answer = solubility of PbCl2 in mol/L = 1.6 x 10-2 M
55Solving Solubility with a Common Ion For the salt AgI at 25C, Ksp = 1.5 x 10-16What is its solubility in 0.05 M NaI?Answer = solubility of AgI in mol/L = 3.0 x M
56Big Idea #5: Spontaneity, Entropy and Free Energy
57Spontaneity, Entropy and Free Energy G = H - TSSpontaneity, Entropy and Free Energy
58Spontaneous Processes and Entropy First Law“Energy can neither be created nor destroyed"The energy of the universe is constantSpontaneous ProcessesProcesses that occur without outside interventionSpontaneous processes may be fast or slowMany forms of combustion are fastConversion of diamond to graphite is slow
59Which of the following reactions is spontaneous? H2(g) + I2(g) ↔ 2HI Kc=49Br2 + Cl2 ↔ 2BrCl Kc=6.9HF + H2O ↔ F- + H Kc=6.8x10-4
60Second Law of Thermodynamics "In any spontaneous process there is always an increase in the entropy of the universe"Ssolid < Sliquid << Sgas
61For reactions at constant temperature: Calculating Free Energy Method #1For reactions at constant temperature:G0 = H0 - TS0
63Calculating Free Energy Method #3 Using standard free energy of formation (Gf0):Gf0 of an element in its standard state is zero
64Free Energy and Equilibrium Equilibrium point occurs at the lowest value of free energy available to the reaction systemAt equilibrium, G = 0 and Q = KG0KG0 = 0K = 1G0 < 0K > 1G0 > 0K < 1
65Bond 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?
68Reaction RateThe change in concentration of a reactant or product per unit of time
69Reaction Rates: 1. Can measure disappearance of reactants 2NO2(g) 2NO(g) + O2(g)Reaction Rates:1. Can measuredisappearance ofreactants2. Can measureappearance ofproducts3. Are proportionalstoichiometrically
70Reaction Rates: 4. Are equal to the slope tangent to that point 2NO2(g) 2NO(g) + O2(g)Reaction Rates:4. Are equal to theslope tangent tothat point5. Change as thereaction proceeds,if the rate isdependent uponconcentration[NO2]t
71Rate LawsDifferential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.The differential rate law is usually just called “the rate law.”Integrated rate laws express (reveal) the relationship between concentration of reactants and time
72Writing a (differential) Rate Law Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction:2 NO(g) + Cl2(g) 2 NOCl(g)Experiment[NO](mol/L)[Cl2]RateMol/L·s10.2501.43 x 10-620.5005.72 x 10-632.86 x 10-6411.4 x 10-6
73Writing a Rate LawPart 1 – Determine the values for the exponents in the rate law:R = k[NO]x[Cl2]yExperiment[NO](mol/L)[Cl2]RateMol/L·s10.2501.43 x 10-620.5005.72 x 10-632.86 x 10-641.14 x 10-5In experiment 1 and 2, [Cl2] is constant while [NO] doubles.The rate quadruples, so the reaction is second order with respect to [NO] R = k[NO]2[Cl2]y
74Writing a Rate LawPart 2 – Determine the value for k, the rate constant, by using any set of experimental data:R = k[NO]2[Cl2]Experiment[NO](mol/L)[Cl2]RateMol/L·s10.2501.43 x 10-6
75Writing a Rate LawPart 3 – Determine the overall order for the reaction.R = k[NO]2[Cl2]2+1= 3 The reaction is 3rd orderOverall order is the sum of the exponents, or orders, of the reactants
76Determining Order with Concentration vs. Time data (the Integrated Rate Law)Zero Order:First Order:Second Order:
77Solving an Integrated Rate Law Time (s)[H2O2] (mol/L)1.001200.913000.786000.5912000.3718000.2224000.1330000.08236000.050Problem: Find the integrated rate law and the value for the rate constant, kA graphing calculator with linear regression analysis greatly simplifies this process!!(Click here to download my Rate Laws program for theTi-83 and Ti-84)
78Time vs. [H2O2] Regression results: y = ax + b a = -2.64 x 10-4 Time (s)[H2O2]1.001200.913000.786000.5912000.3718000.2224000.1330000.08236000.050Regression results:y = ax + ba = x 10-4b = 0.841r2 =r =
79Time vs. ln[H2O2] Regression results: y = ax + b a = -8.35 x 10-4 Time (s)ln[H2O2]12030060012001800-1.5142400-2.043000-2.5013600-2.996Regression results:y = ax + ba = x 10-4b = -.005r2 =r =
80Time vs. 1/[H2O2] Regression results: y = ax + b a = 0.00460 Time (s)1/[H2O2]1.001201.09893001.28216001.694912002.702718004.545524007.6923300012.195360020.000Regression results:y = ax + ba =b =r2 =r =
81And the winner is… Time vs. ln[H2O2] 1. As a result, the reaction is 1st order2. The (differential) rate law is:3. The integrated rate law is:4. But…what is the rate constant, k ?
82Finding the Rate Constant, k Method #2: Obtain k from the linear regresssion analysis.Regression results:y = ax + ba = x 10-4b = -.005r2 =r =Now remember: k = -slopek = 8.35 x 10-4s-1
83Rate Laws Summary Rate = k Rate = k[A] Rate = k[A]2 [A] = -kt + [A]0 Zero OrderFirst OrderSecond OrderRate LawRate = kRate = k[A]Rate = k[A]2Integrated Rate Law[A] = -kt + [A]0ln[A] = -kt + ln[A]0Plot that produces a straight line[A] versus tln[A] versus tRelationship of rate constant to slope of straight lineSlope = -kSlope = kHalf-Life
84Reaction MechanismThe 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 reactionThe mechanism must agree with the experimentally determined rate law
85Rate-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
86Identifying the Rate-Determining Step For the reaction:2H2(g) + 2NO(g) N2(g) + 2H2O(g)The experimental rate law is:R = k[NO]2[H2]Which step in the reaction mechanism is the rate-determining (slowest) step?Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g)Step #2 N2O(g) + H2(g) N2(g) + H2O(g)Step #1 agrees with the experimental rate law
87Identifying Intermediates For the reaction:2H2(g) + 2NO(g) N2(g) + 2H2O(g)Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?)Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g)Step #2 N2O(g) + H2(g) N2(g) + H2O(g)2H2(g) + 2NO(g) N2(g) + 2H2O(g) N2O(g) is an intermediate
88Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?
89Collision ModelCollisions 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.
90Factors Affecting Rate Increasing temperature always increases the rate of a reaction.Particles collide more frequentlyParticles collide more energeticallyIncreasing surface area increases the rate of a reactionIncreasing Concentration USUALLY increases the rate of a reactionPresence of Catalysts, which lower the activation energy by providing alternate pathways
93Relative Magnitudes of Forces The types of bonding forces vary in their strength as measured by average bond energy.StrongestWeakestCovalent 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)
94London Dispersion Forces The temporary separations of charge that lead to the London force attractions are what attract one nonpolar molecule to its neighbors.London forces increase with the size of the molecules.Fritz LondonSynonyms: “Induced dipoles”, “dispersion forces”, and “dispersion-interaction forces”
101Atomic Radius Big Idea #1: Periodic Trends Definition: Half of the distance between nuclei in covalently bonded diatomic moleculeRadius decreases across a periodIncreased effective nuclear charge due to decreased shieldingRadius increases down a groupEach row on the periodic table adds a “shell” or energy level to the atom
104Ionization EnergyDefinition: the energy required to remove an electron from an atomIncreases for successive electrons taken from the same atomTends to increase across a periodElectrons in the same quantum level do not shield as effectively as electrons in inner levelsIrregularities at half filled and filled sublevels due to extra repulsion of electrons paired in orbitals, making them easier to removeTends to decrease down a groupOuter electrons are farther from the nucleus and easier to remove
105Ionization Energy: the energy required to remove an electron from an atom Increases for successive electrons taken from the same atomTends to increase across a periodElectrons in the same quantum level do not shield as effectively as electrons in inner levelsIrregularities at half filled and filled sublevels due to extra repulsion of electrons paired in orbitals, making them easier to removeTends to decrease down a groupOuter electrons are farther from the nucleus
108ElectronegativityDefinition: A measure of the ability of an atom in a chemical compound to attract electronsElectronegativity tends to increase across a periodAs radius decreases, electrons get closer to the bonding atom’s nucleusElectronegativity tends to decrease down a group or remain the sameAs radius increases, electrons are farther from the bonding atom’s nucleus
112Ionic Radii Cations Anions Positively charged ions formed when an atom of a metal loses one ormore electronsCationsSmaller than the correspondingatomNegatively charged ions formedwhen nonmetallic atoms gain oneor more electronsAnionsLarger than the correspondingatom
118Electrochemistry Terminology #1 Oxidation – A process in which an element attains a more positive oxidation stateNa(s) Na+ + e-Reduction – A process in which an element attains a more negative oxidation stateCl2 + 2e- 2Cl-
119Electrochemistry Terminology #2 An old memory device for oxidation and reduction goes like this…LEO says GERLose Electrons = OxidationGain Electrons = Reduction
120Electrochemistry Terminology #4 AnodeThe electrode where oxidation occursCathodeThe electrode where reduction occursMemory device:Reductionat theCathode
122Zn - Cu Galvanic Cell From a table of reduction potentials: Zn2+ + 2e- Zn E = -0.76VCu2+ + 2e- Cu E = +0.34V
123Zn - Cu Galvanic Cell Zn Zn2+ + 2e- E = +0.76V The less positive, or more negative reduction potential becomes the oxidation…Cu2+ + 2e- Cu E = +0.34VZn Zn2+ + 2e E = +0.76VZn + Cu2+ Zn2+ + Cu E0 = V
124Line Notation Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) | || | An abbreviated representation of an electrochemical cellZn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)AnodematerialAnodesolutionCathodesolutionCathodematerial||||
125Calculating G0 for a Cell G0 = -nFE0n = moles of electrons in balanced redox equationF = Faraday constant = 96,485 coulombs/mol e-Zn + Cu2+ Zn2+ + Cu E0 = V
126Both sides have the same components but at different concentrations. ???Concentration CellBoth sides have the same components but at different concentrations.Step 1: Determine which side undergoes oxidation, and which side undergoes reduction.
127Both sides have the same components but at different concentrations. ???Concentration CellBoth sides have the same components but at different concentrations.AnodeCathodeThe 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentrationZn2+ (1.0M) + 2e- Zn (reduction)Zn Zn2+ (0.10M) + 2e- (oxidation)Zn2+ (1.0M) Zn2+ (0.10M)
128Electrolytic Processes Electrolytic processes are NOT spontaneous. They have:A negativecell potential, (-E0)A positive free energy change, (+G)
129Solving 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 Cu2+? (Amp=C/sec)Cu2+ + 2e- Cu10 C1800sec1 mol e-1 mole Cu63.5 g Cu2 mol e-1 mole CuCsec= 5.94g Cu
130Good Luck on the Exam. Try your best 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