F A 7/1/2014. No, I Do Not Drop a Grade!!!!!!! THE IDEAL GAS EQUATION PV = n R T and its APPLICATIONS. PRESSURE (force per unit area) grt VP

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1 7/1/014 Chem 131 Final has 115 points (similar to others) multiple choice questions ( pts each) 30 questions from old material {Chapters questions from new material {Chap & nomenclature questions ( 1 pt each) Key to final exam will be posted in glass case outside room 100 after exam is over A 90 up C B D WHAT TO STUDY? At the end of each chapter Strategies in Chemistry Summary & Key Terms Also Old Exams Web problems & Exams You Will Have Five (5) Grades Exam 1,, 3, 4 + Assignments CHAP 10: Gases (Ideal & Real) I will Average These Five (5) grades {add grades & divide by five} One Exam {1,, or 3} Grade Can be REPLACED By Exam 4 (Final Exam) If Necessary No, I Do Not Drop a Grade!!!!!!! STANDARD CONDITIONS for Gases STP Standard temperature 0C = K Standard pressure 1 atm Measured with 1. Barometers. Manometers Open Closed PRESSURE (force per unit area) P F A THE IDEAL GAS EQUATION PV = n R T and its APPLICATIONS GAS DENSITY Density MOLECULAR WEIGHT grams volume P R grams MW mole MW T grt VP 1

2 7/1/014 Gas Mixtures & Partial Pressures Pi P T n n PV = n R T P i V = n i R T P T V = n T R T or where i is the mole fraction i T P i i ni i n Total P total APPLICATIONS OF THE IDEAL GAS EQUATION 1. Collecting Gases over Water P Total = (P gas + P water ) = (n gas + n water ) R T. Gas Volumes in Chemical Reactions 1 N (gas) + 3 H (gas) = NH 3 (gas) 1 volume reacts with 3 volumes to produce volumes REAL GASES and the van der Waals Equation Chapter 11 Liquids & Solids INTER & INTRA MOLECULAR FORCES P nrt V nb Corrects for molecular volume n a V Corrects for molecular attraction INTRA molecular forces are Chemical Bonds which are formed between elements INTER molecular forces aare the attractive forces between molecules and ions that determine bulk properties of matter types of Intermolecular Forces 1. Ion dipole. Dipole dipole 3. London Forces => Instantaneous induced dipole (dispersion forces) 4. Hydrogen bonds. Some Properties of Liquids Vapor Pressure Boiling Point Viscosity Surface Tension

3 7/1/014 FEATURES OF A PHASE DIAGRAM Triple point: temperature and pressure at which all three phases are in equilibrium. Vaporpressure curve: generally as pressure increases, temperature increases. Critical point: critical temperature and pressure for the gas. Melting point curve: as pressure increases, the solid phase is favored if the solid is more dense than the liquid. Normal melting point: melting point at 1 atm. Supercritical Fluid A state of matter beyond the critical point that is neither liquid nor gas CHAP 1: Properties of Solutions SOLUBILITY RULES Chapter 4 SOLUBILITY factors that affect Pressure (HENRY S LAW) Temperature CONCENTRATION Molecular weight COLLIGATIVE PROPERTIES Molecular solutions Ionic solutions COLLOIDS CONCENTRATION is either based on weight or volume 1. ppm (parts per million) (based on weight). ppb (parts per billion) (based on weight) 3. Mole Fraction. (based on weight) 4. Molarity. (based on volume) 5. Molality.. (based on weight) Solutions, Mixtures, & Colloids (a) A MIXTURE is heterogeneous (b) A SOLUTION is homogeneous (c) A colloidal dispersion is between a solution and a mixture 3

4 7/1/014 COLLIGATIVE PROPERTIES (determination of molecular weight) of non Electrolytes Vapor Pressure LOWERING P = x P 1 Boiling Point ELEVATION Freezint Point LOWERING Osmotic Pressure T b = K b m T f = K f m π = MRT COLLIGATIVE PROPERTIES (determination of molecular weight and/or van t Hoff i facror) of Electrolytes Vapor Pressure LOWERING P = i x P 1 Boiling Point ELEVATION T b = i K b m Freezint Point LOWERING T f = i K f m Osmotic Pressure π = i MRT Chap 14: CHEMICAL KINETICS Part I: The Rates of Reaction Depends On CHEMICAL KINETICS DEALS WITH 1. How FAST {Speed like miles per hour and. By what MECHANISM Does a Reaction Take Place? 1. Nature of Reactants {fixed. Concentration of Reactants {variable 3. Temperature {variable 4. Etc. {variable Catalysts Particle Size Photochemical Determine the rate law for All radioactive decay is 1 st order [ A ] 0 ln k t [ A ] Halflife {t 1/ } is defined as the time required for onehalf of a reactant to react [ A] 0 ln k t [ A] 1 0 1/ 4

5 7/1/014 TEMPERATURE AND RATE AS TEMPERATURE INCREASES, SO DOES THE REACTION RATE. Arrhenius developed a mathematical equation between k (rate constant) and E a (energy of activation) THIS IS BECAUSE k IS TEMPERATURE DEPENDENT. Part II Reaction mechanism. Reactions occur through several discrete steps Each of these is known as an elementary reaction or elementary process The molecularity of a process tells how many molecules are involved in the process The Collision Model Molecules can only react if they collide The following mechanism has been proposed for the gas phase reaction of H with ICl H (g) + ICl (g) HI (g) + HCl HI (g) + ICl I (g) + HCl (g) (a) Write the balanced equation for the reaction (b) Identify intermediates in mechanism (c) Write rate laws for each elementary step (d) If the 1 st step is the slowest what is the rate law for the overall reaction? H (g) + ICl (g) HI (g) + HCl HI (g) + ICl I (g) + HCl (g) (a) H (g) + ICl (g) I (g) + HCl (b) HI (c) Step 1: Rate = k [H ][ICl] Step : Rate = k[hi][icl] (d) Slowest step determines rate law therefore reaction is 1 st order wrt H and ICl and nd order overall EQUILIBRIUM Chapters Molecular Chap 15 Ionic (Weak Acid / Base ) Chap 16 Ionic ( Insoluble Salts) Chap 17 Homogeneous Equilibrium Heterogeneous Equilibrium Equilibrium Calculations Le Châtelier s Principle 5

6 7/1/014 The equilibriumconstant expression depends only on the stoichiometry not on the mechanism Pure Solids and Liquids do not have a concentration so they do not appear in the equilibrium expression Products K Reactants p r The Initial Change Equilibrium method for SO (g) + O (g) = SO 3 (g) moles SO moles O moles SO 3 Initially Change 0.95 ½ (0.95) Equilibrium K eq [ so ] [ so ] [ o ] x Catalysts increase the rate of both the forward and reverse reactions Equilibrium is achieved faster, but the equilibrium composition remains unaltered Therefore Catalysts have No effect on the equilibrium concentrations A Change in Temperature Changes the equilibrium constant Endothermic processes are favored with increase Exothermic processes are favored with decrease Relationship between K c and K p Kp = Kc (RT) n Where n = (moles of gaseous product) (moles of gaseous reactant) Kp = Kc For H (gas) + I (gas) = HI(gas) WHY? Chapter 16 is a continuation of chapter 15 For WEAK Acids & WEAK Bases What is an Acid? What is a Base? 1. ARRHENIUS Acid / Base. BRONSTEDLOWRY Acid / Base 3. LEWIS Acid / Base ph and the ion product constant of water ph = log [H + ] poh = log [OH ] pk = log K w K w = [H + ] [OH ] = 1.0 x10 14 ph + poh = 14 When [H + ] = [OH ] have a neutral solution and ph = poh = 7 6

7 7/1/014 ph of some common fluids Gastric juice in stomach Lemon juice.4 Vinegar 3.0 Grapefruit juice 3. Orange juice 3.5 Urine Saliva Milk 6.5 Blood Tears 7.4 Milk of Magnesia 10.6 Household Ammonia 11.5 ACID K a CONJ. BASE K b HF HNO HCO H (formic) CH 3 CO H (acetic) HCN 7.1 x x x x x Relationship between K a and K b K a F NO HCO CH 3 CO CN x K b = K w 1.4 x x x x x 10 5 Acid Base Properties of Salts SALTS IN WATER EITHER FORM neutral solutions basic solutions or acidic solutions Buffer Solution: is a solution of a Weak acid or a Weak base and its salt BOTH ACID (OR BASE) & SALT MUST BE PRESENT Will the reaction Ba + (aq) + SO 4 (aq) BaSO 4 (s) occur? If [Ba + ][SO ] > K sp it will Moles Ba + = M x 0.050L = 5 x 10 5 Moles SO 4 = M x 0.050L = 5 x 10 6 [Ba + ] = (5 x 10 4 ) {100mL of solution [SO 4 ] =( 5 x 10 5 ) Ba + (aq) + SO 4 (aq) BaSO 4 (s) K sp =[Ba + ][SO ] [Ba + ][SO 4 ] = (5 x 10 4 )(5 x 10 5 ) =.5 x 10 8 From Table: Ksp = 1.1 x10 10 Since the ion product (.5 x 10 8 ) is greater than the Ksp, (1.1 x10 10 ) a precipate of barium sulfate is expected 7

8 7/1/014 The concentration of calcium ion in the blood plasma is M. If the concentration of oxalate ion is 1.0 x 10 7 do you expect calcium oxalate to precipitate? [Ca + ][C O 4 ] = (0.005)(1.0 x 10 7 ) =.5x10 10 Ksp for calcium oxalate is.3 x 10 9 Since the ion product (.5x10 10 ) is smaller than Ksp, (.3 x 10 9 ) do not expect precipitation to occur Henderson Hasselbalch equation. Consider the equilibrium constant expression for the dissociation of a generic weak acid, HA: K a = HA + H O [H 3 O + ] [A ] [HA] = [H 3 O + ] H 3 O + + A [A ] [HA] Taking the negative log of both sides log K a = log [H 3 O + ] + log [A ] [HA] base pk a ph. acid Chap 0 ELECTROCHEMISTRY Two Parts To ELECTROCHEMISTRY Ordinary Chemical Reactions produce Heat Zn(s) + HCl(aq) H (g) + ZnCl (aq) + HEAT Electrochemical Reactions produce Voltage Zn(s) + HCl(aq) H (g) + ZnCl (aq) + VOLTS Part 1 SPONTANEOUS Process A Chemical Reaction Produces Voltage BATTERYS Part NONSpontaneous Process Voltage causes a Chemical Reaction ELECTROLYSIS Two types of batteries Two () electrolytes with a Salt Bridge 1. Two () electrolytes Uses a salt bridge Negative electrode on Zn/ Zn + // H + / H left by convention Zn Zn + + e. One (1) electrolyte Does Not need a salt bridge Pb / H SO 4 (aq) / PbO Pb PbSO 4 PbO PbSO 4 8

9 7/1/014 Zn/ Zn + // H + / H ANODE: Zn (the negative electrode) Zn (s) Zn + (aq) + e CATHODE: H {positive electrode H + (aq) + e H (gas) The overall electrochemical reaction Zn(s) + H + (aq) Zn + (aq) + H (gas) One (1) Electrolyte: Sulfuric Acid No need for a salt bridge ANODE: Pb (the negative electrode) Pb (s) + SO 4 (aq) PbSO 4 (s) + e CATHODE: PbO (lead IV oxide) {positive electrode PbO (s) + SO 4 (aq) + 4H + (aq) + e PbSO 4 (s) + H O(l) The overall electrochemical reaction PbO (s) + Pb (s) + SO 4 (aq) + 4 H + (aq) PbSO 4 (s) + H O(l) ELECTROLYSIS External force causes chemical reactions How Much Ni would be electroplated on the Steel How Much Ni would be electroplated if 1 Faraday was passed through cell? 1 Ni + + e 1 Ni Two () Faradays would produce 1 mole (58.71grams) so one (1) Faraday would produce ½ mole (9.36 gms) How long would it take for 9.36 grams to be deposited? 1 Faraday = 96,500 coulombs Coulombs = Amp seconds Q = I (amps) x t (sec) How long would it take for 9.36 grams of Ni to be electroplated if amps were passed through the cell? 1 Faraday = 96,500 coulombs Coulombs = Amp seconds Q = I (amps) x t (sec) 96,500 coulombs = (amps) x? (sec) Time = ½ (96,500) = 48,50 sec = 804 minutes = 13.4 hours 9

10 7/1/014 Will Fe + (aq) oxidize Al(s)? Or will the following reaction occur? 3 Fe + (aq) + Al (s) 3 Fe (s) + Al +3 (aq) In order to answer that question, calculate E cell e + Fe + (aq) Fe (s) E red = 0.44 Al (s) Al +3 (aq) + 3e E oxid = E cell = E oxid + E red = + Since E cell is positive, YES the reaction will occur End Review Balancing OxidationReduction Reactions In Acidic Solution 1. Write ion half reactions. Balance atoms 3. Add water (if needed) 4. Add electrons In Basic Solution {one more step is added 5. OH is added to equation Example of Balancing Oxidation Reduction Reaction: NO (aq) + Al (s) NH 3 (g) + Al(OH) 4 (aq) in Acidic Solution The two half reactions are NO (aq) NH 3 (g) Al(s) Al(OH) 4 (aq) NO (aq) NH 3 (g) Al(s) Al(OH) 4 (aq) Adding water NO (aq) NH3 (gas) + H O 4 H O+Al (s) Al(OH) 4 (aq) Adding H + 7 H + (aq) + NO (aq) NH3 (gas) + H O 4 H O+Al (s) Al(OH) 4 (aq) + 4 H + (aq) 7 H + (aq) + NO (aq) NH 3 (g) + H O 4 H O +Al(s) Al(OH) 4 (aq) + 4 H + (aq) Adding electrons 6e + 7 H + (aq) + NO (aq) NH 3 (g) + H O 4 H O+Al(s) Al(OH) 4 (aq) + 4 H + (aq) + 3e To balance electrons multiply nd Eq by 8H O +Al (s) Al(OH) 4 (aq) + 8H + (aq) + 6e 6e + 7H + (aq) + NO (aq) NH 3 (aq) + H O 8H O+Al (s) Al(OH) 4 (aq) + 8H + (aq) +6e Adding two half reactions gives: 6 H O+ NO (aq) +Al(s) NH 3 (g)+ Al(OH) 4 (aq)+ H + (aq) Which is Balanced in acidic solution 10

11 7/1/014 Balancing OxidationReduction Reactions in Basic Solution 6H O +NO + Al NH 3 + Al(OH) 4 + H + Add OH (aq) to BOTH sides OH (aq) OH (aq) OH + 5 H O+NO + Al NH 3 + Al(OH) 4 OH (aq) + 5H O+NO (aq) +Al(s) NH 3 (gas) + Al(OH) 4 (aq) The equation is now balanced in basic solution Are you sure? Count elements on both sides of reaction 11

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