101- chem GENERAL CHEMISTRY-1

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1 101- chem GENERAL CHEMISTRY-1 First Mid Term (15 marks) Second Mid Term (15 marks) Lab (30 marks) Mid term (30 marks) Final (100 marks) Course# and Name: Chem-101, General chemistry-i Semester credit hours: 4.0 credit. Second Term Total Contact Hours: 39 hr. theory (Sun,Tue, Thurs 10-11) + 4 hr. lab

2 101-Chem Instructor: Dr. Amal AL-Mohaimeed Office location: Blg. 5, T floor, Room No. 188 Office hours: Sun, Tues, Thurs 9:00-10:00 address: muhemeed@ksu.edu.sa Total Contact Hours: 14 weeks x 3 hrs Theory (Sun, Tues, Thurs 10-11) 14x hrs Lab (Mon 10-1) Mid Term Exam: (First mid-term: Tues, 1/5, 1-1) (Second mid-term: Tues, 7/6,1-1)

3 General Chemistry-1 Book: Chapter Subject 1 Chapter 1: The Mole and Stoichiometry (7 hrs) Chapter : Properties of Gases (6 hrs) 3 Chapter 3: Energy and Thermodynamics (7 hrs) 4 Chapter 4: Chemical Kinetics (4 hrs) 5 Chapter 5: Properties of Solutions (5 hrs) 6 Chapter 6: Chemical Equilibrium (7 hrs) 7 Chapter 7: Acids and Bases (3 hrs) 3

4 Chapter 1 The Mole and Stoichiometry Chemistry: The Molecular Nature of Matter, 6E

5 International System of Units (SI) Standard system of units Metric Seven Base Units 5

6 Your Turn! The SI unit of length is the A. millimeter B. meter C. yard D. centimeter E. foot 6

7 Decimal Multipliers 7

8 Using Decimal Multipliers Use prefixes on SI base units when number is too large or too small for convenient usage Numerical values of multipliers can be interchanged with prefixes Ex. 1 ml = 10 3 L 1 km = 1000 m 1 ng = 10 9 g 1,130,000,000 s = s = 1.13 Gs 8

9 Laboratory Measurements Four common 1. Distance (d). Volume 3. Mass 4. Temperature 9

10 Laboratory Measurements 1. Distance (d) SI Unit is meter (m) Meter too large for most laboratory measurements Commonly use Centimeter (cm) 1 cm = 10 m = 0.01 m Millimeter (mm) 1 mm = 10 3 m = m 10

11 . Volume (V) Dimensions of (dm) 3 SI unit for Volume = m 3 Most laboratory measurements use V in liters (L) 1 L = 1 dm 3 (exactly) Chemistry glassware marked in L or ml 1 L = 1000 ml What is a ml? 1 ml = 1 cm 3 11

12 3. Mass SI unit is kilogram (kg) Frequently use grams (g) in laboratory as more realistic size 1 1 kg = 1000 g 1 g = kg = 1000 g 1

13 A. Celsius scale 4. Temperature Most common for use in science Water freezes at 0 C Water boils at 100 C 100 degree units between melting and boiling points of water 13

14 B. Kelvin scale 4. Temperature SI unit of temperature is kelvin (K) Water freezes at K and boils at K 100 degree units between melting and boiling points Absolute Zero Zero point on Kelvin scale 0 K= C 14

15 Temperature Conversions Must convert to Kelvin scale T K t ( C C) 1 K 1 C Ex. What is the Kelvin temperature of a solution at 5 C? T K (5 C C) 1 K 1 C = 98 K 15

16 Learning Check: T Conversions Ex. Convert 77 K to the Celsius scale. T K t ( C C) 1 K 1 C t C T ( K 73.15K) 1 C 1 K 1 C t C (77 K 73.15K) = 196 C 1 K 16

17 Table. Some Non-SI Metric Units Commonly Used in Chemistry 17

18 Derived SI Units All physical quantities will have units derived from these seven SI base units Ex. Area Derived from SI units based on definition of area length width = area meter meter = area m m = m SI unit for area = square meters = m 18

19 Density density d mass volume m V Units g/ml or g/cm 3 19

20 Learning Check What is the SI unit for velocity? Velocity ( v) distance time Velocity units meters seconds m s What is the SI unit for volume of a cube? Volume (V) = length width height V = meter meter meter V = m 3 0

21 1

22 Stoichiometry Stoichiometric Calculations Conversions from one set of units to another using Dimensional Analysis m= n x MM N= n x N A m: mass, n: amount (mol), MM: molar mass, N: number of particles, N A : Avogadro s no

23 The Mole Number of atoms in exactly 1 grams of 1 C atoms Ex. How many atoms in 1 mole of 1 C? Based on experimental evidence 1 mole of 1 C = atoms = g Avogadro s number = N A 1 mole of X = units of X Number of atoms, molecules or particles in one mole 1 mole Xe = Xe atoms 1 mole NO = NO molecules 3

24 Atoms Moles of Compounds Atomic Mass Mass of atom (from periodic table) 1 mole of atoms = gram atomic mass = atoms Molecules Molecular Mass Sum of atomic masses of all atoms in compound s formula 1 mole of molecule X = gram molecular mass of X = molecules 4

25 Moles of Compounds Ionic compounds Formula Mass Sum of atomic masses of all atoms in ionic compound s formula 1 mole ionic compound X = gram formula mass of X General Molar mass (MM) = formula units Mass of 1 mole of substance (element, molecule, or ionic compound) under consideration 1 mol of X = gram molar mass of X = formula units 5

26 SI Unit for Amount = Mole 1 mole of substance X = gram molar mass of X 1 mole S = 3.06 g S 1 mole NO = g NO Molar mass is our conversion factor between g & moles 1 mole of X = units of X N A is our conversion factor between moles & molecules 1 mole H O = molecules H O 1 mole NaCl = formula units NaCl 6

27 Learning Check: Using Molar Mass Ex. How many moles of iron (Fe) are in g Fe? What do we know? 1 mol Fe = g Fe What do we want to determine? Start g Fe =? Mol Fe End Set up ratio so that what you want is on top & what you start with is on the bottom 1 mol Fe g Fe = mole Fe 55.85g Fe 7

28 Learning Check: Using Molar Mass Ex. If we need mole Ca 3 (PO 4 ) for an experiment, how many grams do we need to weigh out? Calculate MM of Ca 3 (PO 4 ) 3 mass Ca = g = 10.4 g mass P = g = g 8 mass O = g = g 1 mole Ca 3 (PO 4 ) = g Ca 3 (PO 4 ) What do we want to determine? g Ca 3 (PO 4 ) =? Mol Fe Start End 8

29 Learning Check: Using Molar Mass 0.160mol Ca 3 (PO 4 ) g Ca 1 mol Ca 3 3 (PO (PO 4 ) 4 ) = 5.11 g Ca 3 (PO 4 ) 9

30 Your Turn! Ex. How many moles of CO are there in 10.0 g? A mol B mol C mol D mol E. 0.7 mol Molar mass of CO g = 1.01 g C g = 3.00 g O 1 mol CO = g CO n=m/mm 10.0 g CO 1 mol CO 44.01g CO = 0.7 mol CO 30

31 Your Turn! Ex. How many grams of platinum (Pt) are in mole Pt? A. 195 g B g C g D g E. 9.7 g Molar mass of Pt = g/mol g Pt 0.475mol Pt 1 mol Pt = 9.7 g Pt 31

32 Ex. How many silver atoms are in a 85.0 g silver bracelet? What do we know? g Ag = 1 mol Ag 1 mol Ag = Ag atoms What do we want to determine? 85.0 g silver =? atoms silver g Ag mol Ag atoms Ag 85.0 g Ag 1 mol Ag g Ag mol atoms Ag Ag = Ag atoms 3

33 Using Avogadro s Number Ex. What is the mass, in grams, of one molecule of octane, C 8 H 18? Molecules octane mol octane g octane 1. Calculate molar mass of octane Mass C = g = g Mass H = g = g 1 mol octane = 114. g octane. Convert 1 molecule of octane to grams 114.g octane 1 mol octane 1 mol octane molecules octane = g octane 33

34 Learning Check: Mole Conversions Ex. Calculate the number of formula units of Na CO 3 in 1.9 moles of Na CO mol Na CO formula units Na 1 mol NaCO3 = particles Na CO 3 Ex. How many moles of Na CO 3 are there in 1.15 x 10 5 formula units of Na CO 3? CO formula units Na CO 3 1 mol NaCO formula units Na CO 3 = mol Na CO 3 34

35 Your Turn! Ex. How many atoms are in 1.00 x 10 9 g of U? Molar mass U = g/mole. A. 6.0 x atoms B. 4.0 x atoms C..53 x 10 1 atoms D x atoms E..54 x 10 1 atoms 9 1 mol atoms U g U U g U 1 mol U =.53 x 10 1 atoms U 3 35

36 Your Turn! Ex. Calculate the mass in grams of FeCl 3 in formula units. (molar mass = g/mol) A. 16. g B g C g D. 41. g E units FeCl = 41. g FeCl mol FeCl 3 units FeCl g FeCl 1 mol FeCl

37 Mole-to-Mole Conversion Factors In H O there are: mol H 1 mol H O 1 mol O 1 mol H O mol H 1 mol O on atomic scale atom H 1 molecule H O 1 atom O 1 molecule H O atom H 1 molecule O 37

38 Stoichiometric Equivalencies Ex. N O 5 mol N 1 mol N O 5 5 mol O 1 mol N O 5 mol N 5 mol O 38

39 Calculating the Amount of a Compound by Analyzing One Element Ex. sample is found to contain moles of phosphorus P. How many moles of Ca 3 (PO 4 ) are in that sample? What do we want to find? mol P =? mol Ca 3 (PO 4 ) What do we know? Solution mol P 1 mol Ca 3 (PO 4 ) mol P 1 mol Ca 3 (PO mol P 4 ) = 0.43 mol Ca 3 (PO 4 ) 39

40 Your Turn! Ex. Calculate the number of moles of calcium in.53 moles of Ca 3 (PO 4 ) A..53 mol Ca B mol Ca C mol Ca D mol Ca E mol Ca.53 moles of Ca 3 (PO 4 ) =? mol Ca 3 mol Ca 1 mol Ca 3 (PO 4 ) 3 mol Ca.53mol Ca3(PO4) 1 mol Ca3(PO4) = 7.59 mol Ca 40

41 Mass-to-Mass Calculations Ex. Chlorophyll, the green pigment in leaves, has the formula C 55 H 7 MgN 4 O 5. If g of Mg is available to a plant for chlorophyll synthesis, how many grams of carbon will be required to completely use up the magnesium? Analysis g Mg? g C g Mg mol Mg mol C g C Assembling the tools g Mg = 1 mol Mg 1 mol Mg 55 mol C 1 mol C = g C 41

42 Ex. Mass-to-Mass Conversion 1 mol Mg 4.3 g Mg 1 mol C 1.0 g C g Mg mol Mg mol C g C 1 mol Mg 55 mol C g Mg 1 mol Mg 4.3g Mg 55 mol C 1 mol Mg 1.0g C 1 mol C = g C 4

43 Your Turn! Ex. How many g of iron are required to use up all of 5.6 g of oxygen atoms (O) to form Fe O 3? A g B. 9.8 g C g D. 134 g E. 5.4 g 5.6g O = 59.6 g Fe mass O mol O mol Fe mass Fe 5.6 g O? g Fe 3 mol O mol Fe 1 mol O 16.0g O mol Fe 3 mol O g Fe 1 mol Fe 43

44 Percentage Composition Percentage composition tells us mass of each element in g of substance Percentage by Mass: % mass of element % by mass of element 100% mass of sample Ex. Na CO 3 is 43.38% Na 11.33% C 45.9% O What is sum of % by mass? % 44

45 Ex. Percent Composition Ex. A sample of a liquid with a mass of g was decomposed into its elements and gave 5.17 g of carbon, g of hydrogen, and.478 g of oxygen. What is the percentage composition of this compound? 45

46 Ex. % Composition of Compound For C: For H: For O: g C 100% g total g H 100% g total g O 100% g total 5.17 g C 8.657g 0.960g H 8.657g 100% = 60.6% C 100%= 11.11% H.478g O 100% = 8.6% O 8.657g Sum of percentages: 99.99% 46

47 Your Turn! Ex. A sample was analyzed and found to contain g nitrogen and g oxygen. What is the percentage composition of this compound? 1. Calculate total mass of sample Total sample mass = g g = g. Calculate % Composition of N g N g N 100% 100% g total 0.546g 3. Calculate % Composition of O g O 100% g total g O 0.546g 100% = 5.94% N = 74.06% O 47

48 Ex. Using Percent Composition Are the mass percentages 30.54% N & 69.46% O consistent with the formula N O 4? Procedure: 1. Assume 1 mole of compound. Subscripts tell how many moles of each element are present mol N & 4 mol O 3. Use molar masses of elements to determine mass of each element in 1 mole Molar Mass of N O 4 = 9.14 g N O 4 / 1 mol 4. Calculate % by mass of each element 48

49 Ex. Using Percent Composition (cont) mol N 4 mol O 14.07g N 1 mol N 16.00g O 1 mol O = 8.14 g N = g O 8.14 g N %N 100% = 30.54% N in N 9.14g N O O g O %O 100% = 69.46% N in N 9.14g N O O 4 4 The experimental values match the theoretical percentages for the formula N O 4. 49

50 Your Turn Ex. If a sample containing only phosphorous & oxygen has percent composition 56.34% P & 43.66% O, is this P 4 O 10? A. Yes B. No 4 mol P 1 mol P 4 O mol O 1 mol P 4 O 10 4 mol P = g/mol P = 13.9 g P 10 mol O = g/mol O = g O 1 mol P 4 O 10 = 83.9 g P 4 O 10 %P 13.9g P 83.9g P O % = % P %O 160.0g O 83.9g P O % = % O 50

51 Determining Empirical & Molecular Formulas When making or isolating new compounds one must characterize them to determine structure & Molecular Formula Exact composition of one molecule Exact whole # ratio of atoms of each element in molecule Empirical Formula Simplest ratio of atoms of each element in compound Obtained from experimental analysis of compound glucose Empirical formula CH O Molecular formula C 6 H 1 O 6 51

52 Strategy for Determining Empirical Formulas 1. Determine mass in g of each element. Convert mass in g to moles 3. Divide all quantities by smallest number of moles to get smallest ratio of moles 4. Convert any non-integers into integer numbers. If number ends in decimal equivalent of fraction, multiply all quantities by least common denominator Otherwise, round numbers to nearest integers 5

53 1. Empirical Formula from Mass Data Ex. When a g sample of a compound was analyzed, it was found to contain g of C, g of H, and g of N. Calculate the empirical formula of this compound. Step 1: Calculate moles of each substance g C g H g N 1 mol C g C 1 mol H 1.008g H 1 mol N g N mol C mol H mol N 53

54 1. Empirical Formula from Mass Data Step : Select the smallest # of moles. Lowest is 3.7 x 10 3 mole Mole ratio molc C = molc molh H = molc = 1 = moln N = molc = 1 Step 3: Divide all # of moles by the smallest one Empirical formula = CH 5 N Integer ratio 54

55 Empirical Formula from Mass Composition Ex. One of the compounds of iron Fe and oxygen O, When a.448 g sample was analyzed it was found to have g of Fe and g of O. Calculate the empirical formula of this compound. Assembling the tools: 1 mol Fe = g Fe 1 mol O = g O 1. Calculate moles of each substance 1.771g Fe 1 mol Fe g Fe mol Fe g O 1 mol O 16.00g O mol O 55

56 1. Empirical Formula from Mass Data. Divide both by smallest #mol to get smallest whole # ratio mol Fe mol Fe 0.043mol O mol Fe =1.000 Fe =1.33 O 3 = Fe 3 = 3.99 O Or Fe 1 O Fe O (1.003) O(1.333) Fe3O3.99 Fe Empirical Formula = Fe 3 O 4 56

57 . Empirical Formula from % Composition Ex. Calculate the empirical formula of a compound whose % composition data is % P and % O. If the molar mass is determined to be 83.9 g/mol, what is the molecular formula? Step 1: Assume 100 g of compound g P g O 1 mol P = g 1 mol O = g 43.64g P 1 mol P 30.97g P = mol P g O 1 mol O g O = 3.53 mol P 57

58 . Empirical Formula from % Composition Step : Divide by smallest number of moles mol P = mol P 3.53 mol O.500 = mol P Step 3: Multiple by n to get smallest integer ratio Here n = Empirical formula = P O 5 58

59 Determining Molecular Formula Need molecular mass & empirical formula Calculate ratio of molecular mass to mass predicted by empirical formula & round to nearest integer molecular mass n empirical formula mass Ex. Glucose Molecular mass is g/mol Empirical formula = CH O Empirical formula mass = g/mol g n g Molecular formula = C 6 H 1 O 6 59

60 Learning Check The empirical formula of a compound is P O 5. If the molar mass is determined to be 83.9 g/mol, what is the molecular formula? Step 1: Calculate empirical mass empirical mass P 5 O mass P 5 mass O g/mol g/mol g/mol g/molp O 5 Step : Calculate ratio of molecular to empirical mass 83.9 g/ mol n = Molecular formula = P g/mol 4 O 10 60

61 Your Turn! Ex.The empirical formula of hydrazine is NH, and its molecular mass is 3.0. What is its molecular formula? A. NH B. N H 4 C. N 3 H 6 D. N 4 H 8 E. N 1.5 H 3 Molar mass of NH = ( )g + ( 1.008)g = g n = (3.0/16.0) = Atomic Mass: N:14.007; H:1.008; O:

62 Learning Check: Balancing Equations unbalanced AgNO 3 (aq) + Na 3 PO 4 (aq) Ag 3 PO 4 (s) + NaNO 3 (aq) balanced 3AgNO 3 (aq) + Na 3 PO 4 (aq) Ag 3 PO 4 (s) + 3NaNO 3 (aq) unbalanced Zn(s) + HCl(aq) ZnCl (aq) + H (g) Balanced Zn(s) + HCl(aq) ZnCl (aq) + H (g) 6

63 Balance by Inspection C 3 H 8 (g) + O (g) CO (g) + H O(l) Assume 1 in front of C 3 H 8 3C 1C 3 8H H 4 1C 3 H 8 (g) + O (g) 3CO (g) + 4H O(l) O 5 =10 O = (3 ) + 4 = 10 8H H = 4 = 8 1C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O(l) 63

64 Your Turn! Ex. Balance each of the following equations. What are the coefficients in front of each compound? 1 Ba(OH) (aq) + 1 Na SO 4 (aq) 1 BaSO 4 (s) + NaOH(aq) KClO 3 (s) KCl(s) + 3 O (g) H 3 PO 4 (aq) + 3 Ba(OH) (aq) Ba 1 3 (PO 4 ) (s) + H 6 O(l) 64

65 Stoichiometric Ratios Consider the reaction Could be read as: N + 3H NH 3 When 1 molecule of nitrogen reacts with 3 molecules of hydrogen, molecules of ammonia are formed. Molecular relationships 1 molecule N molecule NH 3 3 molecule H molecule NH 3 1 molecule N 3 molecule H 65

66 Stoichiometric Ratios Consider the reaction Could also be read as: N + 3H NH 3 When 1 mole of nitrogen reacts with 3 moles of hydrogen, moles of ammonia are formed. Molar relationships 1 mole N mole NH 3 3 mole H mole NH 3 1 mole N 3 mole H 66

67 Using Stoichiometric Ratios Ex. For the reaction N + 3 H NH 3, how many moles of N are used when.3 moles of NH 3 are produced? Assembling the tools moles NH 3 = 1 mole N.3 mole NH 3 =? moles N 1 mol N.3 mol NH 3 = 1. mol N mol NH 3 67

68 Your Turn! Ex. If mole of CO is produced by the combustion of propane, C 3 H 8, how many moles of oxygen are consumed? A mole B..88 mole C mole D mole E mole mol C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O(g) CO Assembling the tools mole CO =? moles O 3 moles CO = 5 mole O 5 mol O 3 mol CO = mol O 68

69 Using Balanced Equation to Determine Stoichiometry Ex. What mass of O will react with 96.1 g of propane (C 3 H 8 ) gas, to form gaseous carbon dioxide & water? Strategy 1. Write the balanced equation C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O(g). Assemble the tools 96.1 g C 3 H 8 moles C 3 H 8 moles O g O 1 mol C 3 H 8 = 44.1 g C 3 H 8 1 mol O = 3.00 g O 1 mol C 3 H 8 = 5 mol O 69

70 Using Balanced Equation to Determine Stoichiometry Ex. What mass of O will react with 96.1 g of propane in a complete combustion? C 3 H 8 (g) + 5O (g) 3CO (g) + 4H O(g) 3. Assemble conversions so units cancel correctly 1 mol C3H8 5 mol O 96.1 g C3H g C H 1 mol C H g O 1 mol O = 349 g of O are needed 70

71 Your Turn! Ex. How many grams of Al O 3 are produced when 41.5 g Al react? A g B. 157 g C. 314 g D..0 g E g Al(s) + Fe O 3 (s) Al O 3 (s) + Fe(l) 41.5 g Al 1 mol Al 6.98g Al 1 mol AlO mol Al g Al O 1 mol Al O3 3 = 78.4 g Al O 3 71

72 Reactant that is completely used up in the reaction Present in lower # of moles It determines the amount of product produced For this reaction = ethylene Excess reactant Limiting Reactant Reactant that has some amount left over at end Present in higher # of moles For this reaction = water 7

73 Ex. Limiting Reactant Calculation How many grams of NO can form when 30.0 g NH 3 and 40.0 g O react according to: Solution: Step 1 4 NH O 4 NO + 6 H O mass NH 3 mole NH 3 mole O mass O Assembling the tools 1 mol NH 3 = g 1 mol O = 3.00 g Only have 40.0 g O, O limiting reactant 4 mol NH 3 5 mol O 1 mol NH3 5 mol O 30.0 g NH gNH 4 mol NH = 70.5 g O needed g O 1 mol O 73

74 Ex. Limiting Reactant Calculation How many grams of NO can form when 30.0 g NH 3 and 40.0 g O react according to: Solution: Step 4 NH O 4 NO + 6 H O mass O mole O mole NO mass NO Assembling the tools 1 mol O = 3.00 g 1 mol NO = g 5 mol O 4 mol NO 1 mol O 40.0 g O 3.00g O = 30.0 g NO formed 4 mol NO 5 mol O Can only form 30.0 g NO g NO 1 mol NO 74

75 Your Turn! Ex. If 18.1 g NH 3 is reacted with 90.4 g CuO, what is the maximum amount of Cu metal that can be formed? NH 3 (g) + 3CuO(s) N (g) + 3Cu(s) + 3H O(g) (MM) (17.03) (79.55) (8.01) (64.55) (18.0) (g/mol) 18.1gNH 1mol NH gnh 3 mol CuO mol NH g CuO 1mol CuO A. 17 g B. 103 g C. 7. g D. 108 g E g 17 g CuO needed. Only have 90.4g so CuO limiting 1mol CuO 3 mol Cu g Cu 90.4 g CuO g CuO 3 mol CuO 1mol Cu 7. g Cu can be formed 75

76 Theoretical vs. Actual Yield Theoretical Yield Maximum amount of product that must be obtained if no losses occur. Amount of product formed if all of limiting reagent is consumed. Actual Yield Amount of product that is actually isolated at end of reaction. Amount obtained experimentally How much is obtained in mass units or in moles. 76

77 Percentage Yield Useful to calculate % yield. Percent yield Relates the actual yield to the theoretical yield It is calculated as: percentageyield Ex. If a cookie recipe predicts a yield of 36 cookies and yet only 4 are obtained, what is the % yield? percentageyield actual yield theoretical yield 67%

78 Ex. Percentage Yield Calculation When 18.1 g NH 3 and 90.4 g CuO are reacted, the theoretical yield is 7. g Cu. The actual yield is 58.3 g Cu. What is the percent yield? NH 3 (g) + 3CuO(s) N (g) + 3Cu(s) + 3H O(g) 58.3g Cu % yield 100% = 80.7% 7.g Cu 78

79 Learning Check: Percentage Yield Ex. A chemist set up a synthesis of solid phosphorus trichloride PCl 3 by mixing 1.0 g of solid phosphorus with 35.0 g chlorine gas and obtained 4.4 g of solid phosphorus trichloride. Calculate the percentage yield of this compound. Analysis: Write balanced equation P(s) + Cl (g) PCl 3 (s) Determine Limiting Reagent Determine Theoretical Yield Calculate Percentage Yield 79

80 Learning Check: Percentage Yield Assembling the Tools: 1 mol P = g P 1 mol Cl = g Cl 3 mol Cl mol P Solution 1. Determine Limiting Reactant 1 mol P 3 mol Cl 70.90g Cl 1.0 g P = 41. g Cl 30.97g P mol P 1 mol Cl But you only have 35.0 g Cl, so Cl is limiting reactant 80

81 Learning Check: Percentage Yield Solution. Determine Theoretical Yield 1 mol Cl mol PCl g Cl 70.90g Cl 3 mol Cl 137.3g PCl 1 mol PCl 3 3 = 45. g PCl 3 3. Determine Percentage Yield Actual yield = 4.4 g 4.g PCl3 percentageyield g PCl = 93.8 % 3 81

82 Your Turn! Ex. When 6.40 g of CH 3 OH was mixed with 10. g of O and ignited, 6.1 g of CO was obtained. What was the percentage yield of CO? CH 3 OH + 3O CO + 4H O MM(g/mol) (3.04) (3.00) (44.01) (18.0) A. 6.1% 1 mol CH3OH 3 mol CO 6.40 g CH3OH B. 8.79% 3.04g CH3OH mol CH3OH 1 mol C. 100% =9.59 g O needed; CH 3 OH limiting 1 mol CH3OH mol CO D. 14% 6.40 g CH3OH 3.04g CH3OH mol CH3OH 1 mol E. 69.6% = 8.79 g CO in theory 6.1g CH 8.79g CH 3 OHactual 100% OH theory % 3.00g O O 44.01g CO CO 8

83 Molarity Concentration Number of moles of solute per liter of solution. Allows us to express relationship between moles of solute and volume of solution Hence, 0.100M solution of NaCl contains mole NaCl in 1.00 liter of solution Same concentration results if you dissolve mol of NaCl in liter of solution Molarity (M) moles of solute liters of solution mole Volume 0.100mol NaCl 1.00L NaCl soln mol NaCl 0.100L NaCl soln 0.100M NaCl 83

84 Learning Check: Calculating Molarity (from grams and volume) Ex. Calculate the molarity (M) of a solution prepared by dissolving 11.5 g NaOH (40.00 g/mol) solid in enough water to make 1.50 L of solution. g NaOH mol NaOH M NaOH 1 mol NaOH 11.5 g NaOH g NaOH M moles NaOH L so ln 0.88mol 0.88mol NaOH 1.50 L soln NaOH 0.19M NaOH 84

85 Learning Check: Calculating Volume (from Molarity and moles) Ex. How many ml of 0.50 M NaCl solution are needed to obtain mol of NaCl? Use M definition 0.50M NaCl 0.50mol NaCl 1.00 L NaCl soln Given molarity and moles, need volume 1.00 L NaCl soln 0.100mol NaCl 0.50mol NaCl 1000mL NaCl soln 1 L NaCl soln = 400 ml of 0.50 M NaCl solution 85

86 Temperature Insensitive Concentration 1. Percent Concentrations Also called percent by mass or percent by weight percent by mass This is sometimes indicated %(w/w) where w stands for weight The (w/w) is often omitted mass of solute 100% mass of solution 86

87 Ex. Percent by Mass What is the percent by mass of NaCl in a solution consisting of 1.5 g of NaCl and 75.0 g water? percent by mass mass of solute mass of solution 100% 1.5g wt% NaC l 100 ( )g wt% NaC l 14.3% NaCl 87

88 Learning Check Ex. How many grams of sea salt would be needed to prepare 6.5 L of 3.5% w/w solution with density of 1.03 g/ml.? What do we need to find? 6.5 L? g sea salt What do we know? 3.5 g sea salt 100 g solution 1.03 g soln 1.00 ml solution 1000 ml 1.00 L 1000mL 1.03g soln 6.5 L 1 L 1.00mL soln = g sea salt 3.5 g sea salt 100 g soln 88

89 More Temperature Insensitive Concentration Units Molality (m) Number of moles of solute per kilogram solvent molality m mol of solute kg of solvent Also Molal concentration Independent of temperature 89

90 Ex. Concentration Calculation A. If you prepare a solution by dissolving 5.38 g of I in g of water, what is the molality (m) of the solution? What do we need to find? 5.38 g? m What do we know? 53.8 g I 1 mol I m = mol solute/kg solvent g kg Solve it 5.38g I 1 mol I 53.8g I kg water = mol I /kg water = m 90

91 Ex. CalculatingM from m (cont.) B. What is the molarity (M) of this solution? The density of this solution is 1.59 g/ml. What do we need to find? 5.38 g? M What do we know? 53.8 g I 1 mol I M = mol solute/l soln 1.59 g soln 1 ml soln g of soln = g I + g H O = g g Solve it 1 mol I g 5.38g I 53.8g I g soln 1.00 ml = mol I /L soln = M 1000mL 1 L 91

92 Converting between Concentrations Ex. Calculate the molarity and the molality of a 40.0% HBr solution. The density of this solution is 1.38 g/ml g HBr 40.0% HBr wt% g solution If we assume g of solution, then 40.0 g of HBr. m mol HBr kg of O H 40.0g HBr mol HBr 0.494mol HBr g HBr/ mol If 100 g solution, then mass H O = g soln 40.0 g HBr = 60.0 g H O mol HBr 0.494mol HBr m 8. 4m kg H O kg H O 9

93 Converting between Concentrations (cont.) Now Calculate Molarity of 40% HBr Vol Soln M mass density mol HBr L solution 100g = 7.46 ml 1.38g / ml M mol HBr = mol 0.494mol HBr 7.46mL solution 1000mL 1L = 6.8 M 93

94 Your Turn! Ex. What is the molality of 50.0% (w/w) NaOH solution? A m B. 1.5 m C m D. 5 m E. 50 m g soln = 50.0 g NaOH g water 1 mol NaOH 1 m 1000g 50.0g NaOH 40.00g NaOH mol/kg 1 kg g water = 5 m MM(g/mol) H O: 18.0; NaOH:

95 Your Turn! Ex. What is the molarity of the 50%(w/w) solution if its density is 1.59 g/ml? A. 19 M B. 1.5 M C. 1.9 M D M 1 mol NaOH g soln 50 g NaOH x x g ml ml x x = 19 M 100 g soln L or mmol 50 g H O 1.59 g g soln 100 g soln ml 5 x x 19 M 95

96 Other Temperature Insensitive Concentration Units Mole Fraction A # mol Totalmoles of all A components 96

97 Chapter Properties of Gases Chemistry: The Molecular Nature of Matter, 6E

98 Properties of Common Gases Despite wide differences in chemical properties, ALL gases more or less obey the same set of physical properties. Four Physical Properties of Gases 1. Pressure (P). Volume (V) 3. Temperature (T) 4. Amount = moles (n) 98

99 Units of Pressure Pascal = Pa SI unit for Pressure Very small 1atm 101,35 Pa = 101 kpa 1 atm too big for most lab work 1 1torr atm 1 atm 760 mm Hg 760 At sea level 1 torr = 1 mm Hg 99

100 Studied relationship between P and V Work done at constant T as well as constant number of moles (n) T 1 = T As V, P Boyle s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure. 100

101 A plot of V versus P results in a curve A plot of V versus 1/P will be a straight line P a 1/V P x V = constant P 1 x V 1 = P x V P = k (1/V ) This means a plot of P versus 1/V will be a straight line. 101

102 Ex: A sample of chlorine gas occupies a volume of 946 ml at a pressure of 76 mmhg. What is the pressure of the gas (in mmhg) if the volume is reduced at constant temperature to 154 ml? P 1 = 76 mmhg P 1 x V 1 = P x V P =? V 1 = 946 ml V = 154 ml P = P 1 x V 1 V 76 mmhg x 946 ml = = 4460 mmhg 154 ml 10

103 Charles worked on relationship of how V changes with T Kept P and n constant Showed V as T i.e., Charles s Law V T = k The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. V 1 /T 1 = V /T 103

104 Ex. Anesthetic gas is normally given to a patient when the room temperature is 0.0 C and the patient's body temperature is 37.0 C. What would this temperature change do to 1.60 L of gas if the pressure and mass stay the same? V1 T 1 V T V V1T T L K K V = 1.69 L 104

105 Ex: A sample of carbon monoxide gas occupies 3.0 L at 15 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 /T 1 = V /T V 1 = 3.0 L T 1 = K V = 1.54 L T =? T = V x T 1 V L x K = = 19 K 3.0 L 105

106 Gay-Lussac s Law Volume (V) and number of moles (n) are constant P as T V= constant, n= Constant The pressure of a fixed amount of gas at constant volume is directly proportional to its absolute temperature. P T Low T, Low P P 1 T = P T 1 P T (K) High T, High P 106

107 Ratio PV T Combined Gas Law Constant for fixed amount of gas (n) PV C T for fixed amount (moles) PV 1 T 1 1 P V T 107

108 How Other Laws Fit into Combined Gas Law PV 1 T 1 1 P V T Boyle s Law T 1 = T P 1 V 1 = P V Charles Law P 1 = P Gay-Lussac s Law V 1 = V V1 V T1 T P1 P T T 1 108

109 Combined Gas Law Ex. If a sample of air occupies 500. ml at STP*, what is the volume at 85 C and 560 torr? P 760 torr 500.mL K 1 T V 1 1 P T V 560 torr K V *Standard Temperature (73.15K) Standard Pressure (1 atm) = 890 ml 109

110 Ex. Using Combined Gas Law What will be the final pressure of a sample of nitrogen gas with a volume of 950 m 3 at 745 torr and 5.0 C if it is heated to 60.0 C and given a final volume of 1150 m 3? P 1 = 745 torr P =? P V 1 = 950 m 3 V = 1150 m 3 T 1 = 5.0 C = K PV 1 1T T V 1 T = 60.0 C = K 3 745torr 950m K K 1150m P = 688 torr 110

111 Your Turn! Ex. Which units must be used in all gas law calculations? A. K B. atm C. L D. no specific units as long as they cancel 111

112 Relationships between Gas Volumes In reactions in which products and reactants are gases: If T and P are constant Simple relationship among volumes hydrogen + chlorine hydrogen chloride 1 vol 1 vol vol hydrogen + oxygen water (gas) vol 1 vol vol ratios of simple, whole numbers 11

113 Avogadro s Principle When measured at same T and P, equal V's of gas contain equal number of moles Volume of a gas is directly proportional to its number of moles, n V n (at constant P and T) H (g) + Cl (g) HCl (g) Coefficients 1 1 Volumes 1 1 Molecules 1 1 (Avogadro's Principle) Moles

114 Learning Check: Ex. Calculate the volume of ammonia formed by the reaction of 5L of hydrogen with excess nitrogen. N (g) + 3H (g) NH 3 (g) 5 L H 1 L NH 3 L H 3 17 L NH 3 114

115 Learning Check: If 15 L H react with 50L N, what volume of NH 3 can be expected? N (g) + 3H (g) NH 3 (g) 15L H 1 L NH 3 L H L NH 3 50 L N 1 L NH 1 L N 3 100L NH 3 H is limiting reagent 83.3 L 115

116 Learning Check: Ex. How many liters of N (g) at 1.00 atm and 5.0 C are produced by the decomposition of 150. g of NaN 3? NaN 3 (s) Na (s) + 3N (g) 150.g NaN mol N 1 V T V 1 1 V T ; V 1mol NaN g 3.4 L 1mol at STP V1 T T L K 73.15K 1 3 mol N mol NaN L 84.6 L mol N

117 Your Turn! Ex. How many liters of SO 3 will be produced when 5 L of sulfur dioxide reacts with 75 L of oxygen? All gases are at STP. A. 5 L B. 50 L C. 100 L D. 150 L E. 75 L L SO 5 L SO x 5 L SO 3 3 L SO L SO 75 L O x 150 L SO L O 117

118 Ideal Gas Law With Combined Gas Law we saw that PV T With Avogadro s results we see that this is modified to PV T C n R Where R = a new constant = Universal Gas constant PV nrt 118

119 Standard Molar Volume Volume of 1 mole gas must be identical for all gases under same P and T Standard Conditions of Temperature and Pressure STP STP = 1 atm and K (0.0 C) Under these conditions 1 mole gas occupies V =.4 L.4 L standard molar volume 119

120 What is the value of R? Plug in values of T, V, n and P for 1 mole of gas at STP (1 atm and 0.0 C) T = 0.0 C = K P = 1 atm V =.4 L n = 1 mol R PV nt 1 1 atm mol. 4L K R = L atm mol1 K1 10

121 Learning Check: Ex. How many liters of N (g) at 1.00 atm and 5.0 C are produced by the decomposition of 150. g of NaN 3? NaN 3 (s) Na(s) + 3N (g) V =? n mol N 150.g NaN 1 V = nrt/p P = 1 atm T = 5C = K V n = mol N 3 1 mol NaN g 3.461mol N Latm 98.15K molk 1atm 3 mol N mol NaN V=84.6L 3 11

122 Your Turn! Ex. Solid CaCO decomposes to solid CaO and CO when heated. What is the pressure, in atm, of CO in a 50.0 L container at 35 o C when 75.0 g of calcium carbonate decomposes? A atm B atm C atm P = m x R x T / MM x V D atm E. 38 atm g CaO x x x x 308 K g 1 mol CaCO3 K mol 0.38 atm 1 mol CaCO 1 mol CO L atm PV nrt 50.0 L 1

123 Ex. A gas allowed to flow into a 300 ml gas bulb until the pressure was 685 torr. The sample now weighed 1.45 g; its temperature was 7.0 C. What is the molecular mass of this gas? 1L V 300 ml L 1000mL 1atm P 685torr torr n PV RT MM atm atm L ( atm L / mol K ) 300. K mass n 0. T = 7.0 C K = 300. K g mol = mole 13 g/mol 13 Gas = Xe

124 Ex. A gaseous compound with an empirical formula of PF was found to have a density of 5.60 g/l at 3.0 C and 750 torr. Calculate its molecular mass and its molecular formula. T = 3.0 C K = 96. K 1atm P 750torr atm 760torr PV atm L n RT ( atm L / mol K ) 96. K MM mass n g mol mole 138 g/mol 14

125 Ex. Solution (cont) Now to find molecular formula given empirical formula and MM First find mass of empirical formula unit 1 P = 1 31g/mol = 31g/mol F = 19 g/mol = 38 g/mol Mass of PF = 69 g/mol MM molecular mass empirical mass 138g / mol 69g / mol the correct molecular formula is P F 4 15

126 Your Turn! Ex. 7.5 g of a gas with an empirical formula of NO occupies.0 L at a pressure of 1.0 atm and 5 o C. Determine the molar mass and molecular formula of the compound. A g/mol, NO B g/mol, N O 4 C. 7.7 g/mol, NO D g/mol, N O E. Not enough data to determine molar mass 16

127 Your Turn! - Solution MW = L atm 7.5 g x x 98 K K mol 1.0 atm x.0 L 90.0 g/mol g 1 mol NO mol NO 90 x = mol 45.0 g mol Molecular formula is N O 4 17

128 Stoichiometry of Reactions Between Gases Can use stoichiometric coefficients in equations to relate volumes of gases Provided T and P are constant Volume moles V n Ex. Methane burns according to the following equation. CH 4 (g) + O (g) CO (g) + H O (g) 1 vol vol 1 vol vol 18

129 Ex. The combustion of 4.50 L of CH 4 consumes how many liters of O? (Both volumes measured at STP.) P and T are all constant so just look at ratio of stoichiometric coefficients Volume of O L L O 1L CH 4 = 9.00 L O 19

130 Ex. Gas bulb with a volume of 50 ml. How many grams of Na CO 3 (s) would be needed to prepare enough CO (g) to fill this bulb when the pressure is at 738 torr and the temperature is 3 C? The equation is: Na CO 3 (s) + HCl(aq) NaCl(aq) + CO (g) + H O(l) 1L V 50mL 0. 50L 1000mL 1atm P 738torr atm 760torr T = 3.0 C K = 96. K n PV RT atm L ( atm L / mol K ) 96. K 130

131 Ex. Solution (cont) = x 10 3 mole CO mol CO 1mol NaCO3 1mol CO = x 10 3 mol Na CO mol NaCO3 106g NaCO3 1mol Na CO 3 = 1.06 g Na CO 3 131

132 Your Turn! Ex. How many grams of sodium are required to produce 0.0 L of hydrogen gas at 5.0 C, and 750 torr? Na(s) + H O(l ) NaOH(aq) + H (g ) A g B g C g D. 9.6 g E g 13

133 Your Turn! - Solution Moles of H produced: n = 1 atm 750 torr x x 0.0 L 760 torr L atm x 98 K K mol = mol H Grams of sodium required: mol Na 3.0 g g Na = mol H x x = 37.1 g mol H mol Na 133

134 Dalton's Law of Partial Pressure For mixture of non-reacting gases in container Total pressure exerted is sum of the individual partial pressures that each gas would exert alone P total = P a + P b + P c + Where P a, P b, and P c = partial pressures Partial pressure Pressure that particular gas would exert if it were alone in container 134

135 Dalton s Law of Partial Pressures Assuming each gas behaves ideally Partial pressure of each gas can be calculated from Ideal Gas Law n RT RT P a a P V b b P V c c V So Total Pressure is n n RT P total P a P b P nart nbrt V V c ncrt V 135

136 Dalton s Law of Partial Pressures Rearranging Or P P total total ( n n a total n n RT ) V Where n total = n a + n b + n c + n total = sum of # moles of various gases in mixture b RT V c 136

137 Ex. Mixtures of 46 L He at 5 C and 1.0 atm and 1 L O at 5 C and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 5 C. He O P i = 1.0 atm P f = P He P i = 1.0 atm P f = P O V i = 46 L V f = 5.0 L V i = 1 L V f = 5.0 L 137

138 Ex. Solution (cont) First calculate pressure of each gas in 5 L tank (P f ) using combined gas law PV i i 1atm 46L PHe 9. atm V 5L f PV i i 1atm 1L PO 4. atm Vf 5L Then use these partial pressures to calculate total pressure Ptotal PHe PO 9. atm 4. atm atm 138

139 Your Turn! Ex. 50 ml of methane, CH 4, at 35 o C and 0.55 atm and 750 ml of propane, C 3 H 8, at 35 o C and 1.5 atm, were introduced into a 10.0 L container. What is the final pressure, in torr, of the mixture? A torr B. 6.0 x 10 4 torr C. 3.4 x 10 3 torr D. 760 torr E torr 139

140 Your Turn! - Solution 0.55 atm x 0.50 L P(CH 4) 10.0 L = atm 1.5 atm x L P(C3H 8) 10.0 L = 0.11 atm 760 torr P = T atm x atm = 95.6 torr 140

141 141 Mole Fractions and Mole Percents Mole Fraction Ratio of number moles of given component in mixture to total number moles in mixture total A Z C B A A A n n n n n n n X RT V P n A A total A total A A n n P P X total A A P X P

142 Ex. The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O present X A P P A total X O 156torr torr 14

143 Ex. The mole fraction of nitrogen in the air is Calculate the partial pressure of N in air when the atmospheric pressure is 760. torr. P N X N P total P torr 593torr N 143

144 Your Turn! Ex. 50 ml of methane, CH 4, at 35 o C and 0.55 atm and 750 ml of propane, C 3 H 8, at 35 o C and 1.5 atm were introduced into a 10.0 L container. What is the mole fraction of methane in the mixture? A B C D. 0.5 E

145 Your Turn! - Solution 0.55 atm x 0.50 L P(CH 4) 10.0 L = atm 1.5 atm x L P(C3H 8) 10.0 L = 0.11 atm atm X CH = = ( ) atm 145

146 Collecting Gases over Water Water vapor is present because molecules of water escape from surface of liquid and collect in space above liquid Molecules of water return to liquid rate of escape = rate of return Number of water molecules in vapor state remains constant Gas saturated with water vapor = Wet gas Application of Dalton s Law of Partial Pressures 146

147 Vapor Pressure Pressure exerted by vapor present in space above any liquid Constant at constant T When wet gas collected over water, we usually want to know how much dry gas this corresponds to P total = P gas + P water Rearranging P gas = P total P water 147

148 Ex. A sample of oxygen is collected over water at 0.0 C and a pressure of 738 torr. Its volume is 310 ml. The vapor pressure of water at 0 C is torr. a. What is the partial pressure of O? b. What would the volume be when dry at STP? a. P O = P total P water = 738 torr 17.5 torr = 70 torr 148

149 Ex. Solution b. calculate P O at STP P 1 = 70 torr P = 760 torr V 1 = 310 ml V =? T 1 = = 93 K T = K = 73 K V PV 1 1 T 1 PV T V PV 1 1T T P 70torr 310mL 73K V = 74 ml 93K 760torr 1 149

150 Your Turn! Ex. An unknown gas was collected by water displacement. The following data was recorded: T = 7.0 o C; P = 750 torr; V = 37.5 ml; Gas mass = g; P vap (H O) = 6.98 torr Determine the molecular weight of the gas. A. 5.4 g/mol B. 30. g/mol C g/mol D g/mol E g/mol 150

151 Your Turn! - Solution MW MW L atm g x x 300 K grt K mol PV ( )torr x L 60.3 g/mol 151

152 Complete spreading out and intermingling of molecules of one gas into and among those of another gas Ex. Perfume in room Diffusion 15

153 Effusion Movement of gas molecules Through extremely small opening into vacuum Vacuum No other gases present in other half 153

154 Graham's Law of Effusion Rates of effusion of gases are inversely proportional to square roots of their densities, d, when compared at identical pressures and temperatures 1 Effusion Rate (constant P and T) d Effusion Rate ( A) Effusion Rate ( B ) d d B A d d B A Effusion Rate ( A) Effusion Rate ( B ) d d B A M M B A 154

155 Graham's Law of Effusion Ex. Calculate the ratio of the effusion rates of hydrogen gas (H ) and uranium hexafluoride (UF 6 ) - a gas used in the enrichment process to produce fuel for nuclear reactors. First must compute MM's MM (H ) =.016 g/mol MM (UF 6 ) = 35.0 g/mol Effusion Rate (H ) Effusion Rate (UF ) M M UF H 13 Thus the very light H molecules effuse ~13 times as fast as the massive UF 6 molecules. 155

156 Ex. For the series of gases He, Ne, Ar, H, and O what is the order of increasing rate of effusion? substance He Ne Ar H O MM Lightest are fastest So H > He > Ne > O >Ar Heavier gases effuse more slowly Lighter gases effuse more rapidly 156

157 Postulates of Kinetic Theory of Gases 1. Particles are so small compared with distances between them, so volume of individual particles can be assumed to be negligible. V gas ~ 0. Particles are in constant motion Collisions of particles with walls of container are cause of pressure exerted by gas number collisions P gas 157

158 Postulates of Kinetic Theory of Gases 3. Particles are assumed to exert no force on each other Assumed neither to attract nor to repel each other 4. Average kinetic energy of collection of gas particles is assumed to be directly proportional to Kelvin Temperature KE avg T K Root-mean-square speed 158

159 Ex: What are the rms speeds of helium atoms, and nitrogen, hydrogen, and oxygen molecules at 5 C? T = 5 C + 73 = 98 K. R = J/mol K Element Mass (kg) rms speed (m/s) He H N O

160 Kinetic Theory of Gases This means T KE ave Specifically As increase T, KE ave, number collisions with walls, thereby increasing P Kinetic energy For 1 mole of gas is: KE ave 3 RT 160

161 Real Gases Have finite volumes Do exert forces on each other Real Gases deviate Why? PV T constant PV nt R 161

162 Real Gases Deviate from Ideal Gas 1. Gas molecules have finite V's Take up space Less space of kinetic motions V motions < V container particles hit walls of container more often P over ideal Law 16

163 Real Gases. Particles DO attract each other Even weak attractions means they hit walls of container less often P over ideal gas 163

164 van der Waal's equation for Real Gases P n a V V nb nrt corrected P corrected V a and b are van der Waal's constants Obtained by measuring P, V, and T for real gases over wide range of conditions 164

165 van der Waal's equation for Real P Gases n a V corrected P V nb nrt a Pressure correction Indicates some attractions between molecules Large a Means strong attractive forces between molecules Small a Means weak attractive forces between molecules 165

166 van der Waal's equation for Real P Gases n a V V nb nrt corrected V b Volume correction Deals with sizes of molecules Large b Means large molecules Small b Means small molecules Gases that are most easily liquefied have largest van der Waal's constants 166

167 Ex: When will a real gas behave most like an ideal gas? A) at high temperatures and high pressures B) at low temperatures and high pressures C) at low temperatures and low pressures D) at high temperatures and low pressures 167

168 Ex: If mol of an ideal gas were confined to.41 L at 0.0 C, it would exert a pressure of atm. Use the van der Waals equation and the constants in Table 10.3 to estimate the pressure exerted by mol of Cl (g) in.41 L at 0.0 C. Ex.. Use Van de Waal s equation to calculate the pressure exerted by 1.00 molcl confined to a volume of.00 L at 73K. The value of a=6.49l atm/mol and b=0.056 L/mol 168

169 Chapter 3 Energy and Chemical Change Thermodynamics Chemistry: The Molecular Nature of Matter, 6E

170 Thermochemistry Study of energies given off by or absorbed by reactions. Thermodynamics Study of energy transfer (flow) Energy (E) Ability to do work or to transfer heat. Kinetic Energy (KE) Energy of motion KE = ½mv 170

171 Potential Energy (PE) Stored energy Exists in natural attractions and repulsions Gravity Positive and negative charges Springs Chemical Energy PE possessed by chemicals Stored in chemical bonds Breaking bonds requires energy Forming bonds releases energy 171

172 Law of Conservation of Energy Energy can neither be created nor destroyed Can only be converted from one form to another Total Energy of universe is constant Total Energy = Potential + Energy Kinetic Energy 17

173 Temperature vs. Heat Temperature Proportional to average kinetic energy of object s particles Higher average kinetic energy means Higher temperature Faster moving molecules Heat Energy transferred between objects Caused by temperature difference Always passes spontaneously from warmer objects to colder objects Transfers until both are the same temperature 173

174 Heat Transfer hot cold Hot and cold objects placed in contact Molecules in hot object moving faster KE transfers from hotter to colder object average KE of hotter object average KE of colder object Over time Average KEs of both objects becomes the same Temperature of both becomes the same 174

175 Units of Energy Joule (J) 1kg m 1J s value is greater than 1000 J, use kj 1 kj = 1000 J calorie (cal) Energy needed to raise T of 1 g H O by 1 C 1 cal = J (exactly) 175

176 Internal Energy (E) Sum of energies of all particles in system E = Total energy of system E = Potential + Kinetic = PE + KE Change in Internal Energy E = E final E initial means change final initial What we can actually measure Want to know change in E associated with given process 176

177 E, Change in Internal Energy For reaction: reactants products E = E products E reactants Can use to do something useful Work Heat If system absorbs energy during reaction Energy coming into system is positive (+) Final energy > initial energy Ex. Photosynthesis or charging battery As system absorbs energy Increase potential energy Available for later use 177

178 E, Change in Internal Energy E = E products E reactants Energy change can appear entirely as heat Can measure heat Can t measure E product or E reactant Fortunately, we are more interested in E Energy of system depends only on its current condition DOES NOT depend on: How system got it What E for system might be sometime in future 178

179 State Functions Any property that only depends on object s current state or condition Independence from method, path or mechanism by which change occurs is important feature of all state functions Some State functions: Internal energy E = E f E i Pressure P = P f P i Temperature t = t f t i Volume V = V f V i 179

180 Defining the System System What we are interested in studying Reaction in beaker Surroundings Everything else Room in which reaction is run Boundary Separation between system and surroundings Visible Ex. Walls of beaker Invisible Ex. Line separating warm and cold fronts 180

181 Three Types of Systems Open System Open to atmosphere Gain or lose mass and energy across boundary Most reactions done in open systems Closed System Not open to atmosphere Energy can cross boundary, but mass cannot Open system Closed system 181

182 Three Types of Systems Isolated System No energy or matter can cross boundary Energy and mass are constant Ex. Thermos bottle Isolated system 18

183 Your Turn! Ex. A closed system can A.include the surroundings. B.absorb energy and mass. C.not change its temperature. D.not absorb or lose energy and mass. E.absorb or lose energy, but not mass. 183

184 Heat (q) Can t measure heat directly Heat (q) gained or lost by an object Directly proportional to temperature change (t) it undergoes. t =tf -ti Adding heat, increases temperature t>0, q>0 Removing heat, decreases temperature t<0, q<0 Measure changes in temperature to quantify amount of heat transferred C = heat capacity q = C t 184

185 Heat Capacity (C) Amount of heat (q) required to raise temperature of object by 1 C Heat Exchanged = Heat Capacity t q = C t Units = J/ C or J C 1 Depends on two factors 1. Sample size or amount (mass) Doubling amount doubles heat capacity. Identity of substance Water vs. iron 185

186 Learning Check: Heat Capacity Ex. A cup of water is used in an experiment. Its heat capacity is known to be 70 J/ C. How much heat will it absorb if the experimental temperature changed from 19. C to 3.5 C? q C q q t J C C J C C q = J 186

187 Learning Check: Heat Capacity Ex. If it requires J to raise the temperature of 1.00 g of water by 1.00 C, calculate the heat capacity of 1.00 g of water. C q t J C 1.00 g 1.00 C 4.18 J/ C 187

188 Your Turn! Ex. What is the heat capacity of 300 g of water if it requires 510 J to raise the temperature of the water by.00 C? A.4.18 J/ C B.418 J/ C C.837 J/ C D J/ C E J/ C C 510 J 300 g.00 C 188

189 Specific Heat (s) Amount of Heat Energy needed to raise T of 1 g substance by 1 C Units C = s m or J/(g C) or J g 1 C 1 Unique to each substance s m Large specific heat means substance releases large amount of heat as it cools C 189

190 Learning Check Ex. Calculate the specific heat of water if it the heat capacity of 100 g of water is 418 J/ C. s C m s 418 J/ C 100. g 4.18 J/(g C) What is the specific heat of water if heat capacity of 1.00 g of water is 4.18 J/ C? s 4.18 J/ C 1.00 g 4.18 J/(g C) Thus, heat capacity is independent of amount 190

191 Your Turn! Ex. The specific heat of silver 0.35 J g 1 C 1. What is the heat capacity of a 100. g sample of silver? A.0.35 J/ C B..35 J/ C C.3.5 J/ C D.35 J/ C E J/ C C C s m 0.35 g J C 100. g 191

192 19

193 Using Specific Heat Heat Exchanged = (Specific Heat mass) t q = s m t Units = J/(g C) g C = J Substances with high specific heats resist T changes Makes it difficult to change temperature widely Water has unusually high specific heat 193

194 Learning Check: Specific Heat Ex. Calculate the specific heat of a metal if it takes 35 J to raise the temperature of a 3.91 g sample by.53 C. q m s t s q m t 35J 3.91g.53 C s.8 g J C 194

195 Ex. 1 Using Specific Heat Ex. If a 38.6 g of gold absorbs 97 J of heat, what will the final temperature if the initial temperature is 4.5 C? The specific heat of gold is 0.19 J g 1 C 1. Need to find t final t = t f t i First use q = s m t to calculate t q t s m 0.19 J 59.6 C = t f 4.5 C t f = 59.6 C C g 97 J 1 C 1 = 84.1 C = 59.6 C 38.6 g 195

196 Exothermic Reaction Reaction where products have less chemical energy than reactants Ex. Some chemical energy converted to kinetic energy Reaction releases heat to surroundings Heat leaves the system; q negative ( ) Reaction gets warmer (T) CH 4 (g) + O (g) CO (g) + H O(g) + heat 196

197 Endothermic Reaction Reaction where products have more chemical energy than reactants Some kinetic energy converted to chemical energy Reaction absorbs heat from surroundings Heat added to system; q positive (+) Reaction becomes colder (T ) Ex. Photosynthesis 6CO (g) + 6H O(g) + solar energy C 6 H 1 O 6 (s) + 6O (g) 197

198 Work = P V Work Convention P = opposing pressure against which piston pushes V = change in volume of gas during expansion V = V final V initial For Expansion Since V final > V initial V must be positive So expansion work is negative V >0 Work done by system on surrounding, W<0 198

199 Your Turn! Ex. Calculate the work associated with the expansion of a gas from 15.0 L to L at a constant pressure of 17.0 atm. A.69 L atm B. 69 L atm C. 315 L atm D.171 L atm E.315 L atm Work = P V V = L 15.0 L w = 17.0 atm 37.0 L 199

200 First Law of Thermodynamics Energy of system may be transferred as heat or work, but not lost or gained. If we monitor heat transfers (q) of all materials involved and all work processes, can predict that their sum will be zero Some materials gain (have +) energy Others lose (have ) energy By monitoring surroundings, we can predict what is happening to system 00

201 Two Methods of Energy Exchange Between System and Surroundings Heat q Work w E = q + w Conventions of heat and work q + Heat absorbed by system E system q Heat released by system E system w + Work done on system E system w Work done by system E system 01

202 Heat and Work Two ways system can exchange internal energy with surroundings 1. Heat Heat absorbed, System s q Heat lost, System s q. Work Is exchanged when pushing force moves something through distance Ex. Compression of system s gas W>0 expansion of system s gas W<0 0

203 Your Turn! Ex. A gas releases 3.0 J of heat and then performs 1. J of work. What is the change in internal energy of the gas? A. 15. J B.15. J C. 9. J D.9. J E.3.0 J E = q + w E = 3.0 J + ( 1. J) 03

204 Heat at Constant Pressure (q P ) Chemists usually do NOT run reactions at constant V Usually do reactions in open containers Open to atmosphere; constant P Heat of reaction at constant Pressure (q P ) E = q P + w = q P - PV q P = E + PV H = state function At constant Pressure: H = q P H = E + PV 04

205 Enthalpy Change (H) H = state function H = H final H initial H = H products H reactants Significance of sign of H Endothermic reaction System absorbs energy from surroundings H >0 positive Exothermic reaction System loses energy to surroundings H<0 negative 05

206 Enthalpy vs. Internal Energy H = E + PV Rearranging gives H E = PV Difference between H and E is PV Reactions where form or consume gases PV can be large Reactions involving only liquids and solids V negligible V 0 So H E 06

207 Enthalpy Changes in Chemical Focus on systems Reactions Endothermic Reactants + heat products Exothermic Reactants products + heat 07

208 H in Chemical Reactions Standard Conditions for H 5 C and 1 atm Standard Heat of Reaction (H ) Enthalpy change for reaction at 1 atm and 5 C Ex. N (g) + 3H (g) NH 3 (g) mol mol.000 mol When N and H react to form NH 3 at 5 C and 1 atm 9.38 kj released H = 9.38 kj 08

209 Thermochemical Equation Write H immediately after equation N (g) + 3H (g) NH 3 (g) H = 9.38 kj Must give physical states of products and reactants H rxn different for different states CH 4 (g) + O (g) CO (g) + H O(l) H = kj CH 4 (g) + O (g) CO (g) + H O(g) H = 80.3 kj Difference = energy to vaporize water 09

210 Thermochemical Equation Write H immediately after equation N (g) + 3H (g) NH 3 (g) H = 9.38 kj coefficients = # moles 9.38 kj released moles of NH 3 formed If 10 mole of NH 3 formed 5N (g) + 15H (g) 10NH 3 (g) H = kj H rxn = ( kj) = kj Can have fractional coefficients ½N (g) + 3 /H (g) NH 3 (g) H = kj 10

211 State Matters! C 3 H 8 (g) + 5O (g) 3 CO (g) + 4 H O(g) ΔH = 043 kj C 3 H 8 (g) + 5O (g) 3 CO (g) + 4 H O(l) ΔH = 19 kj Note: there is difference in energy because states do not match If H O(l) H O(g) ΔH = 44 kj/mol 4H O(l) 4H O(g) ΔH = 176 kj/mol 4H O(l) 4 H O(l) ΔH= 19 kj+176 kj= 043 kj 11

212 Learning Check: Ex. Consider the following reaction: C H (g) + 5O (g) 4CO (g) + H O(g) ΔH = 511 kj How many kj are released for 1 mol C H? mol C H 511 kj 1 mol C H? kj 511kJ mol C H 1mol C H 1,56 kj 1

213 Learning Check: EX. Consider the reaction 6CO (g) + 6H O C 6 H 1 O 6 (s) + 6O (g) ΔH = 816 kj A) how many kj are required for 44 g CO (MM = g/mol)? 1 mol CO 816kJ 44 g CO 470 kj 44.01g CO 6 mol CO B) If 100. kj are provided, what mass of CO can be converted to glucose? 100kJ 6mol CO 816kJ 44.0gCO 1mol CO 9.38 g 13

214 Your Turn! Ex. Based on the reaction CH 4 (g) + 4Cl (g) CCl 4 (g) + 4HCl(g) H = 434 kj/mol CH 4 What energy change occurs when 1. moles of methane reacts? A kj B kj C kj D kj E kj H = 434 kj/mol 1. mol H = 50.8 kj 14

215 Running Thermochemical Equations in Reverse Consider the reaction CH 4 (g) + O (g) CO (g) + H O(g) Reverse thermochemical equation Must change sign of H H = 80.3 kj CO (g) + H O(g) CH 4 (g) + O (g) H = kj 15

216 Hess s Law Multiple Paths; Same H Path a: Single step C(s) + O (g) CO (g) Path b: Two step Step 1: C(s) + ½O (g) CO(g) H = kj H = kj Step : CO(g) + ½O (g) CO (g) H = 83.0 kj Net Rxn: C(s) + O (g) CO (g) H = kj Chemically and thermochemically, identical results 16

217 Ex. Multiple Paths; Same H Path a: N (g) + O (g) NO (g) Path b: Step 1: N (g) + O (g) NO(g) Step : NO(g) + O (g) NO (g) Net rxn: N (g) + O (g) NO (g) Hess s Law of Heat Summation H = 68 kj H = 180. kj H = 11 kj H = 68 kj For any reaction that can be written into steps, value of H for reactions = sum of H values of each individual step 17

218 Enthalpy Diagrams 18

219 Enthalpy Diagrams 98 kj Ex. H O (l) H O(l) + ½O (g) 86kJ = 188kJ + H rxn H rxn = 86 kj ( 188 kj ) H rxn = 98 kj 19

220 Rules for Manipulating Thermochemical Equations 1. When equation is reversed, sign of H must also be reversed (- H ).. If all coefficients of equation are multiplied or divided by same factor, value of H must likewise be multiplied or divided by that factor 3. Formulas canceled from both sides of equation must be for substance in same physical states 0

221 Ex. Calculate H for C graphite (s) C diamond (s) Given C gr (s) + O (g) CO (g) H = 394 kj 1[ ] C dia (s) + O (g) CO (g) H = 396 kj To get desired equation, must reverse nd equation and add resulting equations C gr (s) + O (g) CO (g) CO (g) C dia (s) + O (g) H = 394 kj H = ( 396 kj) C gr (s) + O (g) + CO (g) C dia (s) + O (g) + CO (g) H = 394 kj kj = + kj 1

222 Learning Check Ex. Calculate H for C gr (s) + H (g) C H (g) Given the following: a. C H (g) + 5 / O (g) CO (g) + H O(l) H = kj b. C gr (s) + O (g) CO (g) H = kj c. H (g) + ½O (g) H O(l) H = 85.8 kj

223 Ex. Calculate for C gr (s) + H (g) C H (g) a CO (g) + H O(l) C H (g) + 5 / O (g) H = ( kj) = kj +b C gr (s) + O (g) CO (g) H =( kj) = kj +c H (g) + ½O (g) H O(l) H = 85.8 kj CO (g) + H O(l) + C gr (s) + O (g) + H (g) + ½O (g) C H (g) + 5 / O (g) + CO (g) + H O(l) C gr (s) + H (g) C H (g) H = +6.8 kj 3

224 Your Turn! Ex. Given the following data: C H (g) + O (g) CO (g) + H O(l) H = kj C(s) + O (g) CO (g) H = 394 kj H (g) + O (g) H O(l) H = 86 kj Calculate for the reaction C(s) + H (g) C H (g) A.6 kj B kj C. 60 kj H = kj + ( 394 kj) + ( 86 kj) D. 6 kj E.60 kj 4

225 Tabulating H values Standard Enthalpy of Formation, H f Amount of heat absorbed or evolved when one mole of substance is formed at 1 atm and 5 C (98 K) from elements in their standard states Standard Heat of Formation H f 5

226 Standard State Most stable form and physical state of element at 1 atm and 5 C (98 K) element O C H Al Ne Standard state O (g) C gr (s) H (g) Al(s) Ne(g) Note: All H f of elements in their standard states = 0 Forming element from itself. 6

227 Uses of Standard Enthalpy (Heat) of Formation, H f 1. From definition of H f, can write balanced equations directly H f {C H 5 OH(l)} C(s, gr) + 3H (g) + ½O (g) C H 5 OH(l) H f = kj/mol H f {Fe O 3 (s)} Fe(s) + 3 /O (g) Fe O 3 (s) H f = 8. kj/mol 7

228 Your Turn! Ex. What is the reaction that corresponds to the standard enthalpy of formation of NaHCO 3 (s), H f = kj/mol? a.na(s) + ½H (g) + 3 /O (g) + C(s, gr) NaHCO 3 (s) b.na + (g) + H + (g) + 3O (g) + C 4+ (g) NaHCO 3 (s) c.na + (aq) + H + (aq) + 3O (aq) + C 4+ (aq) NaHCO 3 (s) d.nahco 3 (s) Na(s) + ½H (g) + 3 /O (g) + C(s, gr) e.na + (aq) + HCO 3 (aq) NaHCO 3 (s) 8

229 Using H f. Way to apply Hess s Law without needing to manipulate thermochemical equations Sum of all H reaction = H f of all of the products Sum of all H f of all of the reactants Consider the reaction: aa + bb cc + dd H reaction = c H f (C) + d H f (D) {a H f (A) + b H f (B)} 9

230 Ex. Calculate H rxn Using H f Ex. Calculate H rxn using H f data for the reaction SO 3 (g) SO (g) + ½O (g) 1.Add H f for each product times its coefficient.subtract H f for each reactant times its coefficient. (SO ) 1 H H g ) H (O g )) H (SO g )) H rxn rxn f ( 97kJ/mol 1 H rxn = 99 kj/mol f ( (0 kj/mol) f 3 ( ( 396kJ/mol) 30

231 Learning Check Ex. Calculate H rxn using H f for the reaction 4NH 3 (g) + 7O (g) 4NO (g) + 6H O(l) H rxn H rxn 4H f 4H (NO H rxn = [ ] kj f H rxn = 1395 kj ( g ) (NH 4 mol(34 kj/mol) 4 mol( ) 3 ( g ) 6H ) f (H 7H kj/mol) f O (O ( ) ) ( g ) 6 mol( 85.9kJ/mol) ) 7 mol(0 kj/mol) 31

232 Check Using Hess s Law 4*[NH 3 (g) ½ N (g) + 3 / H (g)] 4*H (NH f 3, g) 7*[ O (g) O (g) ] 7*H (O f, g) 4*[ O (g) + ½ N (g) NO (g)] 4*H (NO f, g) 6*[ H (g) + ½ O (g) H O(l) ] 6* H (H f O, l) 4NH 3 (g) + 7O (g) 4NO (g) + 6H O(l) H rxn 4H f 4H (NO Same as before f ( g ) (NH ) 3 ( g ) 6H ) f (H 7H f O ( (O )) ( g ) ) 3

233 Other Calculations Ex. Given the following data, what is the value of H f (C H 3 O, aq)? Na + (aq) + C H 3 O (aq) + 3H O(l) NaC H 3 O 3H O(s) H rxn = 19.7 kj/mol H f (Na +, aq) 39.7 kj/mol H f (NaC H 3 O 3H O, s) kj/mol H (H f O, l) 85.9 kj/mol 33

234 Ex. cont. H rxn = H f (NaC H 3 O 3H O, s) H f (Na +, aq) H f (C H 3 O, aq) 3H f (H O, l) Rearranging H f (C H 3 O, aq) = H f (NaC H 3 O 3H O, s) H f (Na +, aq) H rxn 3H f (H O, l) H f (C H 3 O, aq) = 710.4kJ/mol ( 39.7kJ/mol) ( 19.7kJ/mol) 3( 85.9kJ/mol) = kj/mol 34

235 Learning Check Ex. Calculate H for this reaction using H f data. Fe(s) + 6H O(l) Fe(OH) 3 (s) + 3H (g) H f H rxn = *H f (Fe(OH) 3, s) + 3*H f (H, g) * H f (Fe, s) 6*H f (H O, l) H rxn = mol*( kj/mol) + 3*0 *0 6 mol*( 85.8 kj/mol) H rxn = 1393 kj kj H rxn = 31.8 kj 35

236 Learning Check Ex. Calculate H for this reaction using H f data. CO (g) + H O(l) O (g) + CH 4 (g) H f H rxn = *H f (O, g) + H f (CH 4, g) H f (CO, g) * H f (H O, l) H rxn = mol ( 74.8 kj/mol) 1 mol ( kj/mol) mol ( 85.8 kj/mol) H rxn = 74.8 kj kj kj H rxn = kj 36

237 Converting Between E and H For Chemical Reactions When reaction occurs V caused by n of gas Not all reactants and products are gases So redefine as n gas Where n gas = (n gas ) products (n gas ) reactants Substituting into or H H = E + PV E n gas RT 37

238 Ex. Find E for the following reaction at 5 C using data in Table 7.? N O 5 (g) 4 NO (g) + O (g) Step 1: Calculate H using H f data (Table 7.) Recall H H H = (4 mol)(33.8 kj/mol) + (1 mol)(0.0 kj/mol) ( mol)(11 kj/mol) H = 113 kj ( H f ) products ( Hf ) reactants 4Hf (NO ) Hf (O ) Hf (NO5 ) 38

239 Step : Calculate Ex. (cont.) n gas = (n gas ) products (n gas ) reactants n gas = (4 + 1 ) mol = 3 mol Step 3: Calculate E using R = J/K mol E H n RT gas T = 98 K E = 113 kj (3 mol)( J/K mol)(98 K)(1 kj/1000 J) E = 113 kj 7.43 kj = 106 kj 39

240 Learning Check Ex. Consider the following reaction for picric acid: 8O (g) + C 6 H (NO ) 3 OH(l) 3 N (g) + 1CO (g) + 6H O(l) What type of reaction is it? Calculate ΔΗ, ΔΕ 8O (g) + C 6 H (NO ) 3 OH(l) 3N (g) + 1CO (g) + 6H O(l) ΔΗ f (kj/mol) ΔH 0 = 1mol( kj/mol) + 6mol( 41.83kJ/mol) + 6mol(0.00kJ/mol) 8mol(0.00kJ/mol) mol(386.94kj/mol) ΔH 0 = 13,898.9 kj (Exothermic reaction) ΔΕ = ΔH Δn gas RT = ΔH (15 8)mol*98* ΔΕ = 13,898.9 kj 9.0 kj = 13,97.9 kj 40

241 Chapter 4 Chemical Kinetics Chemistry: The Molecular Nature of Matter, 6E

242 Speeds at Which Reactions Occur Kinetics: Study of factors that govern How rapidly reactions occur and How reactants change into products Rate of Reaction: Speed with which reaction occurs How quickly reactants disappear and products form 4

243 Factors that Affect Reaction Rates 1. Chemical nature of reactants What elements, compounds, salts are involved? What bonds must be formed, broken? What are fundamental differences in chemical reactivity? 43

244 Factors that Affect Reaction Rates. Ability of reactants to come in contact If two or more reactants must meet in order to react Gas or solution phase facilitates this Reactants mix and collide with each other easily Homogeneous reaction All reactants in same phase Occurs rapidly Heterogeneous reaction Reactants in different phases Reactants meet only at interface between phases Surface area determines reaction rate area, rate area, rate 44

245 Factors that Affect Reaction Rates 3. Concentrations of reactants Rates of both homogeneous and heterogeneous reactions affected by [X] Collision rate between A and B if we [A] or [B]. Often (but not always) Reaction rate as [X] 45

246 Factors that Affect Reaction Rates 4. Temperature Rates are often very sensitive to T Cooking sugar Raising T usually makes reaction faster for two reasons: a. Faster molecules collide more often and collisions have more Energy b. Most reactions, even exothermic reactions, require Energy to get going 46

247 Factors that Affect Reaction Rates 5. Presence of Catalysts Catalysts Substances that rates of chemical and biochemical reactions without being used up Rate-accelerating agents Speed up rate dramatically 47

248 Measuring Rate of Reaction Rate of Chemical Reaction in [X] of particular species per unit time. reaction rate Always with respect to (WRT) given reactant or product [reactants] w/ time [products] w/ time [ reactant ] time 48

249 Rate of Reaction with Respect to Rate WRT X Concentration in M Time in s Units on rate: Ex. Given Species X [product] by 0.50 mol/l per second rate = 0.50 M/s [reactant] by 0.0 mol/l per second rate = 0.0 M/s [X] mol/l s t [X] t t1 t 1 mol L s M s [X] t WRT: with respect to 49

250 Rate of Reaction Always + Whether something is or in [X]. Reactants Need sign to make rate + Reactant consumed So [X] = Products Produced as reaction goes along So [X] = + Thus Rate = + Rate Rate [reactant] t [product] t 50

251 Rates and Coefficients Relative rates at which reactants are consumed and products are formed Related by coefficients in balanced chemical equation. Know rate with respect to one product or reactant Can use equation to determine rates WRT all other products and reactants. Ex. C 3 H 8 (g) + 5 O (g) 3 CO (g) + 4 H O (g) Rate of Reaction [ C3H8 ] 1 [O ] 1 [CO ] 1 [HO] t 5 t 3 t 4 t 51

252 Rates and Coefficients C 3 H 8 (g) + 5 O (g) 3 CO (g) + 4 H O (g) O reacts 5 times as fast as C 3 H 8 Rate [O t ] 5 [C CO forms 3 times faster than C 3 H 8 consumed Rate t t H O forms 4/5 as fast as O consumed [CO ] 3 H [C 3H 3 t 8 8 ] ] [ HO] 4 [O t 5 t ] 5

253 Rates and Coefficients In general aa + B C + D t D 1 t C 1 t B 1 t A 1 Rate a 53

254 Your Turn! Ex. In the reaction CO (g) + O (g) CO (g), the rate of the reaction of CO is measured to be.0 M/s. What would be the rate of the reaction of O? A. the same =.0 M/s B. twice as great = 4.0 M/s C. half as large = 1.0 M/s D. you cannot tell from the given information 54

255 Change of Reaction Rate with Time Generally reaction rate changes during reaction i.e. Not constant Often initially fast when lots of reactant present Slower at end when reactant used up Why? Rate depends on [reactants] Reactants being used up, so [reactant] is [A] vs. time is curve A is reactant [A] is w/ time 55

256 Measuring Rates Measured in three ways: Instantaneous rate Average rate Initial rate 56

257 [ N O Instantaneous Reaction Rates Instantaneous rate Slope of tangent to curve at any specific time Initial rate Determined at initial time N O a p p e a r a n c e T i m e ( s ) 57

258 [NO] Average Rate of Reaction Slope of line connecting starting and ending coordinates for specified time frame Δ[Product] Δtime rate NO appe arance Time (s) 58

259 Table 14.1 Data at 508 C Ex. HI(g) H (g) + I (g) [HI] (mol/l) Time (s) Initial rate rate between first two data points rate ( (50 ( 0.084M) 50s )M 0)s M / s 59

260 [HI] (mol/l) Time (s) Rate at 300 s HI(g) H (g) + I (g) Rate = tangent of curve at 300 s Rate (0.065 ( M 50s )M 300)s M/ s 60

261 Your Turn! Ex. The concentration of NO was found to be 0.058M at 5 minutes and at 10 minutes the concentration was M. What is the average rate of the reaction between 5 min and 10 min? A. 310 M/min B. 3. x 10-3 M/min C..7 x 10-3 M/min D. 7.1 x 10-3 M/min 0.058M M 3 10 min 5min 3. x 10 M / min 61

262 Concentration and Rate Rate Laws aa + B C + D Homogeneous reaction Rate = k[a] m [B] n Rate Law or Rate expression m and n = exponents found experimentally No necessary connection between stoichiometric coefficients (a, ) and rate exponents (m, n) Usually small integers 1,. Sometimes simple fractions (½, ¾) or zero 6

263 Rate Laws Rate = k[a] m [B] n Exponents tell Order of Reaction with respect to (WRT) each reactant Order of Reaction m = 1 [A] 1 1 st order m = [A] nd order m = 3 [A] 3 3 rd order m = 0 [A] 0 0 th order [A] 0 = 1 means A doesn't affect rate Overall order of reaction sum of orders (m and n) of each reactant in rate law 63

264 Learning Check Ex. The rate law for the reaction A +B 3C is rate= k[a][b] If the concentration of A is 0.M and that of B is 0.3M, and rate constant is M -1 s -1 what will be the reaction rate? rate=0.045 M -1 s -1 [0.][0.3] rate=0.007 M/s M/s 64

265 Ex. 5 Br + BrO 3 + 6H + 3Br + 3H O 3 [ y ] z ] [H BrO ] x k[bro3 ] [Br t x = 1 y = 1 z = 1 st order WRT BrO 3 1 st order WRT Br nd order WRT H + Overall order = = 4 rate k[bro ] [Br ] [H ] 65

266 Ex. Sometimes n and m are coincidentally the same as stoichiometric coefficients HI (g) H (g) + I (g) rate [HI] t nd order WRT HI nd order overall k[hi] 66

267 Your Turn! Ex. The following rate law has been observed: Rate = k[h SeO][I - ] 3 [H + ]. The rate with respect to I - and the overall reaction rate is: A. 6, B., 3 C. 1, 6 D. 3, 6 67

268 Calculating k from Rate Law If we know rate and concentrations, can use rate law to calculate k Ex. at 508 C Rate=.5 x 10 4 M/s [HI] = M rate [HI] t k[hi] k rate [HI] (0.0558M) M / s M 1 s 1 68

269 How To Determine Exponents in Rate Law Experiments Method of Initial Rates If reaction is sufficiently slow or have very fast technique Can measure [A] vs. time at very beginning of reaction before it curves up very much, then [A] 1 [A] 0 initial rate t1 t0 Set up series of experiments, where initial concentrations vary 69

270 Ex. Method of Initial Rates 3A + B products Rate = k[a] m [B] n Expt. # [A] 0, M [B] 0, M Initial Rate, M/s Convenient to set up experiments so [X] of one species is doubled or tripled while [X] of all other species are held constant Tells us effect of [varied species] on initial rate 70

271 Reaction Order and Rate If reaction is 1 st order WRT given species X, Doubling [X] 1 1 Rate doubles If reaction is nd order WRT X, Doubling [X] Rate quadruples If reaction is 0 th order WRT X, Doubling [X] 0 0 Rate doesn't change If reaction is n th order WRT X Doubling [X] n n 71

272 Back to our Example Expt. # [A] 0, M [B] 0, M Initial Rate, M/s Comparing 1 and 4 Doubling [A] Rate Quadruples rate Rate Reaction nd order in A [A] 4 m A B m A B m m n n m k 0.0 m n n m 1 1 k 0.10 Rate k 4 Rate 1 k m = 4 or m = 7

273 Back to our Example Expt. # [A] 0, M [B] 0, M Initial Rate, M/s Comparing and 3 Doubling [B] Rate Rate does not change Rate Reaction 0 th order in B [B] A B 3 m A B m m m n n n 3 k 0.0 n n n n k 0.10 Rate 3 k 1 Rate k n = 1 or n = 0 73

274 Ex. Method of Initial Rates Expt. # [A] 0, M [B] 0, M Initial Rate, M/s Conclusion: rate = k[a] Can use data from any experiment to determine k Let s choose experiment 1 k rate A 0.10M M/s M 1 s 1 74

275 Ex. Method of Initial Rates SO + O SO 3 Rate = k[so ] m [O ] n Expt # [SO ], M [O ], M Initial Rate of SO 3 formation, M s

276 Ex. Compare 1 and [SO ] doubles, [O ] constant, Rate quadruples, 4 Rate Rate 1 Rate Rate m m k k 3 4 SO SO m m m 1 O O n n 1 k k m m 0. 5 m = 4 or m = n n 76

277 Ex. Compare and 4 3 [O ] triples, [SO ] constant Rate triples, 3 1 Rate 4 Rate Rate 4 Rate n n k k SO SO 3 n m 4 m O O n 4 n k k m m n = 3 or n = 1 n n 77

278 Ex. Rate = k[so ] [O ] 1 1 st order WRT O nd order WRT SO 3 rd order overall Can use any experiment to find k k [SO rate ] [O ] M / s M s ( 0. 50M ) ( 0. 90M ) 78

279 Ex. Method of Initial Rates BrO Br + 6H + 3Br + 3H O Rate Expt # [BrO [BrO 3 ], mol/l t 3 ] [Br ], mol/l k [BrO 3 ] [H + ], mol/l m [Br ] n [H ] p Initial Rate, mol/(l s)

280 Ex. Compare 1 and Rate Rate M / M / s s k( 0. 0M ) k( 0. 10M ) m m ( 0. 10M ) ( 0. 10M ) n n ( 0. 10M ) ( 0. 10M ) p p m 0. 0M m. 0 (. 0) m 0. 10M 1 Compare and 3 Rate 3 Rate M / M / n s s k( 0. 0M ) k( 0. 0M ) 0. 0M n. 0 (. 0) n 0. 10M m m ( 0. 0M ) ( 0. 10M ) 1 n n ( 0. 10M ) ( 0. 10M ) p p 80

281 Ex. Compare 1 and 4 Rate 4 Rate M / M / s s k( 0. 10M ) k( 0. 10M ) m m ( 0. 10M ) ( 0. 10M ) n n ( 0. 0M ) ( 0. 10M ) p p p 0. 0M p 4. 0 (. 0) p 0. 10M First order in [BrO 3 ] and [Br ] Second order in [H + ] Overall order = m + n + p = = 4 Rate Law is: Rate = k[bro 3 ][Br ][H + ] 81

282 Your Turn! Ex. Using the following experimental data, determine the order with respect to NO and O. Exp [NO] [O ] Initial Rate of NO t #, M, M formation, M s A., 0 B. 3,1 C., 1 D. 1, 1 8

283 Your Turn! - Solution x 3 1 R x 10 M s M M 3 1 x R1 1.5 x 10 M s 0.1M 0.5M x 1 R3 5. x 10 M s 0.50M 0.50M 3 1 R1 1.5 x 10 M s 0.1M 0.5M y 1 y y y y 83

284 Concentration and Time Rate law tells us how speed of reaction varies with [X]'s. Sometimes want to know [reactants] and [products] at given time during reaction How long for [reactants] to drop below some minimum optimal value Need dependence of Rate on Time 84

285 Concentration vs. Time for 1 st Order [ A] Rate t Reactions k[ A] Corresponding to reactions A products Integrating we get [ A] ln 0 [ A] t kt Rearranging gives ln[ A] t kt ln[ A ] 0 Equation of line y = mx + b 85

286 Plot ln[a] t (y axis) vs. t (x axis) ln[ A] t kt ln[ A ] 0 Slope = k Yields straight line Indicative of 1 st order kinetics slope = k intercept = ln[a] 0 If we don't know already 86

287 First Order Kinetics Graph Plot of [A] vs. time gives an exponential decay [ A] t [ A] o e kt 87

288 Half-lives for 1 st Order Reactions Half-life = t ½ First Order Reactions Set 1 [ A] t [ A] 0 Substituting into Gives ln [ A] 0 [ A] 1 0 [ A] ln 0 [ A] kt 1 t kt Canceling gives ln = kt ½ ln t 1 Rearranging gives k k 1 88

289 Half-life for 1 st Order Reactions Observe: 1. t ½ is independent of [A] o For given reaction (and T) Takes same time for concentration to fall from M to 1 M as from M to M. k 1 has units (time) -1, so t ½ has units (time) t ½ called half-life Time for ½ of sample to decay 89

290 Half-life for 1 st Order Reactions Does this mean that all of sample is gone in two half-lives ( x t ½ )? No! In 1 st t ½, it goes to ½[A] o In nd t ½, it goes to ½(½[A] o ) = ¼[A] o In 3 rd t ½, it goes to ½(¼[A] o ) = ⅛[A] o In n th t ½, it goes to [A] o / n Existence of [X] independent half-life is property of exponential function Property of 1 st order kinetics 90

291 Half-Life Graph 91

292 Ex. Using Half-Life 131 I is used as a metabolic tracer in hospitals. It has a half-life, t ½ = 8.07 days. How long before the activity falls to 1% of the initial value? N N o e kt ln N No 1 kt N ln N ln t o t ln 1 (8.07 days)ln ln days 9

293 Learning Check Ex. The radioactive decay of a new atom occurs so that after 1 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s -1? ln 0 A A 100 ln( ) 33 kt k (1da) k = da -1 k = s -1 93

294 Learning Check Ex. The half-life of I-13 is.95h. What percentage remains after 4 hours? ln() k t 1 k ln.95h 0.30 h 1 = k ln A o A kt ln A o A 0.30h 1 4h 7.48 A / A o =.0711 % 94

295 Your Turn! Ex. Which order has a half-life that is independent of the original amount? A. Zero B. First C. Second D. None depend on the original quantity 95

296 Theories about Reaction Rates Reaction rate depends on [reactants] and T Collision Theory Based on Kinetic Molecular Theory Accounts for both effects on molecular level Central Idea Molecules must collide to react Greater number of collision/sec = greater reaction rate 96

297 Theories about Reaction Rates Collision Theory As [reactants] number of Collisions Reaction rate As T Molecular speed Molecules collide with more force (energy) Reaction rate 97

298 Collision Theory Rate of reaction proportional to number of effective collisions/sec among reactant molecules Effective collision that gives rise to product 1. Molecular Orientation Molecules must be oriented in a certain way during collisions for reaction to occur Ex. NO Cl + Cl NO + Cl Cl must come in pointing directly at another Cl atom for Cl to form 98

299 Molecular Orientation Wrong Orientation Correct Orientation 99

300 . Temperature Over moderate T range, E a unchanged As T, More molecules have E a So more molecules undergo reaction Reaction rate as T 300

301 3. Activation Energy, E a Molecules must possess certain amount of kinetic energy (KE) in order to react Activation Energy, E a Minimum KE needed for reaction to occur Transition State Theory Used to explain details of reactions What happens when reactant molecules collide Potential Energy Diagram To visualize what actually happens during successful collision Relationship between E a and developing Total PE 301

302 Potential Energy Potential Energy Diagram Activation energy (E a ) = hill or barrier between reactants and products heat of reaction (H) = difference in PE between products and reactants H reaction = H products H reactants Products Reaction Coordinate (progress of reaction) 30

303 Exothermic Reaction H reaction < 0 ( ) Exothermic reaction Products lower PE than reactants Exothermic Reaction H = Products E a could be high and reaction slow even if H rxn large and negative E a could be low and reaction rapid 303

304 H reaction > 0 (+) Endothermic Reaction Endothermic Reaction H = + H reaction = H products H reactants 304

305 Ex. NO Cl + Cl NO + Cl As NO Cl and Cl come together Start to form Cl Cl bond Start to break N Cl bond Requires E, as must bring things together In TS N Cl bond ½ broken Cl Cl bond ½ formed After TS Cl Cl bond forms N Cl breaks Releases E as products more stable 305

306 Potential Energy Your Turn! Ex. Examine the potential energy diagram. Which is the Slowest (Rate Determining) Step? A. Step 1 B. Step Has greatest E a C. Can t tell from the given information 1 Reaction Progress 306

307 Measuring Activation Energy Arrhenius Equation Equation expressing T dependence of k k Ae E a RT A = Frequency factor has same units as k R = gas constant in energy units = J mol1 K1 E a = Activation Energy has units of J/mol T = Temperature in K / 307

308 How To Calculate Activation Energy Method 1. Graphically Take natural logarithm of both sides Rearranging Equation for a line E k A a 1 ln ln * R T y = b + m x Arrhenius Plot Plot ln k (y axis) vs. 1/T (x axis) 308

309 Method. van't Hoff Equation van't Hoff Eq uation k E 1 1 ln a k1 R T T 1 309

310 Using van't Hoff Equation Ex. CH 4 + S CS + H S ln E a k (L/mol s) T ( C) T (K) = k = T = k = T a JE / K mol 898K 83K 8.314J / K 1 898K mol 1 83 ln J / mol 310

311 Learning Check Ex. Given that k at 5 C is M/s and that at 50 C it is M/s, what is the activation energy for the reaction? ln( k k 1 ) E R a 1 T 1 T ln( M/s ) M/s Ea 8.314J/(mo l K) 1 33K 1 98K E a = 08 J/mol 311

312 Your Turn! Ex. A reaction has an activation energy of 40 kj/mol. What happens to the rate if you increase the temperature from 70 o C to 80 0 C? A. Rate increases approximately 1.5 times B. Rate increases approximately 5000 times C. Rate does not increase D. Rate increases approximately 3 times 31

313 Your Turn! - Solution Rate is proportional to the rate constant k k J J x(8073)k mol K e J J x(7073)k mol K e

314 Homogeneous Catalyst Same phase as reactants Consider : S (g) + O (g) + H O (g) H SO 4 (g) S (g) + O (g) SO (g) NO (g) + SO (g) NO (g) + SO 3 (g) Catalytic pathway SO 3 (g) + H O (g) H SO 4 (g) NO (g) + ½O (g) NO (g) Regeneration of catalyst Net: S (g) + O (g) + H O (g) H SO 4 (g) What is Catalyst? NO (g) Reactant (used up) in early step Product (regenerated) in later step Which are Intermediates? NO and SO 314

315 Heterogeneous Catalyst Exists in separate phase from reactants Ex. 3 H (g) + N (g) NH 3 (g) H & N approach Fe catalyst H & N bind to Fe & bonds break N H bonds forming N H bonds forming NH 3 formation complete NH 3 dissociates 315

316 Chapter 5 Properties of Solutions Chemistry: The Molecular Nature of Matter, 6E

317 Solution Solutions Homogeneous mixture Composed of solvent and solute(s) Solvent More abundant component of mixture Solute(s) Less abundant or other component(s) of mixture Ex. Lactated Ringer s solution NaCl, KCl, CaCl, NaC 3 H 5 O 3 in water 317

318 Rule of Thumb Like dissolves Like Use polar solvent for polar solute Use Nonpolar solvent for nonpolar solute Polar solutes interact with and dissolve in polar solvents Ex. Ethanol in water Both are polar molecules Both form hydrogen bonds H 3 C CH H O O H H 318

319 miscible Solution Benzene in CCl 4 CCl 4 Nonpolar Benzene, C 6 H 6 Nonpolar Similar in strength to CCl 4 Does dissolve, solution forms Immiscible Solution Solvent and solute are very different No solution forms layers, Don t Mix Benzene in water 319

320 Learning Check Ex. Which of the following are miscible in water? 30

321 Your Turn! Ex. Which of the following molecules is soluble in C 6 H 6? A. NH 3 B. CH 3 NH C. CH 3 OH D. CH 3 CH 3 E. CH 3 Cl 31

322 Solubility Mass of solute that forms saturated solution with given mass of solvent at specified temperature If extra solute added to saturated solution, extra solute will remain as separate phase solubility g solute 100g solvent 3

323 Effect of T on solid Solubility in Liquids Most substances become more soluble as T Amount solubility Varies considerably Depends on substance 33

324 Effect of T on Gas Solubility in Liquids Solubility of gases usually as T Table 13. Solubilities of Common Gases in Water 34

325 Effect of Pressure on Gas Solubility Solubility as P Why? P means V above solution for gas Gas goes into solution Relieves stress on system Conversely, solubility as P Soda in can Gases are more soluble at low temperature and high pressure. 35

326 Effect of Pressure on Gas Solubility A. At some P, equilibrium exists between vapor phase and solution rate in = rate out B. in P puts stress on equilibrium frequency of collisions so rate in > rate out More gas molecules dissolve than are leaving solution C. More gas dissolved Rate out will until Rate out = Rate in and equilibrium restored 36

327 Henry s Law Pressure-Solubility Law Concentration of gas in liquid at any given temperature is directly proportional to partial pressure of gas over solution C gas = k H P gas (T is constant) C gas = concentration of gas P gas = partial pressure of gas k H = Henry's Law constant Unique to each gas Tabulated 37

328 Henry s Law True only at low concentrations and pressures where gases do NOT react with solvent Alternate form C1 P C 1 and P 1 refer to an initial set of conditions C and P refer to a final set of conditions 1 C P 38

329 Ex. Using Henry s Law Calculate the concentration of CO in a soft drink that is bottled with a partial pressure of CO of 5 atm over the liquid at 5 C. The Henry s Law constant for CO in water at this temperature is mol/l atm. C k CO H (CO ) P CO = mol/l atm * 5.0 atm = mol/l 0.16 mol/l When under 5.0 atm pressure 39

330 Ex. Using Henry s Law Calculate the concentration of CO in a soft drink after the bottle is opened and equilibrates at 5 C under a partial pressure of CO of atm. C1 C P C1 C P1 P P1 C mol/L atm 5.0atm C = mol/l When open to air 330

331 Learning Check Ex. What is the concentration of dissolved nitrogen in a solution that is saturated in N at.0 atm? k H = (M / atm) C g =k H P g C g = (M / atm).0 atm C g = M 331

332 Your Turn! Ex. How many grams of oxygen gas at 1.0 atm will dissolve in 10.0 L of water at 5 o C if Henry s constant is 1.3 x 10-3 M atm -1 at this temperature? A. 0.4 g B g C g D. 0.1 g E..4 g 33

333 Colligative Properties Physical properties of solutions Depend mostly on relative populations of particles in mixtures Don t depend on their chemical identities Effects of Solute on Vapor Pressure of Solvents Solutes that can t evaporate from solution are called nonvolatile solutes Fact: All solutions of nonvolatile solutes have lower vapor pressures than their pure solvents 333

334 Raoult's Law Vapor pressure of solution, P soln, equals product of mole fraction of solvent, X solvent, and its vapor pressure when pure, P solvent Applies for dilute solutions P solution X solv ent P solv ent P X P solution solv ent solv ent vapor pressure of the solution mole fraction of the solvent vapor pressure of pure solvent 334

335 Alternate form of Raoult s Law Plot of P soln vs. X solvent should be linear Slope = P solv ent Intercept = 0 Change in vapor pressure can be expressed as P changein P Usually more interested in how solute s mole fraction changes the vapor pressure of solvent P ( P solvent P X solute P solvent solution ) 335

336 Learning Check Ex. The vapor pressure of -methylhexane is torr at 15 C. What would be the pressure of the mixture of 78.0 g -methylhexane and 15 g naphthalene, which is nearly non-volatile at this temperature? P solution = X solvent P o solvent naphthalene C 10 H 8 MM methylhexane C 7 H 16 MM 100. X mole mole - P naphthalene 78.0 g methylhexane 100. g/mol mol ne mol mol g g/mol methy lhexa torr = 33.0 torr = 33 torr mol mol

337 Solutions That Contain Two or More Volatile Components Now vapor contains molecules of both components Partial pressure of each component A and B is given by Raoult s Law P P A B X X Total pressure of solution of components A and B given by Dalton s Law of Partial Pressures A B P P A B P total P A P B X A P A X B P B 337

338 Ex. Benzene and Toluene Consider a mixture of benzene, C 6 H 6, and toluene, C 7 H 8, containing 1.0 mol benzene and.0 mol toluene. At 0 C, the vapor pressures of the pure substances are: P benzene = 75 torr P toluene = torr what is the total pressure above this solution? 338

339 Ex. Benzene and Toluene (cont.) 1. Calculate mole fractions of A and B 1.0mol X benzene 0.33benzene mol.0mol X toluene 0.67 toluene 1.0.0mol. Calculate partial pressures of A and B Pbenzene X benzene Pbenzene torr 5torr Ptoluene Xtoluene Ptoluene 0.67 torr 15torr 3. Calculate total pressure P P P total benzene toluene ( 5 15) torr 40torr 339

340 Learning Check Ex. The vapor pressure of -methylheptane is torr at 55 C. 3-ethylpentane has a vapor pressure of at the same temperature. What would be the pressure of the mixture of 78.0g -methylheptane and 15 g 3-ethylpentane? -methylheptane C 8 H 18 MM g/mol 3-ethylpentane C 7 H 16 MM 100. g/mol P solution = X A P o A + X B P o B 340

341 mole mole X X P Learning Check methylheptane ethylpentane 78.0 g mol g/mol 15 g mol 100.g/mol torr torr P = 30 torr mol (0.6883mol mol) -methy lpentane mol (0.6883mol mol) 3-ethy lpentane

342 Your Turn! Ex. If the vapor pressure of pure hexane is mmhg, and heptane is mm Hg at 5º, which equation is correct if the mixture s vapor pressure is mmhg? A. X(151.8 mmhg) = mmhg B. X(151.8 mmhg) + (X)(45.67 mm Hg) = mmhg C. X(151.8 mmhg) + (1 X)(45.67 mm Hg) = mm Hg D. None of these 34

343 Solutes also Affect Freezing and Boiling Points of Solutions Freezing Point of solution always Lower than pure solvent Boiling Point of solution always Higher than pure solvent Colligative properties Boiling Point Elevation (T b ) in boiling point of solution vs. pure solvent Freezing Point Depression (T f ) in freezing point of solution vs. pure solvent 343

344 Freezing Point Depression (T f ) where T f = K f m T f = (T fp T soln ) m = concentration in Molality K f = molal freezing point depression constant Units of C/molal, Depend on solvent. 344

345 Boiling Point Elevation (T b ) where T b = K b m T b = (T soln T bp ) m = concentration in Molality K b = molal boiling point elevation constant Units of C/m, Depend on solvent. 345

346 Values of K f and K b for solvents 346

347 Ex. Freezing Point Depression Estimate the freezing point of solution contain g ethylene glycol, C H 6 O, (MM = 6.07) and g H O (MM = 18.0). m 1mol CH6O 100.0g CH6O 6.07g CH6O mol solute 1.611mol CH6O kg solvent 0.100kg water T f = K f m = (1.86 C/m) 16.11m T f = (T fp T soln ) = 1.611mol C H 6 O = 16.11m C H 6 O = 30.0 C 30.0 C = 0.0 C T soln T soln = 0.0 C 30.0 C = 30.0 C 347

348 Your Turn! Ex. When 0.5 g of an unknown organic compound is added to 5.0 g of cyclohexane, the freezing point of cyclohexane is lowered by 1.6 o C. K f for the solvent is 0. o C m -1. Determine the molar mass of the unknown. A. 505 g/mol B. 3 g/mol C. 315 g/mol D. 16 g/mol T f K f m C m o o 1.6 C=0. x MW 16 g/mol 0.50 g MW 0.05 kg 348

349 Ex. Boiling Point Elevation A.00 g biomolecule sample was dissolved in 15.0 g of CCl 4. The boiling point of this solution was determined to be C. Calculate the molar mass of the biomolecule. For CCl 4, the K b = 5.07 C/m and BP CCl4 = C. T b T K m m m b b mol solute kg solvent K b mol solute 5.07 C / m 0.684m C g biomoleucle MM biomolecule 497g / mol mole 0.684m kg CCl 3 mol 4 349

350 Osmosis Osmotic Membrane Semipermeable membrane that lets only solvent molecules through Osmosis Net shift of solvent molecules (usually water) through an osmotic membrane Direction of flow in osmosis, Solvent flows from dilute to more concentrated side Flow of solvent molecules across osmotic membrane concentration of solute on dilute side concentration of solute on more concentrated side 350

351 Osmosis and Osmotic Pressure A. Initially, Soln B separated from pure water, A, by osmotic membrane. No osmosis occurred yet B. After a while, volume of fluid in tube higher. Osmosis has occurred. C. Need back pressure to prevent osmosis = osmotic pressure. 351

352 Equation for Osmotic Pressure Assumes dilute solutions = MRT = osmotic pressure M = molarity of solution T = Kelvin Temperature R = Ideal Gas constant = L atm mol 1 K 1 35

353 Ex. Using to determine MM A solution contained 3.50 mg of protein in sufficient H O to form 5.00 ml of solution. The measured osmotic pressure of this solution was 1.54 torr at 5 C. Calculate the molar mass of the protein. M RT 1atm 1.54torr 760torr L atm K K mol mol L mol M L M L mol 3 g g 3 MM g / mol mol mol 353

354 Learning Check Ex. For a typical blood plasma, the osmotic pressure at body temperature (37 C) is 5409 mm Hg. If the dominant solute is serum protein, what is the concentration of serum protein? 5409mm Hg MRT? mol L atm 7.117atm L mol K 1atm 760mm Hg M = 0.80 M K 354

355 Chapter 6 Chemical Equilibrium Chemistry: The Molecular Nature of Matter, 6E

356 Dynamic Equilibrium in Chemical Systems Chemical equilibrium exists when Rates of forward and reverse reactions are equal Reaction appears to stop [reactants] and [products] don't change over time Remain constant Both forward and reverse reaction never cease Equilibrium signified by double arrows ( ) or equal sign (=) 356

357 Dynamic Equilibrium N O 4 NO Initially forward reaction rapid As some reacts [N O 4 ] so rate forward Initially Reverse reaction slow No products As NO forms Reverse rate Ions collide more frequently as [ions] Eventually rate forward = rate reverse Equilibrium 357

358 Dynamic Equilibrium 358

359 Reactants Equilibrium Products N O 4 NO For given overall system composition Always reach same equilibrium concentrations Whether equilibrium is approached from forward or reverse direction 359

360 Equilibrium Simple relationship among [reactants] and [products] for any chemical system at equilibrium Called = mass action expression Derived from thermodynamics Forward reaction: A B Rate = k f [A] k f Reverse reaction: A k r B Rate = k r [B] At equilibrium: A B k f [A] = k r [B] rate forward = rate reverse rearranging: [B] [A] k k f r constant 360

361 Mass Action Expression (MAE) Uses stoichiometric coefficients as exponent for each reactant For reaction: aa + bb cc + dd Reaction quotient Q Numerical value of mass action expression Equals Q at any time, and [C] [A] Equals K only when reaction is known to be at equilibrium c a [D] [B] d b 361

362 Q Mass Action Expression [HI] [H ][I ] = same for all data sets at equilibrium Equilibrium Concentrations (M) Exp t [H ] [I ] [HI] I II III IV Q [HI] [H ][I (0.156) (0.0)(0.0) (0.80) (0.0350)(0.0450) (0.100) (0.0150)(0.0135) (0.311) (0.044)(0.044) ] Average =

363 Equilibrium Law For reaction at equilibrium write the following Equilibrium Law (at 440 C) K c [HI] [H ][I 49.5 Equilibrium constant = K c = constant at given T Use K c since usually working with concentrations in mol/l For chemical equilibrium to exist in reaction mixture, reaction quotient Q must be equal to ] equilibrium constant, K c Q = K c 363

364 Predicting Equilibrium Law For general chemical reaction: dd + ee ff + gg Where D, E, F, and G represent chemical formulas d, e, f, and g are coefficients [F] [G] Mass action expression = d e [D] [E] Note: Exponents in mass action expression are stoichiometric coefficients in balanced equation. f g [F] [G] Equilibrium law is: K c d e [D] [E] f 364 g

365 Ex. Equilibrium Law 3 H (g) + N (g) NH 3 (g) K c = 4.6 x 10 8 at 5 C What is equilibrium law? K c [H [NH ] 3 3 ] [N ]

366 Learning Check Ex. Write mass action expressions for the following: NO (g) N O 4 (g) Q [NO ] [NO 4 ] CO (g) + O (g) CO (g) Q [CO ] [CO] [O ] 366

367 Manipulating Equations for Chemical Equilibria Various operations can be performed on equilibrium expressions 1. When direction of equation is reversed, new equilibrium constant is reciprocal of original A + B C + D K c [C][D] [A][B] C +D A + B K c [A][B] [C][D] 1 K 367 c

368 Ex. Manipulating Equilibria 1 1. When direction of equation is reversed, new equilibrium constant is reciprocal of original K 3 H (g) + N (g) NH 3 (g) at 5 C NH 3 (g) 3 H (g) + N (g) at 5 C c K c [H 3 [H ] [N] [NH ] K [NH ] 3 3 ] [N c ]

369 Manipulating Equilibria. When coefficients in equation are multiplied by a factor, equilibrium constant is raised to a power equal to that factor. A + B C + D K c [C][D] 3A + 3B 3C + 3D [A][B] K 3 3 [C] [D] [C][D] [C][D] [C][D] 3 c K 3 3 c [A] [B] [A][B] [A][B] [A][B] 369

370 Manipulating Equilibria. When coefficients in equation are multiplied by factor, equilibrium constant is raised to power equal to that factor 3 H (g) + N (g) NH 3 (g) at 5 C K c [H [NH ] 3 3 multiply by 3 9 H (g) + 3 N (g) 6 NH 3 (g) K ] [N ] [NH3 ] 3 c K 9 3 c [H ] [N] 370 8

371 Manipulating Equilibria 3 3. When chemical equilibria are added, their equilibrium constants are multiplied A + B C + E A + B + E C + D F + G K c D + F + G [ C ][ D ] 1 [ A][ B ] K c [ F ][ G ] [ C ][ ] E K c [ C ][ D ] [ F ][ G ] [ D ][ F ][ G ] 3 [ A][ B ] [ C ][ E ] [ A][ B ][ E ] 371 K c 1 K c

372 Manipulating Equilibria 3 3. When chemical equilibria are added, their equilibrium constants are multiplied NO (g) NO 3 (g) + NO(g) NO 3 (g) + CO(g) NO (g) + CO(g) [NO][NO [NO ] ] 3 K c [NO K NO (g) + CO (g) [NO NO(g) + CO (g) 3 ][CO ][CO] Therefore 3 K 1 c c ] K c1 K c K c 3 [NO][CO [NO [NO][NO [NO [NO [NO [NO 37 3 ] 3 ] ][CO ][CO] [NO][CO ] ][CO] ] ][CO] ]

373 Learning Check Ex. N (g) + 3 H (g) NH 3 (g) K c = 500 at a particular temperature. What would be K c for following? NH 3 (g) N (g) + 3 H (g) 1 1 K c K c 500 ½ N (g) + 3/ H (g) 0.00 NH 3 (g) 1 K c K c

374 Equilibrium Constant, K c Constant value equal to ratio of product concentrations to reactant concentrations raised to their respective exponents K c Changes with temperature (van t Hoff Equation) Depends on solution concentrations [products] [reactants] Assumes reactants and products are in solution f d 374

375 Equilibrium Constant, K p Based on reactions in which substances are gaseous Assumes gas quantities are expressed in atmospheres in mass action expression Use partial pressures for each gas in place of concentrations Ex. N (g) + 3 H (g) NH 3 (g) K P P P N NH P 3 3 H 375

376 How are K p and K c Related? Start with Ideal Gas Law PV=nRT Substituting P/RT for molar concentration into K c results in pressure-based formula n = moles of gas in product moles of gas in reactant K p K ( RT ) c n For gaseous reactions, use either K P or K C For solution reactions, must use K C 376

377 Learning Check Ex. Consider the reaction: NO (g) N O 4 (g) If K p = for the reaction at 5 C, what is value of K c at same temperature? n = n products n reactants = 1 = 1 K K p c (RT) Δn K c K (RT) p n ( K ) 1 K c =

378 Your Turn! Ex. Consider the reaction A(g) + B(g) 4C(g) If the K c for the reaction is 0.99 at 5ºC, what would be the K p? A B..0 C. 4. D. 400 E. None of these Δn=(4 3)=1 K p = K c (RT) Δn K p = 0.99*( *98.15) 1 K p = 4 378

379 Homogeneous reaction/equilibrium All reactants and products in same phase Can mix freely Heterogeneous reaction/equilibrium Reactants and products in different phases Can t mix freely Solutions are expressed in M Gases are expressed in M Governed by K c 379

380 NaHCO 3 (s) Heterogeneous Equilibria Equilibrium Law = Can write in simpler form Na CO 3 (s) + H O(g) + CO (g) K [ Na 3 ( s ) For any pure liquid or solid, ratio of moles to volume of substance (M) is constant Ex. 1 mol NaHCO 3 occupies 38.9 cm 3 mol NaHCO 3 occupies 77.8 cm 3 CO ][ H [ NaHCO O ( g ) 3 ( s ) ][ CO ] ( g ) ] mol NaHCO M M L mol NaHCO M M L 380

381 NaHCO 3 (s) Heterogeneous Equilibria Na CO 3 (s) + H O(g) + CO (g) concentrations of pure solids and liquids can be incorporated into equilibrium constant, K c Equilibrium law for heterogeneous system written without concentrations of pure solids or liquids K c K [Na CO [NaHCO 3 ( s ) 3 ( s ) ] ] [H O ( g ) ][CO ( g ) ] 381

382 Large K (K>>1) Ex. Interpreting K C Means product rich mixture Reaction goes far toward completion SO (g) + O (g) SO 3 (g) K c = at 5 C K c [SO3] [SO ] [O ]

383 Small K (K<<1) Ex. Means reactant rich mixture Interpreting K C Only very small amounts of product formed H (g) + Br (g) HBr(g) K c = at 5 C [HBr] K c [H ][Br ]

384 K 1 Interpreting K C Means product and reactant concentrations close to equal Reaction goes only ~ halfway 384

385 Size of K gives measure of how reaction proceeds K >> 1 [products] >> [reactants] K = 1 [products] = [reactants] K << 1 [products] << [reactants] 385

386 Learning Check Ex. Consider the reaction of NO (g) N O 4 (g) If K p = at 5 C, does the reaction favor product or reactant? K is small (K < 1) Reaction favors reactant Since K is close to 1, significant amounts of both reactant and product are present 386

387 Equilibrium Positions and Shifts Equilibrium positions Combination of concentrations that allow Q = K Infinite number of possible equilibrium positions Le Châtelier s principle System at equilibrium (Q = K) when upset by disturbance (Q K) will shift to offset stress System said to shift to right when forward reaction is dominant (Q < K) System said to shift to left when reverse direction is dominant (Q > K) 387

388 Relationship Between Q and K Q = K Q < K reaction at equilibrium reactants products Too many reactants Must convert some reactant to product to move reaction toward equilibrium Q > K Too many products reactants products Must convert some product to reactant to move reaction toward equilibrium 388

389 Examples of Le Châtelier s Principle Let s see how this works with changes in 1. Concentration. Pressure and volume 3. Temperature 4. Catalysts 5. Adding inert gas to system at constant volume 389

390 Effect of Change in Concentration Ex. SO (g) + O (g) SO 3 (g) K c =.4 x 10-3 at 700 o C Which direction will the reaction move if 0.15 moles of O is added to an equilibrium mixture? A. Towards the products B. Towards the reactants C. No change will occur 390

391 Effect of Change in Concentration When changing concentrations of reactants or products Equilibrium shifts to remove reactants or products that have been added Equilibrium shifts to replace reactants or products that have been removed 391

392 Effect of Pressure and Volume Changes Ex. Consider gaseous system at constant T and n 3H (g) + N (g) If reduce volume (V) NH 3 (g) Expect Pressure to increase (P) K P To reduce pressure, look at each side of reaction Which has less moles of gas Reactants = = 4 moles gas Products = moles gas Reaction favors products (shifts to right) P P N NH P 3 3 H 39

393 Effect of P and V Changes Consider gaseous system at constant T and n Ex. H (g) + I (g) HI(g) If pressure is increased, what is the effect on equilibrium? n reactant = = n product = K P P P HI H I Predict no change or shift in equilibrium P 393

394 Effect of P and V Changes NaHSO 3 (s) NaSO 3 (s) + H O(g) + SO (g) K P P H OPSO If you decrease volume of reaction, what is the effect on equilibrium? Reactants: no moles gas = all solids Products: moles gas V, causes P Reaction shifts to left (reactants), as this has fewer moles of gas 394

395 Effect of P and V Changes Reducing volume of gaseous reaction mixture causes reaction to decrease number of molecules of gas, if it can Increasing pressure Moderate pressure changes have negligible effect on reactions involving only liquids and/or solids Substances are already almost incompressible Changes in V, P and [X] effect position of equilibrium (Q), but not K 395

396 Effect of Temperature Changes H O(s) H O(l) H =+6 kj (at 0 C) Energy + H O(s) H O(l) Energy is reactant Add heat, shift reaction right 3H (g) + N (g) NH 3 (g) H f = kj 3 H (g) + N (g) NH 3 (g) + energy Energy is product Add heat, shift reaction left 396

397 Effect of Temperature Changes T shifts reaction in direction that produces endothermic (heat absorbing) change T shifts reaction in direction that produces exothermic (heat releasing) change Changes in T change value of mass action expression at equilibrium, so K changed K depends on T T of exothermic reaction makes K smaller More heat (product) forces equilibrium to reactants T of endothermic reaction makes K larger More heat (reactant) forces equilibrium to products 397

398 Catalysts And Equilibrium Catalyst lowers E a for both forward and reverse reaction Change in E a affects rates k r and k f equally Catalysts have no effect on equilibrium 398

399 Inert gas Effect of Adding Inert Gas One that does not react with components of reaction Ex. Argon, Helium, Neon, usually N Adding inert gas to reaction at fixed V (n and T), P of all reactants and products Since it doesn t react with anything No change in concentrations of reactants or products No net effect on reaction 399

400 Your Turn! Ex. The following are equilibrium constants for the reaction of acids in water, K a. Which is the most acid dissociated reaction? A. K a = B. K a = C. K a = D. K a =

401 Calculating K C Given Equilibrium Concentrations Ex. N O 4 (g) NO (g) If you place mol N O 4 in 1 L flask at equilibrium, what is K C? [N O 4 ] eq = 0.09 M [NO ] eq = M K c [ NO] [ N O ] 4 K c [0.0116] [0.09] K C =

402 Your Turn! Ex. For the reaction: A(aq) + B(aq) 3C(aq) the equilibrium concentrations are: A =.0 M, B = 1.0 M and C = 3.0 M. What is the expected value of K c at this temperature? A. 14 B C. 1.5 D K c K c [C] [A] 3 [B] [3.0] [.0] 3 [1.0] 40

403 Calculating K C Given Initial Concentrations and One Final Concentration Ex. SO (g) + O (g) SO 3 (g) mol SO and mol O are placed in a L flask at 1000 K. At equilibrium 0.95 mol SO 3 has formed. Calculate K C for this reaction. 1 st calculate concentrations of each Initial Equilibrium 1.00mol SO] [ O ] 1. 00M 1.00L [ 0.95mol [ SO3] 0. 95M 1.00L 403

404 Set up Concentration Table SO (g) + O (g) SO 3 (g) Initial Conc. (M) Changes in Conc. (M) Equilibrium Conc. (M) K c [SO [SO ] 3 ] [O ] [0.95] K c K [0.075] [0.538] c =.8 10 =

405 Calculate [X] equilibrium from K c and [X] initial Ex. CH 4 (g) + H O(g) CO(g) + 3H (g) At 1500 C, K c = An equilibrium mixture of gases had the following concentrations: [CH 4 ] = M and [H ] = 0.800M and [CO] =0.300M. What is [H O] at equilibrium? 405

406 Calculate [X] equilibrium from K c and [X] initial Ex. CH 4 (g) + H O(g) CO(g) + 3H (g) K c = 5.67 [CH 4 ] = M; [H ] = 0.800M; [CO] =0.300M What is [H O] at equilibrium? First, set up equilibrium K c [CO][H [CH ][H 4 3 ] O] [HO] [CO][H ] [CH4 ]K c Next, plug in equilibrium concentrations and K c 3 [0.300][0.800] [HO] [0.400](5.67) [H O] = M 406

407 Calculate [X] equilibrium from [X] initial and K C Ex. H (g) + I (g) HI(g) at 45 C K C = If one mole each of H and I are placed in a L flask at 45 C, what are the equilibrium concentrations of H, I and HI? Step 1. Write Equilibrium Law K c [ HI ] [ H ][ I ]

408 Ex. Step. Concentration Table Conc (M) H (g) + I (g) HI (g) Initial Change Equil m x x +x.00 x.00 x +x Initial [H ] = [I ] = 1.00 mol/0.500l =.00M Amt of H consumed = Amt of I consumed = x Amt of HI formed = x (.00 (x ) x )(.00 x ) (x ) (.00 x ) 408

409 Ex. Step 3. Solve for x Both sides are squared so we can take square root of both sides to simplify K (x ) (.00 x ) x (.00 x ) 7.459(.00 x ) x x x x x

410 Ex. Step 4. Equilibrium Concentrations Conc (M) H (g) + I (g) HI (g) Initial Change Equil m [H ] equil = [I ] equil = = 0.4 M [HI] equil = x = (1.58) =

411 Your Turn! Ex. N (g) + O (g) NO(g) K c = at 3900 o C If 0.5 moles of N and O are placed in a 50 ml container, what are the equilibrium concentrations of all species? A M, M, M B M, M, M C M M, M D M, M, M 411

412 Your Turn! - Solution Ex. Conc (M) N (g) +O (g) NO (g) Initial Change x x + x Equil 1.00 x 1.00 x + x 0.50 mol [N ] [O ] 1.00M 0.50 L ( x) (1 x ) x x x M [NO] = x = 0.105M 41

413 EX. For the reaction Learning Check A(g) B(g) given that K p = at 5 C, and we place 0. atm A into the container, what will be the pressure of B at equilibrium? A B I 0. 0 atm C x +x E 0. x x x (0.) x = [B]= M 413

414 Calculating K C Given Initial Concentrations and One Final Concentration Ex. H (g) + I (g) 450 C Initially H and I concentrations are 0.00 mol each in.00 L At equilibrium, HI concentration is M. Calculate K C [I ] = [H ] = 0.100M H (g) + I (g) 450 C 0.100M 0.100M 0 -x -x x x x x X=0.08 M x=0.160 [I ] = [H ] = =0.00 M K C = (0.160) /(0.00 ) =64 414

415 Chapter 7 Acids and Bases Chemistry: The Molecular Nature of Matter, 6E

416 Arrhenius Acids and Bases Acid produces H 3 O + in water Base gives OH Acid-base neutralization Acid and base combine to produce water and a salt. Ex. HCl(aq) + NaOH(aq) H O + NaCl(aq) H 3 O + (aq) + Cl (aq) + Na + (aq) + OH (aq) H O + Cl (aq) + Na + (aq) Many reactions resemble this without forming H 3 O + or OH in solution 416

417 Brønsted-Lowry Definition Acid = proton donor Base = proton acceptor Allows for gas phase acid-base reactions Ex. HCl + H O H 3 O + + Cl HCl = acid Donates H + Water = base Accepts H + 417

418 Conjugate Acid-Base Pair Species that differ by H + Ex. HCl + H O H 3 O + + Cl HCl = acid Water = base H 3 O + Conjugate acid of H O Cl Conjugate base of HCl 418

419 Formic Acid is Bronsted Acid Formic acid (HCHO ) is a weak acid Must consider equilibrium HCHO (aq) + H O Focus on forward reaction CHO (aq) + H 3 O + (aq) 419

420 Formate Ion is Bronsted Base Now consider reverse reaction Hydronium ion transfers H + to CHO conjugate pair HCHO + H O H 3 O + + CHO acid base acid base conjugate pair 40

421 Learning Check Identify the Conjugate Partner for Each conjugate base Cl NH 3 C H 3 O CN F conjugate acid HCl NH 4 + HC H 3 O HCN HF 41

422 Learning Check Write a reaction that shows that HCO 3 is a Brønsted acid when reacted with OH HCO 3 (aq) + OH (aq) H O(l) + CO 3 (aq) Write a reaction that shows that HCO 3 is a Brønsted base when reacted with H 3 O + (aq) HCO 3 (aq) + H 3 O + (aq) H CO 3 (aq) + H O(l) 4

423 Your Turn! Ex. In the following reaction, identify the acid/base conjugate pair. (CH 3 ) NH + H SO 4 (CH 3 ) NH + + HSO 4 - A. (CH 3 ) NH / H SO 4 ; (CH 3 ) NH + / HSO - 4 B. (CH 3 ) NH / (CH 3 ) N + ; H SO 4 / SO - 4 C. H SO 4 / HSO - 4 ; (CH 3 ) NH + / (CH 3 ) NH D. H SO 4 / (CH 3 ) NH ; (CH 3 ) NH + / HSO

424 Amphoteric Substances Can act as either acid or base Also called amphiprotic Can be either molecules or ions Ex. hydrogen carbonate ion: Acid: HCO 3 (aq) + OH (aq) CO 3 (aq) + H O(l) Base: HCO 3 (aq) + H 3 O + (aq) H CO 3 (aq) + H O(l) H O(l) + CO (g) 44

425 Your Turn! Ex. Which of the following can act as an amphoteric substance? A. CH 3 COOH B. HCl C. NO - D. HPO

426 Strengths of Acids and Bases Strength of Acid Measure of its ability to transfer H + Strong acids React completely with water Ex. HCl and HNO 3 Weak acids Less than completely ionized Ex. CH 3 COOH and CHOOH Strength of Base classified in similar fashion: Strong bases React completely with water Ex. Oxide ion (O ) and OH Weak bases Undergo incomplete reactions Ex. NH 3 and NRH (NH CH 3, methylamine) 46

427 Reactions of Strong Acids and Bases In water Strongest acid = hydronium ion, H 3 O + If more powerful H + donor added to H O Reacts with H O to produce H 3 O + Similarly, Strongest base is hydroxide ion (OH ) More powerful H + acceptors React with H O to produce OH 47

428 Position of Acid-Base Equilibrium Acetic acid (HC H 3 O ) is weak acid Ionizes only slightly in water HC H 3 O (aq) + H O(l) H 3 O + (aq) + C H 3 O (aq) weaker acid weaker base stronger acid stronger base Hydronium ion Better H + donor than acetic acid Stronger acid Acetate ion Better H + acceptor than water Stronger base Position of equilibrium favors weaker acid and base 48

429 Your Turn! Ex. In the reaction: HCl + H O H 3 O + + Cl - which species is the weakest base? A. HCl B. H O C. H 3 O + D. Cl - 49

430 In General Stronger acids and bases tend to react with each other to produce their weaker conjugates Stronger Brønsted acid has weaker conjugate base Weaker Brønsted acid has stronger conjugate base 430

431 Autoionization of water Trace ionization self-ionization of water H O(l) + H O(l) H 3 O + (aq) + OH (aq) acid base acid base Equilibrium law is: But [H O] pure = K c 1000g 18.0 g/mol 1.00L [H 3 O [H ][OH O] = 55.6 M ] [H O] = constant 431

432 Autoionization of water H O(l) + H O(l) H 3 O + (aq) + OH (aq) Since [H O] = constant Equilibrium law simplifies to Kc [ HO] [ H ][ OH ] Kw Where K w = ion product constant for water Often omit nd H O molecule and write H O(l) H + (aq) + OH (aq) K w [ H ][ OH ] 43

433 Autoionization of water H O (l) H + (aq) + OH (aq) for pure H O at 5 C [H + ] = [OH ] = 1.0 x 10 7 M K w = (1.0 x 10 7 )(1.0 x 10 7 ) = 1.0 x H O auto-ionization occurs in any solution K w = [H + ] [OH ] = 1.0 x at 5 C 433

434 Self-Ionization of Water In aqueous solution, Product of [H + ] and [OH ] equals K w [H + ] and [OH ] may not actually equal each other Solution Classification Neutral [H 3 O + ] = [OH ] Acidic [H 3 O + ] > [OH ] Basic [H 3 O + ] < [OH ] 434

435 Learning Check Ex. In a sample of blood at 5 C, [H + ] = M. Find [OH ] and determine if the solution is acidic, basic or neutral. Kw [ H ][ OH ] 110 So M > M [OH ] > [H 3 O + ] Solution slightly basic Kw [ OH ] [ H ]

436 In general The ph Concept ph log[ H ] [ H ] 10 ph poh log[ OH pk px log X w log Kw. ]

437 Using Logarithms K w [ H ][ OH ] log([ H ][ OH ]) logkw log( ) log[ H ] log[ OH ] logkw ( ) ph poh pkw

438 Redefine Acidic, Basic and Neutral Solutions in terms of ph! As ph, [H + ] ; poh, and [OH ] As ph, [H + ] ; poh, and [OH ] Neutral ph = 7.00 Acidic ph < 7.00 Basic ph >

439 Your Turn! Ex. K w increases with increasing temperature. At 50 o C, K w = x What is the ph of a neutral solution at 50 o C? A B C D [H + ] =[OH - ]=(5.476 x ) 1/ =.34 x 10-7 ph=-log[h + ] = - log.34 x 10-7 =

440 Learning Check Ex. What are [H + ] and [OH ] of ph = 3.00 solution? [H + ] = = M [OH ] = = M Ex. What are [H + ] and [OH ] of ph = 4.00 solution? ph = [H + ] = M 14 [OH ] = = M Or ph 4.00 solution has 10 times less H + than ph 3.00 solution 440

441 Sample ph Calculations Ex. Calculate ph and poh of blood where [H + ] = M [OH ] =. x 10 7 M ph = log(4.6 x 10 8 ) = 7.34 poh = log(. x 10 7 ) = = pk w Or poh = ph = =

442 Sample ph Calculations (cont d) Ex. What is the ph of NaOH solution at 5 C in which the OH concentration is M? [OH ] = M poh = log(0.006) =.59 ph = poh = =

443 Your Turn! Ex. A sample of juice has a ph of Calculate [H + ]. A. 7.6 x 10 3 M B M C M D. 5.9 x 10 9 M E. 1.7 x 10 4 M [H ] 10 ph = = 1.7 x 10 4 M 443

444 Learning Check Ex. What is the [H 3 O + ] and ph of a solution that has [OH ] = M? [H 3 O + ][OH - ] = 1 x [H 3 O + ] = 1 x /3. x 10-3 =3.1 x 10-1 M ph = -log [H 3 O + ] = -log(3.1 x 10-1 )=

445 Your Turn! Ex. What is the [OH ] and ph of a solution that has [H 3 O + ] = M? [H 3 O + ] ph A M 9.40 B M C M 4.60 D M 5.3 E M

446 Learning Check Ex. What is the poh and the [H 3 O + ] of a solution that has a ph of.33? poh = [H 3 O + ]=

447 Your Turn! Ex. What is the ph and the [H 3 O + ] of a solution that has a poh of 1.89? [H 3 O + ] ph A M 1.89 B M C M 10.8 D M 1.11 E M 7.00 poh= 1.89 [OH - ] = Shift log poh [OH - ] =0.019 [H + ]=10-14 / 0.019=7.8 x ph= -log H + = -log 7.8 x ph=

448 Strong Acids: ph of Dilute Solutions Assume 100% dissociated in solution Good ~ if dilute Makes calculating [H + ] and [OH ] easier 1 mole H + for every 1 mole HA So [H + ] = [HA] for strong acids Thus, if M HClO 4 [H + ] = M And ph = log (0.040) = 1.40 Strong Acids HCl HBr HI HNO 3 H SO 4 HClO 3 HClO 4 448

449 ph of Dilute Solutions of Strong Bases Strong Bases NaOH KOH LiOH 1 mole OH for every 1 mole B [OH ] = [B] for strong bases Ca(OH) Ba(OH) mole OH for every 1 mole B [OH ] = *[B] for strong bases Sr(OH) 449

450 Learning Check Ex. Calculate the ph of M Ca(OH). Ca(OH) (s) + H O Ca + (aq) + OH (aq) [OH ] = *[Ca(OH) ] = *0.011M = 0.0M poh = log (0.0) = 1.66 ph = poh = = 1.34 What is this in the [H + ] of the solution? [H + ] = = 4.6 x M 450

451 Learning Check Ex. What is the ph of 0.1M HCl? Assume 100% dissociation HCl(aq) + H O(l) H + (aq) + OH (aq) I 0.1 N/A 0 0 C End 0 N/A ph = log(0.1) = 1 451

452 Learning Check Ex. What is the ph of 0.5M Ca(OH)? Assume 100% dissociation Ca(OH) (aq) Ca + (aq) + OH (aq) I C E poh = -log(1.0) = 0 ph = poh = = 14 45

453 Weak Acids and Bases Incompletely ionized Molecules and ions exist in equilibrium Reaction of a Weak Acid with Water CH 3 COOH(aq) + H O(l) CH 3 COO (aq) + H 3 O + (aq) HSO 3 (aq) + H O(l) SO 3 (aq) + H 3 O + (aq) NH 4 + (aq) + H O (l) NH 3 (aq) + H 3 O + (aq) 453

454 Weak Acid/Base Equilibria Acid + Water Conjugate Base + Hydronium Ion Or generally HA(aq) + H O(l) A (aq) + H 3 O + (aq) [A ][H O ] K 3 c [HA][H O] But [H O] = constant (55.6 M) so rewrite as K c [H O] [A Where K a = acid ionization constant ][H 3 [HA] O ] K a 454

455 Weak Acid/Base Equilibria Often simplify as HA (aq) A (aq) + H + (aq) K a [A ][H [HA] ] pk a log K a p K a 10 K a 455

456 Table 17. Weak Monoprotic Acids at 5 C 456

457 Learning Check Ex. What is the pk a of HOAC if K a = 3.5 x 10 4? HOCN(aq) + H O(l) OCN (aq) + H 3 O + (aq) or HOCN(aq) OCN (aq) + H + (aq) [OCN ][H ] K a [HOCN] = 3.5 x 10 4 pk a = log K a = log(3.5 x 10 4 ) =

458 Reaction of a Weak Base with Water CH 3 COO (aq) + H O(l) CH 3 COOH(aq) + OH (aq) NH 3 (aq) + H O(l) NH 4 + (aq) + OH (aq) Or generally B(aq) + H O(l) BH + (aq) + OH (aq) K c [BH ][OH ] [B][H O] But [H O] = constant so can rewrite as K b [BH ][OH [B] ] Where K b = base ionization constant pk b log K b p K b 10 K b 458

459 Learning Check Ex. What is the pk b of C 5 H 5 Nif K a = 3.5 x 10 4? C 5 H 5 N(aq) + H O(l) C 5 H 5 NH + (aq) + OH (aq) 5 [C5H5NH ][OH ] K b = 1.7 x 10 [C H N] 9 5 pk b = log K b = log(1.7 x 10 9 ) = 8.76 _ 459

460 Table 17.3 Weak Bases at 5 C 460