2. 5 - 1CH110: Chapter 5 CH110 Chapter 5 Chemical Reactions Chemical Changes Chemical Equations Types of Reactions Oxidation-ReductionMoles Energy in Chemical Reactions Reaction Rates

3. 5 - 2CH110: Chapter 5 What is Chemistry? “The study of Matter and its Changes.” Physical Changes Physical Changes = Physical Property Changes in a Physical Property Chemical Changes Chemical Changes = Chemical Property Changes in a Chemical Property Appearance: melting, freezing, evaporation…melting, freezing, evaporation… stretching, molding, cutting…stretching, molding, cutting… Chemical Composition:

4. 5 - 3CH110: Chapter 5 Change in the Chemical Composition Burning of Magnesium Chemical Changes Rusting of Iron Decomposing of wood Souring of Milk Examples:

5. 5 - 4CH110: Chapter 5 Chemical Reactions Color changeColor change Gas formedGas formed Solid precipitate formedSolid precipitate formed Temperature ChangeTemperature Change Gives heat = exothermicGives heat = exothermic Gets cold = endothermicGets cold = endothermic

6. 5 - 5CH110: Chapter 5 Examples Mulching leaves chemicalphysical Which are chemical or physical changes? Tarnishing Silver Fermentation Making ice into water Carbonated Beverage going flat Bleaching a stain

7. 5 - 6CH110: Chapter 5 Mg + O 2  MgO + Energy Chemical Reactions Shows how the Chemical change occurs. Reactants C 3 H 8 + O 2 CO 2 + H 2 O + Energy  Fe + O 2 Fe 2 O 3 ProductsProducts

8. 5 - 7CH110: Chapter 5 Chemical equations Chemist’s shorthand to describe a reaction. ReactantsReactants ProductsProducts The state of all substancesThe state of all substances H 2 + O 2 H 2 O + E (g)(g) (g) Any conditions used in the reactionAny conditions used in the reaction heat Same # & type atoms on each sideSame # & type atoms on each side Law of Conservation of Matter Law of Conservation of Matter 2 2 (g) (l) (s) (aq)

9. 5 - 8CH110: Chapter 5 Balancing Equations WB ___W 10 + ___B 8 ___WB ReactantsReactants ProductsProducts Making Hot dogs: How many packages wieners & buns to buy so none is left over. 4540

10. 5 - 9CH110: Chapter 5 Ca H Cl Balancing Equations CaCl 2 + H 2 Ca + HCl CaCl 2 + H 2 ReactantsReactants ProductsProducts Step 1 Step 1: Count atoms of each element on both sides of equation. 1 1 1 1 2 2

11. 5 - 10CH110: Chapter 5 Balancing Equations CaCl 2 + H 2 Ca + HCl CaCl 2 + H 2 Ca H Cl Ca H Cl ReactantsReactants ProductsProducts 111111 111111 122122 122122 Step 2 Step 2: Determine which atoms are not balanced. - not balanced

12. 5 - 11CH110: Chapter 5 Balancing Equations CaCl 2 + H 2 Ca + HCl CaCl 2 + H 2 ReactantsReactants ProductsProducts 111111 111111 122122 122122 - not balanced Step 3: Step 3: Balance one element at a time with coefficients in front of formulas until all balanced. (Never change the formula!) 2 2 2 Ca H Cl Ca H Cl 2 2

13. 5 - 12CH110: Chapter 5 Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts Step 1 Step 1: Count atoms of each element on both sides of equation. 3 1 4 1 2 1 2 8 3 1

14. 5 - 13CH110: Chapter 5 Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts 3141231412 1283112831 - not balanced Step 2 Step 2: Determine which atoms are not balanced. - not balanced

15. 5 - 14CH110: Chapter 5 Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts 3141231412 1283112831 - not balanced Step 3: Step 3: Balance elements with #’s in front of formulas until all balanced. (Never change the formulas!)

16. 5 - 15CH110: Chapter 5 Hints: Start with a metal in a complex compound, or an element that only appears in one formula. (ie Mg here) Na P O Mg Cl Balancing Equations Mg 3 (PO 4 ) 2 + NaCl Na 3 PO 4 + MgCl 2 Mg 3 (PO 4 ) 2 + NaCl ReactantsReactants ProductsProducts 3141231412 1283112831 - not balanced 6 6 6 6 63 3 3 2 2 6 6 1 8 8 2 6 6

17. 5 - 16CH110: Chapter 5 Hints: Start with an element that only appears in one formula on both sides of the equation. Leave oxygen until last. Balancing Equations CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O ReactantsReactants ProductsProducts CHOCHO

18. 5 - 17CH110: Chapter 5 Balancing Equations CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O CHOCHO CHOCHO ReactantsReactants ProductsProducts Step 1 Step 1: Count atoms of each element on both sides of equation. 262262 2 6 2 123123 1 2 3

19. 5 - 18CH110: Chapter 5 CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O Balancing Equations ReactantsReactants ProductsProducts 262262 262262 123123 123123 Step 2 Step 2: Determine which atoms are not balanced. - not balanced CHOCHO CHOCHO

20. 5 - 19CH110: Chapter 5 262262 262262 - not balanced CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O 2 2 Balancing Equations ReactantsReactants ProductsProducts 123123 123123 - not balanced Step 3: Step 3: Balance one element at a time with coefficients in front of formulas until all balanced. (Never change the formula!) 2 5 5 3 6 6 7 7 3.5 7 7 CHOCHO CHOCHO

21. 5 - 20CH110: Chapter 5 CHOCHO CHOCHO CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O 2 2 Balancing Equations ReactantsReactants ProductsProducts 262262 262262 123123 123123 2 5 5 3 6 6 7 7 3.5 7 7 3.5 O 2 Can’t have 3.5 O 2, so multiply equation by 2!

22. 5 - 21CH110: Chapter 5 3.5 CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O 2 2 Balancing Equations CHOCHO CHOCHO ReactantsReactants ProductsProducts 262262 262262 123123 123123 4 5 5 6 6 6 7 7 7 7 7 3.5 O 2 Can’t have 3.5 O 2, so multiply equation by 2! 2 4 4 4 4 12 14

23. 5 - 22CH110: Chapter 5 Types of Chemical Reactions Complete: C 3 H 8 + 5O 2  3CO 2 + 4H 2 O Combustion Incomplete: 2C 3 H 8 + 7O 2  6CO + 8H 2 O C 3 H 8 + 2O 2  3C + 4H 2 O

24. 5 - 23CH110: Chapter 5 Types of Chemical Reactions Combination Decomposition Single Replacement: Substitution Double Replacement: Metathesis A + BX  B + AX A + B  C C  A + B AX + BY  BX + AY

25. 5 - 24CH110: Chapter 5 Combination Reactions 2H 2 + O 2  2H 2 O Formation of Acid Rain SO 3 + H 2 O  H 2 SO 4 Exposion of Hydrogen Balloon Rusting of Iron 4 Fe + 3 O 2  2 Fe 2 O 3 A + B  C

26. 5 - 25CH110: Chapter 5 Decomposition Reactions Heating Egg Shells CaCO 3  CaO + CO 2 2 H 2 O 2 2 H 2 O + O 2 Blood with peroxide C  A + B

27. 5 - 26CH110: Chapter 5 Single Replacement Reactions Iron Deposits on an Aluminum Pan Al + FeCl 3  Fe + AlCl 3 A + BX  B + AX

28. 5 - 27CH110: Chapter 5 Activity series of metals potassium sodium potassium sodium calcium magnesium aluminum zinc chromium magnesium aluminum zinc chromium iron nickel tin lead iron nickel tin lead copper silver platinum gold copper silver platinum gold Hydrogen Al + Fe +3  Fe + Al +3 Fe + H +  Fe +3 + H 2 increasing reactivity Element give e’s to ion lower on list

29. 5 - 28CH110: Chapter 5 Double Replacement Reaction BaCl 2(aq) + Na 2 SO 4(aq)  BaSO 4(s) + 2NaCl (aq) AX + BY  BX + AY Ba +2 Cl -1 Na +1 SO 4 -2 Insoluble Precipitate Formed

30. 5 - 29CH110: Chapter 5 Predict the products: AgNO 3(aq) MgCl 2(aq) AgNO 3(aq) + MgCl 2(aq) Ag + NO 3 - Mg +2 Cl - AgCl (s) Double Replacement Reaction AX + BY  BX + AY Write correct formulas then balance as needed 22 + Mg(NO 3 ) 2(aq)

31. 5 - 30CH110: Chapter 5 Oxidation and reduction REDOX Where reactants exchange electrons - Examples: All types of batteries All types of batteries alkaline, NiCad, car batteries Rusting and corrosion Rusting and corrosion Metabolism Metabolism Antioxidants (Vit C, E prevent oxidation) Antioxidants (Vit C, E prevent oxidation)

32. 5 - 31CH110: Chapter 5 Oxidation and reduction REDOX Where reactants exchange electrons - LEOGER LEO the lion says GER Reduction Reduction = Gaining electrons GER GER: Gain Electrons Reduction Oxidation Oxidation = Losing electrons LEO LEO: Lose Electrons Oxidation

33. 5 - 32CH110: Chapter 5 Oxidation and reduction 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1-1+ Assign Oxidation States: 0 0 For simple ions, charge. Ox state = charge. For element in natural form 0 Ox State = 0.

34. 5 - 33CH110: Chapter 5 Oxidation and reduction 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1- 1+ Who’s loosing or gaining electrons? 00 Loses 1 e - = LEO Gains 1 e - =GER 1 e - =GER oxidized Na loses e - (LEO) Na gets oxidized reduced Cl gains e - (GER) Cl gets reduced

35. 5 - 34CH110: Chapter 5 Oxidation and reduction Oxidation loses Oxidation - when a reactant loses e - (s). (LEO) NaNa + + e - Na (s) Na + + e - Reduction gains Reduction - when a reactant gains e - (s). (GER) Cl 2 + 2 e - 2 Cl - Cl 2 (g) + 2 e - 2 Cl - half reactions These are half reactions

36. 5 - 35CH110: Chapter 5 Oxidation and reduction 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1- 1+ Who’s loosing or gaining electrons? 00 NaNaoxidized Na loses e - (LEO) so Na gets oxidized Na caused Cl to get reduced Na caused Cl to get reduced Na is the Reducing agent Na is the Reducing agent Loses 1 e - =LEO Gains 1 e - =GER 1 e - =GER

37. 5 - 36CH110: Chapter 5 Oxidation and reduction 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1- 1+ Who’s loosing or gaining electrons? 00 Loses 1 e - =LEO Gains 1 e - =GER 1 e - =GER ClClreducedCl gains e - (GER) so Cl gets reduced Cl caused Na to get oxidizedCl caused Na to get oxidized Cl is the Oxidizing agentCl is the Oxidizing agent

38. 5 - 37CH110: Chapter 5 REDOX reactions 2 H 2 + O 2 2 H 2 O 00 00 1+1+ 2-2- Hydrogen is oxidized and is a reducing agent. Oxygen is reduced and is an oxidizing agent. Lose 1 e - = LEO Gain 1 e - = GER

39. 5 - 38CH110: Chapter 5 Types of chemical reactions Chemical Reactions NonredoxNonredox DoublereplacementDoublereplacement Combination RedoxRedox SinglereplacementSinglereplacement Decomposition

40. 5 - 39CH110: Chapter 5 1 pair = 1 dozen = 1 mole = 1 pair = 1 dozen = 1 mole = The Mole 1 mol eggs___ 6.02 x 10 23 eggs 1 mol Au_______ 6.02 x 10 23 Au atoms _____1 mole H 2 O_____ 6.02 x 10 23 H 2 O molecules 6.02 x 10 23 H 2 O molecules 2 12 6.02 x 10 23 602,000,000,000,000,000,000,000.

41. 5 - 40CH110: Chapter 5 1 car ___ 4 wheels The Mole & Formulas 1 mol cars_ 4 mol wheels 4 mol wheels 1 doz cars 4 doz wheels 1 mole H 2 O 2 mol H 2 mol H 1 mole H 2 O 1 mol O 1 mol O

42. 5 - 41CH110: Chapter 5 = wheels 1 car ___ 4 wheels The Mole & Formulas 1 mol cars_ 4 mol wheels 4 mol wheels 1 doz cars 4 doz wheels 1doz cars 5 doz cars 5 doz cars 1 20 doz 2 mol H 1 mol H 2 O 5 mol H 2 O 5 mol H 2 O 1 10 = mol H

43. 5 - 42CH110: Chapter 5 1 mole = MW in g’s The Mole & Molecular Mass 1 mol S_ 32 g S 32 g S 1 mol S_ 32 g S 32 g S 1 mole S = 32 g S 32 g S 1 mol S 32 g S 1 mol S 1 mol C 12 g C 12 g C 1 mol C 12 g C 12 g C 1 mole C = 12 g C 12 g C 1 mol C 12 g C 1 mol C

44. 5 - 43CH110: Chapter 5 The Mole & Molecular Mass 1 mol H 2 O_ 18.0 g H 2 O 1 mole H 2 O has: 1.0 g H = 1 mol H 2 mol H 2 mol H 1 2.0 g H 16.0 g O = 1 mol O 1 16.0 g O 18.0 g H 2 O 18.0 g H 2 O 1 mol H 2 O 18.0 g

45. 5 - 44CH110: Chapter 5 Molecular Mass Find the MW of Glucose; C 6 H 12 O 6 1.0 g H = 1 mol H 12 mol H 1 12.0 g H 16.0 g O = 1 mol O 6 mol O 1 96.0 g O 180.0 g C 6 H 12 O 6 C 6 H 12 O 6 1 mol C 6 H 12 O 6 12.0 g C = 1 mol C 6 mol C 1 72.0 g C

46. 5 - 45CH110: Chapter 5 1 mol H 2 O = 18 g H 2 O Mass to Mole Conversions How many moles of water are in 36 g H 2 O? What should the answer look like? What is Unique to the problem? 36 g H 2 O 36 g H 2 O 1 mol H 2 O 2.0

47. 5 - 46CH110: Chapter 5 1 mol H 2 O 18 g H 2 O Mass to Mole Conversions How many moles of H are in 36 g H 2 O? What should the answer look like? What is Unique to the problem? 36 g H 2 O 36 g H 2 O 1 mol H 4.0 2 mol H = 1 mol H 2 O

48. 5 - 47CH110: Chapter 5 180 g Gluc = 1 mol Gluc Mole to Mass Conversions How many g’s of Glucose (C 6 H 12 O 6 ) are in 5 mol Glucose? What should the answer look like? What is Unique to the problem? 5 mol Gluc 1 g Glucose 900

49. 5 - 48CH110: Chapter 5 Moles in Chemical Equations 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 2 mol Na 1 mol Cl 2 1 mol Cl 2 2 mol Na 2 mol NaCl Cl 2 1 mol Cl 2 2 mol NaCl Cl 2 are needed to produce 2.5 mol NaCl? How many gs of Cl 2 are needed to produce 2.5 mol NaCl?

50. 5 - 49CH110: Chapter 5 Moles in Chemical Equations 2 Na (s) Cl 2 (g 2 NaCl 2 Na (s) + Cl 2 (g)  2 NaCl 1 mol Cl 2 70.9 g Cl 2 Cl 2 1 mol Cl 2 2 mol NaCl Cl 2 are needed to produce 2.5 mol NaCl? How many gs of Cl 2 are needed to produce 2.5 mol NaCl? Identify any conversion factors. How should the answer look? 2.5 mol NaCl = g What is unique to the problem? 88.625 = 89 g Cl 2 1 mol Cl 2 2 mol NaCl 70.9 g Cl 2 1 mol Cl 2

51. 5 - 50CH110: Chapter 5 H2OH2OH2OH2O H2OH2OH2OH2O more stable Energy in Chemical Reactions Exothermic reaction -  H= heat of reaction Energy Rxn Progress Reactants Products E act = Activation Energy (Gets hot) H 2 + O 2 2H 2 + O 2 2H 2 O + Energy

52. 5 - 51CH110: Chapter 5 Exothermic Reactions Energy Rxn Progress 2Mg + O 2  2MgO + Energy 2H 2 + O 2  2H 2 O + Energy CH 4 + 2O 2  CO 2 + 2H 2 O + 213 kcal

53. 5 - 52CH110: Chapter 5 Exothermic Reactions Energy Rxn Progress Reactants(Water) Products(Ice) Energy is released Products are more stable. -H-H-H-H

54. 5 - 53CH110: Chapter 5 more stable Energy in Chemical Reactions Endothermic reaction  H= heat of reaction Energy Rxn Progress Reactants Products E act = Activation Energy (Gets cold)

55. 5 - 54CH110: Chapter 5 Energy Rxn Progress Endothermic Reactions Energy is required Products are less stable. Products(Water) Reactants(Ice) +H+H+H+H

56. 5 - 55CH110: Chapter 5 Energy Rxn Progress 2H 2(g) + O 2(g)  2H 2 O + Energy Just because something has the potential to react doesn’t mean it will do so immediately. doesn’t mean it will do so immediately. Just because something has the potential to react doesn’t mean it will do so immediately. doesn’t mean it will do so immediately. H 2(g) O 2(g) may stay together for lifetime without reacting to form water. H 2(g) + O 2(g) may stay together for lifetime without reacting to form water. Reaction Rates

57. 5 - 56CH110: Chapter 5 burning Oxidation: Paper burning Paper turning yellow rusting Oxidation: Nails rusting Reaction Rates Fast: Slow:

58. 5 - 57CH110: Chapter 5 They have to have enough E. Reaction Rates For reactants to make products: collideMolecules must collide (solvents really help) alignedThey have to be aligned correctly. (Parked cars don’t collide)

59. 5 - 58CH110: Chapter 5 Rates of Reactions Reaction rates can be affected by : polar vs. nonpolar reactant structure( polar vs. nonpolar ) vapor vs liq physical state of reactants ( vapor vs liq. ) medications concentration of reactants ( medications ) sugar cube vs crystals surface area ( sugar cube vs crystals ) hypothermia & metabolism temperature ( hypothermia & metabolism ) H 2 O 2 & blood catalyst ( H 2 O 2 & blood )

60. 5 - 59CH110: Chapter 5 Reaction Rates Factors that increase reaction rate: 1.More Reactants: More cars  More collisions 8 blocks: 34 surfaces 8 blocks: 24 surfaces More surface area  More collisions

61. 5 - 60CH110: Chapter 5 Reaction Rates Factors that increase reaction rate: 2.Higher Temperature: Faster cars  More collisions More Energy  More collisions

62. 5 - 61CH110: Chapter 5 Reaction Rates Factors that increase reaction rate: 3.Adding a Catalyst: Lower E act  More collisions Uncatalysed reaction

63. 5 - 62CH110: Chapter 5 Alter the reaction mechanism but not change the products Uncatalysed reaction Catalysed reaction Lower activation energy A catalyst will: Enzymes Enzymes are biological catalysts. Catalysts