2. 5 - 1GS105 CHGS105 Chapter 17 Chemical Reactions Chemical Changes Chemical Equations Types of Reactions Oxidation-ReductionMoles Energy in Chemical Reactions Reaction Rates

3. 5 - 2GS105 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 - 3GS105 Change in the Chemical Composition Burning of Magnesium Chemical Changes Rusting of Iron Decomposing of wood Souring of Milk Examples:

5. 5 - 4GS105 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 - 5GS105 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 - 6GS105 Mg + O 2  MgO + Energy Chemical Equations 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 - 7GS105 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 - 8GS105 Law of Conservation of Matter Question:

10. 5 - 9GS105 Law of Conservation of Matter Solution: Fe + O 2  Fe 2 O 3

11. 5 - 10GS105 Law of Conservation of Matter Question:

12. 5 - 11GS105 Law of Conservation of Matter Solution:

13. 5 - 12GS105 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

14. 5 - 13GS105 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

15. 5 - 14GS105 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

16. 5 - 15GS105 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

17. 5 - 16GS105 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

18. 5 - 17GS105 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

19. 5 - 18GS105 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!)

20. 5 - 19GS105 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

21. 5 - 20GS105 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

22. 5 - 21GS105 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

23. 5 - 22GS105 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

24. 5 - 23GS105 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

25. 5 - 24GS105 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!

26. 5 - 25GS105 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

27. 5 - 26GS105 ___O 2  ____O 3 ___Cr + ___O 2  ____Cr 2 O 3 ___HNO 2  ___HNO 3 + ___NO + __ H 2 O ___N 2 H 4 O 3  ___N 2 + ___O 2 + ___H 2 O ___N 2 O  ____N 2 + ____O 2 ___NO + ___O 2  ____NO 2 ___CH 4 + ___O 2 --> CO 2 + ____H 2 O Balancing Equations Practice:

28. 5 - 27GS105 ___O 2  ____O 3 ___Cr + ___O 2  ____Cr 2 O 3 ___HNO 2  ___HNO 3 + ___NO + __ H 2 O _ _N 2 H 4 O 3  _ _N 2 + _ _O 2 + _ _H 2 O ___N 2 O  ____N 2 + ____O 2 ___NO + ___O 2  ____NO 2 ___CH 4 + ___O 2 --> CO 2 + ____H 2 O Balancing Equations Solutions: 32 342 3 2 2 4 2 2 2 22 2

29. 5 - 28GS105 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

30. 5 - 29GS105 Combination Reactions Rusting of Iron 4 Fe + 3 O 2  2 Fe 2 O 3 2H 2 + O 2  2H 2 O Exposion of Hydrogen Balloon A + B  C Decomposition Reactions 2 H 2 O 2 2 H 2 O + O 2 Blood with peroxide C  A + B

31. 5 - 30GS105 Single Replacement Reactions Iron Deposits on an Aluminum Pan Al + FeCl 3  Fe + AlCl 3 A + BX  B + AX Double Replacement Reaction BaCl 2(aq) + Na 2 SO 4(aq)  BaSO 4(s) + 2NaCl (aq) AX + BY  BX + AY Insoluble Precipitate Formed

32. 5 - 31GS105 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.

33. 5 - 32GS105 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

34. 5 - 33GS105 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 __32 g S _ __32 g S _ 6.02 x 10 23 atoms S 6.02 x 10 23 atoms S __32 g S _ __32 g S _ 6.02 x 10 23 atoms S 6.02 x 10 23 atoms S __12 g C _ __12 g C _ 6.02 x 10 23 atoms C 6.02 x 10 23 atoms C __12 g C _ __12 g C _ 6.02 x 10 23 atoms C 6.02 x 10 23 atoms C

35. 5 - 34GS105

36. 5 - 35GS105 Molecular Mass Find the MW (or MM) 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

37. 5 - 36GS105 Molecular Mass Find the MW (MM) of Water; H 2 O 1.0 g H = 1 mol H 2 mol H 1 2.0 g H 2.0 g H 16.0 g O = 1 mol O 1 16.0 g O 18.0 g H 2 O H 2 O 1 mol H 2 O 1 mol H 2 O_ 18.0 g H 2 O

38. 5 - 37GS105 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 1 mol H 2 O_ 18.0 g H 2 O 18.0 g H 2 O 18.0 g H 2 O 1 mol H 2 O

39. 5 - 38GS105 1 mol H 2 O 18 g H 2 O Mass to Mole Conversions How many molecules 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 = molecules H 2 O 1.2 x 10 24 molecules H 2 O 1 mol H 2 O_ 18.0 g H 2 O 1 mol H 2 O_ 6.02x10 23 molecules H 2 O 6.02x10 23 molecules H 2 O 6.02x10 23 molecules H 2 O _ 1 mol H 2 O

40. 5 - 39GS105 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

41. 5 - 40GS105 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 900 180.0 g C 6 H 12 O 6 C 6 H 12 O 6 1 mol C 6 H 12 O 6 1 mol C 6 H 12 O 6 180.0 g C 6 H 12 O 6

42. 5 - 41GS105 Ratios in Chemical Equations C (s) O 2 (g CO 2(g) C (s) + O 2 (g)  CO 2(g) 1 mol C 1 mol O 2 1 mol O 2 1 mol C CO 2 1 mol CO 2 O 2 1 mol O 2 1 mol CO 2 O 2  44 g CO 2 12 gs of C + 32 gs of O 2  44 g CO 2 12 g C 32 g O 2 32 g O 2 12 g C CO 2 44 g CO 2 O 2 32 g O 2 44 g CO 2

43. 5 - 42GS105 Ratios in Chemical Equations C (s) O 2 (g CO 2(g) C (s) + O 2 (g)  CO 2(g) 1 mol C 1 mol O 2 1 mol O 2 1 mol C CO 2 1 mol CO 2 O 2 1 mol O 2 1 mol CO 2 O 2 are needed to produce 11 g CO 2 ? How many gs of O 2 are needed to produce 11 g CO 2 ? 12 g C 32 g O 2 32 g O 2 12 g C CO 2 44 g CO 2 O 2 32 g O 2 44 g CO 2

44. 5 - 43GS105 11 g CO 2 Ratios in Chemical Equations 32 g O 2 44 g CO 2 O 2 1 mol O 2 1 mol CO 2 If it takes 32 g O 2 to make 44 g CO 2 then: 11 g CO 2 = g O 2 8 C (s) O 2 (g CO 2(g) C (s) + O 2 (g)  CO 2(g) O 2 are needed to make 11 g CO 2 ? How many gs of O 2 are needed to make 11 g CO 2 ? 32 g O 2 44 g CO 2 X g O 2 11 g CO 2 = 32 g O 2 44 g CO 2 X g O 2 11 g CO 2 =

45. 5 - 44GS105 Ratios in Chemical Equations 32 g O 2 44 g CO 2 O 2 1 mol O 2 1 mol CO 2 Identify any conversion factors. How should the answer look? 11 g CO 2 = g O 2 What is unique to the problem? 8 O 2 32 g O 2 44 g CO 2 C (s) O 2 (g CO 2(g) C (s) + O 2 (g)  CO 2(g) O 2 are needed to make 11 g CO 2 ? How many gs of O 2 are needed to make 11 g CO 2 ? Factor conversion method:

46. 5 - 45GS105 Ratios in Chemical Equations 11 g CO 2 = g O 2 8 O 2 32 g O 2 44 g CO 2 C (s) O 2 (g CO 2(g) C (s) + O 2 (g)  CO 2(g) O 2 are needed to make 11 g CO 2 ? How many gs of O 2 are needed to make 11 g CO 2 ? O 2  11 g CO 2 ? gs C + 8 gs O 2  11 g CO 2 So: 3 O 2  13 g CO 2 ? 5 gs C + 8 gs O 2  13 g CO 2 ? Then could : NO! There is only enough O 2 to make 11 g’s of CO 2. We would still make 11 g’s of CO 2 but have 2 g’s of C left over

47. 5 - 46GS105 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? 46 g Na 71 g Cl 2 71 g Cl 2 46 g Na 58.5 g NaCl Cl 2 71 g Cl 2 58.5 g NaCl

48. 5 - 47GS105 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? = 89 gCl 2 How many moles of Cl 2 are needed? X mol Cl 2 2.5 mol NaCl Cl 2 1 mol Cl 2 2 mol NaCl = How many grams of Cl 2 are needed? mol Cl 2 X g Cl 2 = 1 mol Cl 2 71 g Cl 2 1.25 1.25 1.25

49. 5 - 48GS105 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 = gCl 2 What is unique to the problem? 88.625 = 89 gCl 2 Cl 2 1 mol Cl 2 2 mol NaCl 70.9 g Cl 2 1 mol Cl 2 Factor conversion method:

50. 5 - 49GS105 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 -486 kJ made

51. 5 - 50GS105 Bond Energy 2H 2 + O 2 2H 2 O + Energy 2H-H + O=O  2 H-O-H 2 (436)4984 (464) 1370 kJ used1856 kJ made-=-486 kJ made

52. 5 - 51GS105 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 + Energy -486 kJ made

53. 5 - 52GS105 Bond Energy CH 4 + 2O 2 CO 2 CO 2 4C-H + 2O=O  O=C=O + 2 H-O-H 4 (414)2(498)2 (799) 2652 kJ used3454 kJ made- =- 802 kJ made + 2H 2 O + Energy + 2H 2 O + Energy 4 (464)

54. 5 - 53GS105 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 - 54GS105 Energy Rxn Progress Endothermic Reactions Energy is required Products are less stable. Products(Water) Reactants(Ice) +H+H+H+H

56. 5 - 55GS105 Exothermic Reactions Energy Rxn Progress Reactants(Water) Products(Ice) Energy is released Products are more stable. -H-H-H-H

57. 5 - 56GS105 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

58. 5 - 57GS105 burning Paper burning Paper yellowing Reaction Rates Fast: Slow: Human Aging Explosion Fruit Ripening rusting Nails rusting

59. 5 - 58GS105 They have to have enough E. Reaction Rates For reactants to make products: collideMolecules must collide (If they don’t meet they can’t react: solvents really help)solvents really help) alignedThey have to be aligned correctly. (Parked cars don’t collide) (Car bumpers meeting don’t make a change)

60. 5 - 59GS105 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 )

61. 5 - 60GS105 Reaction Rates Factors that increase reaction rate: 1.More Reactants (Greater Concentration) : More cars  More collisions 8 blocks: 34 surfaces 8 blocks: 24 surfaces More surface area  More collisions

62. 5 - 61GS105 Reaction Rates Factors that increase reaction rate: 2.Higher Temperature: Faster cars  More collisions More Energy  More collisions Need Energy of Activation (E act )

63. 5 - 62GS105 Reaction Rates Factors that increase reaction rate: 3.Adding a Catalyst: Lower E act  More collisions Uncatalysed reaction

64. 5 - 63GS105 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

65. 5 - 64GS105 Catalysts

66. 5 - 65GS105 Without added energy reactions go to disorder Entropy = The amount of disorder: Entropy Randomness factor Nature tends to move spontaneously from a state of lower probably to one of higher probability. If a beaker with marbles is shaken, which beaker (above) will result?