1. Chemical Reactions Natural Approach to Chemistry Chapter 10 1

2. Learning Objectives Sections 10.1-3 Characterize types of chemical reactions (redox,acid-base, synthesis, and single and double replacement) Apply mole concept & conservation of mass to calculate quantities Distinguish between endothermic and exothermic processes Sections 10.4 Evaluate costs & benefits of resources Discuss technological effects & impacts on environ- mental quality Discuss production and use of natural resources Create and interpret potential energy diagrams 2

3. Be sure you know these terms: Parts of a chemical reaction Synthesis Decomposition Single displacement Double displacement Polymerization Precipitate Polymer, polymerization Exo- and endothermic Enthalpy of reaction/formation Energy barrier Photosynthesis Chemical engineering Biodegradable Hazardous substances Sustainable chemistry Green chemistry 3

4. Chemical Reaction Assignments 10.1 322/1-7,31,37,38,52 10.3 322/8-15,39-41,64,65 10.4 3.22/16-29,43-47,66-71 4

5. A Chemical Equation Represents with symbols and formulas, the identities and relative molecular or molar amounts of the reactants and products in a chemical reaction. 5

6. Word Equation / Formula Equation Methane + oxygen --> carbon dioxide + water CH 4 (g) + 0 2 (g) --> CO 2 (g) + H 2 0(g) Reactants Products The above formula equation is not balanced. 6

7. Chemical Reaction Indications 1.Production of energy as heat and/or light. 2.Production of a gas. 3.Formation of a precipitate – a solid produced as a result of a chemical reaction in solution. A precipitate separates from the solution. 4. A color change. 7

8. Chemical Equation Requirements 1.The equation must represent known facts. 2.The formulas for the reactants and products must be written correctly. DO NOT change subscripts. 3.The law of conservation of mass must be satisfied. The number of atoms of each element must be the same on each side of the yield sign. After a formula is written correctly, place coefficients in front of a formula to show conservation of mass. 8

9. Diatomic Molecules ElementSymbolMolecular FormulaPhysical State at Room Temp. HydrogenHH2H2 Gas NitrogenNN2N2 Gas OxygenOO2O2 Gas FluorineFF2F2 Gas ChlorineClCl 2 Gas BromineBrBr 2 Liquid IodineII2I2 Solid When writing a chemical equation including any of the above elements, they are shown as diatomic molecules as in column 3 above. 9

10. Coefficients When placed in front of a correctly written chemical formula, a coefficient multiplies the number of atoms of each element indicated in the formula 2O 2 means 4 O 2H 2 0 means 4 H and 2 O 10

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12. Practice Write word and balanced chemical equations for: solid calcium reacts with solid sulfur to produce solid calcium sulfide. Include symbols for physical states. Ca(s) + S(s) --> CaS(s) Balanced: 1 Ca on each side 1 S on each side 12

13. Write word and balanced chemical equation for: Hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas. H 2 (g) + F 2 (g) --> HF(g) H 2 (g) + F 2 (g) --> 2HF(g) Solid aluminum metal reacts with aqueous zinc chloride to produce solid zinc metal and aqueous aluminum chloride. Al(s) + ZnCl 2 (aq) --> Zn(s) + AlCl 3 (aq) Al(s) + 3 ZnCl 2 (aq) --> Zn(s) + 2 AlCl 3 (aq) 2 Al(s) + 3 ZnCl 2 (aq) -->3 Zn(s) + 2 AlCl 3 (aq) 13

14. Translate these chemical equations into sentences: CS 2 (l) + 3O 2 (g) --> CO 2 (g) + 2SO 2 (g) Liquid carbon disulfide reacts with oxygen gas to produce carbon dioxide gas and sulfur dioxide gas. NaCl(aq) + AgNO 3 (aq) --> NaNO 3 (aq) + AgCl(s) Aqueous solutions of sodium chloride and silver nitrate react to produce aqueous sodium nitrate and a precipitate of silver chloride. 14

15. Write & Balance: Hydrazine, N 2 H 4, reacts violently with oxygen to produce gaseous nitrogen and water. N 2 H 4 (l) + O 2 (g) --> N 2 (g) + H 2 O(l) Is it balanced? No. There are 4H on reactant side and 2H on product side and O is 2/1. N 2 H 4 (l) + O 2 (g) --> N 2 (g) + 2H 2 O(l) N: 2/2H:4/4O:2/2 15

16. Chemical Equation Indications 1. Coefficients indicate relative amounts of reactants and products (proportions, molecules, moles, grams, ratios). H 2 (g) + Cl 2 (g) --> 2HCl(g) 1 molecule H 2 : 1 molecule Cl 2 : 2 molecules HCl 1 mole H 2 : 1 mole Cl 2 : 2 moles HCl 16

17. 2.Coefficients can be used to determine relative masses of reactants and products H 2 (g) + Cl 2 (g) --> 2HCl(g) 1mol H 2 x 2.02g H 2 = 2.02 g H 2 mol 1 mol Cl 2 x 70.90 g Cl 2 = 70.90 g Cl 2 mol 2 mol HCl x 36.46 g HCl = 72.92 g HCl mol 17

18. 3. The reverse reaction for a chemical equation has the same relative amounts of substances as the forward reactions. 18

19. Balancing Chemical Equations 1.Write equation. 2.Write correct chemical formulae for each compound. 3.Balance according to the Law of Conservation of Mass by adjusting coefficients. 4.Start with the element appearing in the fewest substances. Balance free elements last. 5.Count atoms to be sure the equation is balanced. 19

20. Practice - Write word, formula, and balanced chemical equations for this reaction. 1.Magnesium and hydrochloric acid react to produce magnesium chloride and hydrogen. Word Equation: Magnesium + hydrochloric acid --> magnesium chloride + hydrogen Formula Equation: Mg(s) + HCl(aq) --> MgCl 2 (s) + H 2 (g) Adjust coeffs: Mg(s) + 2HCl(aq) --> MgCl 2 (s) + H 2 (g) Count atoms: Mg: 1/1 H:2/2 Cl:2/2 20

21. Solid sodium combines with chlorine gas to produce solid sodium chloride. Sodium(s) + chlorine(g) --> sodium chloride Na(s) + Cl 2 (g) --> NaCl(s) Balance: Na(s) + Cl 2 (g) --> 2NaCl(s) 2Na(s) + Cl 2 (g) --> 2NaCl(s) 21

22. Types of Chemical Reactions 1.SynthesisA + X --> AX 2.Decomposition AX --> A + X 3.Single-displacementA + BX --> AX + B 4.Double-displacementAX + BY --> AY + BX 5.Combustion – a substance combines with oxygen releasing energy as light and heat. 6.Acid + Base --> Salt + Water 7.Reduction/oxidation (Redox) – covered in a later chapter 22

23. Activity Series – elements organized according to how they react 23 An element can replace any element placed below it BUT It cannot replace any element above it. Zn can replace Cu but Au cannot replace Mg Most active metals: Lireact w/coldCo Do not react w/H 2 0. RbH 2 0 & acidsNi React w/acids, repla- Kreplacing H 2.Sn cing H 2. React w/O 2 BaReact w/O 2 Pb forming oxides. Srforming oxidesH 2 React w/O 2, forming CaSb oxides NaBi Mgreact w/steamCu Al(not cold H 2 0)Hg Mnand acids, repla-Ag Fairly unreactive, Zncing H 2. ReactPt forming oxides only Crwith O 2 formingAu indirectly. Feoxides Cd

24. Nonmetal activity series: Most active F Cl Br I 24

25. Sample Problems – activity series Zn(s) + H 2 O(l) ---> No reaction, water must be 100 o C (steam) at least. Sn(s) + O 2 (g) --> Yes, any metal more active than Ag will react w/O 2 to form an oxide. (Sn is above Ag) 2Sn(s) + O 2 (g) --> 2SnO Cd(s) + Pb(NO 3 ) 2 (aq) --> Yes, Cd is above Pb. Products: Cd(NO 3 ) 2 + Pb Cu(s) + HCl(aq) --> No, Cu is below H 25 50 o C

26. 1. Synthesis Reaction A + X --> AX Samples: Fe(s) + S(s) --> FeS(s) 2Mg(s) + O 2 (g) --> 2MgO(s) H 2 0 + SO 3 --> H 2 SO 4 26

27. Synthesis with Oxides (see handout for more information!!) CaO(s) + H2O(l) --> Ca(OH) 2 (s) Pollution: SO 2 (g) + H 2 O(l) --> H 2 SO 3 (aq) 2H 2 SO 3 (aq) + H 2 0(l) --> 2H 2 SO 4 (aq) Oxides: CaO(s) +SO 2 (g) --> CaSO 3 (s) 27

28. 2. Decomposition Reaction AX --> A + X H 2 0(l) --------> 2H 2 (g) + O 2 (g) (electrolysis) 2HgO(s) ---> 2Hg(l) + O 2 (g) 28 electricity 

29. Decomposition of Metal Oxide CaCO 3 --->CaO + CO 2 Decomp of Metal Hydroxide Ca(OH) 2 ---> CaO + H 2 O Decomp of Metal Chlorate 2KClO 3 ----> 2KCl + 3O 2 Decomp of Acids H 2 CO 3 --> CO 2 + H 2 O (occurs at room temp) H 2 SO 4 ---> SO 3 + H 2 O 29     MnO 2

30. 3. Single Displacement Reaction A + BX --> AX + B Or Y + BX --> BY + X 1. Fe + CuSO 4 --> FeSO 4 + Cu 2. Cu + 2AgN0 3 --> Cu(NO 3 ) 2 + 2Ag 3. CI 2 + 2K I --> 2KCl + I 2 How is 3. different from 1. or 2.? In 1.& 2. metals are being displaced. In 3. a halogen is being displaced. 30

31. Hydrogen displaced by a metal: Mg + 2HCl --> H 2 + MgCl 2 31

32. 4. Double-Displacement Reaction AX + BY --> AY + BX A,X,B, and Y in reactants are ions. AY and BX are ionic or molecular compounds. 1.Formation of a Precipitate 2KI(aq) + Pb(NO 3 ) 2 (aq) --> PbI 2 (s) + 2KNO 3 (aq) 32

33. Formation of a Gas FeS(s) + 2HCl(aq) –> FeCl 2 (aq) + H 2 S(g) Formation of Water HCl(aq) + NaOH(aq) --> NaCl(aq) + H 2 0(l) 33

34. Combustion Reactions 2H 2 (g) + O 2 (g) --> 2H 2 O(g) C 3 H 8 (g) + 5O 2 (g) --> 3CO 2 (g) + 4H 2 0(g) Other products are heat and light. 34

35. Predicting Activity (use the activity series handout) Zn(s) + H 2 0(l) --> ? No, not hot enough. Steam needed (100 o C) Ca(s) + H 2 O(l) --> ? Yes, Ca is above H on the chart. The products are: Ca(OH) 2 + H 2 (g) Pt(s) + O 2 (g) --> ? No. 35

36. Cd(s) + 2HBr(aq) --> Yes, Cd is above H. Products: CdBr + H 2 (g) Mg(s) + steam --> Yes, Products: Mg(OH) 2 + H 2 (g) 36

37. Activity Series Activity of metals:Activity of halogen nonmetals Li<--Most active metalF 2 <-- Most active nonmetal Rb react w/cold Cl 2 K H 2 0 & acids Br 2 Ba replacing H 2. I 2 Sr React w/O 2 Ca forming oxides Na Mgreact w/steam Al(not cold H 2 0) Mnand acids, repla- Zncing H 2. React Crwith O 2 forming Feoxides Cd Co Do not react w/H 2 0. Ni React w/acids, repla- Sn cing H 2. React w/O 2 Pb forming oxides. H 2 Sb Bi React w/O 2, forming Cu oxides Hg Ag Fairly unreactive, Pt forming oxides only Auindirectly. 37

38. Solubility Chart 38

39. There are three key components to a chemical reaction: Reactants Products Energy (in or out)

40. There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions

41. There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Photosynthesis is an endothermic reaction: energy is absorbed

42. There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Photosynthesis is an endothermic reaction: energy is absorbed Cellular respiration is an exothermic reaction: energy is released

43. There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Chapter 9.3 Properties of Solutions

44. There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Chapter 9.3 Properties of Solutions Energy is absorbed from the surroundings so the pack feels cold

45. There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Chapter 9.3 Properties of Solutions Energy is released into the surroundings so the pack feels hot

46. There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Chapter 9.3 Properties of Solutions Change in enthalpy enthalpy: the amount of energy that is released or absorbed during a chemical reaction

47. Reaction Exothermic Endothermic Energy is released is absorbed Enthalpy change (∆H, J/mole) is a negative number ∆H < 0 is a positive number ∆H > 0

48. Chemical equation for the combustion of carbon: Reactants Products Energy thermochemical equation: the equation that gives the chemical reaction and the energy information of the reaction. C(s) + O 2 (g) CO 2 (g)∆H = –393.5 kJ

49. Chemical equation for the combustion of carbon: The reverse chemical reaction involves the same amount of energy, but the energy flow is reversed (“in” instead of “out”): Enthalpy calculations C(s) + O 2 (g) CO 2 (g)∆H = –393.5 kJ CO 2 (g) C(s) + O 2 (g) ∆H = +393.5 kJ

50. Chemical equation for the combustion of carbon: 1 mole 1 mole 1 mole 2 moles 2 moles 2 moles The combustion of twice as much carbon releases twice as much energy: Enthalpy calculations C(s) + O 2 (g) CO 2 (g)∆H = –393.5 kJ 2C(s) + 2O 2 (g) 2CO 2 (g) ∆H = –787.0 kJ

51. Chemical equation for the formation of rust: 2 moles 3/2 moles 1 mole Enthalpy calculations 2Fe(s) + 3/2O 2 (g) Fe 2 O 3 (s) ∆H = –824.2 kJ

52. Chemical equation for the formation of rust: 2 moles 3/2 moles 1 mole Enthalpy calculations 2Fe(s) + 3/2O 2 (g) Fe 2 O 3 (s) ∆H = –824.2 kJ Rewrite the chemical equation using coefficients with the smallest whole numbers possible

53. Chemical equation for the formation of rust: Enthalpy calculations 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) ∆H = ? x 2 2 moles 3/2 moles 1 mole 2Fe(s) + 3/2O 2 (g) Fe 2 O 3 (s) ∆H = –824.2 kJ 4 moles 3 moles 2 moles What is the enthalpy change for this reaction?

54. Chemical equation for the formation of rust: Enthalpy calculations 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) ∆H = –1,648.4 kJ x 2 2 moles 3/2 moles 1 mole 2Fe(s) + 3/2O 2 (g) Fe 2 O 3 (s) ∆H = –824.2 kJ 4 moles 3 moles 2 moles x 2

55. Enthalpy of formation This is also the chemical equation for the formation of CO 2. ∆H reaction = ∆H formation of CO2 = –393.5 kJ Chemical equation for the combustion of carbon: C(s) + O 2 (g) CO 2 (g)∆H = –393.5 kJ The formation of 1 mole of CO 2 releases 393.5 kJ of energy ∆H f (CO 2 ) = –393.5 kJ/mole

56. Enthalpies of formation of some common substances Knowing these values and the following equation, you can calculate unknown enthalpy values: Enthalpy of formation

57. Enthalpy calculations The complete combustion of glucose (C 6 H 12 O 6 ) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose. C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(g)∆H = –2,808 kJ Asked: ∆H f (glucose) = ? Given: From the table of enthalpies of formation Relationships:

58. 10.4 Chemical Reactions and Energy Enthalpy calculations The complete combustion of glucose (C 6 H 12 O 6 ) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose. reactantsproducts Relationships: Formation of glucose C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(g)∆H = –2,808 kJ 6CO 2 (g) + 6H 2 O(g) C 6 H 12 O 6 (s) + 6O 2 (g) ∆H = +2,808 kJ

59. Remember to multiply by the coefficients! reactantsproducts Formation of glucose 6CO 2 (g) + 6H 2 O(g) C 6 H 12 O 6 (s) + 6O 2 (g) ∆H = +2,808 kJ Enthalpy calculations

60. reactantsproducts Formation of glucose 6CO 2 (g) + 6H 2 O(g) C 6 H 12 O 6 (s) + 6O 2 (g) ∆H = +2,808 kJ Enthalpy calculations

61. Formation of glucose 6CO 2 (g) + 6H 2 O(g) C 6 H 12 O 6 (s) + 6O 2 (g) ∆H = +2,808 kJ Answer: ∆H f (glucose) = –1,004 kJ/mole Asked: ∆H f (glucose) = ? 1 mole The complete combustion of glucose (C 6 H 12 O 6 ) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose. Enthalpy calculations

62. AB + AB ∆H = X

63. AB + AB ABAB + ∆H = –X The reverse reaction changes the sign of ∆H

64. AB + AB ∆H = X + ∆H = 3X x 3 A A A B B B AB AB AB If three times more substances are involved, ∆H is three times greater

65. AB + AB ∆H = X ∆H (reaction) = ∆H f (products) – ∆H f (reactants) ABAB ∆H (reaction) =∆H f – ∆H f + ∆H f

66. Reactants Products ∆H = … kJ Energy profile Thermochemical equation: Energy flow during the reaction have stored energy also have stored energy Energy Progress of reaction We can graph the change in energy as the reaction takes place

67. Energy profile Reactants Products ∆H = … kJThermochemical equation: Energy flow during the reaction have stored energy also have stored energy

68. Activation energy The reaction cannot start without this initial input of energy Energy profile

69. C(s) + O 2 (g) CO 2 (g) Combustion of carbon: Wood does not spontaneously light itself up on fire Energy profile

70. Reaction of sodium in water: Na(s) + H 2 O(l) 2NaOH + H 2 (g) Sodium reacts with water immediately (and violently) upon contact Energy profile

71. Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2). Hess’s law ∆H 1 ∆H 2 ∆H 3 RA BA BP ∆H 4 RP

72. Hess’s law: ∆H 4 = ∆H 1 + ∆H 2 + ∆H 3 Hess’s law ∆H 1 ∆H 2 ∆H 3 RA BA BP ∆H 4 12 RP

73. Hess’s law Asked: C gr (s) C d (s)∆H = ? Given that the enthalpy of combustion for graphite (C gr ) and diamond (C d ) are –393.5 kJ/mole and –395.4 kJ/mole, respectively, calculate the enthalpy of formation of diamond from graphite. Given: C gr (s) + O 2 (g) CO 2 (g)∆H = –393.5 kJ/mole C d (s) + O 2 (g) CO 2 (g)∆H = –395.4 kJ/mole Relationships: Hess’s law Strategy: Create a path that leads from C gr to C d.

74. Hess’s law Asked : C gr (s) C d (s)∆H = ? Given : C gr (s) + O 2 (g) CO 2 (g)∆H = –393.5 kJ/mole C d (s) + O 2 (g) CO 2 (g)∆H = –395.4 kJ/mole C d (s) is a product in:C gr (s) C d (s) C d (s) is a reactant in: CO 2 (g) C d (s) + O 2 (g) ∆H = +395.4 kJ/mole Write the reverse reaction so that C d (s) is a product, and adjust  H: C d (s) + O 2 (g) CO 2 (g) ∆H = –395.4 kJ/mole

75. Hess’s law Asked: C gr (s) C d (s)∆H = ? Given: C gr (s) + O 2 (g) CO 2 (g)∆H = –393.5 kJ/mole C d (s) + O 2 (g) CO 2 (g)∆H = –395.4 kJ/mole CO 2 (g) C d (s) + O 2 (g) ∆H = +395.4 kJ/mole C gr (s) + O 2 (g) CO 2 (g) ∆H = –393.5 kJ/mole Write the sum of the two equations:

76. Hess’s law Asked: C gr (s) C d (s)∆H = ? Given: C gr (s) + O 2 (g) CO 2 (g)∆H = –393.5 kJ/mole C d (s) + O 2 (g) CO 2 (g)∆H = –395.4 kJ/mole CO 2 (g) C d (s) + O 2 (g) ∆H = +395.4 kJ/mole C gr (s) + O 2 (g) CO 2 (g) ∆H = –393.5 kJ/mole Write the sum of the two equations:

77. Hess’s law Asked: C gr (s) C d (s)∆H = ? Given: C gr (s) + O 2 (g) CO 2 (g)∆H = –393.5 kJ/mole C d (s) + O 2 (g) CO 2 (g)∆H = –395.4 kJ/mole CO 2 (g) C d (s) + O 2 (g) ∆H = +395.4 kJ/mole C gr (s) + O 2 (g) CO 2 (g) ∆H = –393.5 kJ/mole Write the sum of the two equations:

78. Hess’s law Asked: C gr (s) C d (s)∆H = ? Given: C gr (s) + O 2 (g) CO 2 (g)∆H = –393.5 kJ/mole C d (s) + O 2 (g) CO 2 (g)∆H = –395.4 kJ/mole CO 2 (g) C d (s) + O 2 (g) ∆H = +395.4 kJ/mole C gr (s) + O 2 (g) CO 2 (g) ∆H = –393.5 kJ/mole Write the sum of the two equations: C gr (s) C d (s) ∆H = (–393.5 + 395.4) kJ/mole ∆H = +1.9 kJ/mole

79. Energy profile of a reactionHess’s law

80. MAKE SURE THAT YOU KNOW THE TERMS FROM SLIDE #3!! 80