1. Unit 6 Chemical Reactions General Chemistry Spring 2010

2. NOMENCLATURE REVIEW Section 1

3. Ionic Basic Rules ◦ These have a cation (+) and an anion (-) ◦ Usually a metal and a nonmetal ◦ Sometimes contain polyatomic ion(s)  (back of PT) ◦ Pay attention to charges  Overall charge on ENTIRE formula has to be ZERO  Use subscripts to add charge to make zero  Criss-cross is a shortcut but be careful! ◦ Practice this on the next slide

4. Basic Ionic Formula Give the formulas for the following compounds w/o PAI ◦ Potassium sulfide ◦ Magnesium oxide Give the formulas w/ PAI ◦ (DON’T CHANGE THE FORMULA) ◦ Aluminum nitrate ◦ Calcium phosphate

5. Transition Metals In the name, charge on TM is in () ◦ Ex) Iron (III) chloride ◦ Ex) Tin (II) fluoride  Use that charge to do criss-cross To figure out the TM charge do reverse criss- cross. The charge is on ONE TM! ◦ Ex) CuO ◦ Ex) Cu 2 O

6. Covalent Nomenclature NO CRISS CROSS!!! NO REDUCE!!! The subscript after the symbol = the prefix in the name The prefix in the name = the subscript in the formula (what are these rules?) 1.Ex) NO 2 2.Ex) N 2 O 3.Ex) Sulfur hexafluoride 4.Ex) Triphosphorus pentachloride Diatomic molecules

7. Acids Every acid formula begins with hydrogen The other half is either a halogen or a polyatomic ion ◦ Halogen? Use “hydro” in name  Ex) HCl  Ex) HF ◦ Polyatomic? Change ending to “-ic” (for this class)  Ex) H 3 PO 4  Ex) H 2 SO 4  # of H’s is the charge on the PAI

8. Section 2 Counting Atoms Review

9. Subscript Indicates the number present Only applies to the element it’s with AlCl 3 ◦ One aluminum and three chlorines

10. Subscripts When an element has no subscript ◦ The implied subscript is ONE ◦ Ag 1 = Ag

11. Coefficient Affects anything behind it Multiply everything by the coefficient 3 AlCl 3 ◦ Three aluminums and nine chlorines

12. Section 3 Reaction Basics

13. Chemical Equations The “sentences” of chemistry Show how elements react with each other and what compounds they will form So that’s what happens when sodium and water mix!!!!

14. Parts of a chemical Equation Reactants ◦ Always on the left of the equation ◦ What the reaction STARTS with Products ◦ Always on the right of the equation ◦ What is produced from the reaction (made) Yield ◦ Arrow ◦ Where the reaction actually occurs

15. Can you show me what you just said??? Mg + O 2  MgO ReactantsProducts Yield Sign

16. Symbols in Equations (Table 11.1) SymbolExplanation +Used to separate 2 reactions or 2 products  “Yields,” separates reactions from products Used in place of  for reversible rxns (s)Designates a R or P in the solid state; placed after the formula (l)A R or P in the liquid state; placed after formula (g)“ ” in the gas state; “ ” (aq)Designates an aqueous solution; the substance is dissolved in water; placed after formula heat Indicates that head is supplied to the rxn Pt A formula written above the yield sign indicates its use as a catalyst (in this example, platinum)

17. Section 4 Inventory Reactions

18. Law of Conservation of Mass The reason we balance reactions Matter is neither created nor destroyed ◦ Must account for all elements  Before and after reaction Antoine-Laurent de Lavoisier was the first to state this law.

19. Reaction Anatomy KClO 3  KCl + O 2 K1 Cl1 O3 K1 Cl1 O2 Reactant(s)Products

20. Inventory Compare the number of each atom in the reactants to the products ◦ If equal = “balanced reaction”  Abides by the law of conservation of mass ◦ If not equal = “not balanced”  Does not abide by the law of conservation of mass; MUST BALANCE KClO 3  KCl + O 2 K1 Cl1 O3 K1 Cl1 O2 “Not balanced” Reactant(s)Products

21. Section 5 Balancing Reactions

22. Balance 1. Inventory first 2. Locate one element that does not balance 3. Add a coefficient to make it balance ◦ NEVER TOUCH A SUBSCRIPT ◦ Look to make odd numbers even 4. Re inventory 5. Repeat until balanced KClO 3  KCl + O 2 K1 Cl1 O3 K1 Cl1 O2 2 2 2 6 32 2 2 6 balanced It’s best to leave H and O for the end

23. Balancing When balancing a chemical equation you may only change the coefficient (big number in front of the substance) Balancing example #1: ◦ Mg (s) + O 2 (g)  MgO (s) 1Mg 1 2 O 1 22 2 / / 2

24. Balance Combustion = 2 CHOR Combustion is any reaction with CO 2 + H 2 O for products 2…coefficient of 2 in front of big hydrocarbon C…balance carbons H…balance hydrogen O…balance oxygen R…reduce if possible (divide by a common factor, like 2) C 7 H 16 + O 2  CO 2 + H 2 O2141622 11178 H C

25. Balancing (cont.) When balancing a chemical equation you may only change the coefficient (big number in front of the substance) Balancing example #3: ◦ NaOH(aq) + H 2 SO 4 (aq)  Na 2 SO 4 (aq) + HOH(l) 1Na 2 1 OH 1 2 H 1 1 SO 4 1 22 / 2 2 / Count polyatomic ions as one piece!

26. Practice Problem #1 Balance the following reactions: ◦ P 4 (s) + O 2 (aq)  P 4 O 10 ◦ Zn + HCl  ZnCl 2 + H 2 (g) ◦ Mg (s) + O 2 (g)  MgO (s) ◦ KClO 3  KCl + O 2 (g) 5 2 2 2 223

27. Section 6 Reaction Types

28. Chemical reactions are classified into 5 different types: ◦ This makes it easier to see what is happening in a reaction ◦ This makes it easier to predict products in a reaction Sorting into piles makes it easier to see the similarities and differences

29. Reaction Types (cont.) Five reaction types ◦ 1) Synthesis (Combination) ◦ 2) Decomposition ◦ 3) Single Replacement ◦ 4) Double Replacement ◦ 5) Combustion Now let’s look at each type individually

30. Please Note The following reactions are not balanced The examples are only used to show the type of reaction

31. Synthesis = S (Note: The book uses the term “combination;” it’s the same thing) Two or more substances react to form one new compound Element + element  new compound compound + compound  new compound Examples ◦ H 2 + N 2  NH 3 ◦ Mg + O 2  MgO (magnesium oxide)

32. Decomposition = D One compound breaks into two or more simpler products Compound  element/compound + element/compound Examples ◦ Na 2 O  Na + O 2 ◦ NH 4 NO 3  N 2 O + H 2 O  Ammonium nitrate, when heated, explosively breaks down!

33. Single Replacement = SR Element + compound  new element + new compound Examples ◦ AlCl 3 + O 2  Al 2 O 3 + Cl 2 ◦ H 2 (SO 4 ) + K  K 2 (SO 4 ) + H 2 ◦ Br 2 + NaI  NaBr + I 2 ◦ Br 2 + NaCl  No Reaction  Halogen activity decreases as you go down group

34. Activity Series Whether one metal displaces another depends on upon the reactivities of the metals A reactive metal will replace any metal listed below it on the activity series ◦ Ex) iron will displace copper from a copper compound, but iron does not similarly displace zinc or calcium NameSymbol Decreasing activity LithiumLi PotassiumK CalciumCa SodiumNa MagnesiumMg AluminumAl ZincZn IronFe LeadPb (Hydrogen)(H)* CopperCu MercuryHg SilverAg * Metals from Li to Na will replace H from acids and water; from Mg to Pb they will replace H from acids only

35. Double Replacement = DR Compound + compound  new compound + new compound Examples ◦ Na 2 O + MgBr 2  NaBr + MgO ◦ H(NO 3 ) + Mg(OH) 2  H(OH) + Mg(NO 3 ) 2 Which switch? Think about Paula Abdul when you rewrite the formulas!

36. Combustion = C An element or a compound reacts with oxygen, usually producing heat and light Always involves oxygen as a reactant Reaction with CO 2 + H 2 O for products Examples ◦ C 6 H 12 O 6 + O 2  CO 2 + H 2 O ◦ 2 C 8 H 18 + 25 O 2  16 CO 2 + 18 H 2 O

37. Visual Review of Types of Reactions Single Replacement Double Replacement Combustion Decomposition Synthesis

38. Section 7 Predicting Products (p.338-339)

39. Synthesis (cont.) Steps to predict products: ◦ 1. Combine the two reactants in one product (switchy switchy ◦ 2. Balance Steps in the Irish Countryside

40. Practice Problem #4 Predict the product and balance: ◦ Mg (s) + O 2 (g)  ◦ Be (s) + Br 2 (g)  ◦ Cs (s) + S 2 (g)  MgO (s) BeBr 2 (s) Cs 2 S (s) 2 2 4 2

41. Decomposition (cont.) Steps to predict products: ◦ 1. Break the one reactant into two products  Don’t forget about diatomic molecules H, N, O, F, Cl, Br, I ◦ 2. Balance Ancient Steps in Cancun, Mexico

42. Practice Problem #5 Predict the products and balance: ◦ MgCl 2 (s)  ◦ FeS (s)  ◦ NaI (s)  Mg (s) + Cl 2 (g) Fe (s) + S (s) Na (s) + I 2 (s) 2 2

43. Single Replacement (cont.) Steps to predict products: ◦ 1. Figure out which metal is going to replace which other metal ◦ 2. Write the products:  One metal is now by itself  One metal is now part of a compound (Switchy Switchy) ◦ 3. Balance Steps to the House of the Ñusta at Machu Picchu

44. Practice Problem #2 Predict the products and balance: ◦ K (s) + MgS (aq)  ◦ Ba (s) + Au(C 2 H 3 O 2 ) 3 (aq)  ◦ Zn (s) + HCl (aq)  Mg (s) + K 2 S (aq) Au (s) + Ba(C 2 H 3 O 2 ) 2 (aq) H 2 (s) + ZnCl 2 (aq) 2 3 322 2

45. Double Replacement (cont.) Steps to predict products: ◦ 1. Figure out which metal is going to trade partners with which other metal ◦ 2. Write the products:  One metal is now in a compound with the other anion (switchy switchy) ◦ 3. Balance Steps on a Sand Dune at the edge of the Gobi Desert

46. Practice Problem #3 Predict the products and balance: ◦ FeS (s) + HCl (aq)  ◦ CaCl 2 (aq) + H 2 SO 4 (aq)  ◦ NH 4 I (aq) + AgNO 3 (aq)  FeCl 2 (aq) + H 2 S (g) HCl (aq) + CaSO 4 (s) NH 4 NO 3 (aq) + AgI (s) 2 2

47. Combustion (cont.) Steps to predict products: ◦ 1. Write CO 2 and H 2 O as the products ◦ 2. Balance…2CHOR  1 st balance C  2 nd balance H  3 rd balance O Steps to the Lincoln Memorial

48. Practice Problem #6 Predict the products and balance: ◦ C 3 H 8 (g) + O 2 (g)  ◦ C 5 H 12 O (s) + O 2 (g)  ◦ C 4 H 10 (s) + O 2 (g)  CO 2 (g) + H 2 O (l) 534 2151012 2 13 8 10

49. Section 8 Reaction Rates and Equilibrium

50. Energy Basics For a reaction to proceed… ◦ Reactants’ bonds must break ◦ Bonds must reform to make products Energy is required to break reactants apart (their bonds) Reactions either gain or lose energy, they never stay the same

51. Activation Energy Bonds must break in order to reform Energy required to break bonds = ACTIVATION ENERGY All reactions require this The activation energy is always positive

52. Energy Diagram Reactants  Products NaBr + Li(OH)  LiBr + Na(OH) Energy in Kilojoules (kJ) Time Energy of reactants Energy of products

53. Energy Diagram Reactants  Products NaBr + Li(OH)  LiBr + Na(OH) Energy in Kilojoules (kJ) Time Reactants Products The difference between the energy of reactants and products = The total heat or energy of the reaction OR ΔH (change in heat)

54. Activation Energy Reactants Products (bonds have reformed) NaBr + Li(OH) LiBr + Na(OH) Na +1 Li +1 Br -1 (OH) -1 Bonds are broken Takes Energy The energy required to break the reactants’ bonds = ACTIVATION ENERGY Reactants  Products NaBr + Li(OH)  LiBr + Na(OH)

55. Activated Complex Energy in Kilojoules (kJ) Time Na +1 Li +1 Br -1 (OH) -1 Bonds are broken Reactants  Products NaBr + Li(OH)  LiBr + Na(OH) Reactants Products The point at which all bonds have been broken and products begin to reform = ACTIVATED COMPLEX

56. Summary Energy in Kilojoules (kJ) Time Reactants  Products NaBr + Li(OH)  LiBr + Na(OH) Reactants Products Activated Complex Activation Energy Total heat of reaction or ΔH

57. Rates of Reaction Ways to make reactions happen faster: ◦ Make chemicals at higher concentration ◦ Increase the temperature ◦ Make the particles smaller ◦ Add a catalyst

58. Rates of Reactions (cont.) Catalysts lower the activation energy

59. Heats of Reaction Reactions can either… ◦ Give off heat  Exothermic  The energy level at the end of the reaction is lower than the energy level at the beginning of the reaction  Therefore, the change in heat (∆H) is negative ◦ Take in heat  Endothermic  The energy level at the end of the reaction is higher than the energy level at the beginning of the reaction  Therefore, the change in heat (∆H) is positive Feels Hot Feels Cold

60. Exothermic Reaction ◦ Reaction loses heat ◦ Δ H value is always negative Endothermic Reaction ◦ Reaction gains heat ◦ Δ H value is always positive energy time energy time Exo vs Endothermic Reactions

61. Bond Formation (cont.) Example #1 ◦ Endo or exothermic? ◦ Energy of the Activated complex? ◦ ∆H? Reactants (500 kJ) Products (200 kJ) Act. Energy (150 kJ) Exothermic 650 kJ -300 kJ

62. Bond Formation (cont.) Example #2 ◦ Endo or exothermic? ◦ Activation energy? ◦ Energy of products? Endothermic 750 kJ 600 kJ Reactants (200 kJ) ∆H (400 kJ) Activated Complex (950 kJ)