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Chapter 8

Chapter 8. Chemical Reactions. Indications of a Chemical Reaction. Evolution of heat and light Production of a gas Formation of a precipitate Color change. All chemical reactions. have two parts Reactants - the substances you start with Products - the substances you end up with

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Chapter 8

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  1. Chapter 8 Chemical Reactions

  2. Indications of a Chemical Reaction • Evolution of heat and light • Production of a gas • Formation of a precipitate • Color change

  3. All chemical reactions • have two parts • Reactants - the substances you start with • Products- the substances you end up with • Reactants yield Products

  4. Characteristics of Chemical Equations • Represent the known facts • Contain correct formulas and symbols • Satisfy the law of conservation of mass (they must be balanced using coefficients)

  5. Word and Formula Equations • Copper reacts with oxygen to form copper (II) oxide. • word equation • Copper + oxygen ® copper (II) oxide • formula equation • Cu + O2® CuO • balanced chemical equation • 2Cu + O2® 2CuO

  6. Symbols Used In Chemical Equations •  yields • + and • (s) or (cr) solid • (g) gas • (l) liquid • (aq) aqueous solution

  7. Symbols Used In Chemical Equations • ­ gaseous product • ¯ precipitate • reversible reaction • reactants are heated • A catalyst used to change the rate of the reaction

  8. What is a catalyst? • A substance that speeds up a reaction without being changed by the reaction. • Enzymes are biological or protein catalysts.

  9. What is a reversible reaction? • A chemical reaction in which the products reform the original reactants.

  10. Write the formula equation • Solid iron (III) sulfide reacts with gaseous hydrogen chloride to form solid iron (II) chloride and hydrogen sulfide gas. • Nitric acid dissolved in water reacts with solid sodium carbonate to form liquid water and carbon dioxide gas and sodium nitrate dissolved in water.

  11. Write the word equations • Fe(s) + O2(g) ® Fe2O3(s) • Cu(s) + AgNO3(aq) ® Ag(s) + Cu(NO3)2(aq) • NO2 (g) N2(g) + O2(g)

  12. Balancing Chemical Equations A step by step approach

  13. Balanced Equation • Atoms can’t be created or destroyed • All the atoms we start with we must end up with • A balanced equation has the same number of each element on both sides of the equation.

  14. ® O + C C O O O • Carbon + oxygen yields carbon dioxide • C + O2® CO2 • This equation is already balanced

  15. ® O + C C O O • Carbon + oxygen yields carbon monoxide • C + O2® CO • We need one more oxygen in the products. • Can’t change the formula

  16. C O ® O + C O • Must be another CO • But where did the other C come from? C O

  17. C C O ® O + O • Must have started with two C • 2 C + O2® 2 CO C O C

  18. Rules for balancing • Write the correct formulas for all the reactants and products • Count the number of atoms of each type appearing on both sides • Balance the elements one at a time by adding coefficients (the numbers in front) • Balance polyatomic ions that appear on both sides of the equation as single units

  19. Always • Recount atoms as coefficients are added or changed • Reduce the coefficients to represent the smallest whole number ratio of reactants and products

  20. Never • Add, delete, or change a subscript to balance an equation. • Never put a coefficient in the middle of a formula

  21. Example H2 + O2 ® H2O Make a table to keep track of atoms

  22. Example H2 + O2 ® H2O 2 H 2 2 O 1 Need twice as much O in the product

  23. Example H2 + O2 ® 2 H2O 2 H 2 2 O 1 Changes the O

  24. Example H2 + O2 ® 2 H2O 2 H 2 2 O 1 2 Also changes the H

  25. Example H2 + O2 ® 2 H2O 2 H 2 4 2 O 1 2 Need twice as much H in the reactant

  26. Example 2 H2 + O2 ® 2 H2O 2 H 2 4 2 O 1 2 Recount

  27. Example 2 H2 + O2 ® 2 H2O 4 2 H 2 4 2 O 1 2 The equation is balanced, has the same number of each kind of atom on both sides

  28. Example 2 H2 + O2 ® 2 H2O 4 2 H 2 4 2 O 1 2 This is the answer Not this

  29. Examples • P. 252 – example problem 8-3 • P. 252 – practice problems • P. 253 – example 8-4,8-5 • P. 254 – practice problems • H.W. p. 254 Section Review P. 270 # 18,19, 22 Website for balancing equations http://funbasedlearning.com/chemistry Classic Chembalancer

  30. Types of Reactions5 Basic Types(pages 256-264) Predicting the Products

  31. Synthesis • the combination of 2 or more substances to form a compound • Two reactants one product A + B  AB

  32. Synthesis H2(g) + Cl2(g)  2 HCl(g)

  33. Synthesis • 2 elements, or compounds combine to make one compound. • Three types: • Reaction of elements + oxygen or sulfur • Reaction of metals + halogens • Reactions with oxides

  34. Synthesis • Examples (see pages 256-258)

  35. Decomposition • a compound breaks down into 2 or more simpler substances • one reactant two products AB  A + B

  36. Decomposition • one reactant breaks apart into two or more elements or compounds. • Usually requires heat or electricity • A catalyst may be used to speed up the reaction • The decomposition of a substance by electricity is called electrolysis

  37. Decomposition • Can predict the products easily if it is a binary compound • 2H2O 2H2 + O2 • 2NaCl 2Na + Cl2

  38. Decomposition 2 H2O(l)  2 H2(g) + O2(g)

  39. Decomposition • decomposition of metal carbonates • decomposition of metal hydroxides • decomposition of metal chlorates • decomposition of acids.

  40. Decomposition • Examples (see pp. 259-260) h.w. handout

  41. Single Replacement • one element replaces another in a compound • metal replaces metal (+) and hydrogen • nonmetal replaces nonmetal (-) A + BC  B + AC

  42. Single Replacement Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)

  43. Single Replacement • Examples: (see p. 261-262)

  44. Double Replacement • ions in two compounds “change partners” • cation of one compound combines with anion of the other AB + CD  AD + CB

  45. Double Replacement Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KNO3(aq)

  46. Double Replacement • Formation of a precipitate, gas, or water • Examples ( see pages 262-263) • H.w. handout

  47. Combustion • the burning of any substance in O2 to produce a large amount of energy in the form of heat and light CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

  48. Combustion • Combustion of a hydrocarbon with oxygen • the products will be CO2 and H2O. (may get C and CO under less than perfect conditions) Balance C, H, then O

  49. Examples • C2H6 + O2® • C3H8 + O2® • C4H10 + O2® • C7H16 + O2®

  50. Homework • P. 269 # 1-11 • P. 264 # 1-4 • P. 270 # 20, 21 • P. 271 # 25-29, 33 • P. 272 # 44, 48

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