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CHGS105 Chapter 17 Chemical Reactions

CHGS105 Chapter 17 Chemical Reactions. Chemical Changes Chemical Equations Types of Reactions Oxidation-Reduction Moles Energy in Chemical Reactions Reaction Rates. What is Chemistry?. “The study of Matter and its Changes .”. Physical Changes = Changes in a Physical Property.

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CHGS105 Chapter 17 Chemical Reactions

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  1. CHGS105 Chapter 17Chemical Reactions • Chemical Changes • Chemical Equations • Types of Reactions • Oxidation-Reduction • Moles • Energy in Chemical Reactions • Reaction Rates

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

  3. Chemical Changes • Change in the Chemical Composition Examples: Burning of Magnesium Rusting of Iron Decomposing of wood Souring of Milk

  4. Chemical Reactions • Gas formed • Solid precipitate formed • Color change • Temperature Change • Gives heat = exothermic • Gets cold = endothermic

  5. Chemical Equations • Shows how the Chemical change occurs. Reactants Products  C3H8 + O2 CO2 + H2O + Energy Mg + O2 MgO + Energy Fe + O2 Fe2O3

  6. Chemical equations Chemist’s shorthand to describe a reaction. heat 2 H2 + O2 2 H2O + E (g) (g) (g) • Reactants • Products • The state of all substances (g) (l) (s) (aq) • Any conditions used in the reaction • Same # & type atoms on each side • Law of Conservation of Matter

  7. Law of Conservation of Matter Question:

  8. Law of Conservation of Matter Solution: Fe + O2 Fe2O3

  9. Law of Conservation of Matter Question:

  10. Law of Conservation of Matter Solution:

  11. Balancing Equations Making Hot dogs: How many packages wieners & buns to buy so none is left over. 4 5 40 ___W10 + ___B8 ___WB • Reactants • Products

  12. Balancing Equations Ca + HCl CaCl2 + H2 • Reactants • Products 1 1 1 Ca H Cl 1 2 2 Step 1: Count atoms of each element on both sides of equation.

  13. Balancing Equations Ca + HCl CaCl2 + H2 • Reactants • Products 1 1 1 Ca H Cl 1 2 2 - not balanced - not balanced Step 2: Determine which atoms are not balanced.

  14. Balancing Equations 2 Ca + HCl CaCl2 + H2 • Reactants • Products 1 1 1 Ca H Cl 1 2 2 - not balanced 2 - not balanced 2 Step 3: Balance one element at a time with coefficients in front of formulas until all balanced. (Never change the formula!)

  15. Balancing Equations Na3PO4 + MgCl2Mg3(PO4)2 + NaCl • Reactants • Products 3 1 4 1 2 Na P O Mg Cl 1 2 8 3 1 Step 1: Count atoms of each element on both sides of equation.

  16. Balancing Equations Na3PO4 + MgCl2Mg3(PO4)2 + NaCl • Reactants • Products 3 1 4 1 2 Na P O Mg Cl 1 2 8 3 1 - not balanced - not balanced - not balanced - not balanced - not balanced Step 2: Determine which atoms are not balanced.

  17. Balancing Equations Na3PO4 + MgCl2Mg3(PO4)2 + NaCl • Reactants • Products 3 1 4 1 2 Na P O Mg Cl 1 2 8 3 1 - not balanced - not balanced - not balanced - not balanced - not balanced Step 3: Balance elements with #’s in front of formulas until all balanced. (Never change the formulas!)

  18. Balancing Equations 2 3 1 6 Na3PO4 + MgCl2Mg3(PO4)2 + NaCl • Reactants • Products 3 1 4 1 2 Na P O Mg Cl 1 2 8 3 1 - not balanced 6 6 - not balanced 2 - not balanced 8 - not balanced 3 - not balanced 6 6 • Hints: • Start with a metal in a complex compound, or an element that only appears in one formula. (ie Mg here)

  19. Balancing Equations C2H6 + O2CO2 + H2O • Reactants • Products C H O • Hints: • Start with an element that only appears in one formula on both sides of the equation. • Leave oxygen until last.

  20. Balancing Equations C2H6 + O2CO2 + H2O • Reactants • Products 2 6 2 C H O 1 2 3 Step 1: Count atoms of each element on both sides of equation.

  21. Balancing Equations C2H6 + O2CO2 + H2O • Reactants • Products 2 6 2 C H O 1 2 3 - not balanced - not balanced - not balanced Step 2: Determine which atoms are not balanced.

  22. Balancing Equations 3.5 2 3 C2H6 + O2CO2 + H2O • Reactants • Products 2 6 2 C H O 1 2 3 2 - not balanced - not balanced 6 - not balanced 7 5 7 Step 3: Balance one element at a time with coefficients in front of formulas until all balanced. (Never change the formula!)

  23. Balancing Equations 3.5 2 3 C2H6 + O2CO2 + H2O • Reactants • Products 2 6 2 C H O 1 2 3 2 6 7 5 7 Can’t have 3.5 O2 , so multiply equation by 2!

  24. Balancing Equations 3.5 7 4 6 2 C2H6 + O2CO2 + H2O • Reactants • Products 2 6 2 C H O 1 2 3 4 2 4 12 6 12 14 7 5 7 14 Can’t have 3.5 O2 , so multiply equation by 2!

  25. Balancing Equations Practice: ___O2 ____O3 ___Cr + ___O2 ____Cr2O3 ___HNO2 ___HNO3 + ___NO + __ H2O ___N2H4O3 ___N2 + ___O2 + ___H2O ___N2O  ____N2 + ____O2 ___NO + ___O2 ____NO2 ___CH4 + ___O2 --> CO2 + ____H2O

  26. Balancing Equations Solutions: ___O2 ____O3 ___Cr + ___O2 ____Cr2O3 ___HNO2 ___HNO3 + ___NO + __ H2O __N2H4O3 __N2 + __O2 + __H2O ___N2O  ____N2 + ____O2 ___NO + ___O2 ____NO2 ___CH4 + ___O2 --> CO2 + ____H2O 3 2 4 3 2 3 2 4 2 2 2 2 2 2 2

  27. Types of Chemical Reactions Combustion Complete: C3H8 + 5O2 3CO2 + 4H2O Incomplete: 2C3H8 + 7O2 6CO + 8H2O C3H8 + 2O2 3C + 4H2O

  28. 2 H2O2 2 H2O + O2 Combination Reactions A + B  C Exposion of Hydrogen Balloon 2H2 + O2 2H2O Rusting of Iron 4 Fe + 3 O2 2 Fe2O3 Decomposition Reactions C  A + B Blood with peroxide

  29. Single Replacement Reactions A + BX B + AX Iron Deposits on an Aluminum Pan Al + FeCl3  Fe + AlCl3 Double Replacement Reaction AX + BY  BX + AY BaCl2(aq) + Na2SO4(aq)  BaSO4(s) + 2NaCl(aq) Insoluble Precipitate Formed

  30. The Mole 2 1 pair = 1 dozen = 1 mole = 12 6.02 x 1023 602,000,000,000,000,000,000,000. 1 mol eggs___ 6.02 x 1023eggs 1 mol Au_______ 6.02 x 1023 Au atoms _____1 mole H2O_____ 6.02 x 1023 H2O molecules

  31. The Mole & Formulas 1 car ___ 4 wheels 1 doz cars 4 doz wheels 1 mol cars_ 4 mol wheels 1 mole H2O 2 mol H 1 mole H2O 1 mol O

  32. The Mole & Formulas 1 car ___ 4 wheels 1 doz cars 4 doz wheels 1 mol cars_ 4 mol wheels 5 doz cars 1 4 doz wheels 1doz cars = wheels 20 doz = mol H 10 5 mol H2O 1 2 mol H 1 mol H2O

  33. The Mole & Molecular Mass 1 mole = MW in g’s 1 mole S = 32 g S 1 mol S_ 32 g S 32 g S 1 mol S __32 g S _ 6.02 x 1023 atoms S 1 mole C = 12 g C 1 mol C 12 g C 12 g C 1 mol C __12 g C _ 6.02 x 1023 atoms C

  34. The Mole & Molecular Mass 1 mol H2O_ 18.0 g H2O 1 mole H2O has: 2.0 g H 2 mol H 1 1.0 g H = 1 mol H 18.0 g H2O1 mol H2O 16.0 g O 1 mol O 1 16.0 g O = 1 mol O 18.0 g

  35. Mass to Mole Conversions How many moles of water are in 36 g H2O? What should the answer look like? What is Unique to the problem? mol H2O 2.0 • 1 mol H2O = • 18 g H2O • 36 g H2O • 1 18.0 g H2O1 mol H2O 1 mol H2O_ 18.0 g H2O

  36. Mass to Mole Conversions How many molecules of water are in 36 g H2O? What should the answer look like? What is Unique to the problem? = moleculesH2O 1.2 x 1024 molecules H2O 1 molH2O 18 g H2O 36 g H2O 1 6.02x1023molecules H2O_ 1 mol H2O 1 mol H2O_ 18.0 g H2O 1 mol H2O_ 6.02x1023molecules H2O

  37. Ratios in Chemical Equations C(s)+ O2 (g)CO2(g) 1 molC 1 molO2 1 molC 1 molCO2 1 molO2 1 molCO2 12 g C 32 g O2 12 g C 44 g CO2 32 g O2 44 g CO2 12 gsof C + 32 gs of O2  44 g CO2

  38. Ratios in Chemical Equations C(s)+ O2 (g)CO2(g) 1 molC 1 molO2 1 molC 1 molCO2 1 molO2 1 molCO2 12 g C 32 g O2 12 g C 44 g CO2 32 g O2 44 g CO2 How many gs of O2 are needed to produce 11 g CO2?

  39. Ratios in Chemical Equations C(s)+ O2 (g)CO2(g) How many gs of O2 are needed to make 11 g CO2? • If it takes 32 g O2 to make 44 g CO2 then: 32 g O2 44 g CO2 X g O2 11 g CO2 = 11 g CO2 32 g O2 44 g CO2 X g O2 11 g CO2 11 g CO2 = = g O2 8 1 molO2 1 molCO2 32 g O2 44 g CO2

  40. Ratios in Chemical Equations C(s)+ O2 (g)CO2(g) How many gs of O2 are needed to make 11 g CO2? Factor conversion method: • Identify any conversion factors. • What is unique to the problem? 11 g CO2 32 g O2 44 g CO2 8 = g O2 • How should the answer look? 1 molO2 1 molCO2 32 g O2 44 g CO2

  41. Ratios in Chemical Equations C(s)+ O2 (g)CO2(g) How many gs of O2 are needed to make 11 g CO2? 11 g CO2 32 g O2 44 g CO2 = g O2 8 ? gs C + 8 gsO2  11 g CO2 3 So: Then could : 5 gs C + 8 gsO2  13 g CO2 ? NO! There is only enough O2 to make 11 g’s of CO2. We would still make 11 g’s of CO2 but have 2 g’s of C left over

  42. Moles in Chemical Equations 2 Na(s)+ Cl2 (g)2 NaCl 2 mol Na 1 mol Cl2 2 mol Na 2 mol NaCl 1 mol Cl2 2 mol NaCl 46 g Na 71 g Cl2 46 g Na 58.5 g NaCl 71 g Cl2 58.5 g NaCl How many gs of Cl2 are needed to produce 2.5 mol NaCl?

  43. Moles in Chemical Equations 2 Na(s)+ Cl2 (g)2 NaCl How many gs of Cl2 are needed to produce 2.5 mol NaCl? • How many moles of Cl2 are needed? 1 mol Cl2 2 mol NaCl 1.25 X mol Cl2 2.5 mol NaCl = • How many grams of Cl2 are needed? 1 mol Cl2 71 g Cl2 mol Cl2 X g Cl2 1.25 = = 89 gCl2 1 mol Cl2 2 mol NaCl 1 mol Cl2 70.9 g Cl2

  44. Moles in Chemical Equations 2 Na(s)+ Cl2 (g)2 NaCl How many gs of Cl2 are needed to produce 2.5 mol NaCl? Factor conversion method: • Identify any conversion factors. • What is unique to the problem? 2.5 mol NaCl 1 mol Cl2 2 mol NaCl 70.9 g Cl2 1 mol Cl2 = gCl2 88.625 = 89 gCl2 • How should the answer look? 1 mol Cl2 2 mol NaCl 1 mol Cl2 70.9 g Cl2

  45. Bond Energy 2H2 + O2 2H2O + Energy 2H-H + O=O  2 H-O-H 2 (436) 498 4 (464) 1370 kJ used - 1856 kJ made = -486 kJ made

  46. Energy in Chemical Reactions 2H2 + O2 2H2O + Energy Exothermic reaction Energy Reactants Eact= Activation Energy H2 + O2 -H= heat of reaction (Gets hot) -486 kJ made H2O Products more stable Rxn Progress

  47. Exothermic Reactions 2H2 + O2 2H2O + Energy -486 kJ made 2Mg + O2 2MgO + Energy Energy CH4 + 2O2CO2 + 2H2O +Energy Rxn Progress

  48. Energy in Chemical Reactions Endothermic reaction Eact= Activation Energy Products Energy +H= heat of reaction (Gets cold) Reactants more stable Rxn Progress

  49. Reaction Rates 2H2(g) + O2(g) 2H2O + Energy H2(g)+ O2(g)may stay together for lifetime without reacting to form water. Energy Rxn Progress Just because something has the potential to react doesn’t mean it will do so immediately.

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