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Introductory Chemistry, 3rd Edition Nivaldo Tro Chapter 8 Quantities in Chemical Reactions Car an octane and oxygen molecules and carbon dioxide and water Roy Kennedy Massachusetts Bay Community College Wellesley Hills, Maqqwertd ygoijpk[l 2009, Prentice ‗? Hall Outline • • • • • 8.1 Global Warming : Too much Carbon Dioxide 8.2 Making Pancakes: Relationships between Ingredients 8.3 Making Molecules: Mole to Mole Conversions 8.4 Making Molecule: Mass to Mass Conversions 8.5 More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield • 8.6 Limiting Reactant, Theoretical Yield, and Percent Yield form Initial Reactants • 8.7 Enthalpy: Measure of the Heat Evolved or Absorbed during a Chemical Reaction Tro's ―Introductory Chemistry‖, Chapter 8 2 8.1 Global Warming : Too much Carbon Dioxide Tro's ―Introductory Chemistry‖, Chapter 8 3 Burning of Fossil Fuels Car and molecules from chapter opening 2C8H18 (l) + 25O2(g) Internal combustion engines produe Carbon dioxide by burning fuels. For example octane, C8H18, in gasoline. 16 CO2 (g) + 18H2O(g) Global Warming • Average 0.6 °C rise in atmospheric temperature since 1860. • In the same period atmospheric CO2 levels have risen 25%. • Are the two related? Figure 8.1 and Figure 8.2 Showing global tempertature Rise and greenhouse effect. Tro's ―Introductory Chemistry‖, Chapter 8 5 Increased CO2 • The primary source of the increased CO2 levels is combustion of fossil fuels 1860 = Industrial Revolution in the U.S. and Europe. Methane in natural gas CH4(g) 2 O2(g) CO2(g) 2 H2O(g) Tro's ―Introductory Chemistry‖, Chapter 8 6 Quantities in Chemical Reactions • All substances in a chemical reaction are related. • The numerical relationships between compounds in a chemical reaction is called stoichiometry. Tro's ―Introductory Chemistry‖, Chapter 8 7 8.2 Making Pancakes: Relationships between Ingredients Tro's ―Introductory Chemistry‖, Chapter 8 8 Making Pancakes • Pancakes are made with specific amounts of ingredients. Your book forgot buttermilk! Pancake recipe figure on page 244 1 cup flour + 2 eggs + ½ tsp baking powder +3/4 cup buttermilk 5 pancakes Tro's ―Introductory Chemistry‖, Chapter 8 9 Making Pancakes, Continued • You can scale the recipe up of down. • Notice that for every 2 eggs 5 pancakes • What if you have 8 eggs ? pancakes 5 pancakes 8 eggs 20 pancakes 2 eggs Graphics illustrating the above equation Tro's ―Introductory Chemistry‖, Chapter 8 10 8.2 Making Pancakes: Relationships between Ingredients Tro's ―Introductory Chemistry‖, Chapter 8 11 Making Molecules Mole-to-Mole Conversions • The balanced equation = chemical ―recipe‖ • 3 H2(g) + N2(g) 2 NH3(g) 3 molecules + 1 molecule 2 molcules H2 N2 NH3 H-H H-H H-H + N-N H H-N-H H H-N-H Tro's ―Introductory Chemistry‖, Chapter 8 12 Molecule to Molecule and Mole to Mole conversions • 3 H2(g) + N2(g) 2 NH3(g) • This equation may be scaled up if the ratio of H2:N2:NH3 is kept 3:1:2 • Multiply by 4 • 12 H2(g) + 4N2(g) 8 NH3(g) • Multiply by 6.022 x 1023 • 3x(6.022 x1023) H2(g) + 1x(6.022 x1023)N2(g) 2x(6.022 x1023)NH3(g) • This number of molecules can be expressed as moles. • 3 mol H2(g) + 1 mol N2(g) 2 mol NH3(g) Tro's ―Introductory Chemistry‖, Chapter 8 13 The coefficients in a balanced equation give the number of molecules as well as the number of moles of each substance We can use the ratios of the coefficients to convert between moles of substances in a chemical reaction 8.3 Making Molecules: Mole to Mole Conversions Tro's ―Introductory Chemistry‖, Chapter 8 15 Molar Ratios •3 H2(g) + N2(g) 2 NH3(g) •This equation has 6 molar ratios relating the reactant and products. •For example H2 and N2 are related by 1 mole N2 3 mol H2 or 3 mol H2 1 mole N2 Mole to Mole Conversions • 3 H2(g) + N2(g) 2 NH3(g) • 12 moles of H2 requires how many mole of N2 • The ratio must be H2:N2 is 3:1, so so 12:4 is the same so 4 mole of N2 is needed. • How many moles of N2 are needed for 1.74 mol of H2 Tro's ―Introductory Chemistry‖, Chapter 8 17 Mole to Mole Conversion • • • • 3 H2(g) + N2(g) 2 NH3(g) How many moles of N2 are needed for 1.74 mol of H2? We need to use the molar ratio. mol H2 mole N2 1 mole N2 1.74 mol H2 x = 0.580 mole N2 3 mol H2 Tro's ―Introductory Chemistry‖, Chapter 8 18 8.4 Making Molecule: Mass to Mass Conversions Tro's ―Introductory Chemistry‖, Chapter 8 19 Tro's ―Introductory Chemistry‖, Chapter 8 20 Making Molecules Mass-to-Mass Conversions • We learned previously to convert between moles and grams using the molar mass. • Combining this with moles to moles conversions allows us to related grams of one substrance in an equation to grams of another substance. g of A mol of A Molar mass of A mol of B Coefficients of the Balanced Equation g of B Molar mass of B 21 Example 8.2—How Many Grams of Glucose Can Be Synthesized from 58.5 g of CO2 in Photosynthesis? • Photosynthesis: 6 CO2(g) + 6 H2O(g) C6H12O6(s) + 6 O2(g) g of CO2 mol of CO2 mol of C6H12O6 g of C6H12O6 Molar masses needed (sum masses of all atoms) CO2 = 44.01 g/mol Glucose = 180.16 g/mol Tro's ―Introductory Chemistry‖, Chapter 8 22 Grams moles moles grams 6 CO2(g) + 6 H2O(g) C6H12O6(s) + 6 O2(g) 58.5 g CO2 1 mol CO2 x = 1.33 mol CO2 44.01 g CO2 1.33 mol g CO2 1 mol C6H12O6 x 6 mol CO2 = 0.222 mol C6H12O6 0.222 mol g CO2 180.16 g C6H12O6 x = 1 mol C6H12O6 40.0 g C6H12O6 Another stoichiometry example How many g of Al2(SO4)3 are required to completely react with 24.7 g of Ba(NO3)2? Al2(SO4)3(aq) + 3Ba(NO3)2(aq) 3BaSO4(s) + 2Al(NO3)3(aq) g of Ba(NO3)2 mol of Ba(NO3)2 mol of Al2(SO4)3 molar masses needed Ba(NO3)2 = 261.34g/mol Al2(SO4)3 = 342.15 g/mol g of Al2(SO4)3 Grams moles moles grams Al2(SO4)3(aq) + 3Ba(NO3)2(aq) 24.7 g Ba(NO3)2 3BaSO4(s) +Al(NO3)3(aq) 1 mol Ba(NO3)2 x = 0.0945 mol Ba(NO3)2 261.34 g Ba(NO3)2 0.0945 mol Ba(NO3)2 1 mol Al2(SO4)3 x = 0.0315 mol Al2(SO4)3 3 mol Ba(NO3)2 0.0315 mol g Al2(SO4)3 342.15 g Al2(SO4)3 x = 10.8 g Al2(SO4)3 1 mol Al2(SO4)3 8.5 More Pancakes: Limiting Reactant, Theoretical Yield, and Percent Yield 8.6 Limiting Reactant, Theoretical Yield, and Percent Yield form Initial Reactants Tro's ―Introductory Chemistry‖, Chapter 8 26 Note in these two sections, we are only going to cover percent yield Actual; yield % Yield = x 100 Theoretical yield Actual yield – amount actually obtained Theoretical yield – amount obtained if reaction goes to 100% completion. Tro's ―Introductory Chemistry‖, Chapter 8 27 % yield example A is ran with 100 g of Hg reacting with oxygen. 75.3 grams of HgO is isoloted. What is the % yield? Hg(s) + O2(g) Molar masses: Hg = 200.59 g/mol HgO = 216.59 g/mol HgO(s) 28 % yield example (cont) • First calculate the theoretical yield. • g Hg mol Hg mol HgO g HgO • 100 g Hg 1 mol Hg x = 0.499 mol Hg 200.59 g Hg • 0.499 mol Hg 1 mol HgO x = 0.499 mol HgO 1 mol Hg • 0.499 mol Hg 216.59 g HgO x = 108 g HgO (theoretical 1 mol HgO yield) 29 % yield example 75.0 g HgO % Yield = x 100 108 g HgO = 69.4 % Yield of HgO . Tro's ―Introductory Chemistry‖, Chapter 8 30 Enthalpy Change • We previously described processes as exothermic if they released heat, or endothermic if they absorbed heat. • The enthalpy of reaction is the amount of thermal energy that flows through a process. At constant pressure. DHrxn Tro's ―Introductory Chemistry‖, Chapter 8 31 Sign of Enthalpy Change • For exothermic reactions, the sign of the enthalpy change is negative when: Thermal energy is produced by the reaction. The surroundings get hotter. DH = ─ For the reaction CH4(s) + 2 O2(g) CO2(g) + 2 H2O(l), the DHrxn = −802.3 kJ per mol of CH4. • For endothermic reactions, the sign of the enthalpy change is positive when: Thermal energy is absorbed by the reaction. The surroundings get colder. DH = + For the reaction N2(s) + O2(g) 2 NO(g), the DHrxn = +182.6 kJ per mol of N2. 32 Enthalpy and Stoichiometry • The amount of energy change in a reaction depends on the amount of reactants. You get twice as much heat out when you burn twice as much CH4. • Writing a reaction implies that amount of energy changes for the stoichiometric amount given in the equation. For the reaction C3H8(l) + 5 O2(g) 3 CO2(g) + 4 H2O(g) DHrxn = −2044 kJ So 1 mol C3H8 5 mol O2 3 mol CO2 4 mol H2O −2044 kJ. Tro's ―Introductory Chemistry‖, Chapter 8 33 Enthalpy Calculation • C3H8(l) + 5 O2(g) 3 CO2(g) + 4 H2O(g) DHrxn = −2044 kJ • How much heat is released when 53.8 g of O2 completely reacts with propane? • g O2 mol O2 heat (J) • 53.8 g of O2 1 mol O2 x = 1.68 mol O2 32.00 g O2 • 1.68 mol of O2 -2044 J x = 687 J of heat 5 mol O2 34 Example 8.7—How Much Heat Is Associated with the Complete Combustion of 11.8 x 103 g of C3H8(g)? Given: Find: Solution Map: 11.8 x 103 g C3H8, heat, kJ g C3H8 mol C3H8 1 mol C3H8 44.09 g Relationships: kJ - 2044 kJ 1 mol C3H 8 1 mol C3H8 = -2044 kJ, Molar mass = 44.11 g/mol Solution: 11.8 103 g C3H8 1 mol C3H8 - 2044 kJ 5.47 105 kJ 44.11 g C3H8 1 mol C3H8 Check: The sign is correct and the value is reasonable. Practice—How Much Heat Is Evolved When a 0.483 g Diamond Is Burned? (DHcombustion = −395.4 kJ/mol C) Tro's ―Introductory Chemistry‖, Chapter 8 36