CHEMISTRY Principles and Modern Applications

Transcription

1 Petrucci Harwood Herring Madura EditionGENERAL Ninth CHEMISTRY Principles and Modern Applications Chapter 4: Chemical Reactions Philip Dutton University of Windsor, Canada Slide 1 of 29

2 Contents 4-1 Chemical Reactions and Chemical Equations 4-2 Chemical Equations and Stoichiometry 4-3 Chemical Reactions in Solution 4-4 Determining the Limiting Reactant 4-5 Other Practical Matters in Reaction Stoichiometry Focus On Industrial Chemistry Slide 2 of 29

3 4-1 Chemical Reactions and Chemical Equations As reactants are converted to products we observe: Color change (4-1) Precipitate formation (4-1) Gas evolution (4-2a) Heat absorption or evolution (4-2b) Chemical evidence may be necessary. Slide 3 of 29

4 Chemical Reaction Nitrogen monoxide + oxygen nitrogen dioxide Step 1: Write the reaction using chemical symbols. Step 2: Balance the chemical equation. 2 NO + 1O 2 2NO 2 Slide 4 of 29

5 Molecular Representation Slide 5 of 29

6 Balancing Equations An equation can be balanced only by adjusting the coefficients of formulas. Never introduce extraneous atoms to balance. NO + O 2 NO 2 + O Never change a formula for the purpose of balancing an equation. NO + O 2 NO 3 Slide 6 of 29

7 Balancing Equation Strategy Balance elements that occur in only one compound on each side first. Balance free elements last. Balance unchanged polyatomics (or other groups of atoms) as groups. Fractional coefficients are acceptable and can be cleared at the end by multiplication. Slide 7 of 29

8 EXAMPLE 4-2 Writing and Balancing an Equation: The Combustion of a Carbon-Hydrogen-Oxygen Compound. Liquid triethylene glycol, C 6 H 14 O 4, is used a a solvent and plasticizer for vinyl and polyurethane plastics. Write a balanced chemical equation for its complete combustion. Slide 8 of 29

9 EXAMPLE 4-2 Chemical Equation: C 6 H 14 O O CO H 2 O 1. Balance C. 2. Balance H. 3. Balance O. 4. Multiply by two 2 C 6 H 14 O O 2 12 CO H 2 O and check all elements. Slide 9 of 29

10 4-2 Chemical Equations and Stoichiometry Stoichiometry includes all the quantitative relationships involving: atomic and formula masses chemical formulas. Mole ratio is a central conversion factor. Slide 10 of 29

11 EXAMPLE 4-3 Relating the Numbers of Moles of Reactant and Product How many moles of H 2 O are produced by burning 2.72 mol H 2 in an excess of O 2? Write the Chemical Equation: Balance the Chemical Equation: 2 H 2 + O 2 2 H 2 O Use the stoichiometric factor or mole ratio in an equation: n H2 O = 2.72 mol H 2 2 mol H 2 O 2 mol H 2 = 2.72 mol H 2 O Slide 11 of 29

12 EXAMPLE 4-6 Additional Conversion Factors in a Stoichiometric Calculation: Volume, Density, and Percent Composition. An alloy used in aircraft structures consists of 93.7% Al and 6.3% Cu by mass. The alloy has a density of 2.85 g/cm 3. A cm 3 piece of the alloy reacts with an excess of HCl(aq). If we assume that all the Al but none of the Cu reacts with HCl(aq), what is the mass of H 2 obtained? Slide 12 of 29

13 EXAMPLE 4-6 Write the Chemical Equation: Balance the Chemical Equation: 2 Al + 63 HCl 2AlCl H Slide 13 of 29

14 EXAMPLE Al + 6 HCl 2 AlCl H 2 Plan the strategy: g alloy cm 3 cm 3 alloy g alloy g Al mole Al mol H 2 g H 2 We need 5 conversion factors! Write the Equation g Al g alloy mol Al g Al and Calculate: mol H 2 mol Al m H2 = cm 3 alloy 2.85 g alloy 97.3 g Al 1 cm g alloy g H 2 mol H 2 1 mol Al g Al 3 mol H 2 2 mol Al g H 2 1 mol H 2 = g H 2 Slide 14 of 29

15 4-3 Chemical Reactions in Solution Close contact between atoms, ions and molecules necessary for a reaction to occur. Solvent We will usually use aqueous (aq) solution. Solute A material dissolved by the solvent. Slide 15 of 29

16 Molarity Molarity (M) = Amount of solute (mol solute) Volume of solution (L) If mol of urea is dissolved in enough water to make L of solution the concentration is: c urea = mol urea = M CO(NH2 ) L 2 Slide 16 of 29

17 Preparation of a Solution Weigh the solid sample. Dissolve it in a volumetric flask partially filled with solvent. Carefully fill to the mark. Slide 17 of 29

18 EXAMPLE 4-9 Calculating the Mass of Solute in a Solution of Known Molarity. We want to prepare exactly L (250 ml) of an M K 2 CrO 4 solution in water. What mass of K 2 CrO 4 should we use? Plan strategy: mol L g mol Volume moles mass Write equation and calculate: We need 2 conversion factors! m K2 CrO 4 = L mol g = 12.1 g 1.00 L 1.00 mol Slide 18 of 29

19 Solution Dilution M i V i M f V f M = n V M i V i = n i = n f = M f V f M f = M i V i V = M i i V f V f Slide 19 of 29

20 EXAMPLE 4-10 Preparing a solution by dilution A particular analytical chemistry procedure requires M K 2 CrO 4. What volume of M K 2 CrO 4 should we use to prepare L of M K 2 CrO 4? Plan strategy: M f = M i V i V f V i = V f M f M i Calculate: V K2 CrO 4 = L mol L = L 1.00 L mol Slide 20 of 29

21 4-4 Determining Limiting Reagent The reactant that is completely consumed determines the quantities of the products formed. Slide 21 of 29

22 EXAMPLE 4-12 Determining the Limiting Reactant in a Reaction. Phosphorus trichloride, PCl 3, is a commercially important compound used in the manufacture of pesticides, gasoline additives, and a number of other products. It is made by the direct combination of phosphorus and chlorine P 4 (s) + 6 Cl 2 (g) 4 PCl 3 (l) What mass of PCl 3 forms in the reaction of 125 g P 4 with 323 g Cl 2? Strategy: Compare the actual mole ratio to the required mole ratio. Slide 22 of 29

23 EXAMPLE 4-12 n Cl2 = 323 g Cl 2 1 mol Cl 2 = 4.56 mol Cl g Cl 2 1 mol P n P4 = 125 g P 4 4 = 1.01 mol P g P 4 4 = n n Cl 2 P 4 actual = 4.55 mol Cl 2 /mol P 4 theoretical = 6.00 mol Cl 2 /mol P 4 Chlorine gas is the limiting reagent. Slide 23 of 29

24 EXAMPLE 4-12 Strategy for calculation: mol Cl 2 mol PCl 3 g PCl 3 g Cl 2 mol Cl 2 mol PCl 3 mass Cl 2 moles Cl 2 moles PCl 3 mass PCl 3 Write the Equation and calculate: 1 mol Cl 2 4 mol PCl m PCl = 323.g Cl g PCl g Cl 2 6 mol Cl 2 1 mol PCl 3 = 417 g PCl 3 Slide 24 of 29

25 Other Practical Matters in Reaction Stoichiometry Theoretical yield is the expected yield from a reactant. Actual yield is the amount of product actually produced. Actual yield Percent yield = 100% Theoretical Yield Slide 25 of 29

26 Theoretical, Actual and Percent Yield When actual yield = % yield the reaction is said to be quantitative. Side reactions reduce the percent yield. By-products are formed by side reactions. Slide 26 of 29

27 Consecutive Reactions, Simultaneous Reactions and Overall Reactions Multistep synthesis is often unavoidable. Reactions carried out in sequence are called consecutive reactions. When substances react independently and at the same time the reaction is a simultaneous reaction. Slide 27 of 29

28 Overall Reactions and Intermediates The Overall Reaction is a chemical equation that expresses all the reactions occurring in a single overall equation. An intermediate is a substance produced in one step and consumed in another during a multistep synthesis. Slide 28 of 29

29 End of Chapter Questions Initial problem solving is linear and often based on memorizing solutions for particular situations. Answer c b a Slide 29 of 29