Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 1 of 29 Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 Chapter 4: Chemical Reactions General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 2 of 29 Contents 4-1Chemical Reactions and Chemical Equations 4-2Chemical Equations and Stoichiometry 4-3Chemical Reactions in Solution 4-4Determining the Limiting reagent 4-5Other Practical Matters in Reaction Stoichiometry Focus on Industrial Chemistry

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 3 of Chemical Reactions and Chemical Equations As reactants are converted to products we observe: –Color change –Precipitate formation –Gas evolution –Heat absorption or evolution Chemical evidence may be necessary.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 4 of 29 Chemical Reaction Nitrogen monoxide + oxygen → nitrogen dioxide Step 1: Write the reaction using chemical symbols. NO + O 2 → NO 2 Step 2: Balance the chemical equation. 212

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 5 of 29 Molecular Representation

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 6 of 29 Balancing Equations 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

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 7 of 29 Balancing Equation Strategy Balance elements that occur in only one compound on each side first. Balance free elements last. Balance unchanged polyatomics as groups. Fractional coefficients are acceptable and can be cleared at the end by multiplication.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 8 of 29 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.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 9 of C 6 H 14 O 4 + O 2 → CO 2 + H 2 O6 2. Balance H. 2 C 6 H 14 O O 2 → 12 CO H 2 O 4. Multiply by two Example Balance O. and check all elements. Chemical Equation: 1. Balance C. 6 7

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 10 of Chemical Equations and Stoichiometry Stoichiometry includes all the quantitative relationships involving: –atomic and formula masses –chemical formulas. Mole ratio is a central conversion factor.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 11 of 29 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 ? H 2 + O 2 → H 2 O Write the Chemical Equation: Balance the Chemical Equation: 22 Use the stoichiometric factor or mole ratio in an equation: n H 2 O = 2.72 mol H 2 × = 2.72 mol H 2 O 2 mol H 2 O 2 mol H 2

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 12 of 29 Example 4-6 Additional Conversion Factors ina 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?

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 13 of 29 Al + HCl → AlCl 3 + H 2 Write the Chemical Equation: Example 4-6 Balance the Chemical Equation: 2623

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 14 of 29 2 Al + 6 HCl → 2 AlCl H 2 Example 4-6 Plan the strategy: cm 3 alloy → g alloy → g Al → mol Al → mol H 2 → g H 2 We need 5 conversion factors! × Write the Equation m H 2 = cm 3 alloy × × × 2.85 g alloy 1 cm g Al 100 g alloy 1 mol Al g Al 3 mol H 2 2 mol Al g H 2 1 mol H 2 = g H 2 and Calculate:

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 15 of 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.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 16 of 29 Molarity Molarity (M) = Volume of solution (L) Amount of solute (mol solute) If mol of urea is dissolved in enough water to make L of solution the concentration is: c urea = L mol urea = M CO(NH 2 ) 2

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 17 of 29 Preparation of a Solution Weigh the solid sample. Dissolve it in a volumetric flask partially filled with solvent. Carefully fill to the mark.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 18 of 29 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: Example 4-6 Volume → moles → mass We need 2 conversion factors! Write equation and calculate: m K 2 CrO 4 = L × × = 12.1 g mol 1.00 L g 1.00 mol

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 19 of 29 Solution Dilution M i × V i = n i M i × V i M f × V f = n f = M f × V f M i × V i M f = VfVf = M i ViVi VfVf M = n V

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 20 of 29 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 ? Calculate: V K 2 CrO 4 = L × × = L mol 1.00 L L mol Example 4-10 Plan strategy: M f = MiMi ViVi VfVf V i = VfVf MfMf MiMi

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 21 of Determining Limiting Reagent The reactant that is completely consumed determines the quantities of the products formed.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 22 of 29 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 ? Example 4-12 Strategy:Compare the actual mole ratio to the required mole ratio.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 23 of 29 Example 4-12 n Cl 2 = 323 g Cl 2 × = 4.56 mol Cl 2 1 mol Cl g Cl 2 n P 4 = 125 g P 4 × = 1.01 mol P 4 1 mol P g 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. n n  = P4P4 Cl 2

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 24 of Other Practical Matters in Reaction Stoichiometry Theoretical yield is the expected yield from a reactant. Actual yield is the amount of product actually produced. Percent yield = × 100% Actual yield Theoretical Yield

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 25 of 29 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.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 26 of 29 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.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 27 of 29 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.

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 28 of 29 Focus on Industrial Chemistry

Prentice-Hall © 2002 General Chemistry: Chapter 4 Slide 29 of 29 Chapter 4 Questions 1, 6, 12, 25, 39, 45, 53, 65, 69, 75, 84, 94, 83, 112