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Chapter 1: Matter and Measurement

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Chapter 1 Matter and Measurement Prentice Hall © 2002 General Chemistry C hapter 4 Slide 1 of 29 Philip Dutton University of Windsor, Canada Prentice Hall © 2002 Chapter 4 Chemical Reactions General C[.]

General Chemistry Principles and Modern Applications Petrucci • Harwood • Herring 8th Edition Chapter 4: Chemical Reactions Philip Dutton University of Windsor, Canada Prentice-Hall © 2002 Slide of 29 General Chemistry: C Prentice-Hall © 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 reagent 4-5 Other Practical Matters in Reaction Stoichiometry Focus on Industrial Chemistry Slide of 29 General Chemistry: C Prentice-Hall © 4-1 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 Slide of 29 General Chemistry: C Prentice-Hall © Chemical Reaction Nitrogen monoxide + oxygen → nitrogen dioxide Step 1: Write the reaction using chemical symbols Step 2: Balance the chemical equation NO + O2 → NO2 Slide of 29 General Chemistry: C Prentice-Hall © Molecular Representation Slide of 29 General Chemistry: C Prentice-Hall © Balancing Equations • Never introduce extraneous atoms to balance NO + O2 → NO2 + O • Never change a formula for the purpose of balancing an equation NO + O2 → NO3 Slide of 29 General Chemistry: C Prentice-Hall â Balancing Equation Strategy ã 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 Slide of 29 General Chemistry: C Prentice-Hall © Example 4-2 Writing and Balancing an Equation: The Combustion of a Carbon-Hydrogen-Oxygen Compound Liquid triethylene glycol, C6H14O4, is used a a solvent and plasticizer for vinyl and polyurethane plastics Write a balanced chemical equation for its complete combustion Slide of 29 General Chemistry: C Prentice-Hall © Example 4-2 Chemical Equation: C6H14O4 + 15 O2 → CO2 + H2O Balance C Balance H Balance O Multiply by two C6H14O4 + 15 O2 → 12 CO2 + 14 H2O and check all elements Slide of 29 General Chemistry: C Prentice-Hall © 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 General Chemistry: C Prentice-Hall © 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 General Chemistry: C Prentice-Hall © Molarity Amount of solute (mol solute) Molarity (M) = Volume of solution (L) If 0.444 mol of urea is dissolved in enough water to make 1.000 L of solution the concentration is: curea = Slide 16 of 29 0.444 mol urea = 0.444 M CO(NH2)2 1.000 L General Chemistry: C Prentice-Hall © 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 General Chemistry: C Prentice-Hall © Example 4-6 Calculating the mass of Solute in a solution of Known Molarity We want to prepare exactly 0.2500 L (250 mL) of an 0.250 M K2CrO4 solution in water What mass of K2CrO4 should we use? Plan strategy: Volume → moles → mass We need conversion factors! Write equation and calculate: mK CrO = 0.2500 L × Slide 18 of 29 0.250 mol× 194.02 g= 12.1 g 1.00 mol 1.00 L General Chemistry: C Prentice-Hall © Solution Dilution Mi × V i Mf × V f n M= V Mi × Vi = ni = nf = Mf × Vf Mf = Slide 19 of 29 Mi × Vi Vf = Mi General Chemistry: C Vi Vf Prentice-Hall © Example 4-10 Preparing a solution by dilution A particular analytical chemistry procedure requires 0.0100 M K2CrO4 What volume of 0.250 M K2CrO4 should we use to prepare 0.250 L of 0.0100 M K2CrO4? Plan strategy: M f = Mi Vi Vf Vi = Vf Mf Mi Calculate: VK2CrO4 = 0.2500 L × 0.0100 mol× 1.000 L = 0.0100 L 0.250 mol 1.00 L Slide 20 of 29 General Chemistry: C Prentice-Hall © ... Chemical Reactions and Chemical Equations 4-2 Chemical Equations and Stoichiometry 4-3 Chemical Reactions in Solution 4-4 Determining the Limiting reagent 4-5 Other Practical Matters in Reaction... Fractional coefficients are acceptable and can be cleared at the end by multiplication Slide of 29 General Chemistry: C Prentice-Hall © Example 4-2 Writing and Balancing an Equation: The Combustion... Carbon-Hydrogen-Oxygen Compound Liquid triethylene glycol, C6H14O4, is used a a solvent and plasticizer for vinyl and polyurethane plastics Write a balanced chemical equation for its complete combustion

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