General Chemistry Principles and Modern Applications Petrucci Harwood

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General Chemistry Principles and Modern Applications Petrucci • Harwood • Herring 8 th Edition

General Chemistry Principles and Modern Applications Petrucci • Harwood • Herring 8 th Edition Chapter 4: Chemical Reactions Philip Dutton University of Windsor, Canada N 9 B 3 P 4 Prentice-Hall © 2002 (modified 2003 by Dr. Paul Root and 2005 by Dr. David Tramontozzi) General Chemistry: Chapter 4

Contents 4 -1 4 -2 4 -3 4 -4 4 -5 Chemical Reactions and

Contents 4 -1 4 -2 4 -3 4 -4 4 -5 Chemical Reactions and Chemical Equations and Stoichiometry Chemical Reactions in Solution Determining the Limiting reagent Other Practical Matters in Reaction Stoichiometry Focus on Industrial Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

4 -4 Determining Limiting Reagent • The reactant that is completely consumed determines the

4 -4 Determining Limiting Reagent • The reactant that is completely consumed determines the quantities of the products formed. Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Example 4 -12 Determining the Limiting Reactant in a Reaction. Phosphorus trichloride , PCl

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. Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Example 4 -12 n. Cl = 323 g Cl 2 × 1 mol Cl

Example 4 -12 n. Cl = 323 g Cl 2 × 1 mol Cl 2 = 4. 56 mol Cl 2 2 70. 91 g Cl 2 n. P = 125 g P 4 × 4 = n. Cl n. P actual = 4. 55 mol Cl 2/mol P 4 2 4 Mole ratio 1 mol P 4 = 1. 01 mol P 4 123. 9 g P 4 theoretical = 6. 00 mol Cl 2/mol P 4 Chlorine gas is the limiting reagent. Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Example 4 -12 Since chlorine gas is the limiting reagent, we need to figure

Example 4 -12 Since chlorine gas is the limiting reagent, we need to figure out how much PCl 3 will be produced when all of the chlorine gas reacts. n. Cl = 323 g Cl 2 × 1 mol Cl 2 = 4. 56 mol Cl 2 2 70. 91 g Cl 2 4. 56 mol Cl 2 x = 417 4 mol PCl 3 6 mol Cl 2 x 137. 3 g PCl 3 1 mol PCl 3 g PCl 3 are produced in this reaction. Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Example 4 -12 An alternate pathway involves figuring out how much of the product

Example 4 -12 An alternate pathway involves figuring out how much of the product can be formed by complete reaction of each starting material. 4 mol PCl 3 137. 3 g PCl 3 323 g Cl 2 x 1 mol Cl 2 x x = 417 g PCl 3 70. 91 g Cl 2 6 mol Cl 1 mol PCl 3 2 4 mol PCl 3 137. 3 g PCl 3 125 g P 4 x 1 mol P 4 x x = 554 g PCl 3 123. 9 g P 4 1 mol PCl 3 1 mol P 4 Again, the chlorine gas is the limiting reagent !!! Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

4 -5 Other Practical Matters in Reaction Stoichiometry Theoretical yield is the expected yield

4 -5 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 Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Example 4 -14 – Determining Theoretical, Actual, and Percentage Yields Billions of pounds of

Example 4 -14 – Determining Theoretical, Actual, and Percentage Yields Billions of pounds of urea CO(NO 2)2 are produced annually for use as a fertilizer. The reaction used is: 2 NH 3 + CO 2 CO(NO 2)2 + H 2 O The typical starting reaction mixture has a 3 to 1 mole ratio of NH 3 to CO 2. If 47. 7 g of urea forms per mole of CO 2 that reacts, what is the; (a) theoretical yield (b) actual yield; and (c) percentage yield in this reaction Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Example 4 -14 – Determining Theoretical, Actual, and Percentage Yields a) 2 NH 3

Example 4 -14 – Determining Theoretical, Actual, and Percentage Yields a) 2 NH 3 + CO 2 CO(NH 2)2 + H 2 O - stoichiometric proportions are 2 mol NH 3 : 1 mol CO 2 - since we are given the mole ratio of NH 3 to CO 2 as 3: 1, NH 3 is in excess so CO 2 is the limiting reactant. Because the quantity of urea is given per mol of CO 2, we should base the calculation of 1. 00 mol of CO 2. Theoretical yield = 1. 0 mol of CO 2 x 1. 0 mol CO(NH 2)2 / 1. 0 mol CO 2 = 1. 0 mol CO(NH 2)2 Mass CO(NH 2)2 = 1. 0 mol CO(NH 2)2 x 60. 1 g / 1. 0 mol CO(NH 2)2 = 60. 1 g CO(NH 2)2 Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Example 4 -14 – Determining Theoretical, Actual, and Percentage Yields b) Actual yield =

Example 4 -14 – Determining Theoretical, Actual, and Percentage Yields b) Actual yield = 47. 7 g (given in question) CO(NH 2)2 c) % yield = actual yield / theoretical yield = 47. 7 g / 60. 1 g x 100% = 79. 4 % Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Theoretical, Actual and Percent Yield • When actual yield = % yield the reaction

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. Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Consecutive Reactions, Simultaneous Reactions and Overall Reactions • Multistep synthesis is often unavoidable. •

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. Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Overall Reactions and Intermediates • The Overall Reaction is a chemical equation that expresses

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 multi-step synthesis. Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002

Chapter 4 Questions 1, 6, 12, 21, 23, 25, 27, 35, 39, 45, 53,

Chapter 4 Questions 1, 6, 12, 21, 23, 25, 27, 35, 39, 45, 53, 65, 69, 75, 77, 83, 84, 94, 112. Chemistry 140 Fall 2002 Dutton Prentice-Hall © 2002