Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 3 Slide 1 of 24 PHILIP DUTTON UNIVERSITY OF WINDSOR DEPARTMENT OF CHEMISTRY AND BIOCHEMISTRY TENTH EDITION GENERAL CHEMISTRY Principles and Modern Applications PETRUCCI HERRING MADURA BISSONNETTE Chemical Reactions 4

Chemical Compounds Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 3 Slide 2 of 24 CONTENTS 4-1Chemical Reactions and Chemical Equations 4-2Chemical Equations and Stoichiometry 4-3Chemical Reactions in Solution 4-4Determining the Limiting Reactant 4-5Other Practical Matters in Reaction Stoichiometry

Copyright © 2011 Pearson Canada Inc. 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.

Precipitation of silver chromate Figure 4-1 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 3 Slide 4 of 24

Evidence of a chemical reaction Figure 4-2 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 3 Slide 5 of 24

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 6 of 24 Chemical Reactions 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

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 7 of 24 Molecular Representation

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 8 of 24 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 An equation can be balanced only by adjusting the coefficients of formulas.

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 9 of 24 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.

States of Matter Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 3 Slide 10 of 24 (g) gas (l) liquid (s) solid Thus, the equation for combustion of triethylene glycol can be written as 2 C 6 H 14 O 4 (l) + 15 O 2 (g) 12 CO 2 (g) + 14 H 2 O(l) Another commonly used symbol for reactants or products dissolved in water is (aq) aqueous solution

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 11 of Chemical Equations and Stoichiometry Stoichiometry includes all the quantitative relationships involving: atomic and formula masses chemical formulas. chemical equations Mole ratio or stoichiometric factor is a central conversion factor.

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 3 Slide 12 of 24 KEEP IN MIND that it is important to include units and to work from a balanced chemical equation when solving stoichiometry problems.

A generalized stoichiometry diagram Figure 4-3 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 3 Slide 13 of 24

The reaction of 2 Al(s) 6 HCl(aq)2 AlC l3 (aq) 3 H 2 (g) Figure 4-4 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 3 Slide 14 of 24

Copyright © 2011 Pearson Canada Inc. 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.

Molarity Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 16 of 24 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 C = V n

Solution Dilution Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 17 of 24 Figure 4-6 Visualizing the dilution of a solution M i × V i = n i M i  V i M f  V f = n f = M f × V f M = n V M i × V i M f = VfVf = M i ViVi VfVf

Preparation of M K 2 Cr 2 O 7 —Example 4-9 illustrated Figure 4-5 Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 18 of 24

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 19 of Determining Limiting Reagent The reactant that is completely consumed determines the quantities of the products formed.

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 20 of 24 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 4-5Other Practical Matters in Reaction Stoichiometry Theoretical, Actual and Percent Yield

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 21 of 24 actual yield = theoretical yield quantitative Side reactions reduce the percent yield. By-products are formed by side reactions.

Consecutive Reactions, Simultaneous Reactions and Overall Reactions Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 22 of 24 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.

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 23 of 24 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.

Copyright © 2011 Pearson Canada Inc. General Chemistry: Chapter 4 Slide 24 of 24 End of Chapter Questions Initial problem solving is linear and often based on memorizing solutions for particular situations. Answer a b c