Copyright©2004 by Houghton Mifflin Company. All rights reserved. 1 Introductory Chemistry: A Foundation FIFTH EDITION by Steven S. Zumdahl University of Illinois

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 2 Chemical Reactions: An Introduction Chapter 6

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 3 Chemical Reactions Reactions involve chemical changes in matter resulting in new substances Reactions involve rearrangement and exchange of atoms to produce new molecules –Elements are not transmuted during a reaction Reactants  Products

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 4 Evidence of Chemical Reactions a chemical change occurs when new substances are made visual clues (permanent) –color change, precipitate formation, gas bubbles, flames, heat release, cooling, light other clues –new odor, permanent new state

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 5 Evidence of Chemical Reactions: Color Change

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 6 Evidence of Chemical Reactions

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 7 Chemical Equations Shorthand way of describing a reaction Provides information about the reaction –Formulas of reactants and products –States of reactants and products –Relative numbers of reactant and product molecules that are required –Can be used to determine weights of reactants used and of products that can be made

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 8 Conservation of Mass Matter cannot be created or destroyed In a chemical reaction, all the atoms present at the beginning are still present at the end Therefore the total mass cannot change Therefore the total mass of the reactants will be the same as the total mass of the products

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 9 Combustion of Methane methane gas burns to produce carbon dioxide gas and liquid water –whenever something burns it combines with O 2 (g) CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O(l) H H C H H OO + O O C + O HH 1 C + 4 H + 2 O1 C + 2 O + 2 H + O 1 C + 2 H + 3 O

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 10 Combustion of Methane Balanced to show the reaction obeys the Law of Conservation of Mass it must be balanced CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l) H H C H H OO + O O C + O HH OO + O HH + 1 C + 4 H + 4 O

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 11 Writing Equations Use proper formulas for each reactant and product proper equation should be balanced –obey Law of Conservation of Mass –all elements on reactants side also on product side –equal numbers of atoms of each element on reactant side as on product side balanced equation shows the relationship between the relative numbers of molecules of reactants and products –can be used to determine mass relationships

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 12 Symbols Used in Equations symbols used after chemical formula to indicate physical state –(g) = gas; (l) = liquid; (s) = solid –(aq) = aqueous, dissolved in water

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 13 Sample – Recognizing Reactants and Products when magnesium metal burns in air it produces a white, powdery compound magnesium oxide –burning in air means reacting with O 2 –Metals are solids, except for Hg which is liquid ¬write the equation in words –identify the state of each chemical magnesium(s) + oxygen(g)  magnesium oxide(s) ­write the equation in formulas –identify diatomic elements –identify polyatomic ions –determine formulas Mg(s) + O 2 (g)  MgO(s)

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 14 Balancing Chemical Equations ¬Count atoms of each element apolyatomic ions may be counted as one “element” if it does not change in the reaction Al + FeSO 4  Al 2 (SO 4 ) 3 + Fe 1 SO 4 3 bif an element appears in more than one compound on the same side, count each separately and add CO + O 2  CO O 2

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 15 Balancing Chemical Equations ­Pick an element to balance aavoid elements from 1b ®Find Least Common Multiple and factors needed to make both sides equal ¯Use factors as coefficients in equation aif already a coefficient then multiply by new factor °Recount and Repeat until balanced

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 16 Example 1: Magnesium metal burns in air when magnesium metal burns in air it produces a white, powdery compound magnesium oxide –burning in air means reacting with O 2 ¬write the equation in words magnesium(s) + oxygen(g)  magnesium oxide(s) ­write the equation in formulas - determine formulas Mg(s) + O 2 (g)  MgO(s)

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 17 Example 1: Magnesium metal burns in air ®count the number of atoms of on each side Mg(s) + O 2 (g)  MgO(s) 1  Mg  1 2  O  1

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 18 Example 1: Magnesium metal burns in air ¯pick an element to balance - avoid element in multiple compounds Mg(s) + O 2 (g)  MgO(s) 1  Mg  1 1 x 2  O  1 x 2

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 19 Example 1: Magnesium metal burns in air Mg(s) + O 2 (g)  2 MgO(s) 1  Mg  1 1 x 2  O  1 x 2 Use factors as coefficients in front of compound containing the element 3if coefficient already there, multiply them together

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 20 Example 1: Magnesium metal burns in air ²Recount Mg(s) + O 2 (g)  2 MgO(s) 1  Mg  2 2  O  2 ³Repeat 2 Mg(s) + O 2 (g)  2 MgO(s) 2 x 1  Mg  2 2  O  2

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 21 Example 2 Under appropriate conditions at 1000°C ammonia gas reacts with oxygen gas to produce gaseous nitrogen monoxide and gaseous water write the equation in words ammonia(g) + oxygen(g)  nitrogen monoxide(g) + water(g) ­write the equation in formulas NH 3 (g) + O 2 (g)  NO(g) + H 2 O(g)

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 22 Examples ®count the number of atoms of on each side NH 3 (g) + O 2 (g)  NO(g) + H 2 O(g) 1  N  1 3  H  2 2  O  1 + 1

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 23 Examples ¯pick an element to balance - avoid element in multiple compounds NH 3 (g) + O 2 (g)  NO(g) + H 2 O(g) 1  N  1 2 x 3  H  2 x 3 2  O  1 + 1

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 24 Examples Use factors as coefficients in front of compound containing the element 2 NH 3 (g) + O 2 (g)  NO(g) + 3 H 2 O(g) 1  N  1 2 x 3  H  2 x 3 2  O  1 + 1

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 25 Examples ²Recount 2 NH 3 (g) + O 2 (g)  NO(g) + 3 H 2 O(g) 2  N  1 6  H  6 2  O  ³Repeat 2 NH 3 (g) + O 2 (g)  2 NO(g) + 3 H 2 O(g) 2  N  1 x 2 6  H  6 2  O  1 + 3

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 26 Examples ´Recount 2 NH 3 (g) + O 2 (g)  2 NO(g) + 3 H 2 O(g) 2  N  2 6  H  6 2  O  2 + 3

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 27 Examples µRepeat –A trick of the trade, when you are forced to attack an element that is in 3 or more compounds – find where it is uncombined. You can find a factor to make it any amount you want, even if that factor is a fraction! –We want to make the O on the left equal 5, therefore we will multiply it by NH 3 (g) O 2 (g)  2 NO(g) + 3 H 2 O(g) 2  N  2 6  H  x 2  O  2 + 3

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 28 Examples Multiply all the coefficients by a number to eliminate fractions: 2 x [2 NH 3 (g) O 2 (g)  2 NO(g) + 3 H 2 O(g)] 4 NH 3 (g) + 5 O 2 (g)  4 NO(g) + 6 H 2 O(g) 4  N  4 12  H   O  10