1. 1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 5 Lecture Outline Prepared by Jennifer N. Robertson-Honecker West Virginia University

2. 2 Introduction to Chemical Reactions A chemical change—a chemical reaction—converts one substance into another. Chemical reactions involve: breaking bonds in the reactants (starting materials) forming new bonds in the products CH 4 and O 2 CO 2 and H 2 O

3. 3 Introduction to Chemical Reactions A chemical equation is an expression that uses chemical formulas and other symbols to illustrate what reactants constitute the starting materials in a reaction and what products are formed. The reactants are written on the left. The products are written on the right. Coefficients show the number of molecules of a given element or compound that react or are formed.

4. 4 Introduction to Chemical Reactions The law of conservation of mass states that atoms cannot be created or destroyed in a chemical reaction. Coefficients are used to balance an equation. A balanced equation has the same number of atoms of each element on both sides of the equation.

5. 5 Introduction to Chemical Reactions

6. 6 Balancing Chemical Equations HOW TO Balance a Chemical Equation Example Write a balanced chemical equation for the reaction of propane (C 3 H 8 ) with oxygen (O 2 ) to form carbon dioxide (CO 2 ) and water (H 2 O). Step [1] Write the equation with the correct formulas. C 3 H 8 + O 2 CO 2 + H 2 O The subscripts in a formula can never be changed to balance an equation, because changing a subscript changes the identity of a compound.

7. 7 Balancing Chemical Equations Step [2] HOW TO Balance a Chemical Equation Balance the equation with coefficients one element at a time. Balance the C’s first: Balance the H’s next:

8. 8 Balancing Chemical Equations Step [2] HOW TO Balance a Chemical Equation Balance the equation with coefficients one element at a time. Finally, balance the O’s:

9. 9 Balancing Chemical Equations Step [3] Check to make sure that the smallest set of whole numbers is used. HOW TO Balance a Chemical Equation

10. 10 The Mole and Avogadro’s Number A mole is a quantity that contains 6.02 x 10 23 items. 1 mole of C atoms = 6.02 x 10 23 C atoms 1 mole of H 2 O molecules = 6.02 x 10 23 H 2 O molecules The number 6.02 x 10 23 is Avogadro’s number. It can be used as a conversion factor to relate the number of moles of a substance to the number of atoms or molecules: 1 mol 6.02 x 10 23 atoms or 6.02 x 10 23 atoms 1 mol 1 mole of vitamin C molecules = 6.02 x 10 23 vitamin C molecules

11. 11 The Mole and Avogadro’s Number Sample Problem 5.5 How many molecules are contained in 5.0 moles of carbon dioxide (CO 2 )? Step [1] Identify the original quantity and the desired quantity. 5.0 mol of CO 2 original quantity ? number of molecules of CO 2 desired quantity

12. 12 The Mole and Avogadro’s Number Step [3] Set up and solve the problem. 5.0 molx 6.02 x 10 23 molecules 1 mol = 3.0 x 10 24 molecules CO 2 Unwanted unit cancels. Step [2] Write out the conversion factors. 1 mol 6.02 x 10 23 molecules or 6.02 x 10 23 molecules 1 mol Choose this one to cancel mol.

13. 13 Mass to Mole Conversions The formula weight is the sum of the atomic weights of all the atoms in a compound, reported in atomic mass units (amu). What is the molar mass of nicotine (C 10 H 14 N 2 )? Determine the number of atoms of each element from the subscripts in the chemical formula. C 10 H 14 N 2 contains 10 C atoms, 14 H atoms, and 2 N atoms. Sample Problem 5.7 Step [1]

14. 14 Mass to Mole Conversions Step [2] Multiply the number of atoms of each element by the atomic weight and add the results. 10 C atoms x 12.01 amu = 120.1 amu 14 H atoms x 1.01 amu = 14.14 amu 2 N atoms x 14.01 amu = 28.02 amu Formula weight of C 10 H 14 N 2 = 162.26 amu

15. 15 Mass to Mole Conversions Molar Mass The molar mass is the mass of one mole of any substance, reported in grams. The value of the molar mass of a compound in grams equals the value of its formula weight in amu.

16. 16 Mass to Mole Conversions Relating Grams to Moles The molar mass relates the number of moles to the number of grams of a substance. In this way, molar mass can be used as a conversion factor. Sample Problem 5.9 How many moles are present in 100. g of aspirin (C 9 H 8 O 4, molar mass 180.2 g/mol)? Step [1] Identify the original quantity and the desired quantity. 100. g of aspirin original quantity ? mol of aspirin desired quantity

17. 17 Mass to Mole Conversions Relating Grams to Moles Step [2] Write out the conversion factors. 180.2 g aspirin 1 mol 180.2 g aspirin or Choose this one to cancel g. The conversion factor is the molar mass, and it can be written in two ways. Choose the one that places the unwanted unit, grams, in the denominator so that the units cancel:

18. 18 Mass to Mole Conversions Relating Grams to Moles Step [3] Set up and solve the problem. 100. g aspirinx 1 mol 180.2 g aspirin =0.555 mol aspirin Unwanted unit cancels.

19. 19 Mole Calculations in Chemical Equations A balanced chemical equation also tells us: the number of moles of each reactant that combine the number of moles of each product formed 1 N 2 (g) + 1 O 2 (g)2 NO(g) 1 mole of N 2 1 molecule N 2 1 mole of O 2 1 molecule O 2 2 moles of NO 2 molecules NO (The coefficient “1” has been written for emphasis.)

20. 20 Mole Calculations in Chemical Equations Coefficients are used to form mole ratios, which can serve as conversion factors. N 2 (g) + O 2 (g)2 NO(g) Mole ratios: 1 mol N 2 1 mol O 2 1 mol N 2 2 mol NO 1 mol O 2 2 mol NO Use the mole ratios from the coefficients in the balanced equation to convert moles of one compound (A) into moles of another compound (B).

21. 21 Mole Calculations in Chemical Equations Sample Problem 5.10 Using the balanced chemical equation, how many moles of CO are produced from 3.5 moles of C 2 H 6 ? Step [1] Identify the original and desired quantities. 3.5 mol C 2 H 6 original quantity ? mol CO desired quantity 2 C 2 H 6 (g) + 5 O 2 (g)2 CO(g) + 6 H 2 O(g)

22. 22 Mole Calculations in Chemical Equations Step [2] Write out the conversion factors. 2 mol C 2 H 6 4 mol CO or 4 mol CO 2 mol C 2 H 6 Choose this one to cancel mol C 2 H 6. Step [3] Set up and solve the problem. 3.5 mol C 2 H 6 x 4 mol CO 2 mol C 2 H 6 =7.0 mol CO Unwanted unit cancels.

23. 23 Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Example Using the balanced equation, how many grams of O 3 are formed from 9.0 mol of O 2 ? 3 O 2 (g)2 O 3 (g) Moles of reactant Moles of reactant Grams of product Grams of product mole–mole conversion factor mole–mole conversion factor molar mass conversion factor molar mass conversion factor Moles of product Moles of product [1][2] sunlight

24. 24 Mass Calculations in Chemical Equations Step [1] Convert the number of moles of reactant to the number of moles of product using a mole–mole conversion factor. HOW TO Convert Moles of Reactant to Grams of Product 3 mol O 2 2 mol O 3 or 2 mol O 3 3 mol O 2 Cancel mol O 2 in step [1]. Step [2] Convert the number of moles of product to the number of grams of product using the product’s molar mass. 1 mol O 3 48.0 g O 3 1 mol O 3 or Cancel mol O 3 in step [2].

25. 25 Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Set up and solve the conversion. 9.0 mol O 2 x 2 mol O 3 3 mol O 2 Moles of reactant Moles of reactant Grams of product Grams of product x 48.0 g O 3 1 mol O 3 =290 g O 3 mole–mole conversion factor mole–mole conversion factor molar mass conversion factor molar mass conversion factor Mol O 2 cancel.Mol O 3 cancel.

26. 26 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product Example Ethanol (C 2 H 6 O, molar mass 46.1 g/mol) is synthesized by reacting ethylene (C 2 H 4, molar mass 28.1 g/mol) with water. How many grams of ethanol are formed from 14 g of ethylene?

27. 27 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product Moles of reactant Moles of reactant Grams of product Grams of product mole–mole conversion factor mole–mole conversion factor molar mass conversion factor molar mass conversion factor Moles of product Moles of product Grams of reactant Grams of reactant molar mass conversion factor molar mass conversion factor [1] [2] [3]

28. 28 Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product 14 g C 2 H 4 x 1 mol C 2 H 4 28.1 g C 2 H 4 x 1 mol C 2 H 6 O 1 mol C 2 H 4 x 46.1 g C 2 H 6 O 1 mol C 2 H 6 O Grams of reactant Grams of reactant molar mass conversion factor molar mass conversion factor mole–mole conversion factor mole–mole conversion factor molar mass conversion factor molar mass conversion factor =23 g C 2 H 6 O Grams of product Grams of product Grams C 2 H 4 cancel. Moles C 2 H 4 cancel. Moles C 2 H 6 O cancel.

29. 29 Oxidation and Reduction General Features Oxidation is the loss of electrons from an atom. Reduction is the gain of electrons by an atom. Both processes occur together in a single reaction called an oxidation−reduction or redox reaction. Thus, a redox reaction always has two components, one that is oxidized and one that is reduced. A redox reaction involves the transfer of electrons from one element to another.

30. Oxidation and Reduction General Features A redox reaction occurs when a strip of Zn metal is placed in a solution of Cu 2+ ions.

31. 31 Oxidation and Reduction General Features Zn + Cu 2+ Zn 2+ + Cu Zn loses 2 e – Cu 2+ gains 2 e − Zn loses 2 e − to form Zn 2+, so Zn is oxidized. Cu 2+ gains 2 e − to form Cu, so Cu 2+ is reduced.

32. 32 Oxidation and Reduction General Features Oxidation half reaction:ZnZn 2+ + 2 e − Each of these processes can be written as an individual half reaction: Zn + Cu 2+ Zn 2+ + Cu Zn loses 2 e – Cu 2+ gains 2 e – loss of e − Reduction half reaction:Cu 2+ + 2 e − Cu gain of e −

33. 33 Oxidation and Reduction General Features Zn + Cu 2+ Zn 2+ + Cu Zn acts as a reducing agent because it causes Cu 2+ to gain electrons and become reduced. A compound that is oxidized while causing another compound to be reduced is called a reducing agent. A compound that is reduced while causing another compound to be oxidized is called an oxidizing agent. Cu 2+ acts as an oxidizing agent because it causes Zn to lose electrons and become oxidized. oxidizedreduced

34. 34 Oxidation and Reduction General Features

35. 35 Oxidation and Reduction Examples of Oxidation–Reduction Reactions Iron Rusting 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) Fe 3+ O 2– neutral Feneutral O Fe loses e – and is oxidized. O gains e – and is reduced.

36. 36 2 Li + I 2 2 LiI neutral LiLi + I–I– Li loses e − and is oxidized. I gains e − and is reduced. neutral I Focus on Health & Medicine Pacemakers The Lithium–Iodine Battery in a Pacemaker

37. 37 2 Li + I 2 2 LiI The Lithium–Iodine Battery in a Pacemaker Focus on Health & Medicine Pacemakers

38. 38 Energy Energy Changes in Reactions When molecules come together and react, bonds are broken in the reactants and new bonds are formed in the products. Bond breaking always requires an input of energy. Bond formation always releases energy. Cl To cleave this bond, 58 kcal/mol must be added. To form this bond, 58 kcal/mol is released.

39. 39 Energy Changes in Reactions  H is the energy absorbed or released in a reaction; it is called the heat of reaction or the enthalpy change. When energy is absorbed, the reaction is said to be endothermic and  H is positive (+). When energy is released, the reaction is said to be exothermic and  H is negative (−). Cl To form this bond,  H = −58 kcal/mol. To cleave this bond,  H = +58 kcal/mol.

40. 40 Energy Changes in Reactions Calculations Involving  H Values  H indicates the relative strength of the bonds broken and formed in a reaction. More energy is released in forming bonds than is needed to break the bonds. When  H is negative: The products are lower in energy than the reactants. CH 4 (g) + 2 O 2 (g)CO 2 (g) + 2 H 2 O(l)  H = −213 kcal/mol Heat is released.

41. 41 Energy Changes in Reactions Calculations Involving  H Values More energy is needed, absorbed, to break bonds than is released in the formation of new bonds. When  H is positive: The products are higher in energy than the reactants. 6 CO 2 (g) + 6 H 2 O(l)C 6 H 12 O 6 (aq) + 6 O 2 (g) ΔH = +678 kcal/mol Heat is absorbed.

42. 42 Energy Changes in Reactions Calculations Involving  H Values

43. 43 Energy Diagrams For a reaction to occur, two molecules must collide with enough kinetic energy to break bonds. The energy changes in a reaction are often illustrated on an energy diagram, which plots energy on the vertical axis, and the progress of the reaction on the horizontal axis. Reactants are written on the left side and products on the right side, and a smooth curve that illustrates how energy changes with time connects them.

44. 44 Energy Diagrams E a, the energy of activation, is the difference in energy between the reactants and the transition state.

45. 45 Energy Diagrams The E a is the minimum amount of energy that the reactants must possess for a reaction to occur. E a is called the energy barrier and the height of the barrier determines the reaction rate. When the E a is high, few molecules have enough energy to cross the energy barrier, and the reaction is slow. When the E a is low, many molecules have enough energy to cross the energy barrier, and the reaction is fast.

46. 46 Energy Diagrams The difference in energy between the reactants and the products is the  H. If  H is negative, the reaction is exothermic:

47. 47 Energy Diagrams If  H is positive, the reaction is endothermic:

48. 48 Reaction Rates How Concentration and Temperature Affect Reaction Rate Increasing the concentration of the reactants: increases the number of collisions increases the reaction rate Increasing the temperature: increases the kinetic energy of the molecules increases the reaction rate

49. 49 Reaction Rates Catalysts A catalyst is a substance that speeds up the rate of a reaction. A catalyst is recovered unchanged in a reaction, and does not appear in the product. Catalysts accelerate a reaction by lowering E a without affecting  H.

50. 50 Reaction Rates Catalysts The uncatalyzed reaction (higher E a ) is slower. The catalyzed reaction (lower E a ) is faster.  H is the same for both reactions.

51. 51 Focus on The Environment Catalytic Converters A catalytic converter uses a metal catalyst, rhodium, platinum, or palladium, to catalyze three reactions that clean up the exhaust from an auto engine.