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Chemical Equations Chapter 10 Eugene Passer Chemistry Department Bronx Community College © John Wiley and Sons, Inc Version 1.1.

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Presentation on theme: "Chemical Equations Chapter 10 Eugene Passer Chemistry Department Bronx Community College © John Wiley and Sons, Inc Version 1.1."— Presentation transcript:

1 Chemical Equations Chapter 10 Eugene Passer Chemistry Department Bronx Community College © John Wiley and Sons, Inc Version 1.1

2 Chapter Outline 8.1 The Chemical EquationThe Chemical Equation 8.2 Writing and Balancing EquationsWriting and Balancing Equations 8.3 What Information Does an Equation Tell UsWhat Information Does an Equation Tell Us 8.4 Types of Chemical EquationsTypes of Chemical Equations 8.5 Heat in Chemical ReactionsHeat in Chemical Reactions

3 Chemists use chemical equations to describe reactions they observe in the laboratory or in nature. Chemical equations provide us with the means to 1.summarize the reaction 2.display the substances that are reacting 3.show the products 4.indicate the amounts of all component substances in a reaction.

4 The Chemical Equation

5 Chemical reactions always involve change. Atoms, molecules or ions rearrange to form new substances. The substances entering the reaction are called reactants. The substances formed in the reaction are called products. During reactions, chemical bonds are broken and new bonds are formed.

6 A chemical equation uses the chemical symbols and formulas of the reactants and products and other symbolic terms to represent a chemical reaction. A chemical equation is a shorthand expression for a chemical change or reaction.

7 Coefficients (whole numbers) are placed in front of substances to balance the equation and to indicate the number of units (atoms, molecules, moles, or ions) of each substance that are reacting.

8 Al + Fe 2 O 3  Fe + Al 2 O 3 coefficient 2 2

9 Conditions required to carry out the reaction may be placed above or below the arrow.

10 Al + Fe 2 O 3  Fe + Al 2 O 3 coefficient 2 2   heat

11 The physical state of a substance is indicated by symbols such as (l) for liquid.

12 2Al(s) + Fe 2 O 3 (s)  2Fe(l) + Al 2 O 3 (s) All atoms present in the reactant must also be present in the products. In a chemical reaction atoms are neither created nor destroyed. (s)(s) (l)(l)(s)(s)(s)(s)

13 Symbols Used in Chemical Reactions

14 Al + Fe 2 O 3  Fe + Al 2 O 3 reactantsproducts Al + Fe 2 O 3  Fe + Al 2 O 3 Chemical Equation iron oxygen bonds break aluminum oxygen bonds form

15 (aq) symbol aqueous meaning after formula location

16 placed between substances + symbol plus meaning location

17  symbol heat meaning written above  location

18  symbol gas formation meaning after formula location

19 Writing and Balancing Equations

20 To balance an equation adjust the number of atoms of each element so that they are the same on each side of the equation. Never change a correct formula to balance an equation.

21 Steps for Balancing Equations

22 Step 1 Identify the reaction. Write a description or word equation for the reaction. Mercury (II) oxide decomposes to form mercury and oxygen. mercury(II) oxide → mercury + oxygen

23 HgO  Hg + O 2 –The formulas of the reactants and products must be correct. –The reactants are written to the left of the arrow and the products to the right of the arrow. Step 2 Write the unbalanced (skeleton) equation. The formulas of the reactants and products can never be changed.

24 HgO → Hg + O 2 Step 3a Balance the equation. –There is one mercury atom on the reactant side and one mercury atom on the product side. –Mercury is balanced. Element Reactant Side Product Side Hg 1 1

25 Step 3a Balance the equation. –Count and compare the number of atoms of each element on both sides of the equation. –Determine the elements that require balancing.

26 Element Reactant Side Product Side O 1 2 Step 3a Balance the equation. –There are two oxygen atoms on the product side and there is one oxygen atom on the reactant side. –Oxygen needs to be balanced. HgO  Hg + O 2

27 Step 3b Balance the equation. –Balance each element one at a time, by placing whole numbers (coefficients) in front of the formulas containing the unbalanced element. –A coefficient placed before a formula multiplies every atom in the formula by that coefficient.

28 Element Reactant Side Product Side O 1 2  Oxygen (O) is balanced. Step 3b Balance the equation. Place a 2 in front of HgO to balance O.  There are two oxygen atoms on the reactant side and there are two oxygen atoms on the product side. HgO  Hg + O 2 2 2

29 Step 3c Balance the equation. Check all other elements after each individual element is balanced to see whether, in balancing one element, another element became unbalanced.

30 Element Reactant Side Product Side Hg 2 1 Count and compare the number of mercury (Hg) atoms on both sides of the equation. Step 3c Balance the equation.  Mercury (Hg) is not balanced. 2HgO  Hg + O 2 There are two mercury atoms on the reactant side and there is one mercury atom on the product side.

31 2HgO  Hg + O 2 Step 3c Balance the equation. Place a 2 in front of Hg to balance mercury.  Mercury (Hg) is balanced.  There are two mercury atoms on the reactant side and there are two mercury atoms on the product side. Element Reactant Side Product Side Hg

32 2HgO  2Hg + O 2 Element Reactant Side Product Side Hg 2 2 O 2 2  THE EQUATION IS BALANCED 

33 sulfuric acid + sodium hydroxide → sodium sulfate + water Balance the Equation

34 There is one Na on the reactant side and there are two Na on the product side. Reactant Side Product Side SO 4 11 Na12 O1 1 H3 2 2 H 2 SO 4 (aq) + NaOH(aq) → Na 2 SO 4 (aq) + H 2 O(l) 2 Place a 2 in front of NaOH to balance Na. Balance the Equation 2 4

35 H 2 SO 4 (aq) + NaOH(aq) → Na 2 SO 4 (aq) + H 2 O(l) There are 4 H on the reactant side and two H on the product side. Reactant Side Product Side SO 4 11 Na22 O2 1 H4 2 2 Place a 2 in front of H 2 O to balance H  THE EQUATION IS BALANCED 

36 butane + oxygen → carbon dioxide + water Balance the Equation

37 There are now 26 O on the product side. Reactant Side Product Side C 8 8 H2020 O Place a 13 in front of O 2 to balance O. 26  THE EQUATION IS BALANCED  C 4 H 10 (g) + O 2 (g) → CO 2 (g) + H 2 O(l) 28 10

38 What Information Does an Equation Tell Us

39 The meaning of a formula is context dependent. The formula H 2 O can mean: 1.2H and 1 O atom 2.1 molecule of water 3.1 mol of water x molecules of water g of water

40 In an equation formulas can represent units of individual chemical entities or moles. H2H2 +Cl 2 2HCl→ 1 molecule H 2 1 molecule Cl 2 2 molecules HCl 1 mol H 2 1 mol Cl 2 2 mol HCl

41 Formulas Number of molecules Number of atoms Number of moles Molar masses

42 Types of Chemical Equations

43 Combination Decomposition Single-Displacement Double-Displacement Combustion

44 Combination Reactions

45 A + B  AB Two reactants combine to form one product.

46 2Ca(s) + O 2 (g)  2CaO(s) Metal + Oxygen → Metal Oxide 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s)

47 S(s) + O 2 (g)  SO 2 (g) Nonmetal + Oxygen → Nonmetal Oxide N 2 (g) + O 2 (g)  2NO(g)

48 2K(s) + F 2 (g)  2KF(s) Metal + Nonmetal → Salt 2Al(s) + 3Cl 2 (g)  2AlCl 3 (s)

49 Na 2 O(s) + H 2 O(l)  2NaOH(aq) Metal Oxide + Water → Metal Hydroxide CaO(s) + 2H 2 O(l)  2Ca(OH) 2 (aq)

50 SO 3 (g) + H 2 O(l)  H 2 SO 4 (aq) Nonmetal Oxide + H 2 O(l) → Oxy-acid N 2 O 5 (g) + H 2 O(l)  2HNO 3 (aq)

51 Decomposition Reactions

52 AB  A + B A single substance breaks down to give two or more different substances.

53 Carbonate → CO 2 (g) CaCO 3 (s)  CaO(s) + CO 2 (g) 2NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(g) + CO 2 (g) Hydrogen Carbonate → CO 2 (g)

54 2Ag 2 O(s)  4Ag(s) + O 2 (g) Metal Oxide → Metal + Oxygen Metal Oxide → Metal Oxide + Oxygen 2PbO 2 (s)  2PbO(s) + O 2 (g)

55 Miscellaneous Reactions 2KClO 3 (s)  2KCl(s) + 3O 2 (g) 2NaNO 3 (s)  2NaNO 2 (s) + O 2 (g) 2H 2 O 2 (l)  2H 2 O(l) + O 2 (g)

56 Single Displacement Reactions

57 A + BC  AC + B One element reacts with a compound to replace one the elements of that compound.

58 Mg(s) + 2HCl(aq)  H 2 (g) + MgCl 2 (aq) 2Al(s) + 3H 2 SO 4 (aq)  3H 2 (g) + Al 2 (SO 4 ) 3 (aq) salt Metal + Acid → Hydrogen + Salt salt

59 2Na(s) + 2H 2 O(l)  H 2 (g) + 2NaOH(aq) Ca(s) + 2H 2 O(l)  H 2 (g) + Ca(OH) 2 (aq) Metal + Water → Hydrogen + Metal Hydroxide metal hydroxide

60 Metal + Water → Hydrogen + Metal Oxide metal oxide 3Fe(s) + 4H 2 O(g)  4H 2 (g) + Fe 3 O 4 (s)

61 The Activity Series

62 Metals K Ca Na Mg Al Zn Fe Ni Sn Pb H Cu Ag Hg An atom of an element in the activity series will displace an atom of an element below it from one of its compounds. Sodium (Na) will displace an atom below it from one of its compounds. increasing activity

63 Mg(s) + PbS(s)  MgS(s) + Pb(s) Metal Higher in Activity Series Displacing Metal Below It Magnesium is above lead in the activity series. Metals Mg Al Zn Fe Ni Sn Pb

64 Ag(s) + CuCl 2 (s)  no reaction Metal Lower in Activity Cannot Displace Metal Above It Metals Pb H Cu Ag Hg Silver is below copper in the activity series.

65 Double Displacement Reactions

66 Cl 2 (g) + CaBr 2 (s)  CaCl 2 (aq) + Br 2 (aq) Halogen Higher in Activity Series Displaces Halogen Below It Halogens F 2 Cl 2 Br 2 I 2 Chlorine is above bromine in the activity series.

67 AB + CD  AD + CB Two compounds exchange partners with each other to produce two different compounds. The reaction can be thought of as an exchange of positive and negative groups. A displaces C and combines with D B displaces D and combines with C

68 The Following Accompany Double Displacement Reactions formation of a precipitate release of gas bubbles release of heat formation of water

69 Acid Base Neutralization HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) H 2 SO 4 (aq) + 2NaOH(aq)  Na 2 SO 4 (aq) + 2H 2 O(l) acid + base → salt + water

70 Formation of an Insoluble Precipitate AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) Pb(NO 3 ) 2 (aq) + 2KI(aq)  PbI 2 (s) + 2KNO 3 (aq)

71 Metal Oxide + Acid CuO(s) + 2HNO 3 (aq)  Cu(NO 3 ) 2 (aq) + H 2 O(l) CaO(s) + 2HCl(aq)  CaCl 2 (s) + H 2 O(l) metal oxide + acid → salt + water

72 Formation of a Gas H 2 SO 4 (aq) + 2NaCN(aq)  Na 2 SO 4 (aq) + 2HCN(g) NH 4 Cl(aq) + NaOH(aq)  NaCl(aq) + NH 4 OH(aq) NH 4 OH(aq)  NH 3 (g) + H 2 O(l) indirect gas formation

73 Introduction to Stoichiometry: The Mole-Ratio Method

74 Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and products. Mole Ratio: a ratio between the moles of any two substances involved in a chemical reaction. –The coefficients used in mole ratio expressions are derived from the coefficients used in the balanced equation.

75 1 mol2 mol3 mol N 2 + 3H 2  2NH 3

76 1 mol2 mol3 mol

77 The mole ratio is used to convert the number of moles of one substance to the corresponding number of moles of another substance in a stoichiometry problem. The mole ratio is used in the solution of every type of stoichiometry problem.

78 Identify the starting substance from the data given in the problem statement. Convert the quantity of the starting substance to moles, if it is not already in moles. Step 1 Determine the number of moles of starting substance.

79 The number of moles of each substance in the balanced equation is indicated by the coefficient in front of each substance. Use these coefficients to set up the mole ratio. Step 2 Determine the mole ratio of the desired substance to the starting substance.

80 Multiply the number of moles of starting substance (from Step 1) by the mole ratio to obtain the number of moles of desired substance.

81 Step 3. Calculate the desired substance in the units specified in the problem. If the answer is to be in moles, the calculation is complete If units other than moles are wanted, multiply the moles of the desired substance (from Step 2) by the appropriate factor to convert moles to the units required.

82 Step 3. Calculate the desired substance in the units specified in the problem.

83

84

85 Limiting-Reactant and Yield Calculations

86 It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present. The limiting reactant limits the amount of product that can be formed. The limiting reactant is one of the reactants in a chemical reaction.

87 Steps Used to Determine the Limiting Reactant

88 1.Calculate the amount of product (moles or grams, as needed) formed from each reactant. 2.Determine which reactant is limiting. (The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess. 3.Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. This gives the amount of the that substance that remains unreacted.

89 Reaction Yield

90 The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.

91 Many reactions fail to give a 100% yield of product. This occurs because of side reactions and the fact that many reactions are reversible.

92 The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant. The actual yield is the amount of product finally obtained from a given amount of reactant.

93 The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.


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