Presentation on theme: "Chapter 5 Chemical Reactions"— Presentation transcript:
1 Chapter 5 Chemical Reactions Chemical ChangesChemical EquationsBalancing a Chemical Equation
2 Physical PropertiesThe physical properties of a substance are the characteristics we can observe or measure without changing the substance.
3 Physical and Chemical Change In a physical change,The identity and composition of the substance do not change.The state can change or the material can be torn into smaller pieces.In a chemical change,New substances form with different compositions and properties.A chemical reaction takes place.
6 Learning Check Classify each of the following as a 1) physical change or 2) chemical changeA. ____ Burning a candle.B. ____ Ice melting on the street.C. ____ Toasting a marshmallow.D. ____ Cutting a pizza.E. ____ Polishing a silver bowl.
7 Solution Classify each of the following as a 1) physical change or 2) chemical changeA Burning a candle.B Ice melting on the street.C Toasting a marshmallow.D Cutting a pizza.E Polishing a silver bowl.
8 Chemical ReactionIn a chemical reaction, a chemical change produces one or more new substances.During a reaction, old bonds are broken and newbonds are formed.
9 Chemical ReactionIn a chemical reaction, atoms in the reactants are rearranged to form one or more different substances.In this reaction, Fe and O2 react to form rust (Fe2O3).4Fe + 3O Fe2O3
10 Writing a Chemical Equation Shows the chemical formulas of the reactants to the left of an arrow and the products on the right.Reactants ProductsMgO C CO MgCan be read in words. “Magnesium oxide reacts with carbon to form carbon monoxide and magnesium.”
11 Symbols Used in Equations Symbols used in equations show the states of the reactants and products and the reaction conditions.
12 Quantities in A Chemical Reaction 4 NH O NO H2OFour molecules of NH3 react with five molecules of O2 to produce four molecules of NO and six molecules of H2O.orFour moles of NH3 react with 5 moles of O2 to produce four moles of NO and six moles of H2O.
13 Law of Conservation of Mass In any ordinary chemical reaction, matter is not created nor destroyed.H Cl HClTotal atoms = Total atoms2 H, 2 Cl 2H, 2 ClTotal Mass = Total Mass2(1.0) (35.5) 2(36.5)73.0 g = g
14 Balancing a Chemical Equation A chemical equation is balanced when there are the same numbers of each type of atom on both sides of the equation.Al S Al2S Not Balanced2Al S Al2S3 Balanced
15 Using Coefficients to Balance To balance an equation, place coefficients in front of the appropriate formulas.4 NH O NO H2OCheck the balance by counting the atoms of each element in the reactants and the products.4 N (4 x 1N) = N (4 x 1N)12 H (4 x 3H) = 12 H (6 x 2H)10 O (5 x 2O) = 10 O (4O + 6O)
16 Steps in Balancing an Equation Balance one element at a time.Use only coefficients to balance.Fe3O H Fe H2OFe: Fe3O H Fe H2OO: Fe3O H Fe H2OH: Fe3O H Fe H2O
17 Learning CheckCheck the balance of atoms in the following: Fe3O H Fe H2OA. Number of H atoms in products.1) 2 2) ) 8B. Number of O atoms in reactants.C. Number of Fe atoms in reactants.1) 1 2) ) 4
18 Solution Fe3O4 + 4 H2 3 Fe + 4 H2O A. Number of H atoms in products. B. Number of O atoms in reactants.2) 4 (Fe3O4)C. Number of Fe atoms in reactants.2) 3 (Fe3O4)
19 Balancing with Polyatomic Ions Polyatomic ions can be balanced as a unit when they appear on both sides.Pb(NO3)2 + NaCl NaNO3 + PbCl2Balance NO3- as a unitPb(NO3)2 + NaCl NaNO PbCl22 NO = NO3-Balance Na (or Cl)Pb(NO3)2 + 2NaCl NaNO3 + PbCl22Na = Na+2Cl = Cl-
20 Learning Check Balance each equation. The coefficients in the answers are read from left to right.A. __Mg + __N __Mg3N21) 1, 3, ) 3, 1, ) 3, 1, 1B. __Al __Cl __AlCl31) 3, 3, 2 2) 1, 3, ) 2, 3, 2
21 SolutionA. 3) 3, 1, 13 Mg N Mg3N2B. 3) 2, 3, 22 Al Cl AlCl3
23 Solution A. 2) 2, 3, 4, 3 2 Fe2O3 + 3 C 4 Fe + 3 CO2 B. 1) 2, 3, 3, 1 2 Al FeO Fe Al2O3C. 2) 2, 3, 1, 32 Al H2SO Al2(SO4) H2
24 Chapter 5 Chemical Reactions Types of Reactions
25 Types of ReactionsChemical reactions are classified into general types:CombinationDecompositionSingle ReplacementDouble ReplacementCombustion
26 Combination Reactions In a combination reaction, two or more elements or simple compounds combine to form one product.A B ABExamplesH Cl2 2HCl2S O2 2SO34Fe O2 2Fe2O3
27 Combination Reactions In a combination reaction, magnesium and oxygen react to form magnesium oxide.2Mg + O MgOO2MgOMg
28 Decomposition Reactions In a decomposition reaction, one substance is broken down (split) into two or more simpler substances.AB A + B2HgO 2Hg + O22KClO3 2KCl + 3 O2
29 Learning Check Classify the following reactions as 1) combination or 2) decomposition:___A. H2 + Br HBr___B. Al2(CO3)3 Al2O3 + 3CO2___C. 4 Al + 3C Al4C3
30 Solution Classify the following reactions as 1) combination or 2) decomposition:1 A. H2 + Br HBr2 B. Al2(CO3)3 Al2O3 + 3CO21 C. 4 Al + 3C Al4C3
31 Single ReplacementIn a single replacement, one element takes the place of an element in a reacting compound.A BC AC BZn(s) + HCl(aq) ZnCl2(aq) + H2(g)H2HClZnZnCl2
32 Double ReplacementIn a double replacement, the positive ions in the reacting compounds switch places.AB CD AD CBAgNO3 + NaCl AgCl + NaNO3ZnS HCl ZnCl2 + H2S
33 Example of a Double Replacement When solutions of sodium sulfate and barium chloride are mixed, solid BaSO4 is produced.BaCl Na2SO BaSO NaClBaSO4
34 Learning Check Classify each of the following reactions as a 1) single replacement or 2) double replacement__A. 2Al H2SO Al2(SO4) H2__B. Na2SO4 + 2AgNO Ag2SO4 + 2NaNO3__C. 3C + Fe2O Fe CO
35 Solution Classify each of the following reactions as a 1) single replacement or 2) double replacement1 A. 2Al + 3H2SO Al2(SO4)3 + 3H22 B. Na2SO4 + 2AgNO Ag2SO4 + 2NaNO31 C. 3C + Fe2O Fe + 3CO
36 CombustionIn a combustion reaction, a reactant often containing carbon reacts with oxygen O2.C + O CO2CH4 + 2O CO H2OC3H O CO2 + 4H2OMany combustion reactions utilize fuels that are burned in oxygen to produce CO2, H2O, and energy.
40 Learning Check Identify each reaction as 1) combination 2) decomposition ) combustion4) single replacement 5) double replacementA. 3Ba + N Ba3N2B. 2Ag + H2S Ag2S + H2C. SiO HF SiF H2OD. PbCl2 + K2SO KCl + PbSO4E. K2CO K2O + CO2F. C2H O CO2 + 2H2O
41 Solution Identify each reaction as 1) combination 2) decomposition ) combustion4) single replacement 5) double replacement1 A. 3Ba + N Ba3N24 B. 2Ag + H2S Ag2S + H25 C. SiO HF SiF H2O5 D. PbCl2 + K2SO KCl + PbSO42 E. K2CO K2O + CO23 F. C2H O CO H2O
42 Chapter 5 Chemical Reactions Oxidation-Reduction Reactions
43 Oxidation and Reduction Are an important type of reaction.Provide us with energy from food.Provide electrical energy in batteries.Occur when iron rusts.4Fe + 3O Fe2O3
44 Electron Loss and GainAn oxidation-reduction reaction involves the transfer of electrons from one reactant to another.In oxidation, electrons are lost.Zn Zn2+ + 2e- (loss of electrons)In reduction, electrons are gained.Cu e Cu (gain of electrons)
45 Half-Reactions for Oxidation-Reduction In the oxidation-reduction reaction of zinc and copper(II) sulfate, the zinc is oxidized and the Cu2+ (from Cu2+ SO42-) is reduced.Zn Zn e- oxidationCu e Cu reduction
46 Learning Check Identify each of the following as an 1) oxidation or a 2) reduction:__A. Sn Sn e-__B. Fe e- Fe2+__C. Cl e- 2Cl-
47 Solution Identify each of the following as an 1) oxidation or a 2) reduction:1 A. Sn Sn e-2 B. Fe e- Fe2+2 C. Cl e- 2Cl-
48 Balanced Red-Ox Equations In a balanced oxidation-reduction equation, the loss of electrons is equal to the gain of electrons.Zn + Cu Zn2+ + CuThe loss and gain of two electrons is shown in the separate oxidation and reduction reactions.Zn Zn2+ + 2e- oxidationCu e Cu reduction
49 Learning Check In light-sensitive sunglasses, UV light initiates an oxidation-reduction reaction.uv lightAg+ + Cl Ag ClA. Which reactant is oxidized?1) Ag ) Cl ) AgB. Which reactant is reduced?1) Ag ) Cl ) Cl
50 Solution In light-sensitive sunglasses, UV light initiates an oxidation-reduction reaction.uv lightAg+ + Cl Ag ClA. Which reactant is oxidized2) Cl Cl- Cl + e-B. Which reactant is reduced?1) Ag+ Ag+ + e Ag
51 Learning CheckWrite the separate oxidation and reduction reactions for the following equation.2Cs F CsF
52 SolutionWrite the separate oxidation and reduction reactions for the following equation.2Cs F CsFCs Cs e- oxidationF + 1e- F- reduction
53 Oxidation with OxygenAn early definition of oxidation is the addition of oxygen O2 to a reactant.A metal or nonmetal is oxidized while the O2 is reduced to O2-.4K + O K2OC + O CO22SO2 + O SO3
54 Gain and Loss of Hydrogen In organic and biological reactions, oxidation involves the loss of hydrogen atoms and reduction involves a gain of hydrogen atoms.oxidation = Loss of Hreduction = Gain of HCH3OH H2CO H (loss of H)Methanol Formaldehyde
59 Collection TermsA collection term indicates a specific number of items.For example, 1 dozen doughnuts contains 12 doughnuts.1 ream of paper means 500 sheets.1 case is 24 cans.
60 A MoleA mole contains 6.02 x 1023 particles, which is the number of carbon atoms in g of carbon.1 mole C = x 1023 C atomsThe number 6.02 x 1023 is known as Avogadro’s number.One mole of any element contains Avogadro’s number of atoms.1 mole Na = x 1023 Na atoms1 mole Au = x 1023 Au atoms
61 A Mole of MoleculesAvogadro’s number is also the number of molecules and formula units in one mole of a compound.One mole of a covalent compound contains Avogadro’s number of molecules.1 mole CO2 = x 1023 CO2 molecules1 mole H2O = x 1023 H2O moleculesOne mole of an ionic compound contains Avogadro’s number of formula units.1 mole NaCl = 6.02 x 1023 NaCl formula units
63 Avogadro’s Number Avogadro’s number is written as conversion factors. 6.02 x 1023 particles and 1 mole1 mole x 1023 particlesThe number of molecules in 0.50 mole of CO2 molecules is calculated as0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules1 mole CO2 molecules= 3.0 x 1023 CO2 molecules
64 Learning Check A. Calculate the number of atoms in 2.0 moles of Al. 1) Al atoms2) x 1023 Al atoms3) x 1024 Al atomsB. Calculate the number of moles of S in 1.8 x 1024 S.1) mole S atoms2) mole S atoms3) x 1048 mole S atoms
65 SolutionA. Calculate the number of atoms in 2.0 moles of Al. 3) x 1024 Al atoms2.0 moles Al x x 1023 Al atoms1 mole AlB. Calculate the number of moles of S in 1.8 x 1024 S. 2) 3.0 mole S atoms1.8 x 1024 S atoms x mole S6.02 x 1023 S atoms
66 Molar Mass The mass of one mole is called molar mass. The molar mass of an element is the atomic mass expressed in grams.
67 Learning Check Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms = ________B. 1 mole of Sn atoms = ________
68 Solution Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms = 39.1 gB. 1 mole of Sn atoms = g
69 Molar Mass of CaCl2For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl2 to the nearest 0.1 g as follows.ElementNumber of MolesAtomic MassTotal MassCa140.1 g/mole40.1 gCl2235.5 g/mole71.0 gCaCl2111.1 g
70 Molar Mass of K3PO4 Determine the molar mass of K3PO4 to 0.1 g. K 3 ElementNumber of MolesAtomic MassTotal Mass in K3PO4K339.1 g/mole117.3 gP131.0 g/mole31.0 gO416.0 g/mole64.0 gK3PO4212.3 g
72 Learning Check A. 1 mole of K2O = ______g B. 1 mole of antacid Al(OH)3 = ______g
73 SolutionA. 1 mole of K2O2 moles K (39.1 g/mole) + 1 mole O (16.0 g/mole)78.2 g g = gB. 1 mole of antacid Al(OH)31 mole Al (27.0 g/mole) + 3 moles O (16.0 g/mole)+ 3 moles H (1.0 g/mole)27.0 g g g = g
74 Learning CheckProzac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?1) g/mole2) 262 g/mole3) 309 g/mole
75 SolutionProzac, C17H18F3NO, is a widely used antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac?3) 309 g/mole17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) =
76 Molar Mass FactorsMethane CH4 known as natural gas is used in gas cook tops and gas heaters.1 mole CH4 = gThe molar mass of methane can be written as conversion factors.16.0 g CH and mole CH41 mole CH g CH4
77 Learning CheckAcetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.
78 SolutionAcetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid.1 mole of acetic acid = g acetic acid1 mole acetic acid and g acetic acid60.0 g acetic acid mole acetic acid
79 Calculations with Molar Mass Mole factors are used to convert between the grams of a substance and the number of moles.Mole factorGramsMoles
80 Calculating Grams from Moles Aluminum is often used for the structure oflightweight bicycle frames. How many gramsof Al are in 3.00 moles of Al?3.00 moles Al x g Al = g Al1 mole Almole factor for Al
81 Learning CheckThe artificial sweetener aspartame (Nutri-Sweet) C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many moles of aspartame are present in 225 g of aspartame?
82 Solution Calculate the molar mass of C14H18N2O5. (14 x 12.0) + (18 x 1.0) + (2 x 14.0) + (5 x 16.0) = 294 g/moleSet up the calculation using a mole factor.225 g aspartame x 1 mole aspartame294 g aspartamemole factor(inverted)= mole aspartame
83 Conservation of MassIn a chemical reaction, the mass of the reactants is equal to the mass of the products moles Ag mole S = 1 mole Ag2S 2 (107.9 g) (32.1 g) = 1 (247.9 g)247.9 g reactants = g product
84 Moles in EquationsWe can read the equation in “moles” by placing the word “moles” between each coefficient and formula. 4 Fe O Fe2O3 4 moles Fe moles O moles Fe2O3
85 Writing Mole-Mole Factors A mole-mole factor is a ratio of the coefficients for two substances.4 Fe O Fe2O3Fe and O mole Fe and 3 mole O23 mole O mole FeFe and Fe2O mole Fe and 2 mole Fe2O32 mole Fe2O mole FeO2 and Fe2O mole O and 2 mole Fe2O32 mole Fe2O mole O2
86 Learning Check Consider the following equation: 3 H2 + N2 2 NH3 A. A mole factor for H2 and N2 is1) 3 mole N ) 1 mole N ) 1 mole N21 mole H mole H mole H2B. A mole factor for NH3 and H2 is1) 1 mole H ) 2 mole NH ) 3 mole N22 mole NH mole H mole NH3
87 Solution 3 H2 + N2 2 NH3 A. A mole factor for H2 and N2 is 2) 1 mole N23 mole H2B. A mole factor for NH3 and H2 is2) 2 mole NH3
88 Calculations with Mole Factors Consider the following reaction:4 Fe O Fe2O3How many moles of Fe2O3 are produced when 6.0moles O2 react?Use the appropriate mole factor to determine themoles Fe2O3.6.0 mole O2 x 2 mole Fe2O3 = 4.0 mole Fe2O33 mole O2
89 Learning Check Consider the following reaction: 4 Fe + 3 O2 2 Fe2O3 How many moles of Fe are needed to react with 12.0 moles of O2?1) moles Fe2) moles Fe3) moles Fe
90 Solution 3) 16.0 moles Fe Consider the following reaction: 4 Fe O Fe2O3How many moles of Fe are needed to react with 12.0 moles of O2?12.0 mole O2 x 4 mole Fe = moles Fe3 mole O2
92 Learning Check How many grams of O2 are needed to produce 0.400 mole of Fe2O3?4 Fe O Fe2O31) g O22) g O23) g O2
93 Solution 2) 19.2 g O2 0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2 2 mole Fe2O3 1 mole O2mole factor molar mass= g O2
94 Calculating the Mass of a Reactant The reaction between H2 and O2 produces 13.1 g ofwater. How many grams of O2 reacted?2H O2 2H2O? g gPlan: g H2O mole H2O mole O g O213.1 g H2O x 1 mole H2O x 1 mole O2 x g O g H2O mole H2O mole O2= g O2
95 Learning Check 2 C2H2 + 5 O2 4 CO2 + 2 H2O Acetylene gas C2H2 burns in the oxyactylene torch for welding. How many grams of C2H2 are burned if the reaction produces 75.0 g of CO2?2 C2H O CO H2O1) g C2H22) g C2H23) g C2H2
96 Solution 3) 22.2 g C2H2 2 C2H2 + 5 O2 4 CO2 + 2 H2O = 22.2 g C2H2 75.0 g CO2 x 1 mole CO2 x 2 moles C2H2 x g C2H244.0 g CO moles CO mole C2H2= g C2H2
97 Percent YieldYou prepared cookie dough to make 5 dozen cookies. The phone rings and you answer. While you talk, a sheet of 12 cookies burns. You have to throw them out. The rest of the cookies are okay. The results of our baking can be described as follows:Theoretical yield 60 cookies possibleActual yield cookies to eatPercent yield 48 cookies x 100 = 80% yield cookies
98 Percent YieldThe theoretical yield is the maximum amount of product calculated using the balanced equation.The actual yield is the amount of product obtained when the reaction is run.Percent yield is the ratio of actual yield compared to the theoretical yield. Percent Yield = Actual Yield (g) x Theoretical Yield (g)
99 Sample Exercise % Yield Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide.2C O COWhat is the percent yield if 40.0 g of CO are produced from the reaction of 30.0 g O2?
100 Sample Exercise % Yield (cont.) 1. Calculate theoretical yield of CO.30.0 g O2 x 1 mole O2 x 2 mole CO x g CO32.0 g O mole O mole CO = g CO (theoretical)2. Calculate the percent yield.40.0 g CO (actual) x 100 = % yield52.5 g CO(theoretical)
101 Learning Check In the lab, N2 and 5.0 g of H2 are reacted and produce 16.0 g of NH3. What is thepercent yield for the reaction?N2(g) + 3H2(g) NH3(g)1) %2) %3) %
102 Solution 2) 56.5 % N2(g) + 3H2(g) 2NH3(g) 2) %N2(g) + 3H2(g) NH3(g)5.0 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH32.0 g H moles H mole NH3= g NH3 (theoretical)Percent yield = g NH3 x 100 = %28.3 g NH3