Chapter 5 Chemical Reactions Chemical Changes Chemical Equations Balancing a Chemical Equation

Physical Properties The physical properties of a substance are the characteristics we can observe or measure without changing the substance.

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.

Physical and Chemical Change

Some Examples of Chemical and Physical Changes

Learning Check Classify each of the following as a 1) physical change or 2) chemical change A. ____ Burning a candle. B. ____ Ice melting on the street. C. ____ Toasting a marshmallow. D. ____ Cutting a pizza. E. ____ Polishing a silver bowl.

Solution Classify each of the following as a 1) physical change or 2) chemical change A. 2 Burning a candle. B. 1 Ice melting on the street. C. 2 Toasting a marshmallow. D. 1 Cutting a pizza. E. 2 Polishing a silver bowl.

Chemical Reaction In a chemical reaction, a chemical change produces one or more new substances. During a reaction, old bonds are broken and new bonds are formed.

Chemical Reaction In 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 + 3O2 2Fe2O3

Writing a Chemical Equation Shows the chemical formulas of the reactants to the left of an arrow and the products on the right. Reactants Products MgO + C CO + Mg Can be read in words. “Magnesium oxide reacts with carbon to form carbon monoxide and magnesium.”

Symbols Used in Equations Symbols used in equations show the states of the reactants and products and the reaction conditions.

Quantities in A Chemical Reaction 4 NH3 + 5 O2 4 NO + 6 H2O Four molecules of NH3 react with five molecules of O2 to produce four molecules of NO and six molecules of H2O. or Four moles of NH3 react with 5 moles of O2 to produce four moles of NO and six moles of H2O.

Law of Conservation of Mass In any ordinary chemical reaction, matter is not created nor destroyed. + + H2 + Cl2 2 HCl Total atoms = Total atoms 2 H, 2 Cl 2H, 2 Cl Total Mass = Total Mass 2(1.0) + 2(35.5) 2(36.5) 73.0 g = 73.0 g

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 Al2S3 Not Balanced 2Al + 3S Al2S3 Balanced

Using Coefficients to Balance To balance an equation, place coefficients in front of the appropriate formulas. 4 NH3 + 5 O2 4 NO + 6 H2O Check the balance by counting the atoms of each element in the reactants and the products. 4 N (4 x 1N) = 4 N (4 x 1N) 12 H (4 x 3H) = 12 H (6 x 2H) 10 O (5 x 2O) = 10 O (4O + 6O)

Steps in Balancing an Equation Balance one element at a time. Use only coefficients to balance. Fe3O4 + H2 Fe + H2O Fe: Fe3O4 + H2 3Fe + H2O O: Fe3O4 + H2 3Fe + 4H2O H: Fe3O4 + 4H2 3Fe + 4H2O

Learning Check Check the balance of atoms in the following: Fe3O4 + 4 H2 3 Fe + 4 H2O A. Number of H atoms in products. 1) 2 2) 4 3) 8 B. Number of O atoms in reactants. C. Number of Fe atoms in reactants. 1) 1 2) 3 3) 4

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)

Balancing with Polyatomic Ions Polyatomic ions can be balanced as a unit when they appear on both sides. Pb(NO3)2 + NaCl NaNO3 + PbCl2 Balance NO3- as a unit Pb(NO3)2 + NaCl 2NaNO3 + PbCl2 2 NO3- = 2 NO3- Balance Na (or Cl) Pb(NO3)2 + 2NaCl 2NaNO3 + PbCl2 2Na+ = 2Na+ 2Cl- = 2Cl-

Learning Check Balance each equation. The coefficients in the answers are read from left to right. A. __Mg + __N2 __Mg3N2 1) 1, 3, 2 2) 3, 1, 2 3) 3, 1, 1 B. __Al + __Cl2 __AlCl3 1) 3, 3, 2 2) 1, 3, 1 3) 2, 3, 2

Solution A. 3) 3, 1, 1 3 Mg + 1 N2 1 Mg3N2 B. 3) 2, 3, 2 2 Al + 3 Cl2 2 AlCl3

Learning Check A. __Fe2O3 + __C __Fe + __CO2 1) 2, 3, 2,3 2) 2, 3, 4, 3 3) 1, 1, 2, 3 B. __Al + __FeO __Fe + __Al2O3 1) 2, 3, 3, 1 2) 2, 1, 1, 1 3) 3, 3, 3, 1 C. __Al + __H2SO4 __Al2(SO4)3 + __H2 1) 3, 2, 1, 2 2) 2, 3, 1, 3 3) 2, 3, 2, 3

Solution A. 2) 2, 3, 4, 3 2 Fe2O3 + 3 C 4 Fe + 3 CO2 B. 1) 2, 3, 3, 1 2 Al + 3 FeO 3 Fe + 1 Al2O3 C. 2) 2, 3, 1, 3 2 Al + 3 H2SO4 1 Al2(SO4)3 + 3 H2

Chapter 5 Chemical Reactions Types of Reactions

Types of Reactions Chemical reactions are classified into general types: Combination Decomposition Single Replacement Double Replacement Combustion

Combination Reactions In a combination reaction, two or more elements or simple compounds combine to form one product. A + B AB Examples H2 + Cl2 2HCl 2S + 3O2 2SO3 4Fe + 3O2 2Fe2O3

Combination Reactions In a combination reaction, magnesium and oxygen react to form magnesium oxide. 2Mg + O2 2MgO O2 MgO Mg

Decomposition Reactions In a decomposition reaction, one substance is broken down (split) into two or more simpler substances. AB A + B 2HgO 2Hg + O2 2KClO3 2KCl + 3 O2

Learning Check Classify the following reactions as 1) combination or 2) decomposition: ___A. H2 + Br2 2HBr ___B. Al2(CO3)3 Al2O3 + 3CO2 ___C. 4 Al + 3C Al4C3

Solution Classify the following reactions as 1) combination or 2) decomposition: 1 A. H2 + Br2 2HBr 2 B. Al2(CO3)3 Al2O3 + 3CO2 1 C. 4 Al + 3C Al4C3

Single Replacement In a single replacement, one element takes the place of an element in a reacting compound. A + BC AC + B Zn(s) + HCl(aq) ZnCl2(aq) + H2(g) H2 HCl Zn ZnCl2

Double Replacement In a double replacement, the positive ions in the reacting compounds switch places. AB + CD AD + CB AgNO3 + NaCl AgCl + NaNO3 ZnS + 2HCl ZnCl2 + H2S

Example of a Double Replacement When solutions of sodium sulfate and barium chloride are mixed, solid BaSO4 is produced. BaCl2 + Na2SO4 BaSO4 + 2NaCl BaSO4

Learning Check Classify each of the following reactions as a 1) single replacement or 2) double replacement __A. 2Al + 3H2SO4 Al2(SO4)3 + 3H2 __B. Na2SO4 + 2AgNO3 Ag2SO4 + 2NaNO3 __C. 3C + Fe2O3 2Fe + 3CO

Solution Classify each of the following reactions as a 1) single replacement or 2) double replacement 1 A. 2Al + 3H2SO4 Al2(SO4)3 + 3H2 2 B. Na2SO4 + 2AgNO3 Ag2SO4 + 2NaNO3 1 C. 3C + Fe2O3 2Fe + 3CO

Combustion In a combustion reaction, a reactant often containing carbon reacts with oxygen O2. C + O2 CO2 CH4 + 2O2 CO2 + 2H2O C3H8 + 5O2 3CO2 + 4H2O Many combustion reactions utilize fuels that are burned in oxygen to produce CO2, H2O, and energy.

Learning Check Balance the combustion equation: ___C5H12 + ___O2 ___CO2 + ___H2O

Solution Balance the combustion equation: 1 C5H12 + 8 O2 5 CO2 + 6 H2O

Summary of Reaction Types

Learning Check Identify each reaction as 1) combination 2) decomposition 3) combustion 4) single replacement 5) double replacement A. 3Ba + N2 Ba3N2 B. 2Ag + H2S Ag2S + H2 C. SiO2 + 4HF SiF4 + 2H2O D. PbCl2 + K2SO4 2KCl + PbSO4 E. K2CO3 K2O + CO2 F. C2H4 + 3O2 2CO2 + 2H2O

Solution Identify each reaction as 1) combination 2) decomposition 3) combustion 4) single replacement 5) double replacement 1 A. 3Ba + N2 Ba3N2 4 B. 2Ag + H2S Ag2S + H2 5 C. SiO2 + 4HF SiF4 + 2H2O 5 D. PbCl2 + K2SO4 2KCl + PbSO4 2 E. K2CO3 K2O + CO2 3 F. C2H4 + 3O2 2CO2 + 2H2O

Chapter 5 Chemical Reactions Oxidation-Reduction Reactions

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 + 3O2 2Fe2O3

Electron Loss and Gain An 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. Cu2+ + 2e- Cu (gain of electrons)

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 Zn2+ + 2e- oxidation Cu2+ + 2e- Cu reduction

Learning Check Identify each of the following as an 1) oxidation or a 2) reduction: __A. Sn Sn4+ + 4e- __B. Fe3+ + 1e- Fe2+ __C. Cl2 + 2e- 2Cl-

Solution Identify each of the following as an 1) oxidation or a 2) reduction: 1 A. Sn Sn4+ + 4e- 2 B. Fe3+ + 1e- Fe2+ 2 C. Cl2 + 2e- 2Cl-

Balanced Red-Ox Equations In a balanced oxidation-reduction equation, the loss of electrons is equal to the gain of electrons. Zn + Cu2+ Zn2+ + Cu The loss and gain of two electrons is shown in the separate oxidation and reduction reactions. Zn Zn2+ + 2e- oxidation Cu2+ + 2e- Cu reduction

Learning Check In light-sensitive sunglasses, UV light initiates an oxidation-reduction reaction. uv light Ag+ + Cl- Ag + Cl A. Which reactant is oxidized? 1) Ag+ 2) Cl- 3) Ag B. Which reactant is reduced? 1) Ag+ 2) Cl- 3) Cl

Solution In light-sensitive sunglasses, UV light initiates an oxidation-reduction reaction. uv light Ag+ + Cl- Ag + Cl A. Which reactant is oxidized 2) Cl- Cl- Cl + e- B. Which reactant is reduced? 1) Ag+ Ag+ + e- Ag

Learning Check Write the separate oxidation and reduction reactions for the following equation. 2Cs + F2 2CsF

Solution Write the separate oxidation and reduction reactions for the following equation. 2Cs + F2 2CsF Cs Cs+ + 1e- oxidation F + 1e- F- reduction

Oxidation with Oxygen An 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 + O2 2K2O C + O2 CO2 2SO2 + O2 2SO3

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 H reduction = Gain of H CH3OH H2CO + 2H (loss of H) Methanol Formaldehyde

Summary

Learning Check Identify the substances that are oxidized and reduced in the following reactions. A. 4Fe + 3O2 2Fe2O3 B. 6Na + N2 2Na3N C. 2K + I2 2KI

Solution 3+ 2- A. 4Fe + 3O2 2Fe2O3 Fe oxidized; O2 reduced 1+ 3- 3+ 2- A. 4Fe + 3O2 2Fe2O3 Fe oxidized; O2 reduced 1+ 3- B. 6Na + N2 2Na3N Na oxidized: N2 reduced 1+ 1- C. 2K + I2 2KI K oxidized: I2 reduced

The Mole

Collection Terms A 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.

A Mole A mole contains 6.02 x 1023 particles, which is the number of carbon atoms in 12.01 g of carbon. 1 mole C = 6.02 x 1023 C atoms The 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 = 6.02 x 1023 Na atoms 1 mole Au = 6.02 x 1023 Au atoms

A Mole of Molecules Avogadro’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 = 6.02 x 1023 CO2 molecules 1 mole H2O = 6.02 x 1023 H2O molecules One mole of an ionic compound contains Avogadro’s number of formula units. 1 mole NaCl = 6.02 x 1023 NaCl formula units

Samples of One Mole Quantities

Avogadro’s Number Avogadro’s number is written as conversion factors. 6.02 x 1023 particles and 1 mole 1 mole 6.02 x 1023 particles The number of molecules in 0.50 mole of CO2 molecules is calculated as 0.50 mole CO2 molecules x 6.02 x 1023 CO2 molecules 1 mole CO2 molecules = 3.0 x 1023 CO2 molecules

Learning Check A. Calculate the number of atoms in 2.0 moles of Al. 1) 2.0 Al atoms 2) 3.0 x 1023 Al atoms 3) 1.2 x 1024 Al atoms B. Calculate the number of moles of S in 1.8 x 1024 S. 1) 1.0 mole S atoms 2) 3.0 mole S atoms 3) 1.1 x 1048 mole S atoms

Solution A. Calculate the number of atoms in 2.0 moles of Al. 3) 1.2 x 1024 Al atoms 2.0 moles Al x 6.02 x 1023 Al atoms 1 mole Al B. Calculate the number of moles of S in 1.8 x 1024 S. 2) 3.0 mole S atoms 1.8 x 1024 S atoms x 1 mole S 6.02 x 1023 S atoms

Molar Mass The mass of one mole is called molar mass. The molar mass of an element is the atomic mass expressed in grams.

Learning Check Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms = ________ B. 1 mole of Sn atoms = ________

Solution Give the molar mass to the nearest 0.1 g. A. 1 mole of K atoms = 39.1 g B. 1 mole of Sn atoms = 118.7 g

Molar Mass of CaCl2 For 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. Element Number of Moles Atomic Mass Total Mass Ca 1 40.1 g/mole 40.1 g Cl2 2 35.5 g/mole 71.0 g CaCl2 111.1 g

Molar Mass of K3PO4 Determine the molar mass of K3PO4 to 0.1 g. K 3 Element Number of Moles Atomic Mass Total Mass in K3PO4 K 3 39.1 g/mole 117.3 g P 1 31.0 g/mole 31.0 g O 4 16.0 g/mole 64.0 g K3PO4 212.3 g

One-Mole Quantities 32.1 g 55.9 g 58.5 g 294.2 g 342.3 g

Learning Check A. 1 mole of K2O = ______g B. 1 mole of antacid Al(OH)3 = ______g

Solution A. 1 mole of K2O 2 moles K (39.1 g/mole) + 1 mole O (16.0 g/mole) 78.2 g + 16.0 g = 94.2 g B. 1 mole of antacid Al(OH)3 1 mole Al (27.0 g/mole) + 3 moles O (16.0 g/mole) + 3 moles H (1.0 g/mole) 27.0 g + 48.0 g + 3.0 g = 78.0 g

Learning Check Prozac, C17H18F3NO, is an antidepressant that inhibits the uptake of serotonin by the brain. What is the molar mass of Prozac? 1) 40.0 g/mole 2) 262 g/mole 3) 309 g/mole

Solution Prozac, 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/mole 17C (12.0) + 18H (1.0) + 3F (19.0) + 1N (14.0) + 1 O (16.0) = 204 + 18 + 57.0 + 14.0 + 16.0

Molar Mass Factors Methane CH4 known as natural gas is used in gas cook tops and gas heaters. 1 mole CH4 = 16.0 g The molar mass of methane can be written as conversion factors. 16.0 g CH4 and 1 mole CH4 1 mole CH4 16.0 g CH4

Learning Check Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass conversion factors for acetic acid.

Solution Acetic acid C2H4O2 gives the sour taste to vinegar. Write two molar mass factors for acetic acid. 1 mole of acetic acid = 60.0 g acetic acid 1 mole acetic acid and 60.0 g acetic acid 60.0 g acetic acid 1 mole acetic acid

Calculations with Molar Mass Mole factors are used to convert between the grams of a substance and the number of moles. Mole factor Grams Moles

Calculating Grams from Moles Aluminum is often used for the structure of lightweight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al x 27.0 g Al = 81.0 g Al 1 mole Al mole factor for Al

Learning Check The 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?

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/mole Set up the calculation using a mole factor. 225 g aspartame x 1 mole aspartame 294 g aspartame mole factor(inverted) = 0.765 mole aspartame

Conservation of Mass In a chemical reaction, the mass of the reactants is equal to the mass of the products. 2 moles Ag + 1 mole S = 1 mole Ag2S 2 (107.9 g) + 1(32.1 g) = 1 (247.9 g) 247.9 g reactants = 247.9 g product

Moles in Equations We can read the equation in “moles” by placing the word “moles” between each coefficient and formula. 4 Fe + 3 O2 2 Fe2O3 4 moles Fe + 3 moles O2 2 moles Fe2O3

Writing Mole-Mole Factors A mole-mole factor is a ratio of the coefficients for two substances. 4 Fe + 3 O2 2 Fe2O3 Fe and O2 4 mole Fe and 3 mole O2 3 mole O2 4 mole Fe Fe and Fe2O3 4 mole Fe and 2 mole Fe2O3 2 mole Fe2O3 4 mole Fe O2 and Fe2O3 3 mole O2 and 2 mole Fe2O3 2 mole Fe2O3 3 mole O2

Learning Check Consider the following equation: 3 H2 + N2 2 NH3 A. A mole factor for H2 and N2 is 1) 3 mole N2 2) 1 mole N2 3) 1 mole N2 1 mole H2 3 mole H2 2 mole H2 B. A mole factor for NH3 and H2 is 1) 1 mole H2 2) 2 mole NH3 3) 3 mole N2 2 mole NH3 3 mole H2 2 mole NH3

Solution 3 H2 + N2 2 NH3 A. A mole factor for H2 and N2 is 2) 1 mole N2 3 mole H2 B. A mole factor for NH3 and H2 is 2) 2 mole NH3

Calculations with Mole Factors Consider the following reaction: 4 Fe + 3 O2 2 Fe2O3 How many moles of Fe2O3 are produced when 6.0 moles O2 react? Use the appropriate mole factor to determine the moles Fe2O3. 6.0 mole O2 x 2 mole Fe2O3 = 4.0 mole Fe2O3 3 mole O2

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) 3.00 moles Fe 2) 9.00 moles Fe 3) 16.0 moles Fe

Solution 3) 16.0 moles Fe 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? 12.0 mole O2 x 4 mole Fe = 16.0 moles Fe 3 mole O2

Mass Calculations

Learning Check How many grams of O2 are needed to produce 0.400 mole of Fe2O3? 4 Fe + 3 O2 2 Fe2O3 1) 38.4 g O2 2) 19.2 g O2 3) 1.90 g O2

Solution 2) 19.2 g O2 0.400 mole Fe2O3 x 3 mole O2 x 32.0 g O2 2 mole Fe2O3 1 mole O2 mole factor molar mass = 19.2 g O2

Calculating the Mass of a Reactant The reaction between H2 and O2 produces 13.1 g of water. How many grams of O2 reacted? 2H2 + O2 2H2O ? g 13.1 g Plan: g H2O mole H2O mole O2 g O2 13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2 18.0 g H2O 2 mole H2O 1 mole O2 = 11.6 g O2

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 C2H2 + 5 O2 4 CO2 + 2 H2O 1) 88.6 g C2H2 2) 44.3 g C2H2 3) 22.2 g C2H2

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 26.0 g C2H2 44.0 g CO2 4 moles CO2 1 mole C2H2 = 22.2 g C2H2

Percent Yield You 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 possible Actual yield 48 cookies to eat Percent yield 48 cookies x 100 = 80% yield 60 cookies

Percent Yield The 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 100 Theoretical Yield (g)

Sample Exercise % Yield Without proper ventilation and limited oxygen, the reaction of carbon and oxygen produces carbon monoxide. 2C + O2 2CO What is the percent yield if 40.0 g of CO are produced from the reaction of 30.0 g O2?

Sample Exercise % Yield (cont.) 1. Calculate theoretical yield of CO. 30.0 g O2 x 1 mole O2 x 2 mole CO x 28.0 g CO 32.0 g O2 1 mole O2 1 mole CO = 52.5 g CO (theoretical) 2. Calculate the percent yield. 40.0 g CO (actual) x 100 = 76.2 % yield 52.5 g CO(theoretical)

Learning Check In the lab, N2 and 5.0 g of H2 are reacted and produce 16.0 g of NH3. What is the percent yield for the reaction? N2(g) + 3H2(g) 2NH3(g) 1) 31.3 % 2) 56.5 % 3) 80.0 %

Solution 2) 56.5 % N2(g) + 3H2(g) 2NH3(g) 2) 56.5 % N2(g) + 3H2(g) 2NH3(g) 5.0 g H2 x 1 mole H2 x 2 moles NH3 x 17.0 g NH3 2.0 g H2 3 moles H2 1 mole NH3 = 28.3 g NH3 (theoretical) Percent yield = 16.0 g NH3 x 100 = 56.5 % 28.3 g NH3