Classification and Balancing of Chemical Reactions

Classification and Balancing of Chemical Reactions
Chapter 5 Lecture Fundamentals of General, Organic, and Biological Chemistry 8th Edition McMurry, Ballantine, Hoeger, Peterson Chapter Five Classification and Balancing of Chemical Reactions Christina A. Johnson University of California, San Diego © 2017 Pearson Education, Inc.

Outline 5.1 Chemical Equations 5.2 Balancing Chemical Equations
5.3 Precipitation Reactions and Solubility Guidelines 5.4 Acids, Bases, and Neutralization Reactions 5.5 Redox Reactions 5.6 Recognizing Redox Reactions 5.7 Net Ionic Equations

Concepts to Review Periodic Properties and Ion Formation Section 3.4
H+ and OH– Ions: An Introduction to Acids and Bases Section 3.11

5.1 Chemical Equations A chemical equation is an expression in which symbols and formulas are used to represent a chemical reaction. Learning Objective: Understand the law of conservation of mass and how it applies to chemical equations.

5.1 Chemical Equations Chemical reactions can be thought of as “ “recipes.” All the “ingredients” in a chemical equation and their relative amounts are given. Starting materials and final products are listed, with chemical names replaced with formulas.

5.1 Chemical Equations A reactant is a substance that undergoes change in a chemical reaction and is written on the left side of the reaction arrow in a chemical equation. A product is a substance that is formed in a chemical reaction and is written on the right side of the reaction arrow in a chemical equation. Necessary conditions are written above the arrow.

5.2 Balancing Chemical Equations
A balanced chemical equation indicates the appropriate amounts of reactants needed to generate a given amount of product. Learning Objective: Balance chemical equations.

5.1 Chemical Equations Chemical equations must be balanced—the numbers and kinds of atoms must be the same on both sides of the reaction arrow. Coefficient: A number placed in front of a formula to balance a chemical equation 2 NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g) Heat The coefficient, 2, is placed in front of NaHCO3 in order to balance the number of Na atoms, H atoms, C atoms, and O atoms with the atoms listed as products.

5.1 Chemical Equations Substances that take part in chemical reactions may be solids, liquids, gases, or dissolved. This information is added to an equation by placing symbols after the formulas. (s) = solid (l) = liquid (g) = gas (aq) = dissolved in aqueous solution

5.1 Chemical Equations STEP 1: Write an unbalanced equation, using the correct formulas for all reactants and products. For example, hydrogen and oxygen must be written as H2 and O2 rather than as H and O because both elements exist as diatomic molecules. Subscripts in chemical formulas cannot be changed because doing so would change the identity of the substances in the reaction.

5.1 Chemical Equations STEP 2: Add appropriate coefficients to balance the numbers of atoms of each element. Begin with elements that appear in only one compound or formula on each side of the equation. Leave elements that exist in elemental forms, such as oxygen and hydrogen, until last.

5.1 Chemical Equations If a polyatomic ion appears on both sides of an equation, it is treated as a single unit.

5.1 Chemical Equations STEP 3: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are the same. STEP 4: Make sure the coefficients are reduced to their lowest whole-number values.

Worked Example 5.1 Write a balanced chemical equation for the Haber process, an important industrial reaction in which elemental nitrogen and hydrogen combine to form ammonia.

Worked Example 5.1 Cont. Solution: STEP 1: Write an unbalanced equation, using the correct formulas for all reactants and products. N2(g) + H2(g)  NH3(g) By examination, we see that only two elements, N and H, need to be balanced. Both these elements exist in nature as diatomic gases, as indicated on the reactant side of the unbalanced equation.

Worked Example 5.1 Cont. Solution Continued: STEP 2: Add appropriate coefficients to balance the numbers of atoms of each element. Because there are 2 nitrogen atoms on the left, we must add a coefficient of 2 in front of the NH3 on the right side of the equation to balance the equation with respect to N: N2(g) + H2(g)  2NH3(g)

Worked Example 5.1 Cont. Solution Continued: STEP 2 Continued: Now we see that there are 2 H atoms on the left but 6 H atoms on the right. We can balance the equation with respect to hydrogen by adding a coefficient of 3 in front of the H2(g) on the left side: N2(g) + 3H2(g)  2NH3(g)

Worked Example 5.1 Cont. Solution Continued: STEP 3: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are the same. N2(g) + 3H2(g)  2NH3(g) On the left: (1 × 2) N = 2 N (3 × 2) H = 6 H On the right: (2 × 1) N = 2 N (2 × 3) H = 6 H STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. In this case, the coefficients already represent the lowest whole-number values.

Worked Example 5.2 Natural gas (methane, CH4) burns in oxygen to yield water and carbon dioxide (CO2). Write a balanced chemical equation for the reaction.

Worked Example 5.2 Cont. Solution: STEP 1: Write an unbalanced equation, using the correct formulas, for all reactants and products. CH4 + O2  CO2 + H2O

Worked Example 5.2 Cont. Solution Continued: STEP 2: Because carbon appears in one formula on each side of the arrow, let us begin with that element. In fact, there is only 1 carbon atom in each formula, so the equation is already balanced for that element. Next, note that there are 4 hydrogen atoms on the left (in CH4) and only 2 on the right (in H2O). Placing a coefficient of 2 before H2O gives the same number of hydrogen atoms on both sides: CH4 + O2  CO2 + 2H2O (balanced for C and H)

Worked Example 5.2 Cont. Solution Continued: STEP 2 Continued: Finally, look at the number of oxygen atoms. There are 2 on the left (in O2) but 4 on the right (2 in CO2 and 1 in each H2O). If we place a 2 before the O2, the number of oxygen atoms will be the same on both sides, but the number of the other elements will not change: CH4 + 2O2  CO2 + 2H2O (balanced for C, H, and O)

Worked Example 5.2 Cont. Solution Continued:
STEP 3: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are the same. CH4 + 2O2  CO2 + 2H2O On the left: C H (2 × 2) O = 4 O On the right: 1 C (2 × 2) = 4 H O + 2 O = 4 O STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. In this case, the answer is already correct.

Worked Example 5.3 Sodium chlorate (NaClO3) decomposes when heated to yield sodium chloride and oxygen, a reaction used to provide oxygen for the emergency breathing masks in airliners. Write a balanced chemical equation for the reaction.

Worked Example 5.3 Cont. Solution: STEP 1: Write an unbalanced equation, using the correct formulas for all reactants and products. NaClO3  NaCl + O2

Worked Example 5.3 Cont. Solution Continued: STEP 2: Both the Na and the Cl are already balanced, with only one atom of each on the left and right sides of the equation. There are 3 O atoms on the left but only 2 on the right. The O atoms can be balanced by placing a coefficient of 1½ in front of O2 on the right side of the equation: NaClO3  NaCl + 1½ O2

Worked Example 5.3 Cont. Solution Continued:
STEP 3: Check the equation to make sure the numbers and kinds of atoms on both sides of the equation are the same. NaClO3  NaCl + 1½ O2 On the left: Na Cl O On the right: 1 Na Cl (1½ × 2) O = 3 O

Worked Example 5.3 Cont. On the left: 2 Na 2 Cl (2 × 3) O = 6 O
Solution Continued: STEP 4: Make sure the coefficients are reduced to their lowest whole-number values. In this case, obtaining all coefficients in their smallest whole-number values requires that we multiply all coefficients by 2 to obtain: 2NaClO3  2NaCl + 3O2 On the left: Na Cl (2 × 3) O = 6 O On the right: 2 Na 2 Cl (3 × 2) O = 6 O

5.3 Precipitation Reactions and Solubility Guidelines
Precipitation reactions are processes in which an insoluble solid called a precipitate forms when reactants are combined in an aqueous solution. Learning Objective: Apply solubility rules to predict if a precipitation reaction will occur, and write the appropriate balanced reaction.

Whether a precipitation reaction will occur on mixing aqueous solutions of two ionic compounds depends on the solubilities of the potential products. Solubility is the amount of a compound that will dissolve in a given amount of solvent at a given temperature.

If a substance has a low solubility, then it is likely to precipitate from an aqueous solution. If a substance has a high solubility in water, then no precipitate will form.

General Rules on Solubility 1. A compound is probably soluble if it contains one of the following cations. Group 1A cation: Li+, Na+, K+, Rb+, Cs+ Ammonium cation: NH4+ 2. A compound is probably soluble if it contains one of the following anions. Halide: Cl–, Br–, I–, except Ag+, Hg22+, and Pb2+ compounds Nitrate (NO3–), perchlorate (ClO4–), acetate (CH3CO2–), sulfate (SO42–), except Ba2+, Hg22+, and Pb2+ sulfates

Worked Example 5.4 Will a precipitation reaction occur when aqueous solutions of CdCl2 and (NH4)2S are mixed?

Worked Example 5.4 Cont. Solution: Identify the two potential products, and predict the solubility of each using the solubility guidelines. In this instance, CdCl2 and (NH4)2S might give CdS and NH4Cl. Because the guidelines predict that CdS is insoluble, a precipitation reaction will occur: CdCl2(aq) + (NH4)2S(aq)  CdS(s) + 2NH4Cl(aq)

5.4 Acids, Bases, and Neutralization Reactions
Acid-base neutralization reactions are processes in which an acid reacts with a base to yield water plus an ionic compound called a salt. Learning Objective: Predict the products of an acid-base neutralization reaction.

A neutralization reaction removes H+ and OH– ions from solution and yields neutral H2O. Any ionic compound produced in an acid-base reaction is called a salt.

The most common kind of neutralization reaction occurs between an acid (HA) and a metal hydroxide (MOH) to yield water and a salt. The H+ ion from the acid combines with the OH– ion from the base to give neutral H2O. The anion from the acid combines with the cation from the base to give a salt.

Another kind of neutralization reaction occurs between an acid and a carbonate or bicarbonate to yield water, a salt, and carbon dioxide. The carbonate ion reacts with H+ to yield H2CO3. This is unstable and immediately decomposes to give H2O plus CO2. 2 HA (aq) + M2CO3 (aq) → H2O (l) + 2 MA (aq) + CO2 (g)

Worked Example 5.5 Write an equation for the neutralization reaction of aqueous HBr and aqueous Ba(OH)2.

Worked Example 5.5 Cont. Solution:
The reaction of HBr with Ba(OH)2 involves the combination of a proton (H+) from the acid with OH– from the base to yield water and a salt (BaBr2). 2 HBr(aq) + Ba(OH)2(aq)  2 H2O(l) + BaBr2(aq)

5.5 Redox Reactions Oxidation-reduction reactions or redox reactions are reactions in which electrons are transferred from one atom to another. Learning Objective: Recognize redox reactions, and identify the species being oxidized and reduced.

5.5 Redox Reactions As a result of the electron transfer in a redox reaction, the number of electrons assigned to individual atoms in the various reactants changes. Reactions involving covalent compounds are redox reactions because electrons are rearranged as bonds are broken and new bonds are formed.

5.5 Redox Reactions Historically: The word oxidation referred to the combination of an element with oxygen. Reduction referred to the removal of oxygen from an oxide to yield the element.

5.5 Redox Reactions Today, the words have taken on a much broader meaning. Oxidation is the loss of one or more electrons. Reduction is the gain of one or more electrons.

5.5 Redox Reactions

5.5 Redox Reactions The copper wire reacts with aqueous Ag+ ion and becomes coated with metallic silver. This is an example of a redox reaction. Copper is oxidized, and silver ions are reduced.

5.5 Redox Reactions An iodide ion gives an electron to bromine, forming iodine and bromide ions. An iodide ion is oxidized as its charge increases from –1 to 0. Bromine is reduced as its charge decreases from 0 to –1.

5.5 Redox Reactions Oxidation and reduction always occur together.
When one substance loses an electron (is oxidized), another substance must gain that electron (be reduced). The substance that gives up an electron and causes the reduction is a reducing agent. The substance that gains an electron and causes the oxidation is an oxidizing agent.

5.5 Redox Reactions A reducing agent loses one or more electrons.
Causes reduction Undergoes oxidation Becomes more positive (or less negative) May gain oxygen atoms An oxidizing agent gains one or more electrons. Causes oxidation Undergoes reduction Becomes more negative (or less positive) May lose oxygen atoms

5.5 Redox Reactions Alkali metals and alkaline earth metals are the most powerful reducing agents. They will react with pure water. They have low ionization energy. As ionization energy increases, reducing power decreases.

5.5 Redox Reactions Reactive nonmetals are powerful oxidizing agents.
They have the highest oxidation energies. They have the most favorable electron affinity.

Generalizations about Redox Behavior:
5.5 Redox Reactions Generalizations about Redox Behavior: In reactions involving metals and nonmetals, metals tend to lose electrons while nonmetals tend to gain electrons. The number of electrons lost or gained can often be predicted based on the position of the element in the periodic table.

Generalizations about Redox Behavior (Continued):
5.5 Redox Reactions Generalizations about Redox Behavior (Continued): In reactions involving nonmetals, the “more metallic” element (farther down and/or to the left in the periodic table) tends to lose electrons, and the “less metallic” element (up and/or to the right) tends to gain electrons.

5.5 Redox Reactions Corrosion is the deterioration of a metal by oxidation, such as the rusting of iron in moist air. The economic consequences of rusting are enormous. It has been estimated that up to one-fourth of the iron produced in the United States is used to replace bridges, buildings, and other structures that have been destroyed by corrosion.

5.5 Redox Reactions Combustion is the burning of a fuel by rapid oxidation with oxygen in air. Gasoline, fuel oil, natural gas, wood, paper, and other organic substances of carbon and hydrogen are the most common fuels that burn in air. Some metals will burn in air, including magnesium and calcium.

5.5 Redox Reactions Respiration is the process of breathing and using oxygen for the many biological redox reactions that provide the energy that living organisms need. Energy is released from food molecules slowly and in complex, multistep pathways, but the overall result of respiration is similar to that of combustion reactions.

5.5 Redox Reactions Bleaching makes use of redox reactions to decolorize or lighten colored materials. Dark hair is bleached to turn it blond, clothes are bleached to remove stains, wood pulp is bleached to make white paper. The oxidizing agent used depends on the situation. Hydrogen peroxide is used for hair, sodium hypochlorite (NaOCl) for clothes, and elemental chlorine for wood pulp.

Fe2O3 (s) + 3 CO (g) → 2 Fe (s) + 3 CO2 (g)
5.5 Redox Reactions Metallurgy, the science of extracting and purifying metals from their ores, makes use of numerous redox processes. Worldwide, approximately 800 million tons of iron is produced each year by reduction of the mineral hematite with carbon monoxide. Fe2O3 (s) + 3 CO (g) → 2 Fe (s) + 3 CO2 (g)

Worked Example 5.6 For the following reactions, indicate which atom is oxidized and which is reduced. Identify the oxidizing and reducing agents. Cu (s) + Pt2+ (aq) → Cu2+ (aq) + Pt (s) 2 Mg (s) + CO2 (g) → 2 MgO (s) + C (s) ANALYSIS: The definitions for oxidation include a loss of electrons, an increase in charge, and a gain of oxygen atoms; reduction is defined as a gain of electrons, a decrease in charge, and a loss of oxygen atoms.

In this reaction, the charge on the Cu atom increases from 0 to 2+. This corresponds to a loss of two electrons. The Cu is, therefore, oxidized and acts as the reducing agent. Conversely, the Pt2+ ion undergoes a decrease in charge from 2+ to 0, corresponding to a gain of two electrons for the Pt2+ ion. The Pt2+ is reduced and acts as the oxidizing agent. The Cu is oxidized and acts as the reducing agent.

Worked Example 5.6 Cont. Solution Continued: (b) In this case, the gain or loss of oxygen atoms is the easiest way to identify which atoms are oxidized and reduced. The Mg atom is gaining oxygen to form MgO; therefore, the Mg is being oxidized and acts as the reducing agent. The C atom in CO2 is losing oxygen. Therefore, the C atom in CO2 is being reduced and acts as the oxidizing agent.

Worked Example 5.7 For the following reactions, identify the atom(s) being oxidized and reduced. 2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s) C (s) + 2 Cl2 (g) → CCl4 (l) ANALYSIS: There is no obvious increase or decrease in charge to indicate a gain or loss of electrons. Also, the reactions do not involve a gain or loss of oxygen. We can, however, evaluate the reactions in terms of the typical behavior of metals and nonmetals in reactions.

In this case, we have the reaction of a metal (Al) with a nonmetal (Cl2). Because metals tend to lose electrons and nonmetals tend to gain electrons, we can assume that the Al atom is oxidized (loses electrons) and the Cl2 is reduced (gains electrons).

Worked Example 5.7 Cont. Solution Continued: (b) The carbon atom is the less electronegative element (farther to the left) and is less likely to gain an electron. The more electronegative element (Cl) will tend to gain electrons (be reduced).

Worked Example 5.8 Classify the following as a precipitation, an acid-base neutralization, or a redox reaction. Ca(OH)2(aq) + 2HBr(aq) → 2H2O(l) + CaBr2(aq) Pb(ClO4)2(aq) + 2NaCl(aq) → 2PbCl2(s) + 2NaClO4(aq) 2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq) ANALYSIS: One way to identify the class of reaction is to examine the products that form and match them with the descriptions for the types of reactions.

The products of this reaction are water and an ionic compound, or salt (CaBr2). This is consistent with the description of an acid-base neutralization reaction. This reaction involves two aqueous reactants, Pb(ClO4)2 and NaCl, which combine to form a solid product, PbCl2. This is consistent with a precipitation reaction.

Worked Example 5.8 Cont. Solution Continued: (c) The products of this reaction are a solid, Ag(s), and an aqueous ionic compound, Cu(NO3)2. This does not match the description of a neutralization reaction, which would form water and an ionic compound. One of the products is a solid, but the reactants are not both aqueous compounds; one of the reactants is also a solid (Cu). Therefore, this reaction would not be classified as a precipitation reaction. By the process of elimination, then, it must be a redox reaction.

5.6 Recognizing Redox Reactions
Oxidation number: A number that indicates whether an atom is neutral, electron-rich, or electron-poor. Learning Objective: Determine the oxidation number of an atom in a compound.

How can you tell when a redox reaction is taking place? When ions are involved, determine whether there is a change in charges. For reactions involving metals and nonmetals, predict gain or loss of electrons. Molecular substances can be analyzed in terms of loss and gain of oxygen.

By extending the ideas of oxidation and reduction to an increase or decrease in electron sharing, electronegativity differences can be used. An atom is oxidized when it loses a share in electrons. An atom is reduced when it gains a share in electrons.

A formal system has been devised for keeping track of changes in electron sharing and determining whether atoms are oxidized or reduced in reactions. By comparing the oxidation number of an atom before and after a reaction, we can tell whether the atom has gained or lost shares in electrons.

Rules for Assigning Oxidation Numbers: An atom in its elemental state has an oxidation number of 0.

Rules for Assigning Oxidation Numbers: A monatomic ion has an oxidation number equal to its charge.

Rules for Assigning Oxidation Numbers: In a molecular compound, an atom usually has the same oxidation number it would have if it were a monatomic ion. Hydrogen is usually +1. Oxygen is usually –2. Nitrogen is usually –3. Halogens are usually –1.

Rules for Assigning Oxidation Numbers: For compounds with more than one nonmetal element, the more electronegative element has a negative oxidation number and the less electronegative element has a positive oxidation number.

Rules for Assigning Oxidation Numbers: The sum of the oxidation numbers in a neutral compound is 0.

Worked Example 5.9 What is the oxidation number of the titanium atom in TiCl4? Name the compound using a Roman numeral.

Worked Example 5.9 Cont. Solution: Chlorine, a reactive nonmetal, is more electronegative than titanium and has an oxidation number of –1. Because there are four chlorine atoms in TiCl4, the oxidation number of titanium must be +4. The compound is named titanium (IV) chloride.

Worked Example 5.10 Use oxidation numbers to show that the production of iron metal from its ore (Fe2O3) by reaction with charcoal (C) is a redox reaction. Which reactant has been oxidized, and which has been reduced? Which reactant is the oxidizing agent, and which is the reducing agent? 2 Fe2O3 (s) + 3 C (g) → 4 Fe (s) + 3 CO2 (g)

Worked Example 5.10 Cont. Solution: The idea is to assign oxidation numbers to both reactants and products and see if there has been a change. In the production of iron from Fe2O3, the oxidation number of Fe changes from +3 to 0, and the oxidation number of C changes from 0 to +4. Iron has thus been reduced and carbon has been oxidized. Carbon is the reducing agent, and Fe2O3 is the oxidizing agent.

Ionic equation: An equation in which ions are explicitly shown
5.7 Net Ionic Equations Ionic equation: An equation in which ions are explicitly shown Learning Objective: For ionic reactions, write the molecular, ionic, and net ionic reactions, and identify spectator ions.

5.7 Net Ionic Equations In reactions involving ions, it is more accurate to write the reaction as an ionic equation. Some ions undergo no change during the reaction. They appear on both sides of the reaction but play no role and are called spectator ions.

5.7 Net Ionic Equations The actual reaction can be described more simply by writing a net ionic equation, which includes only the ions that undergo change. A net ionic equation is an equation that does not include spectator ions. A net ionic equation must be balanced both for atoms and for charge, with all coefficients reduced to their lowest whole numbers. All insoluble and molecular compounds are represented by their full formulas.

5.7 Net Ionic Equations Applied to neutralization reactions,
KOH (aq) + HNO3 (aq) → H2O (l) + KNO3 (aq) can be rewritten as an ionic equation, K+ (aq) + OH− (aq) + H+ (aq) + NO3– (aq) → H2O (l) + K+ (aq) + NO3– (aq). Eliminating the spectator ions (K+ and NO3–) yields OH– (aq) + H+ (aq) → H2O (l). This net ionic equation confirms a neutralization reaction.

5.7 Net Ionic Equations Applied to redox reactions,
Cu (s) + 2 AgNO3 (aq) → 2 Ag (s) + Cu(NO3)2 (aq) can be rewritten as an ionic equation, Cu (s) + 2 Ag+ (aq) + 2 NO3– (aq) → 2 Ag (s) + Cu2+ (aq) + 2 NO3– (aq). Eliminating the spectator ion (NO3–) yields Cu (s) + 2 Ag+ (aq) → 2 Ag (s) + Cu2+ (aq). This net ionic equation confirms that copper is oxidized and silver is reduced.

Worked Example 5.11 Write balanced net ionic equations for the following reactions: (a) AgNO3 (aq) + ZnCl2 (aq) → (b) HCl (aq) + Ca(OH)2 (aq) → (c) 6 HCl (aq) + 2 Al (s) → 2 AlCl3 (aq) + 3 H2 (g)

Ag+ (aq) + Cl– (aq) → AgCl (s)
Worked Example 5.11 Cont. Solution: The solubility guidelines predict that a precipitate of insoluble AgCl forms when aqueous solutions of Ag+ and Cl– are mixed. The coefficients can all be divided by 2 to give: Ag+ (aq) + Cl– (aq) → AgCl (s)

Worked Example 5.11 Cont. Solution Continued: (b) Allowing the acid HCl to react with the base Ca(OH)2 leads to a neutralization reaction.

Worked Example 5.11 Cont. Solution Continued: (c) The reaction of Al metal with acid (HCl) is a redox reaction. We write the ionic equation by showing the ions that are formed for each aqueous ionic species.

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Classification and Balancing of Chemical Reactions

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