Spencer L. Seager
Michael R. Slabaugh
www.cengage.com/chemistry/seager
Chapter 5:
Chemical Reactions
Jennifer P. Harris
CHEMICAL EQUATIONS
• Chemical equations are written in terms of reactants and
products.
• A symbol is written in parentheses to the right of each reactant
and product to indicate the state or form in which the substance
exists. Gases are indicated by (g), liquids by (l), solids by (s),
and substances dissolved in water by (aq).
solid sugar = C12H22O11 (s)
and liquid water = H2O (l)
sugar dissolved in water =
C12H22O11 (aq)
CHEMICAL EQUATIONS (continued)
• REACTANTS OF A CHEMICAL EQUATION
• The reactants in a chemical equation are the
substances written on the left side of the arrow.
Reactants
2 H2 (g) + O2 (g) → 2 H2O (l)
Products
• PRODUCTS OF A CHEMICAL EQUATION
• The products in a chemical equation are the substances
written on the right side of the arrow.
• When two or more reactants or products are involved in an
equation, they are separated by a plus sign (+).
BALANCED CHEMICAL EQUATIONS
• A balanced chemical equation is one in which the number
of atoms of each element in the reactants is equal to the
number of atoms of that same element in the products.
• A reaction can be balanced by applying the law of
conservation of matter.
• Coefficients (in red below) are written to the left of each
reactant or product in order to achieve balance.
2 H2 (g) + O2 (g) → 2 H2O (l)
EXAMPLES OF UNBALANCED &
BALANCED EQUATIONS
• Methane, CH4,is the main ingredient in natural gas. It
combines with oxygen, O2, when it burns to form carbon
dioxide, CO2, and water, H2O. This information written in the
form of an unbalanced equation is:
CH4(g) + O2 (g)
CO2 (g) + H2O(g)
• The equation is unbalanced as it is written because the
number of H atoms on the left is 4 and the number on the
right is 2. Also, the number of O atoms on the left is 2 and
the number on the right is 3.
EXAMPLES OF UNBALANCED &
BALANCED EQUATIONS (continued)
• The equation is balanced by inserting appropriate coefficients
on the left of each reactant and product to give the following
balanced equation:
CH4(g) + 2O2 (g)
CO2 (g) + 2H2O
• When atoms are counted, 1C, 4H and 4O atoms are on the
left and the same number are on the right.
BALANCED EQUATION PRACTICE
• Determine if the following equations are balanced.
If not, then add coefficients to balance the
equations.
• NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
• Balanced (1 Na, 1 O, 2 H, and 1 Cl both sides)
• Mg (s) + O2 (g) → MgO (s)
• Unbalanced (2 O vs. 1 O)
• 2 Mg (s) + O2 (g) → 2 MgO (s)
BALANCED EQUATION PRACTICE
(continued)
• Determine if the following equations are balanced.
If not, then add coefficients to balance the
equations.
• Na2CO3 (s) → Na2O (s) + CO2 (g)
• Balanced (2 Na, 1 C, and 3 O both sides)
• Mg (s) + HCl (aq) → MgCl2 (aq) + H2 (g)
• Unbalanced (1 H vs. 2 H and 1 Cl vs. 2 Cl)
• Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)
TYPES OF CHEMICAL REACTIONS
• Chemical reactions are often classified into categories
according to characteristics of the reactions. The following is a
useful classification scheme:
REDOX REACTIONS
• The word redox is a
combination of two words,
reduction and oxidation.
These two words have
multiple meanings when
applied to chemical reactions.
• The concept of oxidation
numbers provides a
convenient way to work with
redox reactions.
OXIDATION NUMBERS
• Oxidation numbers (also called oxidation states) are
positive or negative numbers assigned to elements in
chemical formulas according to a set of rules. The term
oxidation number is abbreviated O.N.
• Rule 1:
The O.N. of any uncombined element is 0.
For example: Fe (0), Cl2 (0), and Ca(0)
• Rule 2:
The O.N. of a simple ion is
equal to the charge on the ion.
For example:
Mg2+(+2), O2-(-2), and Cl-(-1).
OXIDATION NUMBERS (continued)
• Rule 3:
The O.N. of group IA and IIA elements when they are in
compounds are always +1 and +2, respectively.
For example: Na2S (Na = +1) and Ca(NO3)2 (Ca = +2)
• Rule 4:
The O.N. of hydrogen is +1.
For example: HBr (H = +1) and H2SO4(H = +1)
• Rule 5:
The O.N. of oxygen is -2 except in peroxides, where it is -1.
For example: MgO (O = -2), HBrO3 (O = -2), and
H2O2 (O = -1)
A closer look at H2O2 :
2(O.N. of H) + 2(O.N. of O) = 0
2(+1)
+ 2(O.N. of O) = 0
2(O.N. of O) = -2
(O.N. of O) = -1
A closer look at MgO :
(O.N. of Mg) + (O.N. of O) = 0
+2
+ (O.N. of O) = 0
(O.N. of O) = -2
OXIDATION NUMBERS (continued)
• Rule 6:
The algebraic sum of the oxidation numbers of all atoms in a
complete compound equals zero.
For example: MgSO4 (O.N. of Mg = +2 by rule 2, O.N. of O = -2 by
rule 5, and O.N. of S = +6 by algebra and rule 6)
A closer look at MgSO4 :
(O.N. of Mg) + (O.N. of S) + 4(O.N. of O) = 0
+2
+ (O.N. of S) +
4(-2) = 0
+2
+ (O.N. of S) +
-8 = 0
(O.N. of S) = +6
Because there is only one Mg and one S, the total positive
oxidation number is +8. The four O atoms, with an O.N. of -2
each, give a total negative O.N. of -8. Thus, the total positive and
the total negative O.N. values add up to zero.
OXIDATION NUMBERS (continued)
• Rule 7:
The algebraic sum of the O.N. of all the atoms in a
polyatomic ion is equal to the charge on the ion.
For example: HCO3- (O.N. of H is +1 by rule 4, O.N. of O is 2 by rule 5, and O.N. of C is +4 by algebra and rule 7)
A closer look at HCO3- :
(O.N. of H) + (O.N. of C) + 3(O.N. of O) = -1
+1
+ (O.N. of C) +
3(-2) = -1
+1
+ (O.N. of C) +
-6 = -1
(O.N. of S) = +4
Because there is only one H and one C, the total positive
oxidation number is +5. The three O atoms, with an O.N. of
-2 each, give a total negative O.N. of -6. Thus, the total
positive and the total negative O.N. values add up to -1,
which is the charge on the bicarbonate ion.
OXIDIZING & REDUCING AGENTS
• In a redox reaction, the substance that contains an element
that is oxidized during the reaction is called the reducing
agent.
• In a redox reaction, the substance that contains an element
that is reduced during the reaction is called the oxidizing
agent.
OXIDIZING & REDUCING AGENTS EXAMPLE
• Determine the oxidizing and reducing agents in the reaction:
2Na(s) +2H2O(l)
H2 (g) + 2NaOH(aq)
• Solution: First, assign oxidation numbers.
OXIDIZING & REDUCING AGENTS EXAMPLE
(continued)
2Na(s) +2H2O(l)
H2 (g) + 2NaOH(aq)
• Now, determine the oxidizing and reducing agents.
• Na is oxidized; therefore, Na is the reducing agent.
• H is reduced; therefore, H2O is the oxidizing agent.
DECOMPOSITION REACTIONS
• In decomposition reactions, one substance is broken down
into two or more simpler substances. Decomposition
reactions may be either redox or nonredox reactions.
• The general form of the equation for a decomposition
reaction is:
A
B + C.
• An example of a redox decomposition reaction is:
2HI(g)
H2 (g) + I2 (g)
• An example of a nonredox decomposition reaction is:
H2CO3(aq)
CO2(g) + H2O(l)
DECOMPOSITION REACTIONS
(continued)
• A pictorial representation of a decomposition reaction:
2 H2O (l) → 2 H2 (g) + O2 (g)
COMBINATION REACTIONS
• In combination reactions two or more substances react to
form a single substance. Combination reactions may be
either redox or nonredox reactions.
• The general form of the equation for a combination reaction
is:
A+B
C
• An example of a redox combination reaction is:
S(s) + O2 (g)
SO2 (g)
• An example of a nonredox combination reaction is:
N2O5(g) + H2O(l)
2HNO3(aq)
COMBINATION REACTIONS
(continued)
• A pictorial representation of a combination reaction:
2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s)
SINGLE-REPLACEMENT REACTIONS
• Single-replacement reactions are always redox reactions
because they occur when one element reacts with a
compound, displaces one of the elements from the
compound, and becomes part of a new compound.
• The general form of the equation for a single replacement
reaction is:
A + BX
B + AX
In this equation, A and B represent elements and AX and
BX are compounds.
• An example of a single replacement reaction is:
Zn(s) + CuSO4(aq)
Cu(s) + ZnSO4(aq)
SINGLE-REPLACEMENT REACTIONS
(continued)
• A pictorial representation of a single replacement
reaction:
Cu (s) + 2 AgNO3 (aq)→ 2 Ag (s) + Cu(NO3)2 (aq)
DOUBLE-REPLACEMENT REACTIONS
• Double-replacement reactions are never redox reactions.
These reactions often take place between substances
dissolved in water. In typical reactions, two dissolved
compounds react and exchange partners to form two new
compounds.
• The following general form of the equation for double
replacement reactions shows the partner-swapping
characteristic of the reactions:
AX + BY
BX + AY
• An example of a double-replacement reaction is:
Ba(NO3)2(aq) + Na2S(aq)
BaS(s) +2NaNO3(aq)
DOUBLE-REPLACEMENT REACTIONS
(continued)
• A pictorial representation of a double-replacement reaction:
NaCl (aq) + AgNO3 (aq) → NaNO3 (aq) + AgCl (s)
REACTION CLASSIFICATION
• Classify each of the following equations as redox or
nonredox reactions and as combination,
decomposition, single-replacement, or
double-replacement reactions.
• NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
• Nonredox reaction; double-replacement reaction
• 2 Mg (s) + O2 (g) → 2 MgO (s)
• Redox reaction; combination
REACTION CLASSIFICATION (continued)
• Classify each of the following equations as redox or
nonredox reactions and as combination,
decomposition, single-replacement, or doublereplacement reactions.
• Na2CO3 (s)→ Na2O (s) + CO2 (g)
• Nonredox reaction; decomposition
• Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)
• Redox reaction; single-replacement reaction
IONIC EQUATIONS
• Many reactions take place between compounds or elements
that are dissolved in water. Ionic compounds and some polar
covalent compounds break apart (dissociate) when they
dissolve in water and form ions.
• The equations for reactions that occur between dissolved
materials can be written in three ways called molecular
equations, total ionic equations, and net ionic equations.
IONIC EQUATIONS (continued)
• MOLECULAR EQUATIONS
• In a molecular equation, each compound is represented by
its formula.
• TOTAL IONIC EQUATIONS
• In a total ionic equation, all soluble ionic substances are
represented by the ions they form in solution. Substances
that do not dissolve or that dissolve but do not dissociate into
ions are represented by their formulas.
NaCl (aq) = Na+ (aq) + Cl− (aq)
Na2S (aq) = 2 Na+ (aq) + S2− (aq)
Na 3PO4 (aq) = 3 Na+ (aq) + PO43− (aq)
IONIC EQUATIONS (continued)
• NET IONIC EQUATIONS
• In a net ionic equation, only unionized or insoluble materials
and ions that undergo changes as the reaction proceeds are
represented.
• Any ions that appeared on both the left and right side of the
total ionic equation are called spectator ions and are not
included in the net ionic equation.
NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)
Na+ (aq) + Ag+ (aq)
+ Na+ (aq)
→ AgCl (s)
+ Cl− (aq) + NO3− (aq)
+ NO3− (aq)
Cl− (aq) + Ag+ (aq)
→ AgCl (s)
IONIC EQUATIONS EXAMPLE
• Write the following molecular equation in total ionic and net ionic
forms. Soluble substances are indicated by (aq) after their formulas
and insoluble solids are indicated by (s) after their formulas.
BaCl2 (aq) + Na2S(aq)
BaS(s) + 2NaCl(aq)
• In total ionic form, all substances except the insoluble BaS will be
written in the form of the ions they form:
Ba2+(aq) + 2Cl-(aq)
BaS(s)
+ 2Na+(aq) + S2-(aq)
+ 2Na+(aq) + 2Cl-(aq)
• In net ionic form, all spectator ions are dropped. Both the Na+ and
Cl- ions are spectator ions because they appear on both sides of the
equation. The net ionic equation is:
Ba2+(aq) + S2-(aq)
BaS(s)
ENERGY AND REACTIONS
• In addition to changes in chemical composition, all chemical
reactions are also accompanied by changes in energy. That
is, all reactions either absorb or give up energy as they
proceed.
• The energy involved in chemical reactions can take
numerous forms such as the electrical energy released by the
chemical reactions of an ordinary cell phone battery. Often,
all or most of the energy takes the form of heat.
• Chemical reactions that release heat as a product are called
exothermic reactions. Ordinary combustion of a log in a
fireplace is an example of an exothermic reaction.
• While it is a physical process and not chemical, a familiar
example of an endothermic process is the melting of
ordinary ice. As the ice melts, heat is absorbed from the air
surrounding the ice.
THE MOLE AND CHEMICAL EQUATIONS
• The mole concept can be applied to balanced chemical
equations and used to calculate mass relationships in chemical
reactions.
• Balanced equations can be interpreted in terms of the mole
concept and the results used to provide factors for use in factorunit solutions to numerical problems.
THE MOLE AND CHEMICAL EQUATIONS
EXAMPLE
• Consider the following balanced reaction equation:
2H2S(g) + 3O2(g)
2SO2 (g) + 2H2O(l)
• The mole concept can be used to write useful statements that
will be the source of factors needed to solve numerical
problems. The following are three of the possible statements:
2 mol H2S + 3 mol O2
2(6.02x1023) molecules H2S
+3(6.02x1023) molecules O2
68.2 g H2S + 96.0 g O2
2 mol SO2 + 2 mol H2O
2(6.02x1023) molecules SO2
2(6.02x1023) molecules H2O
128.2 g SO2 + 36.0 g H2O
FACTOR-UNIT METHOD REVIEW
• Step 1: Write down the known or given quantity. Include
both the numerical value and units of the quantity.
• Step 2: Leave some working space and set the known
quantity equal to the units of the unknown quantity.
• Step 3: Multiply the known quantity by one or more factors,
such that the units of the factor cancel the units of the known
quantity and generate the units of the unknown quantity.
• Step 4: After you generate the desired units, do the
necessary arithmetic to produce the final answer.
FACTOR-UNIT METHOD EXAMPLE
• Calculate the number of moles of H2S that would need to
react with excess O2 in order to produce 115 g of SO2.
• Solution: Note that the factor used was obtained from the
two statements given earlier.
2 mol H2S
115 g SO2 
 1.79 mol H2S
128.2 g SO2
• Note that the “g SO2 ” units in the denominator of the
factor cancel the “g SO2 ” units of the known quantity, and
the “mol H2S” units of the numerator of the factor generate
the needed “mol H2S” units of the answer.
THE LIMITING REACTANT
• The limiting reactant present in a mixture of reactants is
the reactant that will run out first, and thus, it determines the
amount of product that can be produced.
• A useful approach to solving limiting reactant problems is
to calculate the amount of product that could be produced
by each of the quantities of reactant that are available. The
reactant that gives the least amount of product is then the
limiting reactant.
THE LIMITING REACTANT EXAMPLE
• Calculate the maximum amount of SO2 that could be produced
by reacting 55.2 g of O2 with 50.8 g of H2S.
• Solution:
• The amounts of SO2 that could be produced from
55.2 g of O2 reacting with excess H2S as well as from
50.8 g of H2S reacting with excess O2 will be calculated.
• The reactant giving the least amount of SO2 will be the
limiting reactant.
• The amount of SO2 produced by the limiting reactant is
the amount the reaction would produce.
THE LIMITING REACTANT EXAMPLE
• Calculation for SO2 produced from 55.2 g O2 and excess
H2S:
128.2 g SO2
55.2 g O 2 
 73.7 g SO2
96.0 g O 2
• Calculation for SO2 produced from 50.8 g H2S and excess
O 2:
128.2 g SO2
50.8 g H2 S 
 95.5g SO2
68.2 g H2S
• According to the calculations, the 55.2 g of O2 gives the
smallest amount of SO2, so O2 is the limiting reactant
and the reaction will produce 73.7 g of SO2.
REACTION YIELDS
• The amounts of product calculated in the last three examples
are not the amounts that would be produced if the reactions
were actually done in the laboratory.
• In each case, less product would be obtained than was
calculated. There are numerous causes. Some materials
are lost during transfers from one container to another and
side reactions take place that are different from the one that
is intended to take place.
• The amount of product calculated in the examples is called
the theoretical yield. The amount of product actually
produced is called the actual yield. These two quantities are
used to calculate the percentage yield using the following
equation:
actual yield
% yield 
 100
theoretica l yeild
REACTION YIELDS EXAMPLE
• Suppose the mixture of reactants calculated earlier to give
73.7 g SO2 was done in the laboratory and only 42.7 g of SO2
was collected. What is the percentage yield of the reaction?
• Solution:
42.7 g SO2
% yield 
 100  57.9%
7.37 g SO2