Chapter 4: The Major Classes of

1
Chapter 4 The Major Classes of
Chemical Reactions
4.1 The Role of Water as a Solvent 4.2
Precipitation Reactions and Acid-Base Reactions
2
The Role of Water as a Solvent The Solubility
of Ionic Compounds
Electrical conductivity - The flow of electrical
current in a solution is a
measure of the solubility of ionic
compounds or a
measurement of the presence of ions in
solution. Electrolyte - A substance that
conducts a current when dissolved in
water. Soluble ionic compound
dissociate completely and
may conduct a large current, and are called
strong Eeectrolytes.
NaCl(s) H2O(l)
Na(aq) Cl -(aq)
When sodium chloride dissolves into water the
ions become solvated, and are surrounded by water
molecules. These ions are called aqueous and
are free to move through out the solution, and
are conducting electricity, or helping electrons
to move through out the solution
3
The Electrical Conductivity of Ionic Solutions
Fig. 4.1
4
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - I
Problem How many moles of each ion are in each
of the following a) 4.0 moles of sodium
carbonate dissolved in water b) 46.5 g of
rubidium fluoride dissolved in water c) 5.14 x
1021 formula units of iron (III) chloride
dissolved in water d) 75.0 ml of 0.56M
scandium bromide dissolved in water e) 7.8
moles of ammonium sulfate dissolved in
water a) Na2CO3 (s)
2 Na(aq) CO3-2(aq)
H2O
5
Determining Moles of Ions in Aqueous
Solutions of Ionic Compounds - I
Problem How many moles of each ion are in each
of the following a) 4.0 moles of sodium
carbonate dissolved in water b) 46.5 g of
rubidium fluoride dissolved in water c) 5.14 x
1021 formula units of iron (III) chloride
dissolved in water d) 75.0 ml of 0.56M
scandium bromide dissolved in water e) 7.8
moles of ammonium sulfate dissolved in
water a) Na2CO3 (s)
2 Na(aq) CO3-2(aq)
moles of Na 4.0 moles Na2CO3 x
8.0 moles Na and 4.0 moles of
CO3-2 are present
H2O
2 mol Na 1 mol Na2CO3
6
Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - II
H2O
b) RbF(s)
Rb(aq) F -(aq)
7
Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - II
H2O
b) RbF(s)
Rb(aq) F -(aq)
1 mol RbF 104.47 g RbF
moles of RbF 46.5 g RbF x
0.445 moles RbF
thus, 0.445 mol Rb and 0.445 mol F - are present

8
Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - II
H2O
b) RbF(s)
Rb(aq) F -(aq)
1 mol RbF 104.47 g RbF
moles of RbF 46.5 g RbF x
0.445 moles RbF
thus, 0.445 mol Rb and 0.445 mol F - are present
H2O
c) FeCl3 (s)
Fe3(aq) 3 Cl -(aq)
moles of FeCl3 9.32 x 1021 formula units
1 mol FeCl3 6.022 x 1023 formula
units FeCl3
x
0.0155 mol FeCl3
3 mol Cl - 1 mol FeCl3
moles of Cl - 0.0155 mol FeCl3 x
0.0465 mol Cl -
and 0.0155 mol Fe3 are also present.
9
Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - III
H2O
d) ScBr3 (s)
Sc3(aq) 3 Br -(aq)
Converting from volume to moles
1 L 103 ml
0.56 mol ScBr3 1 L
Moles of ScBr3 75.0 ml x x
0.042 mol ScBr3
3 mol Br - 1 mol ScBr3
Moles of Br - 0.042 mol ScBr3 x
0.126 mol Br -
0.042 mol Sc3 are also present
H2O
e) (NH4)2SO4 (s)
2 NH4(aq) SO4- 2(aq)
2 mol NH4 1 mol(NH4)2SO4
Moles of NH4 7.8 moles (NH4)2SO4 x
15.6 mol NH4
and 7.8 mol SO4- 2 are also present.
10
Determining Moles of Ions in Aqueous Solutions
of Ionic Compounds - III
H2O
d) ScBr3 (s)
Sc3(aq) 3 Br -(aq)
Converting from volume to moles
1 L 103 ml
0.56 mol ScBr3 1 L
Moles of ScBr3 75.0 ml x x
0.042 mol ScBr3
3 mol Br - 1 mol ScBr3
Moles of Br - 0.042 mol ScBr3 x
0.126 mol Br -
0.042 mol Sc3 are also present
H2O
e) (NH4)2SO4 (s)
2 NH4(aq) SO4- 2(aq)
2 mol NH4 1 mol(NH4)2SO4
Moles of NH4 7.8 moles (NH4)2SO4 x
15.6 mol NH4
and 7.8 mol SO4- 2 are also present.
11
Fig. 4.2
12
Fig. 4.3
13
The Solubility of Ionic Compounds in Water
The solubility of ionic compounds in water
depends upon the relative strengths of the
electrostatic forces between ions in the ionic
compound and the attractive forces between the
ions and water molecules in the solvent. There
is a tremendous range in the solubility of ionic
compounds in water! The solubility of so called
insoluble compounds may be several orders of
magnitude less than ones that are
called soluble in water, for example
Solubility of NaCl in water at 20oC 365
g/L Solubility of MgCl2 in water at 20oC 542.5
g/L Solubility of AlCl3 in water at 20oC 699
g/L Solubility of PbCl2 in water at 20oC 9.9
g/L Solubility of AgCl in water at 20oC 0.009
g/L Solubility of CuCl in water at 20oC 0.0062
g/L
14
The Solubility of Covalent Compounds in Water
The covalent compounds that are very soluble in
water are the ones with -OH group in them and
are called polar and can have strong polar
(electrostatic)interactions with water. Examples
are compound such as table sugar, sucrose
(C12H22O11) beverage alcohol, ethanol
(C2H5-OH) and ethylene glycol (C2H6O2) in
antifreeze.
H
Methanol Methyl Alcohol
C
H
O H
H
Other covalent compounds that do not contain a
polar center, or the -OH group are considered
nonpolar , and have little or no interactions
with water molecules. Examples are the
hydrocarbons in gasoline and oil. This leads to
the obvious problems in oil spills, where the oil
will not mix with the water and forms a layer on
the surface!
Octane C8H18 and / or Benzene C6H6
15
Acids - A Group of Covalent Molecules Which Lose
Hydrogen Ions to Water Molecules in
Solution
When gaseous hydrogen iodide dissolves in water,
the attraction of the oxygen atom of the water
molecule for the hydrogen atom in HI is greater
that the attraction of the of the iodide ion for
the hydrogen atom, and it is lost to the water
molecule to form an hydronium ion and an iodide
ion in solution. We can write the hydrogen atom
in solution as either H(aq) or as H3O(aq) they
mean the same thing in solution. The presence of
a hydrogen atom that is easily lost in solution
is an Acid and is called an acidic solution.
The water (H2O) could also be written above the
arrow indicating that the solvent was water in
which the HI was dissolved.
HI(g) H2O(L) H(aq) I
-(aq)
HI(g) H2O(L) H3O(aq) I
-(aq)
H2O
HI(g) H(aq) I
-(aq)
16
Fig. 4.4
17
Strong Acids and the Molarity of H Ions in
Aqueous Solutions of Acids
Problem In aqueous solutions, each molecule of
sulfuric acid will loose two protons to yield
two hydronium ions, and one sulfate ion. What
is the molarity of the sulfate and hydronium ions
in a solution prepared by dissolving 155g of
concentrate sulfuric acid into sufficient water
to produce 2.30 Liters of acid solution? Plan
Determine the number of moles of sulfuric acid,
divide the moles by the volume to get the
molarity of the acid and the sulfate ion. The
hydronium ions concentration will be twice the
acid molarity. Solution Two moles of H are
released for every mole of acid
H2SO4 (l) 2 H2O(l)
2 H3O(aq) SO4- 2(aq)
18
Strong Acids and the Molarity of H Ions in
Aqueous Solutions of Acids
Problem In aqueous solutions, each molecule of
sulfuric acid will loose two protons to yield
two hydronium ions, and one sulfate ion. What
is the molarity of the sulfate and hydronium ions
in a solution prepared by dissolving 155g of
concentrate sulfuric acid into sufficient water
to produce 2.30 Liters of acid solution? Plan
Determine the number of moles of sulfuric acid,
divide the moles by the volume to get the
molarity of the acid and the sulfate ion. The
hydronium ions concentration will be twice the
acid molarity. Solution Two moles of H are
released for every mole of acid
H2SO4 (l) 2 H2O(l)
2 H3O(aq) SO4- 2(aq)
1 mole H2SO4 98.09 g H2SO4
Moles H2SO4
1.58 moles H2SO4
155 g H2SO4
x

19
Strong Acids and the Molarity of H Ions in
Aqueous Solutions of Acids
Problem In aqueous solutions, each molecule of
sulfuric acid will loose two protons to yield
two hydronium ions, and one sulfate ion. What
is the molarity of the sulfate and hydronium ions
in a solution prepared by dissolving 155g of
concentrate sulfuric acid into sufficient water
to produce 2.30 Liters of acid solution? Plan
Determine the number of moles of sulfuric acid,
divide the moles by the volume to get the
molarity of the acid and the sulfate ion. The
hydronium ions concentration will be twice the
acid molarity. Solution Two moles of H are
released for every mole of acid
H2SO4 (l) 2 H2O(l)
2 H3O(aq) SO4- 2(aq)
1 mole H2SO4 98.09 g H2SO4
Moles H2SO4
1.58 moles H2SO4
155 g H2SO4
x
1.58 mol SO4-2 2.30 l solution
Molarity of SO4- 2
0.687 Molar in SO4- 2
20
Strong Acids and the Molarity of H Ions in
Aqueous Solutions of Acids
Problem In aqueous solutions, each molecule of
sulfuric acid will loose two protons to yield
two hydronium ions, and one sulfate ion. What
is the molarity of the sulfate and hydronium ions
in a solution prepared by dissolving 155g of
concentrate sulfuric acid into sufficient water
to produce 2.30 Liters of acid solution? Plan
Determine the number of moles of sulfuric acid,
divide the moles by the volume to get the
molarity of the acid and the sulfate ion. The
hydronium ions concentration will be twice the
acid molarity. Solution Two moles of H are
released for every mole of acid
H2SO4 (l) 2 H2O(l)
2 H3O(aq) SO4- 2(aq)
1 mole H2SO4 98.09 g H2SO4
Moles H2SO4
1.58 moles H2SO4
155 g H2SO4
x
1.58 mol SO4-2 2.30 l solution
Molarity of SO4- 2
0.687 Molar in SO4- 2
Molarity of H 2 x 1.58 mol H / 2.30 liters
1.38 Molar in H
21
Fig. 4.5
22
Precipitation Reactions A Solid Product is Formed
When ever two aqueous solutions are mixed, there
is the possibility of forming an insoluble
compound. Let us look at some examples to see
how we can predict the result of adding two
different solutions together.
Pb(NO3) (aq) NaI(aq) Pb2(aq) 2
NO3-(aq) Na(aq) I-(aq)
When we add these two solutions together, the
ions can combine in the way they came into the
solution, or they can exchange partners. In this
case we could have lead nitrate and sodium
iodide, or lead iodide and sodium nitrate formed,
to determine which will happen we must look
at the solubility table (p.141) to determine what
could form. The table indicates that lead iodide
will be insoluble, so a precipitate will form.
Pb(NO3)2 (aq) 2 NaI(aq)
PbI2 (s) 2 NaNO3 (aq)
23
Precipitation Reactions Will a Precipitate Form?
If we add a solution containing potassium
chloride to a solution containing ammonium
nitrate, will we get a precipitate?
KCl(aq) NH4NO3 (aq) K(aq)
Cl-(aq) NH4(aq) NO3-(aq)
By exchanging cations and anions we see that we
could have potassium chloride and ammonium
nitrate, or potassium nitrate and
ammonium chloride. In looking at the solubility
table it shows all possible products as soluble,
so there is no net reaction!
KCl(aq) NH4NO3 (aq) No Reaction!
If we mix a solution of sodium sulfate with a
solution of barium nitrate, will we get a
precipitate? From the solubility table it shows
that barium sulfate is insoluble, therefore we
will get a precipitate!
Na2SO4 (aq) Ba(NO3)2 (aq)
BaSO4 (s) 2 NaNO3 (aq)
24
Fig. 4.6
25
Solubility Rules for Ionic Compounds in Water
Soluble Ionic Compounds
Insoluble Ionic Compounds
1) All common compounds of Group 1A(I) ions
(Na,K, etc.) and ammonium(NH4) are
soluble. 2) All common nitrates (NO3-),
acetates (CH3COO-), and most perchlorates
(ClO4-) are soluble. 3) All common chlorides
(Cl-), bromides (Br -), and iodides (I-) are
soluble, except those of Ag, Pb2, Cu2,
and Hg22. 4) All common sulfates (SO42-) are
soluble, except those of Ca2, Sr2, Ba2,
and Pb2.
1) All common metal hydroxides are insoluble,
except those of Group 1A (1) and the larger
members of Group 2A(2) (beginning with
Ca2). 2) All common carbonates (CO32-) and
phosphates (PO43-) are insoluble, except those
of Group 1A(1) and NH4. 3) All common
sulfides are insoluble, except those of
Group 1A(1), Group 2A(2), and NH4.
26
Predicting Whether a Precipitation Reaction
Occurs Writing Equations
a) Calcium Nitrate and Sodium Sulfate solutions
are added together.
Molecular Equation
Total Ionic Equation
Net Ionic Equation
-
b) Ammonium Sulfate and Magnesium Chloride are
added together.
27
Predicting Whether a Precipitation Reaction
Occurs Writing Equations
a) Calcium Nitrate and Sodium Sulfate solutions
are added together.
Molecular Equation
Ca(NO3)2 (aq) Na2SO4 (aq)
CaSO4 (s) NaNO3 (aq)
Total Ionic Equation
Net Ionic Equation
b) Ammonium Sulfate and Magnesium Chloride are
added together.
28
Predicting Whether a Precipitation Reaction
Occurs Writing Equations
a) Calcium Nitrate and Sodium Sulfate solutions
are added together.
Molecular Equation
Ca(NO3)2 (aq) Na2SO4 (aq)
CaSO4 (s) NaNO3 (aq)
Total Ionic Equation
Ca2(aq)2 NO3-(aq) 2 Na(aq) SO4-2(aq)
CaSO4 (s) 2 Na(aq) 2 NO3-(aq)
Net Ionic Equation
b) Ammonium Sulfate and Magnesium Chloride are
added together.
29
Predicting Whether a Precipitation Reaction
Occurs Writing Equations
a) Calcium Nitrate and Sodium Sulfate solutions
are added together.
Molecular Equation
Ca(NO3)2 (aq) Na2SO4 (aq)
CaSO4 (s) NaNO3 (aq)
Total Ionic Equation
Ca2(aq)2 NO3-(aq) 2 Na(aq) SO4-2(aq)
CaSO4 (s) 2 Na(aq) 2 NO3-(aq)
Net Ionic Equation
Ca2(aq) SO4-2(aq)
CaSO4 (s)
Spectator Ions are Na and NO3-
b) Ammonium Sulfate and Magnesium Chloride are
added together.
30
Predicting Whether a Precipitation Reaction
Occurs Writing Equations
a) Calcium Nitrate and Sodium Sulfate solutions
are added together.
Molecular Equation
Ca(NO3)2 (aq) Na2SO4 (aq)
CaSO4 (s) NaNO3 (aq)
Total Ionic Equation
Ca2(aq)2 NO3-(aq) 2 Na(aq) SO4-2(aq)
CaSO4 (s) 2 Na(aq) 2 NO3-(aq)
Net Ionic Equation
Ca2(aq) SO-4(aq)
CaSO4 (s)
Spectator Ions are Na and NO3-
b) Ammonium Sulfate and Magnesium Chloride are
added together.
In exchanging ions, no precipitates will be
formed, so there will be no chemical reactions
occurring! All ions are spectator ions!
31
Acid - Base Reactions Neutralization Rxns.
An Acid is a substance that produces H (H3O)
ions when dissolved in water. A Base is a
substance that produces OH - ions when dissolved
in water. Acids and bases are electrolytes, and
their strength is categorized in terms of their
degree of dissociation in water to make hydronium
or hydroxide ions. Strong acids and bases
dissociate completely, and are strong
electrolytes. Weak acids and bases dissociate
weakly and are weak electrolytes.
The generalized reaction between an Acid and a
Base is
HX(aq) MOH(aq)
MX(aq) H2O(L)
Acid Base
Salt Water
32
Fig. 4.7
33
Selected Acids and Bases
Acids
Bases
Strong
Strong Hydrochloric acid, HCl
Sodium hydroxide, NaOH Hydrobromic
acid, HBr
Potassium hydroxide, KOH Hydriodoic acid, HI
Calcium
hydroxide, Ca(OH)2 Nitric acid, HNO3
Strontium
hydroxide, Sr(OH)2 Sulfuric acid, H2SO4
Barium hydroxide,
Ba(OH)2 Perchloric acid, HClO4 Weak
Weak
Hydrofluoric acid, HF
Ammonia, NH3 Phosphoric acid, H3PO4
Acidic acid, CH3COOH (or HC2H3O2)
Table 4.2
34
Writing Balanced Equations forNeutralization
Reactions - I
Problem Write balanced chemical reactions
(molecular, total ionic, and
net ionic) for the following chemical reactions
a) calcium hydroxide(aq) and hydroiodic
acid(aq) b) lithium hydroxide(aq) and
nitric acid(aq) c) barium hydroxide(aq)
and sulfuric acid(aq) Plan These are all strong
acids and bases, therefore they will make
water and the corresponding salts. Solution
35
Writing Balanced Equations forNeutralization
Reactions - I
Problem Write balanced chemical reactions
(molecular, total ionic, and
net ionic) for the following chemical reactions
a) calcium hydroxide(aq) and hydroiodic
acid(aq) b) lithium hydroxide(aq) and
nitric acid(aq) c) barium hydroxide(aq)
and sulfuric acid(aq) Plan These are all strong
acids and bases, therefore they will make
water and the corresponding salts. Solution
a) Ca(OH)2 (aq) 2HI(aq)
CaI2 (aq) 2H2O(l)
36
Writing Balanced Equations forNeutralization
Reactions - I
Problem Write balanced chemical reactions
(molecular, total ionic, and
net ionic) for the following chemical reactions
a) calcium hydroxide(aq) and hydroiodic
acid(aq) b) lithium hydroxide(aq) and
nitric acid(aq) c) barium hydroxide(aq)
and sulfuric acid(aq) Plan These are all strong
acids and bases, therefore they will make
water and the corresponding salts. Solution
a) Ca(OH)2 (aq) 2HI(aq)
CaI2 (aq) 2H2O(l) Ca2(aq)
2 OH -(aq) 2 H(aq) 2 I -(aq)

Ca2(aq)
2 I -(aq) 2 H2O(l)
37
Writing Balanced Equations forNeutralization
Reactions - I
Problem Write balanced chemical reactions
(molecular, total ionic, and
net ionic) for the following chemical reactions
a) calcium hydroxide(aq) and hydroiodic
acid(aq) b) lithium hydroxide(aq) and
nitric acid(aq) c) barium hydroxide(aq)
and sulfuric acid(aq) Plan These are all strong
acids and bases, therefore they will make
water and the corresponding salts. Solution
a) Ca(OH)2 (aq) 2HI(aq)
CaI2 (aq) 2H2O(l) Ca2(aq)
2 OH -(aq) 2 H(aq) 2 I -(aq)

Ca2(aq)
2 I -(aq) 2 H2O(l) 2 OH -(aq) 2
H(aq) 2 H2O(l)
38
Writing Balanced Equations for Neutralization
Reactions - II
b) LiOH(aq) HNO3 (aq)
LiNO3 (aq) H2O(l)
39
Writing Balanced Equations for Neutralization
Reactions - II
b) LiOH(aq) HNO3 (aq)
LiNO3 (aq) H2O(l) Li(aq) OH
-(aq) H(aq) NO3-(aq)

Li(aq) NO3-(aq) H2O(l)
OH -(aq) H(aq)
H2O(l)
40
Writing Balanced Equations for Neutralization
Reactions - II
b) LiOH(aq) HNO3 (aq)
LiNO3 (aq) H2O(l) Li(aq) OH
-(aq) H(aq) NO3-(aq)

Li(aq) NO3-(aq) H2O(l)
OH -(aq) H(aq)
H2O(l)
c) Ba(OH)2 (aq) H2SO4 (aq)
BaSO4 (s) 2 H2O(l)

41
Writing Balanced Equations for Neutralization
Reactions - II
b) LiOH(aq) HNO3 (aq)
LiNO3 (aq) H2O(l) Li(aq) OH
-(aq) H(aq) NO3-(aq)

Li(aq) NO3-(aq) H2O(l)
OH -(aq) H(aq)
H2O(l)
c) Ba(OH)2 (aq) H2SO4 (aq)
BaSO4 (s) 2 H2O(l) Ba2(aq) 2
OH -(aq) 2 H(aq) SO42-(aq) BaSO4
(s) 2 H2O(l)

Ba2(aq) 2 OH -(aq) 2 H(aq) SO42-(aq)
BaSO4 (s) 2 H2O(l)
42
An Acid-Base Titration
Fig. 4.8
43
Finding the Concentration of Acid from an Acid -
Base Titration
Volume (L) of base (difference in buret readings)
M (mol/L) of base
Moles of base
molar ratio
Moles of acid
volume (L) of acid
M (mol/L) of acid
44
Finding the Concentration of Base from an Acid -
Base Titration - I
Problem A titration is performed between sodium
hydroxide and potassium hydrogenphthalate (KHP)
to standardize the base solution, by placing
50.00 mg of solid potassium hydrogenphthalate in
a flask with a few drops of an indicator. A
buret is filled with the base, and the initial
buret reading is 0.55 ml at the end of the
titration the buret reading is 33.87 ml. What is
the concentration of the base? Plan Use the
molar mass of KHP (204.2 g/mol) to calculate the
number of moles of the acid, from the balanced
chemical equation, the reaction is equal molar,
so we know the moles of base, and from the
difference in the buret readings, we can
calculate the molarity of the base. Solution
HKC8H4O4 (aq) OH -(aq)
KC8H4O4-(aq) H2O(aq)
45
Potassium Hydrogenphthalate KHC8H4O4
O
C
K
O
O
C
H
O
46
Finding the Concentration of Base from an Acid -
Base Titration - II
50.00 mg KHP 204.2 g KHP 1 mol KHP
1.00 g 1000 mg
moles KHP x
0.00024486 mol KHP
Volume of base Final buret reading - Initial
buret reading 33.87
ml - 0.55 ml 33.32 ml of base
one mole of acid one mole of base therefore
0.00024486 moles of acid will yield 0.00024486
moles of base in a volume of 33.32 ml.
0.00024486 moles 0.03332 L
molarity of base
0.07348679 moles per liter
molarity of base 0.07349 M
47
An Aqueous Strong Acid-Strong BaseReaction on
the Atomic Scale
Acid Base Salt H2O
Fig. 4.9
48
An Acid-Base Reaction That Formsa Gaseous
Product The reaction of acid with
carbonates or bicarbonates will produce carbon
dioxide gas that is released from solution as
a gas in the form of bubbles that leave the
solution.
Fig. 4.10
49
Types of Chemical Reactions - I
I) Combination reactions that are redox
reactions a) A metal and a nonmetal form an
Ionic compound b) Two nonmetals form a
covalent compound c) Combination of an
element and a compound II) Combination reactions
that are not redox reactions a) A metal
oxide and a nonmetal form an ionic compound with
a polyatomic anion b) Metal oxides
and water form bases c) Nonmetal oxides and
water form acids III) Decomposition reactions
a) Thermal decomposition i) Many
ionic compounds with oxoanions form a metal
oxide and a gaseous nonmetal
ii) Many metal oxides, chlorates, and
perchlorates release oxygen
b) Electrolytic decomposition
50
Types of Chemical Reactions - II
IV) Displacement Reactions a)
Single-Displacement Reactions - Activity Series
of the Metals i) A metal displaces
hydrogen from water or an acid ii)
A metal displaces another metal ion from
solution iii) A halogen displaces a
halide ion from solution b)
Double-Displacement Reactions -
i) In precipitation reactions A precipitate
forms ii) In acid-base reactions
Acid Base form a salt water V) Combustion
Reactions - All are Redox Processes a)
Combustion of an element with oxygen to form
oxides b) Combustion of hydrocarbons to
yield water carbon dioxide
Reactants
Products
51
Fig. 4.15
52
Two Nonmetals Combine to Form a Binary
Covalent Compound - Redox Rxns
Halogens of nonmetals
P4 (s) 6 F2 (g)
4 PF3 (g)
I2 (s) 5 F2 (g)
2 IF5 (l)
S8 (s) 4 Br2 (l)
4 S2Br2 (l)
N2 (g) 3 Cl2 (g)
2 NCl3 (g)
Nitrides and sulfides
P4 (s) 10 N2 (g)
4 P3N5 (s)
S8 (s) 2 N2 (g)
2 S4N2 (s)
8 P4 (s) 5 S8 (s)
8 P4S5 (s)
53
Other Elements Combine with Oxygen to Form Oxides
Metals combining with xygen
4 Na(s) O2 (g) 2
Na2O(s)
2 Ca(s) O2 (g) 2
CaO(s)
4 Al(s) 3 O2 (g) 2
Al2O3 (s)
Ti(s) O2 (g) TiO2
(s)
Nonmetals with oxygen
N2 (g) O2 (g) 2 NO (g)
P4 (s) 5 O2 (g)
P4O10 (s)
S8 (s) 8 O2 (g) 8
SO2 (g)
2 F2 (g) O2 (g)
2 OF2 (g)
54
Nonmetal Oxides Halides React with Additional
Oxygen Halogens to Form Higher Oxides and
Halides
1) Nonmetal oxides with oxygen
2 NO(g) O2 (g)
2 NO2 (g)
P4O6 (s) 2 O2 (g)
P4O10 (s)
2 CO(g) O2 (g) 2
CO2 (g)
2) Nonmetal halides with halogens
ClF(g) F2 (g)
ClF3 (g)
BrF3 (g) F2 (g)
BrF5 (l)
IF3 (g) F2 (g)
IF5 (l)
IF5 (l) F2 (g)
IF7 (l)
55
Combination of a Compound and an
Element Nonmetal Oxides Halides React with
Additional Oxygen and Halogens to Form Higher
Oxides and Halides
1) Nonmetal oxides with oxygen
2 NO(g) O2 (g)
2 NO2 (g)
P4O6 (s) 2 O2 (g)
P4O10 (s)
2 CO(g) O2 (g) 2
CO2 (g)
2) Nonmetal halides with halogens
ClF(g) F2 (g)
ClF3 (g)
PF3 (g) F2 (g)
PF5 (l)
IF3 (g) F2 (g)
IF5 (l)
IF5 (l) F2 (g)
IF7 (l)
56
Combination of Two Compounds - I
1) Metal oxide with a nonmetal oxide to form an
ionic compound with a polyatomic anion.
Na2O(s) CO2 (g)
Na2CO3 (s)
K2O(s) SO2 (g)
K2SO3 (s)
CaO(s) SO3 (g)
CaSO4 (s)
2) Metal oxides react with water to form
hydroxides
Na2O(s) H2O(l)
2 NaOH(aq)
BaO(s) H2O(l)
Ba(OH)2 (s)
2 Sc2O3 (s) 6 H2O(l)
4 Sc(OH)3 (s)
FeO(s) H2O(l)
Fe(OH)2 (s)
57
Combination of Two Compounds - II
3) Nonmetal oxides react with water to form acids.
CO2 (g) H2O(l)
H2CO3 (aq)
SO2 (g) H2O(l)
H2SO3 (aq)
3 NO2 (g) H2O(l)
2 HNO3 (aq) NO(g)
4) Hydrates result from the reaction of anhydrous
(without water) compounds with water.
.
CuSO4 (s) 5 H2O(l)
CuSO4 5H2O (s)
.

MgSO4 (s) 7 H2O(l)
MgSO4 7H2O(s)
.
Na2CO3 (s) 10 H2O(l)
Na2CO3 10H2O(s)
5) Addition reactions with carbon compounds.
C2H4 (g) Cl2 (g)
C2H4Cl2 (g)
Ethylene Chlorine
dichloroethane
58
Fig. 4.16
59
Thermal Decomposition Reactions - I
Carbonates
Oxides and carbon dioxide
Na2CO3 (s) Na2O(s)
CO2 (g)
MgCO3 (s) MgO(s)
CO2 (g)
Sulfites
Oxides and sulfur dioxide
MgSO3 (s) MgO(s)
SO2 (g)
K2SO3 (s) K2O(s)
SO2 (g)
Metal oxides, chlorates, and perchlorates
Oxygen
2 Na2O(s) 4 Na(s)
O2 (g)
KClO4 (s) KCl(s)
2 O2 (g)
60
Thermal Decomposition Reactions - II
Hydroxides, hydrates, and some oxoacids
Water
Ca(OH)2 (s) CaO(s)
H2O(g)
Na2CO3 10H2O (s)
Na2CO3 (s) 10 H2O(g)
H2SO3 (l) SO2 (g)
H2O(g)
2 Fe(OH)3 (s) Fe2O3
(s) 3 H2O(g)
MgSO4 7H2O(s) MgSO4
(s) 7 H2O(g)
H2CO3 (aq) CO2
(g) H2O(g)
2 NaOH(s) Na2O(s)
H2O(g)
61
Fig. 4.17
62
Fig. 4.20
63
Metals Displace Hydrogen from Water
Metals that will displace ydrogen from cold water
2 Cs(s) 2 H2O
H2 (g) 2 CsOH(aq)
Ba(s) 2 H2O(l) H2
(g) Ba(OH)2 (aq)
2 Na(s) 2 H2O(l)
H2 (g) 2 NaOH(aq)
Metals that will displace hydrogen from steam
Mg(s) 2 H2O(g) H2
(g) Mg(OH)2 (s)
2 Cr(s) 6 H2O(g) 3
H2 (g) 2 Cr(OH)3 (s)
Zn(s) 2 H2O(g) H2
(g) Zn(OH)2 (s)
64
Metals Displace Hydrogen from Acids
Reactions of metals above hydrogen in the
activity series
Mg(s) 2 HCl(aq) MgCl2
(aq) H2 (g)
Zn(s) H2SO4 (aq) ZnSO4
(aq) H2 (g)
2 Al(s) 6 HCl(aq) 2
AlCl3 (aq) 3 H2 (g)
Cd(s) 2 HBr(aq)
CdBr2 (aq) H2 (g)
Sn(s) 2 H2SO4 (aq)
Sn(SO4)2 (aq) 2 H2 (g)
Reactions of metals below hydrogen in the
activity series
Cu(s) HCl(aq) No
Reaction!
Au(s) H2SO4 (aq) No
Reaction!
65
Fig. 4.19
66
Single-Displacement Reactions - Metals
Replace Metal Ions from Solution
Ba(s) Co(NO3)2 (aq)
Co(s) Ba(NO3)2 (aq)
Cd(s) AgNO3 (aq)
2 Ag(s) Cd(NO3)2 (aq)
Mg(s) Pb(NO3)2 (aq)
Pb(s) Mg(NO3)2 (aq)
Al(s) Ba(NO3)2 (aq)
No Reaction
4 Cr(s) 3 PtCl4 (aq)
3 Pt(s) 4 CrCl3 (aq)
Ca(s) Hg(NO3)2 (aq)
Hg(l) Ca(NO3)2 (aq)
2 Li(s) Na2SO4 (aq) H2O(l) 2
NaOH(aq) H2 (g) Li2SO4 (aq)
67
Double-Displacement Reactions
1) Precipitation reactions - an insoluble product
is formed
Pb(NO3)2 (aq) 2 NaI(aq)
PbI2 (s) 2 NaNO3 (aq)
Ba(NO3)2 (aq) Na2SO4 (aq)
BaSO4 (s) 2 NaNO3 (aq)
2) Acid-base neutralization reactions - water is
formed
HCl(aq) NaOH(aq)
NaCl(aq) H2O(l)
H2SO4 (aq) Ca(OH)2 (s)
CaSO4 (s) 2 H2O(l)
3) A carbonate or sulfite reacts with acid to
form a gas
Na2CO3 (aq) 2 HBr(aq) 2
NaBr(aq) H2O(l) CO2 (g)
K2SO3 (aq) 2 HI(aq)
2 KI(aq) H2O(l) SO2 (g)
68
Combustion Reactions Redox Reactions
Elements combining with oxygen to form oxides
2 Al(s) 3 O2 (g) 2
Al2O3 (s)
S8 (s) 8 O2 (g) 8
SO2 (g)
C(s) O2 (g) CO2 (g)
Compounds combining with oxygen to form oxides
2 Fe2S3 (s) 9 O2 (g) 2
Fe2O3 (s) 6 SO2 (g)
2 Ca3N2 (s) 7 O2 (g)
6 CaO(s) 4 NO2 (g)
Hydrocarbons combining with oxygen to form CO2
and H2O
CH4 (g) 2 O2 (g) CO2
(g) 2 H2O(g)
2 C4H10 18 O2 (g) 8 CO2
(g) 10 H2O(g)
69
Identifying the Type of Chemical Reaction - I
Problem Classify the following as combination,
decomposition, or displacement reactions,
identify the underlying chemical process as
precipitation, acid-base, or redox, and write a
balanced molecular equation for each. For redox,
identify the oxidizing and reducing agents. a)
Barium chloride(aq) and ammonium sulfate(aq) b)
Manganese metal and tin IV chloride (aq) c)
Strontium hydroxide and bromic acid d) Cobalt
metal and nitrogen gas to yield cobalt II
nitride e) Sodium peroxide to give sodium oxide
and oxygen gas Plan Identify the reaction type,
and identify the products and equation. Solution
a) Displacement (metathesis) two substances form
two others, one of which is a precipitate.
BaCl2 (aq) (NH4)2SO4 (aq)
BaSO4 (s) 2 NH4Cl(aq)
70
Identifying the Type of Chemical Reaction -II
b) Displacement reaction(single) This redox
reaction occurs when the more active
manganese displaces the less reactive tin.
2 Mn(s) SnCl4 (aq) 2
MnCl2 (aq) Sn(s)
Mn is the reducing agent, and SnCl4 is the
oxidizing agent.
c) Displacement reaction(metathesis) Two
substances form two others. This is an
acid-base reaction. The base Sr(OH)2 reacts with
twice as much bromic acid to give water and
the salt, strontium bromate dissolved in
water.
Sr(OH)2 (aq) 2 HBrO3 (aq)
Sr(BrO3)2 (aq) 2 H2O(L)
71
Identifying the Type of Chemical Reaction -III
d) Combination Reaction This redox reaction
occurs when cobalt metal is heated in
nitrogen gas and forms the solid compound
cobalt II nitride.
3 Co(s) N2 (g) Co3N2 (s)
Nitrogen is the oxidizing agent, and is reduced
cobalt is the reducing agent, and is oxidized.
e) Decomposition reaction One substance forms
two substances. This is a redox reaction
that forms oxygen gas.
2 Na2O2 (s) 2 Na2O(s)
O2 (g)
Sodium peroxide is the reducing and oxidizing
agent, as oxygen is both oxidized, and reduced.