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University of Lusaka
BSPH 111 - Refresher Chemistry
2015
1
Introduction
Chemistry is the study of the composition, structure, properties and reactions of matter,
especially of atomic molecular systems. This course takes particular interest in the use of
chemistry in the public health sector. The course is not your ordinary chemistry class; it
highlights the relevance of both organic and inorganic chemistry and places an emphasis on
man’s use of chemical substances and the effects of chemicals on the environment. It pushes
the minds of student’s to apply theory to the practical world to impact decision making in
both local and international communities as public health specialists.
Module Learning Outcomes
By the end of the course, students should be able to:

Foster a lasting interest in Chemistry so that they find studying or applying
knowledge from these subjects enjoyable and satisfying;

Develop transferable life-long skills that are relevant to public health and the
increasingly technological world in which we live;

Develop chemistry abilities, skills, and attitudes relevant to public health and
scientific inquiry.
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Course Outline
1. Atoms, Molecules, Stoichiometry; Atomic structure;
2. Chemical bonding
3. States of matter
4. Chemical energetics
5. Electrochemistry
6. Equilibria
7. Reaction kinetics
8. Inorganic chemistry
9. Periodic table
10. Organic chemistry
11. Applications of analytical chemistry.
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Module Descriptor:
Purpose or overall objective: The course will refresh and introduce students to basic
concepts in Chemistry and their relevance in public health.
Course Code: BSPH 111
Refresher Chemistry
Lecturer:
Total Programme Duration: One Semester
4
Contents
BSPH 111 - Refresher Chemistry ........................................................................................................... 1
Introduction ............................................................................................................................................. 2
Module Learning Outcomes ................................................................................................................... 2
Course Outline ........................................................................................................................................ 3
Module Descriptor: ................................................................................................................................. 4
Contents .................................................................................................................................................. 5
Unit 1: Atoms, Molecules, Stoichiometry; Atomic structure ................................................................. 9
Stoichiometry .................................................................................................................................... 16
Terms ............................................................................................................................................ 16
Stoichiometric Calculations .......................................................................................................... 17
Unit 2: Chemical bonding and Molecular Structure ............................................................................. 36
What Are Chemical Bonds, and Why Do They Form? .................................................................... 36
Types of Chemical Bonds ............................................................................................................. 37
Drawing Lewis Structures............................................................................................................. 38
Exceptions to Regular Lewis Structures—Resonance Structures ................................................. 40
Unit Summary ................................................................................................................................... 42
Unit 3: States of matter ......................................................................................................................... 43
Solids ................................................................................................................................................ 43
Ionic solids .................................................................................................................................... 44
Molecular solids ............................................................................................................................ 44
Covalent-network .......................................................................................................................... 44
Metallic solids ............................................................................................................................... 45
Liquids .............................................................................................................................................. 45
Gases ................................................................................................................................................. 45
Phase Changes .................................................................................................................................. 45
Specific Heat ................................................................................................................................. 47
Unit Summary ................................................................................................................................... 49
Unit 4. Chemical energetics .................................................................................................................. 50
Definitions......................................................................................................................................... 50
Hess' Law .......................................................................................................................................... 50
Entropy.............................................................................................................................................. 52
Spontaneity of a reaction- Gibbs Free Energy .............................................................................. 52
5
Gibbs free energy equation ........................................................................................................... 55
Unit Summary ................................................................................................................................... 59
Unit 5: Electrochemistry and Redox Reactions .................................................................................... 60
Electrochemistry: .............................................................................................................................. 60
Oxidation-Reduction Reactions ........................................................................................................ 60
Oxidation: ..................................................................................................................................... 60
Reduction: ..................................................................................................................................... 60
Oxidation number: ........................................................................................................................ 60
Rules for Assigning Oxidation States ........................................................................................... 60
Voltaic (or Galvanic) Cells ............................................................................................................... 63
Standard Reduction Potentials ...................................................................................................... 64
Electrolytic Cells ........................................................................................................................... 66
Unit Summary ................................................................................................................................... 67
Unit 6: Equilibria .................................................................................................................................. 68
Dynamic Equilibrium........................................................................................................................ 68
The position of equilibrium........................................................................................................... 69
The equilibrium constant .............................................................................................................. 69
Effect of Temperature ................................................................................................................... 69
Effect of Concentration ................................................................................................................. 69
Effect of Pressure .......................................................................................................................... 69
Effect of catalysts on equilibrium ................................................................................................. 70
Unit Summary ................................................................................................................................... 71
Unit 7: Reaction kinetics ....................................................................................................................... 72
Rates of reaction ............................................................................................................................... 72
Collision Theory ............................................................................................................................... 72
Activation energy .............................................................................................................................. 72
Reaction Mechanisms ....................................................................................................................... 73
Unit Summary ................................................................................................................................... 76
Unit 8: Inorganic chemistry .................................................................................................................. 77
Metals, Non-metals and Metalloids .................................................................................................. 77
Metals............................................................................................................................................ 77
Non-metals .................................................................................................................................... 78
Metalloids ..................................................................................................................................... 79
Trends in Metallic and Non-metallic Character ............................................................................ 80
6
Uses of Transition Metals ............................................................................................................. 85
Uses of non-transition metals ........................................................................................................ 89
Different steels for different uses: ................................................................................................ 92
Unit Summary ................................................................................................................................... 94
Unit 9: The Periodic Table .................................................................................................................... 95
Trends ............................................................................................................................................... 95
Atomic Size (Atomic Radius) ....................................................................................................... 95
Moving Across a Period................................................................................................................ 95
Moving Down a Group ................................................................................................................. 95
Cations and Anions ....................................................................................................................... 96
Ionization Energy and Electron Affinity ........................................................................................... 96
Ionization Energy .......................................................................................................................... 96
Electron Affinity ........................................................................................................................... 97
Electronegativity ........................................................................................................................... 98
Electron Configuration and Valence Electrons ................................................................................. 99
Electron Configuration.................................................................................................................. 99
Valency and Valence Electrons .................................................................................................. 100
Diamagnetism and Paramagnetism ............................................................................................. 102
Unit Summary ................................................................................................................................. 103
Unit 10: Organic chemistry ................................................................................................................. 104
Why Is Organic Chemistry Important? ........................................................................................... 104
What Does an Organic Chemist Do? .......................................................................................... 104
Alkanes ........................................................................................................................................... 104
Isomerism and branching ............................................................................................................ 106
Alkenes ........................................................................................................................................... 107
Alcohols .......................................................................................................................................... 108
Halogenoalkanes ............................................................................................................................. 110
Reactions of halogenoalkanes ..................................................................................................... 111
Chemical properties of the different functional groups .................................................................. 111
Unit Summary ................................................................................................................................. 115
Unit 11: Applications of analytical chemistry .................................................................................... 116
Application in All Areas of Chemistry ........................................................................................... 116
References ........................................................................................................................................... 118
7
8
Unit 1: Atoms, Molecules, Stoichiometry; Atomic structure
An atom consists of a nucleus of protons and neutrons, surrounded by electrons. Each of the
elements in the periodic table is classified according to its atomic number, which is the
number of protons in that element's nucleus. Protons have a charge of +1, electrons have a
charge of -1, and neutrons have no charge. Neutral atoms have the same number of electrons
and protons, but they can have a varying number of neutrons. Within a given element, atoms
with different numbers of neutrons are isotopes of that element. Isotopes typically exhibit
similar chemical behaviour to each other. Isotopes are atoms of the same element with the
same number of protons but different number of neutrons.
Electrons have such little mass that they exhibit properties of both particles and waves; in.
We further know from Heisenberg's Uncertainty Principle that it is impossible to know the
precise location of an electron. Despite this limitation, there are regions around the atom
where the electron has a high probability of being found. Such regions are referred to as
atomic orbitals.
Exercise
-Describe what an Atom is; use the periodic table to illustrate examples of different atoms
giving the relative atomic mass, and the proton number.
-What is an isotope?
Atomic Orbitals and Quantum Numbers
The relation of a particular electron to the nucleus can be described through a series of four
numbers, called the Quantum Numbers. The first three of these numbers describe the energy
(Principle quantum number), shape (Angular momentum quantum number), and orientation
of the orbital (magnetic quantum number). The fourth number represents the "spin" of the
electron (spin quantum number). The four quantum numbers are described below.
Principle Quantum Number (n)
The principle quantum number indicates how the distance of the orbital from the nucleus.
Electrons are farther away for higher values of n . Electrons are negatively charged, so
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electrons that are closer to the positively charged nucleus are more powerfully attracted and
tightly bound than those that are farther away. Electrons that are closer to the nucleus are thus
more stable, and less likely to be lost by the atom. In other words, as n increases, so does the
energy of the electron and the likelihood of that electron being lost by the atom. In a given
atom, all the atomic orbitals with the same n are collectively known as a shell. n can take on
integer values of 1 or higher (ex. 1, 2, 3, etc.).
Angular Momentum Quantum Number (l)
The angular momentum quantum number describes the shape of the orbital. The angular
momentum number (or subshell) can be represented either by a number (any integer from 0
up to n-1) or by a letter (s, p, d, f, g, and then up the alphabet), with 0 corresponding to s, 1 to
p,
2
to
d,
and
so
on.
For
example:
when n = 1, l can only equal 0; meaning that shell n = 1 has only an s orbital (l= 0).
when n = 3, l can equal 0, 1, or 2; meaning that shell n = 3 has s, p, and dorbitals.
s orbitals are spherical, whereas p orbitals are dumbbell-shaped. d orbitals and beyond are
much harder to visually represent.
Figure %: s and p atomic orbital shapes
Magnetic Quantum Number (m)
Gives the orientation of the orbital in space; in other words, the value of mdescribes whether
an orbital lies along the x-, y-, or z-axis on a three-dimensional graph, with the nucleus of the
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atom at the origin. m can take on any value from -l to l. For our purposes, it is only important
that this quantum number tells us that for each value of n there may be up to one s -orbital,
three p -orbitals,
five d -orbitals,
and
so
on.
For
example:
The s orbital (l = 0) has one orbital, since m can only equal 0. That orbital is spherically
symmetrical about the nucleus.
Figure %: s orbital
The p orbital (l = 1) has three orbitals, since m = -1, 0, and 1. These three orbitals lie along
the x -, y -, and z -axes.
11
Figure %: p orbitals
The d orbital (l = 2) has five orbitals, since m = -2, -1, 0, 1, and 2. It is far more difficult to
describe the orientation of d orbitals, as you can see:
12
Figure %: d orbitals
Spin Quantum Number (s):
The spin quantum number tells whether a given electron is spin up (+1/2) or spin down (1/2). An orbital contains two electrons, and each of those electrons must have different spins.
Orbital Energy Diagrams
It is often convenient to depict orbitals in an orbital energy diagram, as seen below in . Such
diagrams show the orbitals and their electron occupancies, as well as any orbital interactions
that exist. In this case we have the orbitals of the hydrogen atom with electrons omitted. The
first electron shell (n = 1) contains just the 1s orbital. The second shell (n = 2) holds a
2s orbital and three 2p orbitals. The third shell (n = 3) holds one 3s orbital, three 3p orbitals,
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and five 3d orbitals, and so forth. Note that the relative spacing between orbitals becomes
smaller for larger n. In fact, as n gets large the spacing becomes infinitesimally small.
Figure %: Energy diagram of the unoccupied atomic orbitals of hydrogen. Potential energy
is on the y-axis.
You will see such energy diagrams quite often in your continuing study of chemistry. Notice
that all orbitals with the same n have the same energy. Orbitals with identical energies are
said to be degenerate (not in the moral sense!). Electrons in higher-level orbitals have more
potential energy and are more reactive, i.e. more likely to undergo chemical reactions.
Multi-electron atoms
When an atom only contains a single electron, its orbital energies depend only on the
principle quantum numbers: a 2s orbital would be degenerate with a 2p orbital. However, this
degeneracy is broken when an atom has more than one electron. This is due to the fact that
the attractive nuclear force any electron feels is shielded by the other electrons. s-orbitals tend
to be closer to the nucleus than p-orbitals and don't get as much shielding, and hence become
lower in energy. This process of breaking degeneracies within a shell is known as splitting. In
general s orbitals are lowest in energy, followed by p orbitals, d orbitals, and so forth.
Figure %: Splitting of orbital energies in multi-electron systems
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Electron Energy
The energy diagram of imply a further fact about the energy of electrons. Note that the
energy levels in these diagrams do not follow a continuous line: an atom is either in one
energy subshell or it is in another. There is no in between. In this way, the diagram perfectly
represents the quantized nature of electrons, meaning that electrons can only exist at specific
and defined energy levels. The energy level of an electron in a particular energy shell can be
determined according to the following equation:
E n = /frac-2.178x10-18joulesn 2
where n is the principal quantum number and E n is the energy level at that quantum number.
When an electron absorbs a specific quanta of energy it can jump to a higher energy level. It
can also give off a specific quanta and fall back to a lower energy level. An atom whose
electrons are at their lowest energy levels is said to be in the ground state. The discovery of
the quantum nature of energy and electrons, first formulated by Max Planck in 1900, led to
the creation of an entirely new field, quantum mechanics.
Exercise
-What is Spin quantum number?
-What is the Principle quantum number?
-What is Electron energy?
-What do orbital energy diagrams illustrate?
-What is the magnetic quantum number?
-What is angular momentum quantum number?
15
Stoichiometry
Terms
Actual yield - The amount of product that is actually produced in a chemical reaction.
Endothermic reaction - A reaction that requires heat energy to be carried to completion.
Enthalpy - The amount of heat a substance has at a given temperature and pressure.
Excess reagent - The reactant that is in excess in a chemical reaction. There is more than
enough of it to react with the other reactant(s).
Exothermic reaction - A reaction that releases energy in the form of heat.
Heat of reaction - The change in enthalpy in a chemical reaction.
Limiting reagent - The reactant that limits or determines the amount of product that can be
formed in a chemical reaction.
Percent yield - The decimal percentage resulting from dividing actual yield by theoretical
yield.
Standard heat of formation - Unique to every substance, it is a measure of the change in
enthalpy in the reaction that produces 1 mole of the substance from its constituent elements.
Theoretical yield - The calculated or expected amount of product to be formed in a
chemical reaction.
Thermochemical reaction - Chemical reactions that include the amount of heat energy
produced or absorbed.
Formulas
Percent Yield =
Formula for Percent Yield
δH = δH f (products) - δH f (reactants)
δH
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Conversion Factor between kJ and
kcal
Stoichiometric Calculations
Applying Conversion Factors to Stoichiometry
Now you're ready to use what you know about conversion factors to solve some
stoichiometric problems in chemistry. Almost all stoichiometric problems can be solved in
just four simple steps:
1.
Balance the equation.
2.
Convert units of a given substance to moles.
3.
Using the mole ratio, calculate the moles of substance yielded by the reaction.
4.
Convert moles of wanted substance to desired units.
These "simple" steps probably look complicated at first glance, but relax, they will all
become clear.
Let's begin our tour of stoichiometry by looking at the equation for how iron rusts:
Fe + O2→Fe2O3
Step 1. Balancing the Equation
The constituent parts of a chemical equation are never destroyed or lost: the yield of a
reaction must exactly correspond to the original reagents. This fact holds not just for the type
of elements in the yield, but also the number. Given our unbalanced equation:
Fe + O2→Fe2O3
This equation states that 1 iron (Fe) atom will react with two oxygen (O) atoms to yield 2 iron
atoms and 3 oxygen atoms. (The subscript number, such as the two in O2 describe how many
atoms of an element are in a molecule.) This unbalanced reaction can't possibly represent a
real reaction because it describes a reaction in which one Fe atom magically becomes two Fe
atoms.
17
Therefore, we must balance the equation by placing coefficients before the various molecules
and atoms to ensure that the number of atoms on the left side of the arrow corresponds
exactly to the number of elements on the right.
4Fe +3O2→2Fe2O3
Let's count up the atoms in this new, balanced version of the reaction. On the left of the arrow
we have 4 atoms of iron and 6 atoms of oxygen (since 3×2 = 6 ). On the right we also have 4
iron (since 2×2 = 4 ) and 6 oxygen ( 2×3 = 6 ). The atoms on both sides of the equation
match.
The process of balancing an equation is basically trial and error. It gets easier and easier with
practice. You will likely start to balance equations almost automatically in your mind.
Step 2. Converting Given Units of a Substance to Moles
The process of converting given units into moles involves conversion factors. Below we will
provide the most common and important conversion factors to convert between moles and
grams, moles and volumes of gases, moles and molecules, and moles and solutions. Note also
that though these conversion factors focus on converting from some other unit to moles, they
can also be turned around, allowing you to convert from moles to some other unit.
Converting from Grams to Moles
The gram formula mass of a compound (or element) can be defined as the mass of one mole
of the compound. As the definition suggests, it is measured in grams/mole and is found by
summing the atomic weights of every atom in the compound. Atomic weights on the periodic
table are given in terms of amu (atomic mass units), but, by design, amu correspond to the
gram formula mass. In other words, a mole of a 12 amu carbon atom will weigh 12 grams.
The gram formula mass can be used as a conversion factor in stoichiometric calculations
through the following equation:
Moles =
18
Gram formula mass is also known as GFM. You may also see the term gram molecular mass,
abbreviated GMM. This term is often used instead of GFM when the substance is molecular
and not ionic. However, only the terminology is different, GMM is used in the same way as
GFM. Therefore, I will use the catch- all term GFM in this study guide.
Converting between Volume of a Gas and Moles
The Ideal Gas law, discussed at length in the Sparknote on Gases, provides a handy means of
converting between moles and a gas, provided you know certain qualities of that gas. The
Ideal Gas Law is PV = nRT , with n representing the number of moles. If we rearrange the
equation to solve for n, we get:
n=
with P representing pressure in atm, V representing volume in liters, T representing
temperature in Kelvins, and R the gas constant, which equals .0821 L-atm/mol-K.
Given P , V , and T , you can calculate the number of moles of substance in a gas.
In those instances when a problem specifies that the calculations are to be made at STP
(Standard Temperature and Pressure; P = 1 atm, T = 273 K)), the problem becomes even
simpler. At STP, a mole of gas will always occupy 22.4 L of volume. If you are given a
volume of a gas at STP, you can calculate the moles in that gas by calculating the volume you
are given as a fraction of 22.4 L. At STP, 11.2 L of a gas will be .5 moles; 89.6 L of gas will
be 4 moles.
Converting between Individual Particles and Moles
Avogadro's Number provides the conversion factor for moving from number of particles to
moles. There are 6.02×1023 formula units of particles in every mole of substance, with
formula unit describing the substance we are looking at, whether it is a compound, molecule,
atom, or ion. A formula unit is the smallest unit of a substance that still retains that
substance's properties and is the simplest way to write the formula of the substance without
coefficients. Some representative formula units are listed below.

Compounds: Cu2S , NaCl

Molecules: N2 , H2

Atoms: Fe, Na
19

Ions: Na+(aq) , Cl-(aq)
Since 1 mole = 6.02×1023 formula units, the conversion from formula units to moles is
simple:
Moles =
Converting between Solutions and Moles
Solutions are discussed in much greater detail in the series of Solutions SparkNotes. But it is
possible, and fairly easy to convert between the measures of solution (molarity and molality)
and moles.
Molarity is defined as the number of moles of solute divided by the number of liters of
solvent. Rearranging the equation to solve for moles yields:
Moles = molarity × liters of solution
MolaLity is defined as the number of moles of solute divided by the number of kilograms of
solvent. Rearranging the equation to solve for moles yields:
Moles = molality × kilograms of solution
Using the Mole Ratio to Calulate Yield
Before demonstrating how to calculate how much yield a reaction will produce, we must first
explain what the mole ratio is.
The Mole Ratio
Let's look once again at our balanced demonstration reaction:
4Fe +3O2→2Fe2O3
The coefficients in front of iron, oxygen, and iron (III) oxide are ratios that govern the
reaction; in other words, these numbers do not demand that the reaction can only take place
with the presence of exactly 4 moles of iron and 3 moles of oxygen, producing 2 moles of
iron (III) oxide. Instead, these numbers state the ratio of the reaction: the amount of iron and
oxygen reaction together will follow a ratio of 4 to 3. The mole ratio describes exactly what
its name suggests, the molar ratio at which a reaction will proceed. For example, 2 moles of
20
Fe will react with 1.5 moles of O2 to yield 1 mole of Fe2O3 . Alternatively, 20 moles of Fe
will react with 15 moles of O2 to yield 10 moles of Fe2O3. Each of these examples of the
reaction follow the 4:3:2 ratio described by the coefficients.
Now, with a balanced equation, the given units converted to moles, and our understanding of
the mole ration, which will allow us to see the ratio of reactants to each other and to their
product, we can calculate the yield of a reaction in moles. Step 4 demands that we be able to
convert from moles to back to the units requested in a specific problem, but that only
involves turning backwards the specific converstion factors described above.
Sample Exercise
Problem: Given the following equation at STP:
N2(g) + H2(g)→NH3(g)
Determine what volume of H2(g) is needed to produce 224 L of NH3(g).
Solution:
Step 1: Balance the equation.
N2(g) + 3H2(g)→2NH3(g)
Step 2: Convert the given quantity to moles. Note in this step, 22.4 L is on the denominator
of the conversion factors since we want to convert from liters to moles. Remember your
conversion factors must always be arranged so that the units cancel.
= 10 moles of NH3(g)
Step 3: mole ratio.
= 15 moles H2(g)
Step 4: convert to desired units:
= 336 L H2(g)
21
Now for a more challenging problem:
Given the following reaction:
2H2S(g) + O2(g)→SO2(g) + 2H2O(s)
How many atoms of oxygen do I need in order to get 18 g of ice?
Solution
Step 1: The equation is partially balanced, but let's finish the job.
2H2S(g) +3O2(g)→2SO2(g) + 2H2O(s)
Step 2: convert to moles:
1 formula unit of H2O has 2 atoms of H and 1 atom of O;
The atomic mass of H is 1 gram/mole;
Atomic mass of O = 16 grams/mole.
GFM of H2O(s) =
+
grams / mole
×1 mole = 1 mole of H2O(s)
Step 3: mole ratio:
22
= 18
×3 moles O2(g) = 1.5 moles O2(g)
Step 4: convert to desired units:
= 9.03×1023 molecules O2(g)
Is this the answer? No. The question asks for ATOMS of oxygen. There are two atoms of
oxygen in each molecule of O2(g).
×2 atoms O = 1.806×1024 atoms O
Exercise
Place a tick in the correct box
1
Questions
Zn(s)
1
+
2
4
refer
HCl(aq)
to
the
------>
chemical
ZnCl2(aq)
equation
+
H2(g)
How many moles of HCl are required for complete reaction of 0.40
moles of zinc?
0.40 moles
0.80 moles
0.20 moles
None of the previous answers.
2
Zn(s)
+
2
HCl(aq)
------->
ZnCl2(aq)
+
H2(g)
For the reaction above, how many moles of hydrogen are produced from
the reaction of 0.40 moles of zinc with an excess of HCl?
23
0.40 moles
0.80 moles
0.20 moles
None of the previous answers.
3
Zn(s)
+
2
HCl(aq)-------->
ZnCl2(aq)
+
H2(g)
For the reaction above, how many moles of hydrogen are produced from
the reaction of 0.40 moles of HCl with an excess of zinc?
0.40 moles
0.80 moles
0.20 moles
None of the previous answers.
4
Zn(s)
+
2
HCl(aq)
--------->
ZnCl2(aq)
+
H2(g)
For the reaction above, how many moles of hydrogen are produced from
the reaction of 0.40 moles of Zn with an 0.40 moles of HCl?
0.40 moles
0.80 moles
0.20 moles
None of the previous answers.
5
Questions 5 - 8 refer to the chemical equation:
C3H8(g) + 5 O2(g) = 3 CO2(g) + 4 H2O(g)
How many moles of O2 are required for complete reaction of 1.5 moles
of propane (C3H8)?
1.5 moles
0.30 moles
7.5 moles
24
None of the previous answers.
6
C3H8(g)
+
5
O2(g)
=
3
CO2(g)
+
4
H2O(g)
For the reaction above, how many moles of carbon dioxide are produced
from the reaction of 1.5 moles of propane with an excess of oxygen?
4.5 moles
0.50 moles
1.5 moles
None of the previous answers.
7
C3H8(g)
+
5
O2(g)
=
3
CO2(g)
+
4
H2O(g)
For the reaction above, how many moles of propane are burned when
0.60 moles of carbon dioxide are produced?
1.8 moles
0.2 moles
0.60 moles
None of the previous answers.
8
C3H8(g)
+
5
O2(g)
=
3
CO2(g)
+
4
H2O(g)
For the reaction above, how many moles of carbon dioxide are produced
from the reaction of 0.20 moles of propane with 1.2 moles of oxygen?
0.72 moles
0.067 moles
2.0 moles
0.60 moles
None of the previous answers.
9
Questions
4
FeS2(s)
9
+
11
13
refer
O2(g)
---->
to
the
2
25
chemical
Fe2O3(s)
+
equation
8
SO2(g)
How many moles of O2 are required for complete reaction of 4.0x103
moles of FeS2?
1.5x10-3 moles
1.1x10-2 moles
4.0x10-3 moles
None of the previous answers
10 4
FeS2(s)
+
11
O2(g)
------>
2
Fe2O3(s)
+
8
SO2(g)
For the reaction above, how many moles of Fe2O3 should be produced
from the reaction of 4.0x10-3 moles of FeS2 with an excess of oxygen?
8.0x10-3 moles
4.0x10-3 moles
2.0x10-3 moles
None of the previous answers.
11 4
FeS2(s)
+
11
O2(g)
----->
2
Fe2O3(s)
+
8
SO2(g)
For the reaction above, how many moles of SO2 should be produced
from the reaction of 4.0x10-3 moles of FeS2 with an excess of oxygen?
8.0x10-3 moles
4.0x10-3 moles
2.0x10-3 moles
None of the previous answers.
12 4
FeS2(s)
+
11
O2(g)
------>
2
Fe2O3(s)
+
8
SO2(g)
For the reaction above, how many moles of SO2 should be produced
from the reaction of 4.0x10-3 moles of FeS2 with 8.0x10-3 moles of
oxygen?
1.6x10-2 moles
4.0x10-3 moles
26
1.1x10-2 moles
5.8x10-3 moles
None of the previous answers.
13 4
FeS2(s)
+
11
O2(g)
-------->
2
Fe2O3(s)
+
8
SO2(g)
For the reaction above, how many moles of FeS2 with an excess of
oxygen were used to produce 3.0x10-2 moles of Fe2O3?
6.0x10-2 moles
3.0x10-2 moles
1.50x10-2 moles
None of the previous answers.
14 Questions
CH4(g)
14
+
2
17
O2(g)
refer
to
-------->
the
CO2(g)
chemical
equation
+
H2O(g)
2
How many grams of O2 are required for complete reaction of 1.6 grams
of methane (CH4)?
3.2 g
1.6 g
6.4 g
0.80 g
None of the previous answers.
15 CH4(g)
+
2
O2(g)
-------->
CO2(g)
+
2
H2O(g)
For the above reaction, how many grams of water should be produced
from the reaction of 1.6 grams of methane with an excess of oxygen?
3.6 g
3.2 g
1.6 g
27
0.90 g
None of the previous answers.
16 CH4(g)
+
2
O2(g)
------>
CO2(g)
+
2
H2O(g)
For the above reaction, how many grams of carbon dioxide should be
produced from the reaction of 1.6 grams of methane with an excess of
oxygen?
44 g
4.4 g
1.6 g
None of the previous answers.
17 CH4(g)
+
2
O2(g)
--------->
CO2(g)
+
2
H2O(g)
For the above reaction, how many grams of methane were burned in an
excess of oxygen to produce 7.2 grams of water?
13 g
3.6 g
3.2 g
None of the previous answers.
18 Questions
2
H2S(g)
18
+
22
SO2(g)
refer
to
the
-------->
3S(s)
chemical
+
equation
2
H2O(l)
How many grams of SO2 are required for complete reaction of 1.7
grams of hydrogen sulphide?
0.85 g
6.4 g
3.2 g
1.6 g
28
None of the previous answers
19.
2H2S(g)
+
SO2(g)
------->
3
S(s)
+
2
H2O(l)
How many grams of S are produced from the reaction of 1.7 grams of
hydrogen sulphide with an excess of sulphur dioxide?
2.4 g
1.8 g
1.1 g
2.6 g
None of the previous answers
20.
2H2S(g)
+
SO2(g)
-------->
3S(s)
+
2H2O(l)
How many grams of SO2 with an excess of hydrogen sulphide were
used to produce 0.75 grams of sulphur?
0.25 g
1.5 g
0.50 g
4.5 g
None of the previous answers
21.
2 H2S(g) + SO2(g) --------> 3 S(s) + 2 H2O(l)
How many grams of S are produced from the reaction of 1.7 grams of
hydrogen sulfide with 1.2 g of sulfur dioxide?
2.4 g
1.8 g
2.9 g
1.2 g
None of the previous answers.
29
22.
2 H2S(g) + SO2(g) = 3 S(s) + 2 H2O(l)
Assume the amounts indicated in problem 21 are used and 1.2 grams of
S are obtained. What is the percent yield of sulphur?
100%
50%
67%
41%
None of the previous answers.
23. Questions 23 - 26 refer to the chemical equation:
2 Al(s) + 6 HCl(aq) = 2 AlCl3(aq) + 3 H2(g)
How many grams of HCl are required for complete reaction of 5.4
grams of aluminium?
16 g
2.4 g
7.3 g
5.4 g
None of the previous answers
24.
2 Al(s) + 6 HCl(aq) = 2 AlCl3(aq) + 3 H2(g)
How many grams of hydrogen gas are produced from the reaction of 5.4
grams of aluminum with an excess of HCl?
0.60 g
0.30 g
8.1 g
0.27 g
None of the previous answers.
25.
2 Al(s) + 6 HCl(aq) --------> 2 AlCl3(aq) + 3 H2(g)
30
How many grams of hydrogen gas are produced from the reaction of 5.4
grams of aluminum with 25 g of HCl?
0.60 g
0.30 g
8.1 g
0.69 g
None of the previous answers.
26.
2 Al(s) + 6 HCl(aq) = 2 AlCl3(aq) + 3 H2(g)
Assuming the amounts of reactants given in number 25 are used and
0.50 g of hydrogen are obtained, what is the percent yield of hydrogen
gas?
100%
83%
72%
None of the previous answers.
27. Questions 27 and 28 refer to the chemical equation
CaCl2(aq) + 2 AgNO3(aq) = 2 AgCl(s) + Ca(NO3)2(aq)
How many grams of AgNO3 are required for complete reaction of 0.55
grams of CaCl2?
1.7 g
1.1 g
0.42 g
0.55 g
None of the previous answers.
28. CaCl2(aq) + 2 AgNO3(aq) = 2 AgCl(s) + Ca(NO3)2(aq)
A 2.0 grams mixture of sodium nitrate and silver chloride is reacted
31
with an excess of calcium chloride and 1.00 grams of silver chloride are
obtained. Determine the mass percent of silver nitrate in the original
mixture.
50%
59%
118%
30%
None of the previous answers.
29. The Haber process for the preparation of ammonia involves a
combination reaction between nitrogen and hydrogen. Write a balanced
molecular equation for the reaction and determine the hydrogen to
nitrogen mole ratio for the reaction.
1 mol H2/1 mol N2
1 mol H2/3 moles N2
3 moles H2/1 mol N2
3 moles H2/2 moles N2
None of the previous answers.
30. Gasoline is a mixture of many compounds including isomers of octane
(C8H18). The products of the complete combustion of gasoline are water
and carbon dioxide. Write a balanced reaction for the combustion of
octane and determine the optimum octane to oxygen mole ratio for the
reaction.
12 moles O2/1 mole C8H18
25 moles O2/2 moles C8H18
25 moles O2/1 mole C8H18
None of the previous answers.
32
31. The chemical equation for the reaction
----->
is best represented by:
6 H2(g) + 2 O2(g) ----> 4 H2O(g) + 2 H2(g)
2 H2(g) + O2(g) ----> 2 H2O(g)
H2(g) + O2(g) ----> H2O2(g)
N2(g) + 2 O2(g) ----> 2 NO2(g)
None of the previous answers.
32. For the reaction in number 31, what was the limiting reagent?
Hydrogen
Oxygen
Water
Neither, stoichiometric amounts were used.
None of the previous answers.
33. The
chemical
equation
for
is best represented by:
N2(g) + O2(g) = 2 NO(g)
H2(g) + O2(g) = H2O2(g)
33
the
reaction
2 H2(g) + O2(g) = 2 H2O(g)
H2(g) + Cl2(g) = 2 HCl(g)
None of the previous answers
34. For the reaction in number 33, what was the limiting reagent?
Hydrogen
Chlorine
hydrogen chloride
Neither, they were used in stoichiometric amounts.
None of the previous answers.
35. The
chemical
equation
for
the
reaction
is best represented by:
H2O2(g) = H2(g) + O2(g)
4 H2O2(g) + 2 H2O(g) = 6 H2O(g) + 2 O2(g)
2 H2O2(g) = 2 H2O(g) + O2(g)
2 H2O(g) = 2 H2(g) + O2(g)
None of the previous answers.
Balancing equations exercise:
Write balanced equations for each of the following reactions.
Then classify each by type of reaction.
1.) Fe + O2 -----> Fe2O3
34
2.) KClO3 -----> KCl + O2
3.) Ca(OH)2 + H2SO4 ----> HOH + CaSO4
4.) HgO -------> Hg + O2
5.) Cu + AgNO3 -------> Cu(NO3)2 + Ag
6.) C4H10 + O2 --------->
7.) C9H20 + O2 -------->
8.) CrCl3 + NaOH--------> Cr2O3 + HCl + NaCl
9.) NaClO3 -------> NaCl + O2
10.) C23H48 + O2 ------>
Unit Summary
All substances contain atoms, the simplest form of particles. Atoms have electrons, protons
and neutrons. Isotopes of the same element consist of the same proton number but different
mass number or number of neutrons. Electrons in atoms exist and are arranged in their
respective shells and orbitals. Removing an electron in a shell closest to the nucleus requires
more energy than removing an electron in an atom’s outermost shell. A mole is an amount of
substance and stoichiometric calculations make it possible for us to analyze chemicals
numerically using moles, mass, volume, and molality calculations to deepen our
understanding of quantitative chemistry.
35
Unit 2: Chemical bonding and Molecular Structure
What Are Chemical Bonds, and Why Do They Form?
A chemical bond is the result of an attraction between atoms or ions. The types of bonds that
a molecule contains will determine its physical properties, such as melting point, hardness,
electrical and thermal conductivity, and solubility. How do chemical bonds occur? As we
mentioned before, only the outermost, or valence, electrons of an atom are involved in
chemical bonds. Let’s begin our discussion by looking at the simplest element, hydrogen.
When two hydrogen atoms approach each other, electron-electron repulsion and protonproton repulsion both act to try to keep the atoms apart. However, proton-electron attraction
can counterbalance this, pulling the two hydrogen atoms together so that a bond is formed.
Look at the energy diagram below for the formation of an H–H bond.
As you’ll see throughout our discussion, atoms will often gain, lose, or share electrons in
order to possess the same number of electrons as the noble gas that’s nearest them on the
periodic table. All of the noble gases have eight valence electrons (s2p6) and are very
chemically stable, so this phenomenon is known as the octet rule. There are, however,
certain exceptions to the octet rule. One group of exceptions is atoms with fewer than eight
electrons—hydrogen (H) has just one electron. In BeH2, there are only four valence electrons
around Be: Beryllium contributes two electrons and each hydrogen contributes one. The
second exception to the octet rule is seen in elements in periods 4 and higher. Atoms of these
elements can be surrounded by more than four valence pairs in certain compounds.
36
Types of Chemical Bonds
You’ll need to be familiar with three types of chemical bonds for the exam: ionic bonds,
covalent bonds, and metallic bonds.
Ionic bonds are the result of an electrostatic attraction between ions that have opposite
charges; in other words, cations and anions. Ionic bonds usually form between metals and
nonmetals; elements that participate in ionic bonds are often from opposite ends of the
periodic table and have an electronegativity difference greater than 1.67. Ionic bonds are very
strong, so compounds that contain these types of bonds have high melting points and exist in
a solid state under standard conditions. Finally, remember that in an ionic bond, an electron is
actually transferred from the less electronegative atom to the more electronegative element.
One example of a molecule that contains an ionic bond is table salt, NaCl.
Covalent bonds form when electrons are shared between atoms rather than transferred from
one atom to another. However, this sharing rarely occurs equally because of course no two
atoms have the same electronegativity value. (The obvious exception is in a bond between
two atoms of the same element.) We say that covalent bonds are nonpolar if the
electronegativity difference between the two atoms involved falls between 0 and 0.4. We say
they are polar if the electronegativity difference falls between 0.4 and 1.67. In both nonpolar
and polar covalent bonds, the element with the higher electronegativity attracts the electron
pair more strongly. The two bonds in a molecule of carbon dioxide, CO2, are covalent bonds.
Covalent bonds can be single, double, or triple. If only one pair of electrons is shared,
a single bond is formed. This single bond is a sigma bond (s), in which the electron density
is concentrated along the line that represents the bond joining the two atoms.
However, double and triple bonds occur frequently (especially among carbon, nitrogen,
oxygen, phosphorus, and sulfur atoms) and come about when atoms can achieve a complete
octet by sharing more than one pair of electrons between them. If two electron pairs are
shared between the two atoms, a double bond forms, where one of the bonds is a sigma
bond, and the other is a pi bond (p). A pi bond is a bond in which the electron density is
concentrated above and below the line that represents the bond joining the two atoms. If three
electron pairs are shared between the two nuclei, a triple bond forms. In a triple bond, the
first bond to form is a single, sigma bond and the next two to form are both pi.
37
Multiple bonds increase electron density between two nuclei: they decrease nuclear repulsion
while enhancing the nucleus-to-electron density attractions. The nuclei move closer together,
which means that double bonds are shorter than single bonds and triple bonds are shortest of
all.
Metallic bonds exist only in metals, such as aluminum, gold, copper, and iron. In metals,
each atom is bonded to several other metal atoms, and their electrons are free to move
throughout the metal structure. This special situation is responsible for the unique properties
of metals, such as their high conductivity.
Drawing Lewis Structures
Here are some rules to follow when drawing Lewis structures—you should follow these
simple steps for every Lewis structure you draw, and soon enough you’ll find that you’ve
memorized them. While you will not specifically be asked to draw Lewis structures on the
test, you will be asked to predict molecular shapes, and in order to do this you need to be able
to draw the Lewis structure—so memorize these rules! To predict arrangement of atoms
within the molecule
-
Find the total number of valence electrons by adding up group numbers of the
elements. For anions, add the appropriate number of electrons, and for cations,
subtract the appropriate number of electrons. Divide by 2 to get the number of
electron pairs.
-
Determine which is the central atom—in situations where the central atom has a
group of other atoms bonded to it, the central atom is usually written first. For
example, in CCl4, the carbon atom is the central atom. You should also note that the
central atom is usually less electronegative than the ones that surround it, so you can
use this fact to determine which is the central atom in cases that seem more
ambiguous.
-
Place one pair of electrons between each pair of bonded atoms and subtract the
number of electrons used for each bond (2) from your total.
38
-
Place lone pairs about each terminal atom (except H, which can only have two
electrons) to satisfy the octet rule. Leftover pairs should be assigned to the central
atom. If the central atom is from the third or higher period, it can accommodate more
than four electron pairs since it has d orbitals in which to place them.
-
If the central atom is not yet surrounded by four electron pairs, convert one or more
terminal atom lone pairs to double bonds. Remember that not all elements form
double bonds: only C, N, O, P, and S!
Example
Which one of the following molecules contains a triple bond: PF3, NF3, C2H2, H2CO, or
HOF?
Explanation
The answer is C2H2, which is also known as ethyne. When drawing this structure, remember
the rules. Find the total number of valence electrons in the molecule by adding the group
numbers of its constituent atoms. So for C2H2, this would mean C = 4
2 (since there are
two carbons) = 8. Add to this the group number of H, which is 1, times 2 because there are
two hydrogens = a total of 10 valence electrons. Next, the carbons are clearly acting as the
central atoms since hydrogen can only have two electrons and thus can’t form more than one
bond. So your molecule looks like this: H—C—C—H. So far you’ve used up six electrons in
three bonds. Hydrogen can’t support any more electrons, though: both H’s have their
maximum number! So your first thought might be to add the remaining electrons to the
central carbons—but there is no way of spreading out the remaining four electrons to satisfy
the octets of both carbon atoms except to draw a triple bond between the two carbons.
For practice, try drawing the structures of the other four compounds listed.
Example
39
How many sigma (s) bonds and how many pi (p) bonds does the molecule ethene, C2H4,
contain?
Explanation
First draw the Lewis structure for this compound, and you’ll see that it contains one double
bond (between the two carbons) and four single bonds. Each single bond is a sigma bond, and
the double bond is made up of one sigma bond and one pi bond, so there are five sigma bonds
and one pi bond.
Exceptions to Regular Lewis Structures—Resonance Structures
Sometimes you’ll come across a structure that can’t be determined by following the Lewis
dot structure rules. For example, ozone (O3) contains two bonds of equal bond length, which
seems to indicate that there are an equal number of bonding pairs on each side of the central
O atom. But try drawing the Lewis structure for ozone, and this is what you get:
We have drawn the molecule with one double bond and one single bond, but since we know
that the bond lengths in the molecule are equal, ozone can’t have one double and one single
bond—the double bond would be much shorter than the single one. Think about it again,
though—we could also draw the structure as below, with the double bond on the other side:
Together, our two drawings of ozone are resonance structures for the molecule.
Resonance structures are two or more Lewis structures that describe a molecule: their
composite represents a true structure for the molecule. We use the double-directional arrows
to indicate resonance and also bracket the structures or simply draw a single, composite.
40
Let’s look at another example of resonance, in the carbonate ion CO32-
:
Notice that resonance structures differ only in electron pair positions, not atom positions!
Example
Draw the Lewis structures for the following molecules: HF, N2, NH3, CH4, CF4, and NO+.
Explanation
41
Unit Summary
A chemical bond is the result of an attraction between atoms or ions. The types of bonds that
a molecule contains will determine its physical properties, such as melting point, hardness,
electrical and thermal conductivity, and solubility. Cations are positively charged ions, anions
are negatively charged ions. Lewis dot structures give an illustration bonding and allow us to
view the electrons in more detail as they surround the nucleus. The electrons that participate
in chemical bonding are the valence electrons (the electrons in the outermost shell).
42
Unit 3: States of matter
Now that you know a bit about chemical bonding, let’s talk about the different forms that
groups of molecules can take. In other words, let’s talk about the states of matter. The states
of matter that you’ll need to know for the Chemistry tests are solid, liquid, and gas.
You might wonder why there are different states of matter at all. After all, molecules only
bond together in one way, right? The answer lies in the type of intramolecular and
intermolecular forces that exist both within and between molecules of substances.
Solids
As we mentioned above, the molecules that make up solids are generally held together by
ionic or strong covalent bonding, and the attractive forces between the atoms, ions, or
molecules in solids are very strong. In fact, these forces are so strong that particles in a solid
are held in fixed positions and have very little freedom of movement. Solids have definite
shapes and definite volumes and are not compressible to any extent. There are a few types of
solids that you should be familiar with, and we’ve listed them below. However, we will start
by saying that there are two main categories of solids—crystalline solids and amorphous
solids. Crystalline solids are those in which the atoms, ions, or molecules that make up the
solid exist in a regular, well-defined arrangement. The smallest repeating pattern of
crystalline solids is known as the unit cell, and unit cells are like bricks in a wall—they are
all identical and repeating. The other main type of solids is called the amorphous
solids. Amorphous solids do not have much order in their structures. Though their molecules
are close together and have little freedom to move, they are not arranged in a regular order as
are those in crystalline solids. Common examples of this type of solid are glass and plastics.
43
There are four types of crystalline solids, all of which you should be familiar with for this
course:
Ionic solids—Made up of positive and negative ions and held together by electrostatic
attractions. They’re characterized by very high melting points and brittleness and are poor
conductors in the solid state. An example of an ionic solid is table salt, NaCl.
Molecular solids—Made up of atoms or molecules held together by London dispersion forces,
dipole-dipole forces, or hydrogen bonds. Characterized by low melting points and flexibility
and are poor conductors. An example of a molecular solid is sucrose.
Covalent-network (also called atomic) solids—Made up of atoms connected by covalent
bonds; the intermolecular forces are covalent bonds as well. Characterized as being very hard
with very high melting points and being poor conductors. Examples of this type of solid are
diamond and graphite, and the fullerenes. As you can see below, graphite has only 2-D
hexagonal structure and therefore is not hard like diamond. The sheets of graphite are held
together by only weak London forces!
44
Metallic solids—Made up of metal atoms that are held together by metallic bonds.
Characterized by high melting points, can range from soft and malleable to very hard, and are
good conductors of electricity.
Liquids
Liquids are generally made up of molecules that contain covalent bonds and have strong
intermolecular attractive forces. The atoms and molecules that make up liquids have more
freedom of movement than do those in solids. Also, liquids have no definite shape but do
have a definite volume, and they are not easily compressible. We will discuss liquids in more
detail in the section on solutions in this chapter.
Gases
Gases generally consist of atoms and molecules that are covalently bonded, and their
intermolecular forces are very weak. The molecules of a gas are highly separated, so we say
that gases are mostly empty space. A gas has no definite shape—it will take the shape of the
container that holds it, and gases are easily compressible.
Phase Changes
In order for a substance to move between the states of matter; for example, to turn from a
solid into a liquid, which is called fusion, or from a gas to a liquid (vaporization), energy
must be gained or lost. The heat of fusion (symbolized Hfus) of a substance is the amount of
energy that must be put into the substance for it to melt. For example, the heat of fusion of
water is 6.01 kJ/mol, or in other terms, 80 cal/g. The heat of vaporization, not surprisingly,
is the amount of energy needed to cause the transition from liquid to gas, and it is
symbolized Hvap. You will not be required to memorize heat of fusion or vaporization values
for the exam.
Changes in the states of matter are often shown on phase diagrams. Let’s start with the phase
diagram for water. The phase diagram for water is a graph of pressure versus temperature.
Each of the lines on the graph represents an equilibrium position, at which the substance is
present in two states at once. For example, anywhere along the line that separates ice and
water, melting and freezing are occurring simultaneously.
45
The intersection of all three lines is known as the triple point (represented by a dot and
a T on the figure). At this point, all three phases of matter are in equilibrium with each other.
Point X represents the critical point, and at the critical point and beyond, the substance is
forever in the vapor phase.
This diagram allows us to explain strange phenomena, such as why water boils at a lower
temperature at higher altitudes, for example. At higher altitudes, the air pressure is lower, and
this means that water can reach the boiling point at a lower temperature. Interestingly enough,
water would boil at room temperature if the pressure was low enough!
One final note: If we put a liquid into a closed container, the evaporation of the liquid will
cause an initial increase in the total pressure of the system, and then the pressure of the
system will become a constant. The value of this final pressure is unique to each liquid and is
known as the liquid’s vapor pressure. Water has a relatively low vapor pressure because it
takes a lot of energy to break the hydrogen bonds so that molecules enter the gas phase.
Water and other liquids that have low vapor pressures are said to be nonvolatile. Substances
like rubbing alcohol and gasoline, which have relatively high vapor pressures, are said to
be volatile.
Example
What happens to water when the pressure remains constant at 1 atm but the temperature
changes from -10ºC to 75ºC?
46
Explanation
Looking at the phase change diagram for water and following the dashed line at 1 atm, you
can see that water would begin as a solid (ice) and melt at 0ºC. All of the water would be in
liquid form by the time the temperature reached 75ºC.
The second type of phase change graph you might see is called a heating curve. This is a
graph of the change in temperature of a substance as energy is added in the form of heat. The
pressure of the system is assumed to be held constant, at normal pressure (1 atm). As you can
see from the graph below, at normal pressure water freezes at 0ºC and boils at 100ºC.
The plateaus on this diagram represent the points where water is being converted from one
phase to another; at these stages the temperature remains constant since all the heat energy
added is being used to break the attractions between the water molecules.
Specific Heat
In our lectures you will see a diagram that looks something like this one, and you might come
across a question that asks you to calculate the amount of energy needed to take a particular
47
substance through a phase change. This would be one of the most difficult questions on the
exam, but you might see something like it, or at least part of it. If you were asked to do this,
you would need to use the following equation:
Energy (in calories) = mCp DT
-where m = the mass of the substance (in grams)
-Cp = the specific heat of the substance (in cal/g ºC)
-DT = the change in temperature of the substance (in either Kelvins or ºC, but make sure all
your units are compatible!)
As you can see, this requires that you know the specific heat of the substance. A
substance’s specific heat refers to the heat required to raise the temperature of 1 g of a
substance by 1ºC. You will not be required to remember any specific heat values for the
exam.
Work through the example below to get a feel for how to use this equation.
Example
If you had a 10.0 g piece of ice at -10ºC, under constant pressure of 1 atm, how much energy
would be needed to melt this ice and raise the temperature to 25.0ºC?
Explanation
First, the temperature of the ice would need to be raised from -10ºC to 0ºC. This would
require the following calculation. The specific heat for ice is 0.485 cal/g ºC. Substituting in
the formula
energy = mCp DT; energy = (10.0 g) (0.485 cal/g ºC) (10.0ºC) = 48.5 cal
So 48.5 calories are needed to raise temperature.
Next, we must calculate the heat of fusion of this ice: we must determine how much energy is
needed to completely melt the 10 g of it.
energy = mH fus
energy = (10.0 g) (80 cal/g) = 800 cal
48
So 800 cal of energy are needed to completely melt this sample of ice.
Next, we need to see how much energy would be needed to raise the temperature of water
from 0ºC to 25ºC. The specific heat for liquid water is 1.00 cal/g ºC. So again use
energy = mCp DT to get energy = (10.0 g) (1.00 cal/g ºC) (25.0ºC) = 250 cal
Finally, add together all of the energies to get the total: 48.5 + 800 + 250 = about 1100
calories are needed to convert the ice to water at these given temperatures.
Exercise
-Define and show the structural differences of solids, liquids and gases.
-What is a phase diagram?
-What is Specific Heat?
-Highlight the difference between a triple point and critical point.
-Give an account of the different forms of solids and their characteristics.
Unit Summary
The intramolecular and intermolecular forces that exist both within and between molecules of
substances dictate whether they will be solid, liquid or gas.
49
Unit 4. Chemical energetics
Definitions
-Standard enthalpy changes of reaction
Standard state -- Pressure 101 kPa, temperature 298 K (or 1 atm, 25 degrees celsius). The
standard state of an element or compound is the form in which it exists under standard
conditions (not to be confused with STP)
Standard enthalpy change of formation -- The enthalpy change when 1 mol of a substance is
made from its elements in their standard states
Example:
C(graphite) + 2H2(g) -> CH4(g)
Hess' Law
The energy difference between two states is independent of the route between them.
This allows calculation of energy changes from other data by the manipulation of chemical
equations. Equations may be treated using the four rules of number (doing the same operation
on the energy value).
Example: Calculate the enthalpy of formation of methane.
Data:

ΔH combustion (carbon) = -394kj

ΔH combustion (hydrogen) =-242kj

ΔHcombustion (methane) =-891kj
ΔHf methane is represented by the equation:
C(s) + H2(g)
CH4 (g)
This equation can be constructed using the equations for combustion of the reactants (carbon
and hydrogen) and the product (methane)
50
C(s) + O2(g)
-
CO2(g)
H2(g) + 1/2O2(g)
394kj
-
H2O(l)
242kj
enthalpy 1
enthalpy 2
multiply this equation by 2 to get
2H2(g) + O2(g)
-
2H2O(l)
484kj
enthalpy 3
Add the first and third equation together to get:
C(s) + 2O2(g) + 2H2(g)
CO2(g) + 2H2O(l)
-
enthalpy (1 +
878kj 3) = 4
Now take away the equation for the combustion of methane
CH4 (g) + 2O2(g)
CO2(g) + 2H2O(l)
891kj
enthalpy 5
And after rearrangement ( take the CH4 to the right hand
side) the result is the equation for the formation of methane
C(s) + H2(g)
CH4 (g)
+13kj
enthalpy ( 4 5)
Born Haber cycles are the application of Hess' law to ionic systems. An ionic solid consists of
a giant structure of ions held together in a giant lattice.
Application of Hess law tells us that the enthalpy of formation of an ionic crystal is equal to
the sum of the energies of formation of the ions plus the enthalpy of the lattice. It is a several
step process that is best represented by a diagram showing the individual steps as
endothermic upwards and exothermic downwards.
51
Entropy
Chemistry is concerned with the statistical likelihood of a process occurring. This is related to
the number of possible states which the particles can adopt. This can be regarded as the
degree of disorder or Entropy of the system. Factors which increase disorder in a system are:

Increased number of particles (when more gas particles are produced, this is far more
important than the other factors)

Mixing of particles

Change of state to greater distance between particles (solid->liquid or liquid->gas)

Increased particle movement (temperature)
Δ-S is positive when entropy increases (more disorder) and negative when entropy decreases
(less disorder).
Calculations can be carried out using absolute entropy values in the same way as enthalpy
values in Hess' law. The change in entropy form one state to another will always be the same
regardless of the route taken.
The standard entropy change can be calculated by subtracting the absolute entropy of the
reactants from that of the products.
ΔS (products) - ΔS (reactants) = standard entropy change for a reaction
Spontaneity of a reaction- Gibbs Free Energy
Reactions which release heat (and so increase stability) tend to occur. Reactions which
increase entropy (ΔS is positive) tend to occur, but neither can be used to accurately predict
spontaneity alone.
Gibbs free energy (G) is defined as a measure of the total entropy of the universe. Hence the
change in Gibbs free energy (ΔG) is the change in the total entropy of the universe. The total
entropy of the universe must increase for any process to occur.
When heat is released in a reaction (exothermic change) this energy heats up the universe and
effectively increases its entropy (there are a greater number of possible energy states that the
particles in the universe can adopt).
52
The total entropy of the universe must increase and consequently exothermic reactions are
favourable.
If the entropy of a reaction mixture increases then this is also favourable as the total entropy
of the universe also increases.
Gibbs free energy change = ΔH - TΔS
If Gibbs free energy change is negative (convention) then the total entropy of the universe
increases and the reaction is spontaneous. Why is the sign negative?
When ΔG is negative, the reaction is spontaneous, when it's positive, the reaction is not.
Gibbs free energy calculations
Enthalpy changes can be calculated indirectly by summing the enthalpy values of related
equations using Hess' law. Entropy changes can be calculated in the same way. It follows
then that Gibbs free energy changes can be calculated from a knowledge of Gibbs free energy
values in related equations.
Spontaneity of reaction
Determined by the relationship
ΔG = ΔH - Temperature(in kelvin) x ΔS
Enthalpy
Entropy
change
change
positive
positive
Gibbs free energy
Spontaneity
depends on T, may be + yes, if the temperature is high
or -
enough
53
negative
positive
negative
negative
positive
negative
always negative
always spontaneous
depends on T, may be + yes, if the temperature is low
or -
enough
always positive
never spontaneous
Side Notes:
Electrostatic attraction
This is the most important force in chemistry. It depends on two factors:
1. The size of the charged particles (eg ionic radius)
2. The magnitude of the charge Z1 and Z2
The electrostatic attractive force may be considered proportional to the product of the charge
magnitudes divided by the distance between them. Smaller charged particles can approach
closer and therefore exert a greater force of attraction.
Examples
NaCl - the charges on the ions are single plus and single minus respectively
MgCl2 - in this case the magnesium has a double positive charge and consequently exerts a
much larger force than the sodium ion in sodium chloride. The MgCl2 lattice is much stronger
than NaCl- higher mp and greater lattice enthalpy
54
LiCl - In this case the Li+ ion has the same charge as the Na+ ion but it is much smaller and
can get closer to the chloride ion exerting a greater force. The lattice is stronger than NaCl
and has a higher mp
Gibbs free energy equation
Why is the sign negative?
This is because of the convention adopted for enthalpy. If the enthalpy is negative it's an
exothermic process and the entropy of the universe increases as explained above. The second
term in the equation is a negative term (-TΔS) and if the entropy increases it makes Gibbs
free energy more negative.
Exercise
1.
The table below contains some mean bond enthalpy data.
Bond
C=C
Mean bond enthalpy / kJ mol–1
(a)
436
348
612
944
388
Explain the term mean bond enthalpy.
...............................................................................................................................
...............................................................................................................................
...............................................................................................................................
...............................................................................................................................
(2)
(b)
(i)
Write an equation for the formation of one mole of ammonia, NH3, from its
elements.
.....................................................................................................................
(ii)
Use data from the table above to calculate a value for the enthalpy of
55
formation of ammonia.
.......................................................................................................................
.......................................................................................................................
.......................................................................................................................
.......................................................................................................................
(4)
(c)
Use the following equation and data from the table above to calculate a value for
the
C–H bond enthalpy in ethane.
H
H
C
C + H
H
H
H
H
H
H
C
C
H
H
H
H = –136 kJ mol–1
...............................................................................................................................
...............................................................................................................................
...............................................................................................................................
...............................................................................................................................
(3)
(Total 9 marks)
56
2.
(a)
The table below contains some mean bond enthalpy data.
Bond
H–O
O–O
O=O
Mean bond enthalpy/kJ mol–1
463
146
496
The bonding in hydrogen peroxide, H2O2, can be represented by H–O–O–H. Use
these data to calculate the enthalpy change for the following reaction.
H2O2(g)
2O(g)
+
1
2
O2(g)
.....................…………………………………………………………………………
.
.....................…………………………………………………………………………
.
.....................…………………………………………………………………………
.
.....................…………………………………………………………………………
.
(3)
(b)
The standard enthalpy of
Hf
for methane, is –74.9 kJ mol–1. Write
an equation, including state symbols, for the reaction to which this enthalpy
change applies.
.....................…………………………………………………………………………
.
(2)
(c)
The enthalpy changes for the formation of atomic hydrogen and atomic carbon
from their respective elements in their standard states are as follows.
1
2
H = +218 kJ mol–1
H2
H = +715 kJ mol–1
(i)
By reference to its structure, suggest why a large amount of heat energy is
57
required to produce free carbon atoms from solid carbon.
...........………………………………………………………………………
….
...........………………………………………………………………………
….
58
(ii)
Parts (b) and (c) give enthalpy data for the formation of CH4(g), H(g) and
C(g).
Use these data and Hess’s Law to calculate the value of the enthalpy change
for the following reaction.
CH4(g)
...........………………………………………………………………………
….
...........………………………………………………………………………
….
...........………………………………………………………………………
….
...........………………………………………………………………………
….
...........………………………………………………………………………
….
(iii)
Use your answer from part (c)(ii) to calculate a value for the mean bond
enthalpy of a C–H bond in methane.
...........………………………………………………………………………
….
Unit Summary
Understand Hess’s Law which posits that the energy difference between two states is
independent of the route between them. Entropy is disorder in a system; enthalpy and Gibb’s
free energy are all part of chemical energetics. Let us not forget how to interpret data tables!!
59
Unit 5: Electrochemistry and Redox Reactions
Electrochemistry: The study of the interchange of chemical and electrical energy.
Oxidation-reduction (redox) reactions are another important type of reaction that you will see
throughout this course. You will be expected to be able to identify elements that are oxidized
and reduced, know their oxidation numbers, identify half-cells, and balance redox reactions.
The following is a brief overview of the basics.
Oxidation-Reduction Reactions
Oxidation-reduction reactions involve the transfer of electrons between substances. They take
place simultaneously, which makes sense because if one substance loses electrons, another
must gain them. Many of the reactions we’ve encountered thus far fall into this category. For
example, all single-replacement reactions are redox reactions. Before we go on, let’s review
some important terms you’ll need to be familiar with.
Oxidation: The loss of electrons. Since electrons are negative, this will appear as an increase
in the charge (e.g., Zn loses two electrons; its charge goes from 0 to +2). Metals are oxidized.
Oxidizing agent (OA): The species that is reduced and thus causes oxidation.
Reduction: The gain of electrons. When an element gains electrons, the charge on the element
appears to decrease, so we say it has a reduction of charge (e.g., Cl gains one electron and
goes from an oxidation number of 0 to -1). Non-metals are reduced.
Reducing agent (RA): The species that is oxidized and thus causes reduction.
Oxidation number: The assigned charge on an atom. You’ve been using these numbers to
balance formulas.
Half-reaction: An equation that shows either oxidation or reduction alone.
Rules for Assigning Oxidation States
60
A reaction is considered a redox reaction if the oxidation numbers of the elements in the
reaction change in the course of the reaction. We can determine which elements undergo a
change in oxidation state by keeping track of the oxidation numbers as the reaction
progresses. You can use the following rules to assign oxidation states to the components of
oxidation-reduction reactions:
-
The oxidation state of an element is zero, including all elemental forms of the
elements (e.g., N2, P4, S8, O3).
-
The oxidation state of a monatomic ion is the same as its charge.
-
In compounds, fluorine is always assigned an oxidation state of -1.
-
Oxygen is usually assigned an oxidation state of -2 in its covalent compounds.
Exceptions to this rule include peroxides (compounds containing the
group), where
each oxygen is assigned an oxidation state of -1, as in hydrogen peroxide (H2O2).
-
Hydrogen is assigned an oxidation state of +1. Metal hydrides are an exception: in
metal hydrides, H has an oxidation state of -1.
-
The sum of the oxidation states must be zero for an electrically neutral compound.
-
For a polyatomic ion, the sum of the oxidation states must equal the charge of the ion.
Now try applying these rules to a problem.
Exercise
Assign oxidation numbers to each element in the following:
1.
H2S
61
2.
MgF2
3.
Explanation
1.
The sum of the oxidation numbers in this compound must be zero since the compound
has no net charge. H has an oxidation state of +1, and since there are two H atoms, +1 times 2
atoms = +2 total charge on H. The sulfur S must have a charge of -2 since there is only one
atom of sulfur, and +2 - 2 = 0, which equals no charge.
2.
F is assigned an oxidation state of -1 (according to rule 3), and there are two atoms of
F, so this gives F a total charge of -2. Mg must have a +2 oxidation state since +2 - 2 = 0 and
the compound is electrically neutral.
3.
This time the net charge is equal to -3 (the charge of the polyatomic ion—according
to rule 7). Oxygen is assigned a -2 oxidation state (rule 4). Multiply the oxidation number by
its subscript: -2
4 = -8. Since there is only 1 phosphorus, just use those algebra skills: P +
-8 = -3. Phosphorus must have a +5 charge.
Exercise
When powdered zinc metal is mixed with iodine crystals and a drop of water is added, the
resulting reaction produces a great deal of energy. The mixture bursts into flames, and a
purple smoke made up of I2 vapor is produced from the excess iodine. The equation for the
reaction is
Zn (s) + I2(s)
ZnI 2(s) + energy
Identify the elements that are oxidized and reduced, and determine the oxidizing and reducing
agents.
Explanation
1.
Assign oxidation numbers to each species. Zn and I2 are both assigned values of 0
(rule 1). For zinc iodide, I has an oxidation number of -1 (group 7A—most common charge),
which means that for zinc, the oxidation number is +2.
62
2.
Evaluate the changes that are taking place. Zn goes from 0 to +2 (electrons are lost
and Zn is oxidized). The half-reaction would look like this:
Zn 0
Zn 2+ + 2e -
And I2 goes from 0 to -1 (it gains electrons and so is reduced). This half-reaction would look
like this:
Here, zinc metal is the reducing agent—it causes the reduction to take place by donating
electrons—while iodine solid is the oxidizing agent; iodine solid accepts electrons.
Voltaic (or Galvanic) Cells
Redox reactions release energy, and this energy can be used to do work if the reactions take
place in a voltaic cell. In a voltaic cell (sometimes called a galvanic cell), the transfer of
electrons occurs through an external pathway instead of directly between the two elements.
The figure below shows a typical voltaic cell (this one contains the redox reaction between
zinc and copper):
As you can see, the anode is the electrode at which oxidation occurs; you can remember this
if you remember the phrase “an ox”—“oxidation occurs at the anode.” Reduction takes place
at the cathode, and you can remember this with the phrase “red cat”—“reduction occurs at
the cathode.” An important component of the voltaic cell is the salt bridge, which is a device
used to maintain electrical neutrality; it may be filled with agar, which contains a neutral salt,
63
or be replaced with a porous cup. Remember that electron flow always occurs from anode to
cathode, through the wire that connects the two half-cells, and a voltmeter is used to measure
the cell potential in volts.
Batteries are cells that are connected in series; the potentials add to give a total voltage. One
common example is the lead storage battery (car battery), which has a Pb anode, a
PbO2 cathode, and H2SO4 electrolyte is their salt bridge.
Standard Reduction Potentials
The potential of a voltaic cell as a whole will depend on the half-cells that are involved. Each
half-cell has a known potential, called its standard reduction potential (Eº). The cell
potential is a measure of the difference between the two electrode potentials, and the potential
at each electrode is calculated as the potential for reduction at the electrode. That’s why
they’re standard reduction potentials, not standard oxidation potentials. Here is the chart:
64
On this reduction potential chart, the elements that have the most positive reduction potentials
are easily reduced and would be good oxidizing agents (in general, the non-metals), while the
elements that have the least positive reduction potentials are easily oxidized and would be
good reducing agents (in general, metals). Let’s try a quick problem.
Exercise
Which of the following elements would be most easily oxidized: Ca, Cu, Fe, Li, or Au?
Explanation
65
Use the reduction potential chart: nonmetals are at the top and are most easily reduced.
Metals are at the bottom and are most easily oxidized. Lithium is at the bottom of the chart—
it’s the most easily oxidized of all. So the order, from most easily oxidized to least easily
oxidized, is Au, Fe, Cu, Ca, Li.
Exercise
Which one of the following would be the best oxidizing agent: Ba, Na, Cl, F, or Br?
Explanation
Using the reduction potential chart and the fact that oxidizing agents are the elements that are
most easily reduced, we determine fluorine is the best oxidizing agent.
Electrolytic Cells
While voltaic cells harness the energy from redox reactions, electrolytic cells can be used to
drive nonspontaneous redox reactions, which are also called electrolysis reactions.
Electrolytic cells are used to produce pure forms of an element; for example, they’re used to
separate ores, in electroplating metals (such as applying gold to a less expensive metal), and
to charge batteries (such as car batteries). These types of cells rely on a battery or any DC
source—in other words, whereas the voltaic cell is a battery, the electrolytic cell needs a
battery. Also unlike voltaic cells, which are made up of two containers, electrolytic cells have
just one container. However, like in voltaic cells, in electrolytic cells electrons still flow from
the anode to the cathode. An electrolytic cell is shown below.
66
Unit Summary
Oxidation-reduction reactions involve the transfer of electrons between substances.
Electrochemistry is the study of the interchange of chemical and electrical energy.
67
Unit 6: Equilibria
Dynamic Equilibrium
In all reactions, there are in fact two processes occurring, a forward reaction where the
reactants produce the products, and a reverse reaction where the products react to form the
reactants.
In some reactions, this reverse reaction is insignificant, but in others there comes a point
where the rate of the two reactions is exactly equal and consequently the reactants and
products remain in equal proportions, though both are continually being used up and
produced at the same time.
68
The position of equilibrium
The equilibrium constant
Kc is a constant which represents how far the reaction will proceed at a given temperature.
When Kc is greater than 1, products exceed reactants (at equilibrium). When much greater
than 1, the reaction goes almost to completion. When Kc is less than 1, reactants exceed
products. When much less than 1 (Kc can never be negative...so when it is close to zero) the
reaction hardly occurs at all.
The only thing which can change the value of Kc for a given reaction is a change in
temperature. The position of equilibrium, however, can change without a change in the value
of Kc.
Effect of Temperature
The effect of a change of temperature on a reaction will depend on whether the reaction is
exothermic or endothermic. When the temperature increases, Le Chatelier's principle says the
reaction will proceed in such a way as to counteract this change, ie lower the temperature.
Therefore, endothermic reactions will move forward, and exothermic reactions will move
backwards (thus becoming endothermic). The reverse is true for a lowering of temperature.
Effect of Concentration
When the concentration of a product is increased, the reaction proceeds in reverse to decrease
the concentration of the products. When the concentration of a reactant is increased, the
reaction proceeds forward to decrease the concentration of reactants.
Effect of Pressure
In reactions where gases are produced (or there are more mols of gas on the left), and
increase in pressure will force the reaction to move to the left (in reverse). If pressure is
decreased, the reaction will proceed forward to increase pressure. If there are more mols of
gas on the left of the equation, this is all reversed.
69
Effect of catalysts on equilibrium
A catalyst does not have effects either Kc or the position of equilibrium, it only effects the
rate of reaction. As the rate of forward reaction and reverse reaction is affected equally then
the equilibrium cannot be affected.
The Haber process
N2(g) + 3H2(g)
2NH3(g)
Δ-H = -92.4 kJ mol-1
There are more moles of gas on the left than the right, so a greater yield will be produced at
high pressure. (The equilibrium position will lie further to the right)
The reaction is exothermic, therefore it will give a greater yield at low temperatures. (The
equilibrium position lies further to the right)
In practice, if low temperatures are used the time taken for the reaction to attain equilibrium
becomes unfeasably long. An intermediate temperature is chosen (450ºC) which allows the
reaction to get to an equilibrium in a reasonable time and still has enough of the products in
the equilibrium mixture.
A catalyst of finely divided iron is also used to help speed the reaction (finely divided to
maximise the surface area).
70
To make the process more efficient the ammonia produced at equilibrium is removed by first
cooling the mixture when the ammonia turns into a liquid which can be tapped off. The
unreacted gases in the process are then mixed with fresh reactants and returned to the reaction
chamber to re-establish the equilibrium again and the cycle is repeated continuously.
Unit Summary
The equilibrium constant Kc represents how far the reaction will proceed at a given
temperature. When Kc is greater than 1, products exceed reactants (at equilibrium). When
much greater than 1, the reaction goes almost to completion. When Kc is less than 1,
reactants exceed products. When much less than 1 (Kc can never be negative...so when it is
close to zero) the reaction hardly occurs at all.
71
Unit 7: Reaction kinetics
Rates of reaction
Rate of reaction is concerned with how quickly a reaction reaches a certain point. It can be
defined as the decrease in concentration of the reactants per unit time or the increase in
concentration of the products per unit time.
A graph may be plotted of concentration against time, with time on the x-axis and some
measure of how far the reaction has gone (ie concentration, volume, mass loss etc) on the yaxis. This will produce a curve and the rate at any given point is the gradient of the tangent to
this curve.
Collision Theory
Collision theory -- reactions take place as a result of particles (atoms or molecules) colliding
and then undergoing a reaction. Not all collisions cause reaction, however, even in a system
where the reaction is spontaneous. The particles must have sufficient kinetic energy, and the
correct orientation with respect to each other for them to react.
Activation energy
This is the minimum energy that particles colliding must have in order to produce successful
reaction. It is given the symbol Ea (Energy of Activation). The energy of particles is
expressed by their speed.
72
Changing the conditions
Increasing the temperature of a substance increases the average speed (Energy) of the
particles and consequently the number of particles colliding with sufficient energy (Ea) to
react. At higher temperatures there are more successful collisions and therefore a faster
reaction.
At higher concentrations there are more collisions and consequently a faster reaction.
Catalysts lower the activation energy by providing an alternative mechanism for the reaction/
greater probability of proper orientation. This results in a faster reaction.
In hetrogeneous reactions (where the reactants are in different states) the size of the particles
of a solid may change reaction rate, since the surface is where the reaction takes place, and
the surface area is increased when the particles are more finely divided (therefore smaller
solid particles in a heterogeneous reaction tend to produce a faster reaction).
Reaction Mechanisms
Most reactions involve several steps, which can be individually slow of fast, and which, all
together, make up the complete reaction. The slowest of these steps is called the rate
determining step, as is determines how fast the reaction will go. It is also not necessary that
all the reactants are involved in ever step, and so the rate determining step may not involve all
the reactants. As a result, increasing the concentration (for example) of a reactant which is
not involved in the rate determining step will not change the overall reaction rate.
Exercise
Exercise: To answer the questions external research may be needed so feel free to consult text
books and academic sources concerning rates of reactions.
1.
Explain what is meant by the term "rate of reaction".
2. The initial rates of the reaction 2A + B  2C + D at various concentrations of A and B are
given below:
73
[A] moldm-3
[B] moldm-3
Initial rate /moldm-3s-1
0.01
0.20
0.10
0.02
0.20
0.20
0.01
0.40
0.40
a) What is the order of reaction with respect to A and B?
b) What is the overall order of reaction?
c) What is the rate constant?
d) What will be the rate of the reaction if the concentrations of A and B are both 0.01
moldm-3?
3. For the reaction 2NO(g) + H2(g)  N2O(g) + H2O(g), the following rate data were
collected:
Initial [NO]/M
Initial [H2]/M
Initial rate/Ms-1
0.60
0.37
3.0 x 10-3
1.20
0.37
1.2 x 10-2
1.20
0.74
1.2 x 10-2
What is the rate constant for the reaction?
What can you deduce about the rate-determining step of the reaction?
74
4. For the reaction PCl3 + Cl2  PCl5, the following data were obtained:
Experiment No.
[PCl3]/M
[Cl2]/M
Rate /Ms-1
1
0.36
1.26
6.0 x 10-4
2
0.36
0.63
1.5 x 10-4
3
0.72
2.52
4.8 x 10-3
Deduce the rate equation and the rate constant.
5. Two compounds, X and Y, are known to undergo the reaction
X + 3Y  XY3
Using the experimental results in the table below:
EXPERIMENT
Initial concentration Initial concentration Initial
of X/moldm-3
of Y/moldm-3
rate
formation
of
of
-3 -1
XY3/moldm s
1
0.100
0.100
0.00200
2
0.100
0.200
0.00798
3
0.100
0.300
0.01805
4
0.200
0.100
0.00399
5
0.300
0.100
0.00601
Find the rate constant.
6. The data in the table below relates to the reaction between hydrogen and nitrogen
monoxide at 673K. 2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
Experiment number
Initial concentration Initial concentration Initial
of H2 /moldm-3
of NO /moldm-3
rate
production of N2 /
moldm-3s-1
1
2.0 x 10-3
6.0 x 10-3
6.0 x 10-3
2
3.0 x 10-3
6.0 x 10-3
9.0 x 10-3
75
of
3
6.0 x 10-3
1.0 x 10-3
0.5 x 10-3
Deduce the rate equation and calculate the rate constant.
Unit Summary
Condition
Effect on rate
Temperature
increasing
Explanation
the Two
temperature
reasons:
increases 1. There are more particles with sufficient energy
the rate of a reaction
to react (most important) - more successful
collisions
2. There are more collisions
Concentration
Increasing
the There are more collisions as there are more
concentration
of
a particles in closer proximity
reactant increases the
rate of the reaction
(usually)
Particle size
The
smaller
the Collisions occur at the surface of particles. The
particles the faster the larger the particle size the smaller the surface area
reaction.
solute
(note:
the and the fewer collisions can occur.
particles
solutions
have
in
the
smallest particle size
possible.
and
so
solutions react fastest)
Catalysts
The
presence
of
a Catalysts provide an alternative mechanism with a
catalyst increases the lower activation energy
rate of a reaction
76
Unit 8: Inorganic chemistry
Metals, Non-metals and Metalloids
Characteristic properties of metallic and non-metallic elements:
Metallic Elements
Non-metallic elements
Distinguishing luster (shine)
Non-lustrous, various colours
Malleable and ductile (flexible) as solids
Brittle, hard or soft
Conduct heat and electricity
Poor conductors
Metallic oxides are basic, ionic
Nonmetallic oxides are acidic, compounds
Cations in aqueous solution
Anions, oxyanions in aqueous solution
Metals

Most metals are malleable (can be pounded into thin sheets; a sugar cube chunk of
gold can be pounded into a thin sheet which will cover a football field), and ductile
(can be drawn out into a thin wire).

All are solids at room temp (except Mercury, which is a liquid)

Metals tend to have low ionization energies, and typically lose electrons (i.e.
are oxidized) when they undergo chemical reactions
o
Alkali metals are always 1+ (lose the electron in s subshell)
o
Alkaline earth metals are always 2+ (lose both electrons in s subshell)
o
Transition metal ions do not follow an obvious pattern, 2+ is common, and
1+ and 3+ are also observed

Compounds of metals with non-metals tend to be ionic in nature

Most metal oxides are basic oxides; those that dissolve in water react to form metal
hydroxides:
Metal oxide + water -> metal hydroxide
Na2O(s) + H2O(l) -> 2NaOH(aq)
CaO(s) + H2O(l) -> Ca(OH)2(aq)
77

Metal oxides exhibit their basic chemical
nature by reacting with acids to
form salts and water:
Metal oxide + acid -> salt + water
MgO(s) + HCl(aq) -> MgCl2(aq) + H2O(l)
NiO(s) + H2SO4(aq) -> NiSO4(aq) + H2O(l)

What is the chemical formula for aluminum oxide?
Al has 3+ charge, the oxide ion is O2-, thus Al2O3

Would you expect it to be solid, liquid or gas at room temp?
Oxides of metals are characteristically solid at room temp

Write the balanced chemical equation for the reaction of aluminum oxide with nitric
acid:
Metal oxide + acid ----> salt + water
Al2O3(s) + 6 HNO3(aq) ---> 2Al(NO3)3(aq) + 3H2O(l)
Non-metals

Vary greatly in appearance

Non-lustrous

Poor conductors of heat and electricity

The melting points of non-metals are generally lower than metals

Seven non-metals exist under standard conditions as diatomic molecules:
1. H2(g)
2. N2(g)
78
3. O2(g)
4. F2(g)
5. Cl2(g)
6. Br2(l)
7. I2(l) (volatile liquid - evaporates readily)

Nonmetals, when reacting with metals, tend to gain electrons (typically attaining
noble gas electron configuration) and become anions:
Nonmetal + Metal ----> Salt
For example, the reaction between aluminium and bromine
3Br2(l) + 2Al(s) ----> 2AlBr3(s)

Compounds composed entirely of non-metals are molecular substances (not ionic)

Most non-metal oxides are acidic oxides. Those that dissolve in water react to form
acids:
Non-metal oxide + water ----> acid
CO2(g) + H2O(l) ----> H2CO3(aq) [carbonic acid]
(carbonated water is slightly acidic)

Non-metal oxides can combine with bases to form salts
Non-metal oxide + base -----> salt
CO2(g) + 2NaOH(aq) -> Na2CO3(aq) + H2O(l)
Metalloids
Properties intermediate between the metals and non-metals.
Silicon for example appears lustrous, but is not malleable or ductile (it is brittle - a
characteristic of some non-metals). It is a much poorer conductor of heat and electricity than
the metals. Metalloids are useful in the semiconductor industry.
79
Trends in Metallic and Non-metallic Character

Metallic character is strongest for the elements in the leftmost part of the periodic
table, and tends to decrease as we move to the right in any period (nonmetallic
character increases with increasing ionization values)

Within any group of elements (columns), the metallic character increases from top
to bottom (the ionization values generally decrease as we move down a group). This
general trend is not necessarily observed with the transition metals.
80
Transition Metals: Known for their ability to form coloured compounds and ions in solution
They tend to be much less reactive than the Alkali Metals. They do not react as
quickly with water or oxygen so do not corrode as quickly. Transition metals tend
to form more coloured compounds more than other elements either in solid form
or dissolved in a solvent like water. Examples of the colours of some transition
metal salts in aqueous solution are shown below (grey = colourless in the diagrams). These
coloured ions/compounds often have quite a complex structure and indeed called complexes.
1
2
3
4
5
6
7
8
9
10
1. Sc - scandium salts, such as the chloride, ScCl3, are colourless and are not typical
of transition metals
2. Ti - titanium(III) chloride, TiCl3, is purple
3. V - vanadium(III) chloride, VCl3, is green
4. Cr - chromium(III) sulphate, Cr2(SO4)3, is dark green (chromate(VI) salts are
yellow, dichromate(VI) salts are orange)
5. Mn - manganese compound - potassium manganate(VII), KMnO4, is purple
(manganese(II) salts eg MnCl2 are pale pink)
6. Fe - iron(III) chloride, FeCl3, is yellow-orange-brown.
o
Iron(II) compounds are usually light green and iron(III) compounds
orange/brown.
7. Co - cobalt sulphate, CoSO4, is pinkish
8. Ni - nickel chloride, NiCl2, is green
9. Cu - copper(II) sulphate, CuSO4, is blue.
o
Most common copper compounds are blue in their crystals or solution and
sometimes green.
o
The blue aqueous copper ion, Cu2+(aq), actually has a more complicated
structure:

*[Cu(H2O)6]2+(aq) and when excess ammonia solution is added,
81

after the initial gelatinous blue copper(II) hydroxide precipitate is
formed, Cu(OH)2,

it dissolves to form the deep royal blue ion: *[Cu(H2O)2(NH3)4]2+(aq).

*are called complex ions and when coloured are typical of transition
metal chemistry.
o
Copper(II) oxide, CuO, black insoluble solid, readily dissolving in acids to
give soluble blue salts e.g.

copper(II) sulphate, CuSO4, from dilute sulphuric acid,

copper(II) nitrate, Cu(NO3)2, from dilute nitric acid

and greeny-blue copper(II) chloride, CuCl2, from dilute hydrochloric
acid.
o
Copper(II) hydroxide, Cu(OH)2, blue gelatinous precipitate formed when
alkali added to copper salt solutions.
o
Copper(II) carbonate, CuCO3, is turquoise-green insoluble solid, readily
dissolving in acids, evolving carbon dioxide, to give soluble blue salts (see
above)
o
Copper's valency or combining power is usually two e.g. compounds
containing the Cu2+ ion.

However there are copper(I) compounds where the valency is one.

This variable valency, hence compounds of the same elements, but
with
different
formulae,
is
typical
of
transition
metal
compounds e.g.

copper(I) oxide, Cu2O, an insoluble red-brown solid (CuO is black),

or copper(I) sulphate, Cu2SO4, a white solid.
10. Zn - zinc salts such as zinc sulphate, ZnSO4, are usually colourless and are not
typical of transition metals.

Many of the transition metal carbonates are unstable on heating and readily
undergo thermal decomposition.
o
metal carbonate ==> metal oxide + carbon dioxide
o
e.g.
o
copper(II) carbonate ==> copper(II) oxide + carbon dioxide

o
CuCO3(s) ==> CuO(s) + CO2(g)
or
82
o
zinc carbonate ==> zinc oxide + carbon dioxide

ZnCO3(s) ==> ZnO(s) + CO2(g)
o
In general the equation is ...
o
MCO3(s) ==> MO(s) + CO2(g) where M could be Fe, Cu, Mn or Zn
o
The carbon dioxide can be confirmed by giving a white milky precipitate with
limewater.
o

Sometimes the two solids show a colour change eg

for M = Cu: turquoise green carbonate ==> black copper(II) oxide

for M = Zn: white carbonate ==> white zinc oxide, but yellow hot
Many transition metal ions (e.g. in soluble salt solutions) give
hydroxide precipitates when mixed with aqueous sodium hydroxide
solution.
o
These reactions can be used to help identify transition metal ions in
solution

transition metal salt solution + sodium hydroxide ==> solid hydroxide precipitate +
sodium salt

ionically the precipitation reaction is :
o
metal ion + hydroxide ion ==> hydroxide precipitate
o
M2+(aq) + 2OH-(aq) ==> M(OH)2(s)

M can be for the ...

iron(II) ion Fe2+, pale green in aqueous solution,

which gives a dark grey-green gelatinous precipitate of
iron(II) hydroxide with sodium hydroxide solution

iron(II) sulfate + sodium hydroxide ==> iron(II) hydroxide
+ sodium sulfate

FeSO4(aq)
+
2NaOH(aq)
==>
Fe(OH)2(s)
+
Na2SO4(aq)

or

iron(II) chloride + sodium hydroxide ==> iron(II)
hydroxide + sodium chloride


FeCl2(aq) + 2NaOH(aq) ==> Fe(OH)2(s) + 2NaCl(aq)
For these reactions the ionic equation is ..

Fe2+(aq) + 2OH-(aq) ==> Fe(OH)2(s)
83

or for the copper(II) ion Cu2+, blue in aqueous solution,

which gives a blue copper(II) hydroxide precipitate with
sodium hydroxide solution.

e.g.

copper(II) sulphate + sodium hydroxide ==> copper(II)
hydroxide + sodium sulphate

CuSO4(aq)
+
2NaOH(aq)
==>
Cu(OH)2(s)
+
Na2SO4(aq)

or

copper(II) chloride + sodium hydroxide ==> copper(II)
hydroxide + sodium chloride

CuCl2(aq)
+
2NaOH(aq)
==>
Cu(OH)2(s)
+
2NaCl(aq)

For these reactions the ionic equation is ..


Cu2+(aq) + 2OH-(aq) ==> Cu(OH)2(s)
Note that the copper ion can be also detected by its flame
colour of green-blue.

The flame test is conducted by dipping a nichrome
(cheap)/platinum (expensive) wire into a copper salt
solution and placing the end of the wire plus drop into
the hottest part of a roaring bunsen flame when you see
flashes of blue and green colour.
o
and for the iron(III) ion Fe3+:

giving a brown iron(III) hydroxide precipitate with sodium
hydroxide solution,

e.g.

iron(III) chloride + sodium hydroxide ==> iron(III) hydroxide +
sodium chloride


the ionic equation is ...

o
FeCl3(aq) + 3NaOH(aq) ==> Fe(OH)3(s) + 3NaCl(aq)
Fe3+(aq) + 3OH-(aq) ==> Fe(OH)3(s)
These hydroxide precipitates are basically solids, but of a somewhat
gelatinous nature because they incorporate lots of water in their structure.
84

Also note that iron has two valencies or combining power giving
different
compound
formulae. Multiple valency, hence multiple
compound formation, is another characteristic (but not unique) feature of
transition metal chemistry.
o
The valency of chlorine is 1 and iron can have a combining power of 2 (II) or
3 (III).
o
FeCl2 iron(II) chloride (once called ferrous chloride)
o
FeCl3 iron(III) chloride (once called ferric chloride)

The number in Roman numerals is the valency or combining power
e.g.

oxygen's valency is 2 and copper, another transition element, has a
valency of 1 (I) or 2 (II)

so we have Cu2O copper(I) oxide (once called cuprous oxide)

and CuO copper(II) oxide (once called cupric oxide).

hence the need for the valency of the metal of the
metal to be shown in Roman numerals.
Uses of Transition Metals
Many transition metals are used directly as catalysts in industrial chemical processes and in
the anti-pollution catalytic converters in car exhausts.

For example iron is used in the HABER PROCESS for the synthesis of
ammonia:

o
Nitrogen + Hydrogen ==> Ammonia (via a catalyst of Fe atoms)
o
or N2(g) + 3H2(g) ==> 2NH3(g)
Platinum and rhodium (in other transition series below Sc-Zn) are used in the
catalytic converters in car exhausts to reduce the emission of carbon monoxide
and nitrogen monoxide, which are converted to the non-polluting gases nitrogen
and carbon dioxide.

2NO(g) + 2CO(g) ==> N2(g) + 2CO2(g)

Nickel is the catalyst for 'hydrogenation' in the margarine industry. It catalyses the
addition of hydrogen to an alkene carbon=carbon double bond (>C=C< +
85
H2 ==> (>CH-CH<)n. This process converts unsaturated vegetable oils into higher
melting saturated fats which are more 'spreadable' with a knife!

As well as the metals, the compounds of transition metals also act as catalysts.
EXAMPLES
o
For example manganese dioxide (or manganese(IV) oxide), MnO2, a black
powder, readily decomposes an aqueous solution of hydrogen peroxide:

Hydrogen peroxide ==> water + oxygen


o
2H2O2(aq) ==> 2H2O(l) + O2(g)
A useful reaction in the laboratory for preparing oxygen gas.
Vanadium(V) oxide (vanadium pentoxide, V2O5) is used as the catalyst in
converting sulphur dioxide into sulphur trioxide as a stage in the manufacture
of sulphuric acid in the CONTACT PROCESS.

2SO2(g) + O2(g) ==> 2SO3(g) (via V2O5 catalyst)

A very important industrial process because sulphuric acid is a widely
used chemical in industry.

Transition metals are extremely useful metals on account of their physical or
chemical properties eg lack of corrosion and greater strength compared to the Group 1
Alkali Metals.

Many are used in alloys, with a wide range of applications and uses.
o
An ALLOY is mixture of metal with at least one other metallic or nonmetallic substances - usually other elements.
o
By mixing metal with metal (and sometimes non-metals) together to
make alloys you can improve the metal's properties to better suit a particular
purpose.

For catalysts; their strength and hardness makes them very useful as structural
materials.
86
IRON, Fe

Cast iron is used for man-hole covers because it is so hard wearing but it is brittle
due to a high carbon content.

When alloyed with 1% carbon iron forms mild steel which is not brittle, but is more
malleable and corrosion resistant than cast iron. Mild steel is used for food cans, car
bodies (but galvanising and several coats of paint help it to last!) and machinery etc.

Steel is an alloy based on iron mixed with carbon and usually other metals added
too. There are huge number of steel 'recipes' which can be made to suit particular
purposes by changing the % carbon and adding other metals e.g. titanium steel for
armour plating.

CHROMIUM, Cr
o
Chromium steel (stainless steel, mixing and melting together Fe + Cr and
maybe Ni too) with good anti-corrosion properties, used for cutlery and
chemical plant reactors.
COPPER, Cu

The alloy BRASS is a mixture copper and zinc. It is a much more hard wearing
metal than copper (too soft) and zinc (too brittle) but is more malleable than bronze
for 'stamping' or 'cutting' it into shape.

Copper is used in electrical wiring because it is a good conductor of
electricity but for safety it is insulated by using poorly electrical
conductors like PVC plastic.

Copper is used in domestic hot water pipes because it is relatively
unreactive to water and therefore doesn't corrode easily.

Copper is used for cooking pans because it is relatively unreactive to water and
therefore doesn't corrode easily, readily beaten or pressed into shape but strong
enough, it is high melting and a good conductor of heat.
87

Copper is also used as a roof covering and weathers to a green colour as a surface
coating of a basic carbonate is formed on corrosion.

The alloy BRONZE is a mixture of copper and tin (Sn) and is stronger than copper
and just as corrosion resistant, e.g. used for sculptures.

Iron and steel are used for boilers because of their good heat conduction properties
and high melting point.

Copper compounds are used in fungicides and pesticides e.g. a traditional recipe is
copper sulphate solution plus lime is used to kill greenfly.

Copper is alloyed with nickel to give 'cupro-nickel', an attractive hard wearing
'silvery' metal for coins.

Steel, iron or copper are used for cooking pans because they are malleable, good
heat conductors and high melting.

NICKEL is alloyed with copper to give 'cupro-nickel', an attractive hard wearing
'silvery' metal for coins.
ZINC

Zinc is used to galvanise (coat) iron or steel to sacrificially protect them from
corrosion.
The
zinc
layer
can
be
put
on
the
iron/steel
object
by
chemical (see electroplating and below) or physically dipping it into a bath of molten
zinc.

Zinc sulphate solution can be used as the electrolyte for electroplating/galvanising
objects with a zinc layer.

Zinc is used as a sacrificed electrode in a zinc-carbon battery. It slowly reacts with a
weakly acidic ammonium chloride paste, converting chemical energy into electrical
energy.

The alloy BRASS is a mixture copper and zinc. It is a much more hard wearing
metal than copper (too soft) and zinc (too brittle) but is more malleable than bronze
for 'stamping' or 'cutting' it into shape.
88
Transition
metal
compounds
(often
oxides) of copper, iron, chromium and cobalt are
used
topigments for artwork, and give bright colours to stained
glass and ceramic/pottery glazes e.g.

Paint pigments: chromium oxide Cr2O3 green, iron oxide
(haematite) Fe2O3 red-brown, manganese oxide MnO2 black, copper hydroxidecarbonates (malachite-green, azurite-blue) and titanium dioxide TiO2 white.

Stained glass: cobalt oxide CoO blue, iron oxide/carbonate green, Cu metal red,
CuO turquoise.

NICHROME is an alloy of chromium and nickel. It has a high melting point and a
high electrical resistance and so it is used for electrical heating element wires.

NITINIOL: Titanium and nickel are the main components of Nitinol 'smart' alloys
which are very useful intermetallic compounds. Nitinol belongs to a group of shape
memory alloys (SMA) which can 'remember their original shape'. For example they
can regain there original shape on heating (e.g. used in thermostats in cookers , coffer
makers etc.) or after release of a physical stress (e.g. used in 'bendable' eyeglass
frames, very handy if you tread on them!). The other main metal used in these

TUNGSTEN is used as the filament in light bulbs because its melting point is so
high.

Transition metals like platinum and rhodium are used as metal catalysts in the
catalytic converters used in car vehicle exhausts to reduce carbon monoxide and
nitrogen oxide polluting emissions.

Bright, shiny and relatively unreactive copper, silver and gold are used in jewellery.

There is a note about the bonding in metals and structure of alloys on another
page.
Uses of non-transition metals

Aluminium is NOT a transition metal!
o
e.g. it does not form coloured compounds, it does not act as a catalyst etc.
o
BUT it is high melting, of low density and one of the most used and useful
non-transition metals.
89
o
It
is
rather
weak
BUT
when alloyed
with
copper,
manganese
and magnesium and it forms a much stronger alloy called duralumin.
o
It does not readily corrode due to a permanent Al2O3 aluminium oxide layer
on the surface which does not flake off and protects the aluminium from
further oxidation.
o
Because of its alloyed strength, lightness and anti-corrosion properties it is
used in aircraft construction, window frames, hifi chassis etc.
o
It is a good conductor of heat and can be used in radiators.
o
It is quite a good conductor of electricity, and also because its light, it is used
in conjunction with copper (excellent electrical conductor) in overhead power
lines (don't want them too heavy when iced up!). The cables however do have
a steel core for strength!

Poorly electrical conducting ceramic materials are used to insulate the
wires from the pylons and the ground.

A mixture of tin and lead is mixed to give the alloy SOLDER which has a relatively
low melting solid for electrical connections.

Tin is an unreactive metal and is used to coat more corrodible metals like iron-steel.
A 'tin can' is actually made of steel with a fine protective coating of tin metal over the
surface of it.

Lead is a soft, very malleable relatively unreactive metal used in roofing. 'Flashings'
are used to seal sections of roofs e.g. between walls and the ends of layers of tiles or
slates. Electrical cables can be encased in it. It is used with lead oxide in the
manufacture of electrodes of road vehicle car batteries. Because of its high density it
is used as a shield from dangerous alpha/beta/gamma radiation from radioactive
materials and X-rays, so it is used in nuclear processing facilities etc. and
radiographers wear a lead apron when you go for an X-ray on your bones.

PEWTER is an alloy of mainly tin plus small amounts of copper, bismuth (Bi)
and antimony (Sb), it is stronger than tin but is easy to etch and engrave.

DENTAL AMALGAM ALLOY is a mixture of tin, mercury and silver (a
transition metal). When first prepared its soft and malleable before hardening to that
undesired filling! It has good anti-corrosion properties and resists the attack of acidic
products produced by bacteria in the mouth. An amalgam is an alloy metal compound
90
made from a mixture of mercury and other metals which may be liquid and set to a
solid after preparation.
More on Al and Fe
Aluminium (Al) can be made more resistant to corrosion by a process called
anodising. Iron (Fe) can be made more useful by mixing it with other substances to make
various types of steel. Many metals can be given a coating of a different metal to protect
them or to improve their appearance.

Aluminium is a reactive metal but it is resistant to corrosion. This is because
aluminium reacts in air to form a layer of aluminium oxide which then protects the
aluminium from further attack.
o
This is why it appears to be less reactive than its position in the reactivity
series of metals would predict.

For some uses of aluminium it is desirable to increase artificially the thickness of the
protective oxide layer in a process is called anodising.
o
This involves removing the oxide layer by treating the aluminium sheet with
sodium hydroxide solution.
o
The aluminium is then placed in dilute sulphuric acid and is made the positive
electrode (anode) used in the electrolysis of the acid.
o
Oxygen forms on the surface of the aluminium and reacts with the aluminium
metal to form a thicker protective oxide layer.

Aluminium can be alloyed to make 'Duralumin' by adding copper (and smaller
amounts of magnesium, silicon and iron),to make a stronger alloy used in aircraft
components (low density = 'lighter'!), greenhouse and window frames (good anticorrosion properties), overhead power lines (quite a good conductor and 'light'), but
steel strands are included to make the 'line' stronger and poorly electrical conducting
ceramic materials are used to insulate the wires from the pylons and the ground.
o

There is a note about the bonding and structure of alloys on another page.
The properties of iron can be altered by adding small quantities of other metals or
carbon to make steel.
91

Steels are alloys since they are mixtures of iron with other metals or with non-metals
like carbon or silicon.

Making Steel:
o
(1) Molten iron from the blast furnace iron extraction process is mixed with
recycled scrap iron
o
(2) Then pure oxygen is passed into the mixture and the non-metal impurities
such as silicon or phosphorus are then converted into acidic oxides (oxidation
process) ..

o
e.g. Si + O2 ==> SiO2, or 4P + 5O2 ==> P4O10
(3) Calcium carbonate (a base) is then added to remove the acidic oxide
impurities (in an acid-base reaction). The salts produced by this reaction form
a slag which can be tapped off separately.

e.g. CaCO3 + SiO2 ==> CaSiO3 + CO2 (calcium silicate slag)
o
Reactions (1)-(3) produce pure iron.
o
Calculated quantities of carbon and/or other metallic elements such as
titanium, manganese or chromium are then added to make a wide range of
steels with particular properties.
o
Because of the high temperatures the mixture is stirred by bubbling in
unreactive argon gas!
o
Economics of recycling scrap steel or ion: Most steel consists of >25%
recycled iron/steel and you do have the 'scrap' collection costs and problems
with varying steel composition* BUT you save enormously because there is
no mining cost or overseas transport costs AND less junk lying around!
(NOTE: * some companies send their own scrap to be mixed with the next
batch of 'specialised' steel they order, this saves both companies money!)
Different steels for different uses:
o
High % carbon steel is strong but brittle.
o
Low carbon steel or mild steel is softer and is easily shaped and pressed e.g.
into a motor car body.
o
Stainless steel alloys contain chromium and nickel and are tougher and more
resistant to corrosion.
92
o
Very strong steels can be made by alloying the iron with titanium or
manganese metal.


There is a note about the bonding and structure of alloys on another page.
Steel can be galvanised by coating in zinc, this is physically done by dipping the
object into a bath of molten zinc. On removal and cooling a thin layer of zinc is left
on. The zinc chemically bonds to the iron via the free electrons of both metals - its all
the same atoms to them! It can also be done by electroplating (details below).

Steel (and most metals) can be electroplated.
o
The steel object to be plated is made the negative electrode (cathode) and
placed in a solution containing ions of the plating metal.
o
The positive electrode (anode) is made of the pure plating metal (which
dissolves and forms the fresh deposit on the negative electrode).
o
Nickel, zinc, copper, silver and gold are examples of plating metals.
o
The details of copper purification amount to copper plating, so all you have
to do is swap the pure negative copper cathode with the metal you want to coat
(e.g. Ni, Ag or Au or any material with a conducting surface). Swap the
impure positive copper anode with a pure block of the metal you want to form
the coating layer. The electrodes dip into a salt solution of nickel, zinc, copper,
silver or gold ions etc. and a low d.c. voltage passed through. If M = Ni, Cu,
Zn ....

At the positive (+) anode, the process is an oxidation, electron loss, as
the metal atoms dissolve to form metal(II) ions.

M(s) ==> M2+(aq) + 2e-

at the negative (-) cathode, the process is a reduction, two electron
gain by the attracted metal(II) ions to form neutral metal atoms on the
surface of the metal being coated.

M2+(aq) + 2e- ==> M(s)

For silver plating it is Ag+, Ag and a single electron change.

Any conducting (usually metal) object can be electroplated with
copper or silver for aesthetic reasons orsteel with zinc or
chromium as anti-corrosion protective layer.
93

Many other metals have countless uses e.g. zinc
o
Zinc is used to make the outer casing of zinc-carbon-weak acid batteries.
o
Zinc is alloyed with copper to make the useful metal brass (electrical plug
pins). Brass alloy is stronger and more hardwearing than copper AND not as
brittle as zinc.
Exercise: The use of text books and academic references for a deeper understanding are
permitted to effectively answer the following questions.
1.
2.
a)
Explain why Sc and Zn are not classified as transition metals
b)
Explain how transition metals can form complex ions
d)
Explain why complex ions are often coloured
e)
Explain why Cu+ is not coloured
Explain how the colour of solutions containing transition metals can be used to
determine their concentration.
Unit Summary
Inorganic Chemistry is concerned with metals, non-metals and metalloids. The characteristic
properties of metallic and non-metallic elements are:
Metallic Elements
Non-metallic elements
Distinguishing luster (shine)
Non-lustrous, various colors
Malleable and ductile (flexible) as solids
Brittle, hard or soft
Conduct heat and electricity
Poor conductors
Metallic oxides are basic, ionic
Nonmetallic oxides are acidic, compounds
Cations in aqueous solution
Anions, oxyanions in aqueous solution
Transition metals are known for their ability to form colours in solution. They are also known
to have differing oxidation numbers.
94
Unit 9: The Periodic Table
Trends
Before getting into these trends, we should engage a quick review and establish some
terminology. As seen in the previous section on the octet rule, atoms tend to lose or gain
electrons in order to attain a full valence shell and the stability a full valence shell imparts.
Because electrons are negatively charged, an atom becomes positively or negatively charged
as it loses or gains an electron, respectively. Any atom or group of atoms with a net charge
(whether positive or negative) is called an ion. A positively charged ion is a cation while a
negatively charged ion is an anion.
Now we are ready to discuss the periodic trends of atomic size, ionization energy, electron
affinity, and electronegativity.
Atomic Size (Atomic Radius)
The atomic size of an atom, also called the atomic radius, refers to the distance between an
atom's nucleus and its valence electrons. Remember, the closer an electron is to the nucleus,
the lower its energy and the more tightly it is held.
Moving Across a Period
Moving from left to right across a period, the atomic radius decreases. The nucleus of the
atom gains protons moving from left to right, increasing the positive charge of the nucleus
and increasing the attractive force of the nucleus upon the electrons. True, electrons are also
added as the elements move from left to right across a period, but these electrons reside in the
same energy shell and do not offer increased shielding.
Moving Down a Group
The atomic radius increases moving down a group. Once again protons are added moving
down a group, but so are new energy shells of electrons. The new energy shells provide
shielding, allowing the valence electrons to experience only a minimal amount of the protons'
positive charge.
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Cations and Anions
Cations and anions do not actually represent a periodic trend in terms of atomic radius, but
they do affect atomic radius, and so we will discuss them here.
A cation is positively charged, meaning that it is an atom that has lost an electron or
electrons. The positive charge of the nucleus is thus distributed over a smaller number of
electrons and electron-electron repulsion is decreased, meaning that the electrons are held
more tightly and the atomic radius is smaller than in the normal neutral atom. Anions,
conversely, are negatively charged ions: atoms that have gained electrons. In anions,
electron-electron repulsion increases and the positive charge of the nucleus is distributed over
a large number of electrons. Anions have a greater atomic radius than the neutral atom from
which they derive.
Ionization Energy and Electron Affinity
The process of gaining or losing an electron requires energy. There are two common ways to
measure this energy change: ionization energy and electron affinity.
Ionization Energy
The ionization energy is the energy it takes to fully remove an electron from the atom. When
several electrons are removed from an atom, the energy that it takes to remove the first
electron is called the first ionization energy, the energy it takes to remove the second electron
is the second ionization energy, and so on. In general, the second ionization energy is greater
than first ionization energy. This is because the first electron removed feels the effect of
shielding by the second electron and is therefore less strongly attracted to the nucleus. If
particular ionization energy follows a previous electron loss that emptied a subshell, the next
96
ionization energy will take a rather large leap, rather than follow its normal gently increasing
trend. This fact helps to show that just as electrons are more stable when they have a full
valence shell, they are also relatively more stable when they at least have a full subshell.
Ionization Energy Across a Period
Ionization energy predictably increases moving across the periodic table from left to right.
Just as we described in the case of atomic size, moving from left to right, the number of
protons increases. The electrons also increase in number, but without adding new shells or
shielding. From left to right, the electrons therefore become more tightly held meaning it
takes more energy to pry them loose. This fact gives a physical basis to the octet rule, which
states that elements with few valence electrons (those on the left of the periodic table) readily
give those electrons up in order to attain a full octet within their inner shells, while those with
many valence electrons tend to gain electrons. The electrons on the left tend to lose electrons
since their ionization energy is so low (it takes such little energy to remove an electron) while
those on the right tend to gain electrons since their nucleus has a powerful positive force and
their ionization energy is high. Note that ionization energy does show a sensitivity to the
filling of subshells; in moving from group 12 to group 13 for example, after the d shell has
been filled, ionization energy actually drops. In general, though, the trend is of increasing
ionziation energy from left to right.
Ionization Energy Down a Group
Ionization energy decreases moving down a group for the same reason atomic size increases:
electrons add new shells creating extra shielding that supersedes the addition of protons. The
atomic radius increases, as does the energy of the valence electrons. This means it takes less
energy to remove an electron, which is what ionization energy measures.
Electron Affinity
An atom's electron affinity is the energy change in an atom when that atom gains an electron.
The sign of the electron affinity can be confusing. When an atom gains an electron and
becomes more stable, its potential energy decreases: upon gaining an electron the atom gives
off energy and the electron affinity is negative. When an atom becomes less stable upon
gaining an electron, its potential energy increases, which implies that the atom gains energy
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as it acquires the electron. In such a case, the atom's electron affinity is positive. An atom
with a negative electron affinity is far more likely to gain electrons.
Electron Affinities Across a Period
Electron affinities becoming increasingly negative from left to right. Just as in ionization
energy, this trend conforms to and helps explain the octet rule. The octet rule states that
atoms with close to full valence shells will tend to gain electrons. Such atoms are located on
the right of the periodic table and have very negative electron affinities, meaning they give
off a great deal of energy upon gaining an electron and become more stable. Be careful,
though: the nobel gases, located in the extreme right hand column of the periodic table do not
conform to this trend. Noble gases have full valence shells, are very stable, and do not want
to add more electrons: noble gas electron affinities are positive. Similarly, atoms with full
subshells also have more positive electron affinities (are less attractive of electrons) than the
elements around them.
Electron Affinities Down a Group
Electron affinities change little moving down a group, though they do generally become
slightly more positive (less attractive toward electrons). The biggest exception to this rule are
the third period elements, which often have more negative electron affinities than the
corresponding elements in the second period. For this reason, Chlorine, Cl, (group VIIa and
period 3) has the most negative electron affinity.
Electronegativity
Electronegativity refers to the ability of an atom to attract the electrons of another atom to it
when those two atoms are associated through a bond. Electronegativity is based on an atom's
98
ionization energy and electron affinity. For that reason, electronegativity follows similar
trends as its two constituent measures.
Electronegativity generally increases moving across a period and decreases moving down a
group. Flourine (F), in group VIIa and period 2, is the most powerfully electronegative of the
elements. Electronegativity plays a very large role in the processes of Chemical Bonding.
Electron Configuration and Valence Electrons
Electron Configuration
The electrons in an atom fill up its atomic orbitals according to the Aufbau Principle;
"Aufbau," in German, means "building up." The Aufbau Principle, which incorporates the
Pauli Exclusion Principle and Hund's Rule prescribes a few simple rules to determine the
order in which electrons fill atomic orbitals:
1.
Electrons always fill orbitals of lower energy first. 1s is filled before 2s, and 2s before
2p.
2.
The Pauli Exclusion Principle states no two electrons within a particular atom can
have identical quantum numbers. In function, this principle means that if two electrons
occupy the same orbital, they must have opposite spin.
3.
Hund's Rule states that when an electron joins an atom and has to choose between two
or more orbitals of the same energy, the electron will prefer to enter an empty orbital
rather than one already occupied. As more electrons are added to the atom, these
electrons tend to half-fill orbitals of the same energy before pairing with existing
electrons to fill orbitals.
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Figure %: The ground state electron configuration of carbon, which has a total of six
electrons. The configuration is determined by applying the rules of the Aufbau Principle.
Valency and Valence Electrons
The outermost orbital shell of an atom is called its valence shell, and the electrons in the
valence shell are valence electrons. Valence electrons are the highest energy electrons in an
atom and are therefore the most reactive. While inner electrons (those not in the valence
shell) typically don't participate in chemical bonding and reactions, valence electrons can be
gained, lost, or shared to form chemical bonds. For this reason, elements with the same
number of valence electrons tend to have similar chemical properties, since they tend to gain,
lose, or share valence electrons in the same way. The Periodic Table was designed with this
feature in mind. Each element has a number of valence electrons equal to its group number
on the Periodic Table.
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Figure %: The periodicity of valence electrons
This table illustrates a number of interesting, and complicating, features of electron
configuration.
First, as electrons become higher in energy, a shift takes place. Up until now, we have said
that as the principle quantum number, increases, so does the energy level of the orbital. And,
as we stated above in the Aufbau principle, electrons fill lower energy orbitals before filling
higher energy orbitals. However, the diagram above clearly shows that the 4s orbital is filled
before the 3d orbital. In other words, once we get to principle quantum number 3, the highest
subshells of the lower quantum numbers eclipse in energy the lowest subshells of higher
quantum numbers: 3d is of higher energy than 4s.
Second, the above indicates a method of describing an element according to its electron
configuration. As you move from left to right across the periodic table, the above diagram
shows the order in which orbitals are filled. If we were the actually break down the above
diagram into groups rather than the blocks we have, it would show how exactly how many
electrons each element has. For example, the element of hydrogen, located in the uppermost
left-hand corner of the periodic table, is described as 1s 1, with the s describing which orbital
contains electrons and the 1 describing how many electrons reside in that orbital. Lithium,
which resides on the periodic table just below hydrogen, would be described as 1s 22s 1. The
electron configurations of the first ten elements are shown below (note that the valence
electrons are the electron in highest energy shell, not just the electrons in the highest energy
subshell).
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The Octet Rule
Our discussion of valence electron configurations leads us to one of the cardinal tenets of
chemical bonding, the octet rule. The octet rule states that atoms become especially stable
when their valence shells gain a full complement of valence electrons. For example, in above,
Helium (He) and Neon (Ne) have outer valence shells that are completely filled, so neither
has a tendency to gain or lose electrons. Therefore, Helium and Neon, two of the so-called
Noble gases, exist in free atomic form and do not usually form chemical bonds with other
atoms.
Most elements, however, do not have a full outer shell and are too unstable to exist as free
atoms. Instead they seek to fill their outer electron shells by forming chemical bonds with
other atoms and thereby attain Noble Gas configuration. An element will tend to take the
shortest path to achieving Noble Gas configuration, whether that means gaining or losing one
electron. For example, sodium (Na), which has a single electron in its outer 3s orbital, can
lose that electron to attain the electron configuration of neon. Chlorine, with seven valence
electrons, can gain one electron to attain the configuration of argon. When two different
elements have the same electron configuration, they are called isoelectronic.
Diamagnetism and Paramagnetism
The electron configuration of an atom also has consequences on its behaviour in relation to
magnetic fields. Such behaviour is dependent on the number of electrons an atom has that are
spin paired. Remember that Hund's Rule and the Pauli Exclusion Principle combine to dictate
that an atom's orbitals will all half-fill before beginning to completely fill, and that when they
completely fill with two electrons, those two electrons will have opposite spins.
An atom with all of its orbitals filled, and therefore all of its electrons paired with an electron
of opposite spin, will be very little affected by magnetic fields. Such atoms are
102
called diagmetic. Conversely, paramagnetic atoms do not have all of their electrons spinpaired and are affected by magnetic fields. There are degrees of paramagnetism, since an
atom might have one unpaired electron, or it might have four.
Exercise
1.
2.
Give the electronic configuration of the following atoms:
a)
V
b)
Cr
c)
Co
d)
Cu
e)
Zn
Give the electronic configuration of the following ions:
a)
Co2+
b)
Cu+
c)
V3+
d)
Cr3+
e)
Fe3+
3. Define Ionisation energy, electron afiinity and electronegativity.
Unit Summary
The atomic size of an atom, also called the atomic, the closer an electron is to the nucleus, the
lower its energy and the more tightly it is held. Moving from left to right across a period, the
atomic radius decreases. The nucleus of the atom gains protons moving from left to right,
increasing the positive charge of the nucleus and increasing the attractive force of the nucleus
upon the electrons. The atomic radius increases moving down a group. Once again protons
are added moving down a group, but so are new energy shells of electrons. The new energy
shells provide shielding, allowing the valence electrons to experience only a minimal amount
of the protons' positive charge.
103
Unit 10: Organic chemistry
Organic chemistry is the study of carbon and the study of the chemistry of life. Since not all
carbon reactions are organic, another way to look at organic chemistry would be to consider it
the study of molecules containing the carbon-hydrogen (C-H) bond and their reactions.
Why Is Organic Chemistry Important?
Organic chemistry is important because it is the study of life and all of the chemical reactions
related to life. Several careers apply an understanding of organic chemistry, such as doctors,
veterinarians, dentists, pharmacologists, chemical engineers, and chemists. Organic chemistry
plays a part in the development of common household chemicals, foods, plastics, drugs,
fuels... really most of the chemicals part of daily life.
What Does an Organic Chemist Do?
An organic chemist is a chemist with a college degree in chemistry. Typically this
would be a doctorate or master's degree in organic chemistry, though a bachelor's
degree in chemistry may be sufficient for some entry level positions. Organic
chemists usually conduct research and development in a laboratory setting. Projects
that would use organic chemists would include development of a better painkilling
drug, formulating a shampoo that would result in silkier hair, making a stain resistant
carpet, or finding a non-toxic insect repellent.
Alkanes
The boiling points of alkanes increase as the chains get longer (increased number of electrons
causes increased Van de Waal's forces), increasing rapidly initially but flattening off. Click
diagram to enlarge
104
Alkanes
Compounds containing only hydrogen and carbon. There are three types alkanes, alkenes and
alkynes.
Alkanes have a CH3 group at each end (except methane has only one CH4) and fill out the
required number with CH2 groups.
Nomenclature (Naming system)
methane ethane
propane
butane
105
Isomerism and branching
Any structure that can be drawn can exist providing the fundamental rules have been
fulfilled.
1. Each carbon forms four bonds (may be all single, one double and two singles, etc)
2. Each hydrogen forms one bond
3. Each oxygen forms two bonds
This means that one molecular formula can have other possible structures. This is called
isomerism. The chains formed by the alkanes and other organic molecules do not have to be
straight (actually zigzag) but may be branched i.e having "branches" of carbon atoms
attached to the main unbroken chain.
Example:
The alkane C4H10 exists in two isomeric forms - a straight chain form and a branched form
Butane
methyl propane
106
Alkenes
These structures will be similar to those of the alkanes except two hydrogens on adjacent
carbons are replaced by a double bond between those carbons. The number '1' in the names
refers to the position of the carbon starting the double bond. No numbering is needed in the
first two members as there can be no ambiguity.
ethene
propene
but-1-ene
pent-1-ene
Combustion
107
This is effectively a technical word for burning. Most organic compounds burn with the
exception of chlorinated (halogenated ) hydrocarbons.
Complete combustion produces CO2 and H2O, incomplete combustion produces CO, C and
H2O (usually occurs with unsaturated compounds, where there is a limited supply of oxygen).
C produces a 'dirty' flame leaving carbon deposits on everything, CO is toxic and CO2 is a
greenhouse gas. Incomplete combustion is where the carbon is not completely oxidised.
The combustion of hydrocarbons is an exothermic process (otherwise there wouldn't be much
point in burning them to produce energy for fuel and heat). This is because the O-H bond is
stronger than the C-H bond, and the C=O bond is stronger than the C-C. This means that, the
C-C and C-H bonds breaking requires energy, but this is more than made up for by the energy
released by the formation of the C=O and O-H bonds.
Alcohols
homologous
functional
series name (old group
naming system
description
name)
Has a carbonyl C=O at the
Alkanal
(aldehyde)
-CHO
ends with -anal
end of a carbon chain, with
the carbon also attached to a
hydrogen
Has a carbonyl group C=O
Alkanone
(ketone)
-CO-
ends with -anone
but it is in the middle of a
carbon chain and the carbon
has no hydrogens attached
ends with -anol (may also go These have an -OH group (or
Alkanol (alcohol) -OH
at the start of the name as more than one in the case of
hydroxy- if there is another diols,
108
triols
etc)
at
any
more important group in the position in the chain.
molecule)
The group must go at the end
of a carbon chain as it has a
carbon attached to an oxygen
Alkanoic
acid
(carboxylic acid)
by a double bond and also an
-COOH
ends with -anoic acid
-OH
group
(leaving
the
carbon with only one other
bond which it uses to attach
to the rest of the carbon
chain)
ends with -ylamine or starts
the
Amine
-NH2
name
with
Amino-
(depending on whether there
is a more important functional
group int he molecule)
The NH2 group can go
anywhere
chain.
on
It
the
carbon
infers
basic
to
the
characteristics
molecule. (ability to accept a
proton from an acid)
The
amine
group
is
associated with a carbonyl
group
Amide
-CONH2
ends with -anamide
inferring
characteristics
different
to
the
molecule. It can also be a
linking
group
-CONH-
between two alkyl chains
(proteins and nylon)
The halogen atom may be
becomes a prefix: chloro- added into any position in the
Halogenoalkanes
-X
bromo- iodo- folowed by the chain. If there is more than
rest of the name
one then the prefixes di- tritetra- etc are used.
Esters
-COO-
name is derived from the acid Esters
109
are
linkage
and alcohol that were used in compoounds
formed
by
making the ester. The alkyl condensation (esterification)
group ending with the -COO reactions between carboxylic
was
the
acid
part
and acids (or compounds deriving
becomes -(name)anoate and from them) and alcohols.
the part after the -COO group They also occur naturally and
starts the name. For example: may be polymers (polyester)
ethyl ethanoate
see amide linkage above
derives from the shortest These are linkage compounds
alkyl chain ending in -oxy where the oxygen bridges
Ethers
-O-
followed by the longest alkyl two carbon chains. They are
chain
eg.
methoxyethane of little importance except as
CH3-O-C2H5
solvents (ethoxyethane)
Halogenoalkanes
Halogenoalkanes (also called haloalkanes) have a halogen atom attached to a hydrocarbon
chain. The halogen atom may be at the end of the chain or on any of the carbons.
1-chloropropane
2-chloropropane
Halogenoalkanes may be classified as primary, secondarty or tertiary, depending on the
number of carbon atoms attached to the carbon holding the halogen.
110
1-chloropropane is a primary halgenoalkane, as there is only one carbon attached to the
carbon holding the halogen, whereas 2-chloropropane is a secondary halogenoalkane as there
are two carbon atoms attached to the carbon holding the halogen.
Reactions of halogenoalkanes

Nucleophilic substitution

Elimination reactions (not SL)
Substitution involves removal of the halogen and replacing it with another ion, or group (just
like substitution in football). The nucleophilic refers to the mechanism of the reaction and
says that the attacking species must be looking for a positive charge (nucleophile = 'positive
seeking'), i.e. it has a lone pair of electrons and may be negatively charged.
The simplest reaction is with dilute sodium hydroxide which contains free hydroxide ions
OH-(aq).
CH3CH2-Br + NaOH
CH3CH2-OH + NaBr
The mechanism of the reaction depends on the nature (1º, 2º or 3º) of the halogenoalkane.

1º Halogenoalkanes react via sN2 mechanism

3º Halogenoalkanes react via sN1 mechanism

2º probabaly use a mixture of both.
Chemical properties of the different functional groups
The following table summarises the chemical properties of the functional groups studied at
this level.
homologous
Important
series
types
Combustion
reaction
reagent and conditions
product (s)
air/oxygen, heat
carbon
dioxide
water
Alkanes
free
radical halogen and UV light
111
halogenoalkanes
and
substitution
of
(mixture)
halogens
Combustion
air/oxygen
carbon dioxide and water
heat
Alkenes
Addition
Br2, Br2(aq), dibromoalkane, bromoalcohol,
H2
alkane
Polymerization
catalyst
polyalkene
Combustion
air/oxygen, heat
carbon dioxide and
water
Alcohols
dehydration
phosphoric
or alkene
(elimination)
sulphuric acid
Oxidation
sodium
alkanal, alkanone or
dichromate/sulphuric
carboxylic acid
acid
substitution of the - phosphorus
halogenoalkane,
OH group
pentachloride, sodium
sodium alkoxide
Esterification
carboxylic acid
ester
Addition
ammonia
compound containing
hydroxy and amine
Oxidation
sodium dichromate/dil. carboxylic acid
sulphuric acid
aldehydes
(alkanals)
Reduction
lithium
aluminium 1º alcohol
hydride
addition - elimination amines,
(condensation)
hydroxylamine,
112
depends on reagent
acid - base reactions
base
(eg
sodium salt and water
hydroxide)
carboxylic acids
Reduction
lithium
aluminium alkanal (or 1º alcohol)
hydride
Esterification
alcohol/conc. sulphuric ester
acid
substitution
haloalkanes
of
the sodium
halogen
hydroxide, alcohol, amine
ammonia
(halogenoalkanes) dehydrohalogenation
ethanolic NaOH/ reflux alkene
(elimination)
reaction as a base with mineral (strong) acid
amine salt
an acid
Amines
condensation (addition carbonyl compounds
depends on reagent
- elimination)
hydrolysis
(breaking dilute acid/heat
amine and carboxylic
apart in solution)
acid
hydrolysis
amine
Amides
(breaking dilute base/heat
and
sodium
apart in solution)
carboxylate
hydrolysis
alcohol and carboxylic
(breaking dilute acid/heat
apart in solution)
acid
hydrolysis
alcohol and sodium
Esters
(breaking dilute base/heat
apart in solution)
carboxylate
113
Exercise
1. Given two molecular formulae C4H10 and C4H8;
Select a compound which could be:
a) an alkane
b) a cycloalkane
c) an alkene
In each case, draw one possible structure to show how your choice is correct.
2. State the class of organic molecule to which the following compounds belong:
a)
CH3CH2CH3
b)
CH3CH=CH2
114
c)
CH3CH2CH2Br
Unit Summary
Organic chemistry is important because it is the study of life and all of the chemical reactions
related to life. Several careers apply an understanding of organic chemistry, such as doctors,
veterinarians, dentists, pharmacologists, chemical engineers, and chemists. Organic chemistry
plays a part in the development of common household chemicals. Organic chemistry is the
study of carbon and the study of the chemistry of life. Since not all carbon reactions are
organic, another way to look at organic chemistry would be to consider it the study of
molecules containing the carbon-hydrogen (C-H) bond and their reactions.
115
Unit 11: Applications of analytical chemistry
Analytical chemistry is the science of obtaining, processing, and communicating information about
the composition and structure of matter. In other words, it is the art and science of determining what
matter is and how much of it exists.
Application in All Areas of Chemistry
Analytical chemists perform qualitative and quantitative analysis; use the science of sampling,
defining, isolating, concentrating, and preserving samples; set error limits; validate and verify results
through calibration and standardization; perform separations based on differential chemical
properties; create new ways to make measurements; interpret data in proper context; and
communicate results. They use their knowledge of chemistry, instrumentation, computers, and
statistics to solve problems in almost all areas of chemistry. For example, their measurements are used
to assure compliance with environmental and other regulations; to assure the safety and quality of
food, pharmaceuticals, and water; to support the legal process; to help physicians diagnose disease;
and to provide chemical measurements essential to trade and commerce.
Analytical chemists are employed in all aspects of chemical research in industry, academia, and
government. They do basic laboratory research, develop processes and products, design instruments
used in analytical analysis, teach, and work in marketing and law. Analytical chemistry is a
challenging profession that makes significant contributions to many fields of science.
Analytical methods using robots and instrumentation specifically designed to prepare and analyze
samples have been automated. In addition, increasingly powerful personal computers and
workstations are enabling the development and use of increasingly sophisticated techniques and
methods of interpreting instrumental data. So, in some cases, because the instrumentation does more,
fewer chemists are required to prepare the sample and measure and interpret the data. On the other
hand, the demand for new and increasingly sophisticated analytical techniques, new instrumentation,
automation and computerization, and regulatory requirements have opened up new opportunities for
analytical chemists in other areas. For example, quality assurance specialists help validate that
analytical laboratories and the chemists working there follow documented and approved procedures;
new instrumentation and laboratory information management systems have opened up opportunities
for chemists with solid technical and computer skills; and corporate downsizings have provided the
impetus for entrepreneurial analytical chemists to start their own businesses.
116
Good oral and written communication skills are essential, particularly when oral presentations,
reports, and memos are required to defend a measurement and its interpretation. In addition,
familiarity with the various roles analytical chemists play in different industries and exposure to
business and management practices are valuable assets that will allow growth into management,
manufacturing, sales, and marketing positions.
Exercise
-In your 4 groups discuss and research on the applications of Analytical Chemistry in
Zambia. How has it been incorporated in the business world?
What technology can be integrated to improve its application for the masses to improve
health procedures and increase awareness of the importance of Analytical Chemistry in
Zambia today?
117
References
Texts:
AS-Level Chemistry AQA Complete Revision & Practice Richard Parsons (Paperback - Jul
14, 2008)
A2-Level Biology OCR Complete Revision & Practice (A2 Level Aqa Re…Richard Parsons
(Paperback - Sep 1, 2009)
AS/A2 Level Chemistry AQA Complete Revision & Practice Richard Parsons (Paperback Jan 20, 2010)
A2-Level Chemistry AQA Complete Revision & Practice (A2 Level Aqa …Richard Parsons
(Paperback - Jul 16, 2009)
AQA Chemistry AS: Student's Book Ted Lister, Janet Renshaw (Paperback - May 12, 2008)
AS Level Chemistry: The Complete Course for AQA Richard Parsons (Paperback - Jan 20,
2012)
Websites:
Analytical Chemisry, ACS Chemistry for Life,
<http://portal.acs.org/portal/acs/corg/content?_nfpb=true&_pageLabel=PP_ARTICLEMAIN
&node_id=1188&content_id=CTP_003375&use_sec=true&sec_url_var=region1&__uuid=9e
214550-3dc3-48b8-b077-699bd6556d55> [Accessed on 18 October 2012]
Energetics, < http://www.alevelchemistry.co.uk/AQA%20AS%20Chemistry/Unit%202/2.1%20Energetics/2.1%20home
.htm> [Accessed on 18 October 2012]
118
Transition Metals, < http://www.docbrown.info/page04/4_75trans.htm> [Accesson on 18
October 2012].
119