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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

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ContentsBSPH 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

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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

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.

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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

Formula for Percent Yield

Percent Yield =

δH

δH = δH f (products) - δH f (reactants)

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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.

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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 =

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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

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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

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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)

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http://www.sparknotes.com/chemistry/stoichiometry/stoichiometriccalculations/section1.rhtml#conversionfactors

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) = + = 18 grams / mole

×1 mole = 1 mole of H2O(s)

Step 3: mole ratio:

×3 moles O2(g) = 1.5 moles O2(g)

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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 1 - 4 refer to the chemical equation

Zn(s) + 2 HCl(aq) ------> ZnCl2(aq) + 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


3 Zn(s) + 2 HCl(aq)--------> ZnCl2(aq) + H2(g)


from the reaction of 0.40 moles of HCl with an excess of zinc?

0.40 moles

0.80 moles

0.20 moles


4 Zn(s) + 2 HCl(aq) ---------> ZnCl2(aq) + H2(g)


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


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

24

0.30 moles

7.5 moles


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


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


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

25

0.067 moles

2.0 moles

0.60 moles



4 FeS2(s) + 11 O2(g) ----> 2 Fe2O3(s) + 8 SO2(g)

How many moles of O2 are required for complete reaction of 4.0x10-

3 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


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

26

4.0x10-3 moles

2.0x10-3 moles


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

1.1x10-2 moles

5.8x10-3 moles


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



CH4(g) + 2 O2(g) --------> CO2(g) + 2 H2O(g)

How many grams of O2 are required for complete reaction of 1.6 grams

of methane (CH4)?

3.2 g

27

1.6 g

6.4 g

0.80 g


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

0.90 g


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


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

28

3.6 g

3.2 g



2 H2S(g) + SO2(g) --------> 3S(s) + 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


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


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

29

1.5 g

0.50 g

4.5 g


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


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%


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

30

2.4 g

7.3 g

5.4 g


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


25. 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 25 g of HCl?

0.60 g

0.30 g

8.1 g

0.69 g


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%

31

83%

72%


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


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

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%


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

32

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


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


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)

33


32. For the reaction in number 31, what was the limiting reagent?

Hydrogen

Oxygen

Water

Neither, stoichiometric amounts were used.




N2(g) + O2(g) = 2 NO(g)

H2(g) + O2(g) = H2O2(g)

2 H2(g) + O2(g) = 2 H2O(g)

H2(g) + Cl2(g) = 2 HCl(g)


34. For the reaction in number 33, what was the limiting reagent?

Hydrogen

Chlorine

hydrogen chloride

Neither, they were used in stoichiometric amounts.

34




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)


Balancing equations exercise:

Write balanced equations for each of the following reactions.

Then classify each by type of reaction.

1.) Fe + O2 -----> Fe2O3

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

35

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.

36

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 proton-

proton 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.

37

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.

38

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.

39

- 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

40

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.

41

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

42

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).

43

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.

44

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!

45

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.

46

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?

47

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

48

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

49

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.

50

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

ΔH0f methane is represented by the equation:

C(s) + H2(g) CH4 (g)

51

This equation can be constructed using the equations for combustion of the reactants (carbon

and hydrogen) and the product (methane)

C(s) + O2(g) CO2(g)

-

394kjenthalpy 1

H2(g) + 1/2O2(g) H2O(l)-

242kjenthalpy 2

multiply this equation by 2 to get

2H2(g) + O2(g) 2H2O(l)-

484kjenthalpy 3

Add the first and third equation together to get:

C(s) + 2O2(g) + 2H2(g) CO2(g) + 2H2O(l)-

878kj

enthalpy (1 +

3) = 4

Now take away the equation for the combustion of

methane

CH4 (g) + 2O2(g) CO2(g) + 2H2O(l)-

891kjenthalpy 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) +13kjenthalpy ( 4 -

5)

52

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.

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

53

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).

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

54

http://ibchem.com/IB/ibnotes/brief/ene-hl.htm#gib

http://ibchem.com/IB/ibnotes/brief/ene-hl.htm#gib

ΔG = ΔH - Temperature(in kelvin) x ΔS

Enthalpy

change

Entropy

changeGibbs free energy Spontaneity

positive positivedepends on T, may be +

or -

yes, if the temperature is high

enough

negative positive always negative always spontaneous

negative negativedepends on T, may be +

or -

yes, if the temperature is low

enough

positive negative 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.

55

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

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 HH CC C=C NN NH

Mean bond enthalpy / kJ mol–1 436 348 612 944 388

(a) Explain the term mean bond enthalpy.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

56

(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

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 HH H

C CHHC C H H = – 1 3 6 k J m o lH+

H HH H

– 1

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

(3)

(Total 9 marks)

57

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) H2O(g) + 21 O2(g)

.....................

………………………………………………………………………….

.....................

………………………………………………………………………….

.....................

………………………………………………………………………….

.....................

………………………………………………………………………….

(3)

(b) The standard enthalpy of formation, 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.

21 H2(g) H(g) H = +218 kJ mol–1

C(s) C(g) H = +715 kJ mol–1

58

(i) By reference to its structure, suggest why a large amount of heat energy is

required to produce free carbon atoms from solid carbon.

...........

…………………………………………………………………………

.

...........

…………………………………………………………………………

.

59

(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) C(g) + 4H(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.

...........

…………………………………………………………………………

.

60

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!!

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.

61

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

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.

62

Now try applying these rules to a problem.

Exercise

Assign oxidation numbers to each element in the following:

1. H2S

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) ZnI2(s) + energy

63

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.

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:

Zn0 Zn2+ + 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):

64

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,

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:

65

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

66

Use the reduction potential chart: non-metals 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.

67

Unit Summary

Oxidation-reduction reactions involve the transfer of electrons between substances.

Electrochemistry is the study of the interchange of chemical and electrical energy.

68

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.

69

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 i.e. 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 moles 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 moles of

gas on the left of the equation, this is all reversed.

70

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 unfeasibly 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).

71

http://ibchem.com/IB/ibfiles/equilibrium/equ_img/haber2.gif

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.

72

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 y-

axis. 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.

73

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:

74

[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?

75

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

of X/moldm-3

Initial concentration

of Y/moldm-3

Initial rate of

formation of

XY3/moldm-3s-1

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

of H2 /moldm-3

Initial concentration

of NO /moldm-3

Initial rate of

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

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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 Explanation

Temperature increasing the

temperature increases

the rate of a reaction

Two reasons:

1. There are more particles with sufficient energy

to react (most important) - more successful

collisions

2. There are more collisions

Concentration Increasing the

concentration of a

reactant increases the

rate of the reaction

(usually)

There are more collisions as there are more

particles in closer proximity

Particle size The smaller the

particles the faster the

reaction. (note: the

solute particles in

solutions have the

smallest particle size

possible. and so

solutions react fastest)

Collisions occur at the surface of particles. The

larger the particle size the smaller the surface area

and the fewer collisions can occur.

Catalysts The presence of a

catalyst increases the

rate of a reaction

Catalysts provide an alternative mechanism with a

lower activation energy

77

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)

78

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)

3. O2(g)

4. F2(g)

79

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.

80

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 (non-metallic

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.

Transition Metals: Known for their ability to form coloured compounds and ions in solution

81

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,

after the initial gelatinous blue copper(II) hydroxide precipitate is

formed, Cu(OH)2,

82

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

CuCO3(s) ==> CuO(s) + CO2(g)

o or

o zinc carbonate ==> zinc oxide + carbon dioxide

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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)

or for the copper(II) ion Cu2+, blue in aqueous solution,

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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

FeCl3(aq) + 3NaOH(aq) ==> Fe(OH)3(s) + 3NaCl(aq)

the ionic equation is ...

Fe3+(aq) + 3OH-(aq) ==> Fe(OH)3(s)

o These hydroxide precipitates are basically solids, but of a somewhat

gelatinous nature because they incorporate lots of water in their structure.

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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< +

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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

2H2O2(aq) ==> 2H2O(l) + O2(g)

A useful reaction in the laboratory for preparing oxygen gas.

o 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 non-

metallic 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.

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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.

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.

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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.

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Transition metal compounds (often oxides) of copper, iron, chromium and cobalt are

used to pigments 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 hydroxide-carbonates (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.

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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

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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 anti-

corrosion 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.

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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) ..

e.g. Si + O2 ==> SiO2, or 4P + 5O2 ==> P4O10

o (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.

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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.

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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. 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

2. 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.

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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.

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.

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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

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.

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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

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

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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

ionization energy and electron affinity. For that reason, electronegativity follows similar

trends as its two constituent measures.

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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.

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.

101

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).

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

102

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

called diagmetic. Conversely, paramagnetic atoms do not have all of their electrons spin-

paired 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. Give the electronic configuration of the following atoms:

a) V

b) Cr

c) Co

d) Cu

e) Zn

2. 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.

103

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.

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.

104

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

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

http://ibchem.com/IB/ibfiles/organic/org_img/bp(alkanes)big.gif

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 CO 2 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

series name (old

name)

functional

groupnaming system description

Alkanal

(aldehyde)-CHO ends with -anal

Has a carbonyl C=O at the

end of a carbon chain, with

the carbon also attached to a

hydrogen

Alkanone

(ketone)-CO- ends with -anone

Has a carbonyl group C=O

but it is in the middle of a

carbon chain and the carbon

has no hydrogens attached

Alkanol (alcohol) -OH ends with -anol (may also go

at the start of the name as

hydroxy- if there is another

These have an -OH group (or

more than one in the case of

diols, triols etc) at any

108

more important group in the

molecule)position in the chain.

Alkanoic acid

(carboxylic acid)-COOH ends with -anoic acid

The group must go at the end

of a carbon chain as it has a

carbon attached to an oxygen

by a double bond and also an

-OH group (leaving the

carbon with only one other

bond which it uses to attach

to the rest of the carbon

chain)

Amine -NH2

ends with -ylamine or starts

the name with Amino-

(depending on whether there

is a more important

functional group int he

molecule)

The NH2 group can go

anywhere on the carbon

chain. It infers basic

characteristics to the

molecule. (ability to accept a

proton from an acid)

Amide -CONH2 ends with -anamide

The amine group is

associated with a carbonyl

group inferring different

characteristics to the

molecule. It can also be a

linking group -CONH-

between two alkyl chains

(proteins and nylon)

Halogenoalkanes -X

becomes a prefix: chloro-

bromo- iodo- folowed by the

rest of the name

The halogen atom may be

added into any position in the

chain. If there is more than

one then the prefixes di- tri-

tetra- etc are used.

Esters -COO- name is derived from the acid Esters are linkage

109

and alcohol that were used in

making the ester. The alkyl

group ending with the -COO

was the acid part and

becomes -(name)anoate and

the part after the -COO group

starts the name. For example:

ethyl ethanoate

compoounds formed by

condensation (esterification)

reactions between carboxylic

acids (or compounds

deriving from them) and

alcohols. They also occur

naturally and may be

polymers (polyester) see

amide linkage above

Ethers -O-

derives from the shortest

alkyl chain ending in -oxy

followed by the longest alkyl

chain eg. methoxyethane

CH3-O-C2H5

These are linkage

compounds where the

oxygen bridges two carbon

chains. They are of little

importance except as

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

series

Important reaction

typesreagent and conditions product (s)

Alkanes Combustion air/oxygen, heat carbon dioxide and

water

free radical halogen and UV light halogenoalkanes

111

substitution of

halogens

(mixture)

Alkenes

Combustion air/oxygen

heat

carbon dioxide and water

Addition Br2, Br2(aq),

H2

dibromoalkane, bromoalcohol,

alkane

Polymerization catalyst polyalkene

Alcohols

Combustion air/oxygen, heat carbon dioxide and

water

dehydration

(elimination)

phosphoric or

sulphuric acid

alkene

Oxidation sodium

dichromate/sulphuric

acid

alkanal, alkanone or

carboxylic acid

substitution of the -

OH group

phosphorus

pentachloride, sodium

halogenoalkane,

sodium alkoxide

Esterification carboxylic acid ester

aldehydes

(alkanals)

Addition ammonia compound containing

hydroxy and amine

Oxidation sodium dichromate/dil.

sulphuric acid

carboxylic acid

Reduction lithium aluminium

hydride

1º alcohol

addition - elimination

(condensation)

amines,

hydroxylamine,

depends on reagent

112

carboxylic acids

acid - base reactions base (eg sodium

hydroxide)

salt and water

Reduction lithium aluminium

hydride

alkanal (or 1º alcohol)

Esterification alcohol/conc. sulphuric

acid

ester

haloalkanes

(halogenoalkanes

)

substitution of the

halogen

sodium hydroxide,

ammonia

alcohol, amine

dehydrohalogenation

(elimination)

ethanolic NaOH/

reflux

alkene

Amines

reaction as a base

with an acid

mineral (strong) acid amine salt

condensation

(addition -

elimination)

carbonyl compounds depends on reagent

Amides

hydrolysis (breaking

apart in solution)

dilute acid/heat amine and carboxylic

acid


apart in solution)

dilute base/heat amine and sodium

carboxylate

Esters


apart in solution)

dilute acid/heat alcohol and

carboxylic acid


apart in solution)

dilute base/heat alcohol and sodium

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 c) CH3CH2CH2Br

114

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=9e214550-3dc3-48b8-b077-699bd6556d55>

[Accessed on 18 October 2012]

Energetics, < http://www.a-levelchemistry.co.uk/AQA%20AS%20Chemistry/Unit

%202/2.1%20Energetics/2.1%20home.htm> [Accessed on 18 October 2012]

118

http://www.a-levelchemistry.co.uk/AQA%20AS%20Chemistry/Unit%202/2.1%20Energetics/2.1%20home.htm

http://www.a-levelchemistry.co.uk/AQA%20AS%20Chemistry/Unit%202/2.1%20Energetics/2.1%20home.htm

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=9e214550-3dc3-48b8-b077-699bd6556d55



Transition Metals, < http://www.docbrown.info/page04/4_75trans.htm> [Accesson on 18

October 2012].

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http://www.docbrown.info/page04/4_75trans.htm

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