82
A DEDICATION This manual is dedicated to the late Professor Max L. McGlashan, a graduate of Canterbury College of the University of New Zealand, and Professor of Chemistry first at the University of Exeter, UK, and then as Head of Department at University College, London. His belief in, and writings on, the importance of absolute rigour and logic in the presentation of chemistry, especially in all areas involving measurement were far reaching, and of great importance to the proper teaching of chemistry. Unfortunately much teaching on measurement and of chemical calculation at the elementary level is still woefully inaccurate and illogical leading to unnecessary difficulties for students. Sections 7, 8 and 10 of this manual, which are an attempt to rectify this situation, owe much to his inspiration. INTRODUCTION TO LET’S TALK CHEMISTRY WHAT IS CHEMISTRY? Chemistry may be defined as the study of matter, its composition and properties, and the changes it undergoes. WHAT IS MEANT BY THE LANGUAGE AND VOCABULARY OF CHEMISTRY? Over time as chemists have developed basic ideas and concepts a distinctive chemical language has evolved. Chemists often use common words in a very special context (e.g. amount of substance). They use symbols for the elements and combine these with numbers to write formulae to represent different substances. They represent chemical reactions (the interaction of substances to give new substances) by writing chemical equations. To show the structure of various species they expand basic formulae to show the necessary detail. In short, chemists have developed a distinct vocabulary and language. As with other languages it has evolved and grown as new needs arise, and consequently is not always logical, or unambiguous. WHAT IS THIS MANUAL ABOUT? This manual is an introduction to the basic concepts, vocabulary and language of chemistry. As in studying a foreign language a sound grasp of this vocabulary and language is essential to the understanding of the subject. Many student problems arise from not knowing the basic language, or from non-rigorous use of by teachers. It takes time and practice to acquire a good appreciation and command of it. WHO IS IT FOR? It is written for all students who need or wish to study chemistry for its own sake or need it as the basis of another discipline (biology, medicine, physics, engineering) at first year university level. It is of special importance for those who have not had the opportunity to study chemistry over their formative years, or who have not been fortunate enough to have experienced good teaching. However, it is also useful to those who have successfully studied chemistry over several years at school to reinforce their basic knowledge and understanding of its all important language. It is also for teachers with limited background of chemistry but who are called on to teach the subject. In teaching it is so important that basic terms and definitions are precise and accurate. HOW IS IT USED? There are several ways this manual could be used.

Lets Talk Chemistry

Embed Size (px)

Citation preview

Page 1: Lets Talk Chemistry

A DEDICATION This manual is dedicated to the late Professor Max L. McGlashan, a graduate of Canterbury College of the University of New Zealand, and Professor of Chemistry first at the University of Exeter, UK, and then as Head of Department at University College, London. His belief in, and writings on, the importance of absolute rigour and logic in the presentation of chemistry, especially in all areas involving measurement were far reaching, and of great importance to the proper teaching of chemistry. Unfortunately much teaching on measurement and of chemical calculation at the elementary level is still woefully inaccurate and illogical leading to unnecessary difficulties for students. Sections 7, 8 and 10 of this manual, which are an attempt to rectify this situation, owe much to his inspiration.

INTRODUCTION TO LET’S TALK CHEMISTRY

WHAT IS CHEMISTRY? Chemistry may be defined as the study of matter, its composition and properties, and the changes it undergoes. WHAT IS MEANT BY THE LANGUAGE AND VOCABULARY OF CHEMISTRY? Over time as chemists have developed basic ideas and concepts a distinctive chemical language has evolved. Chemists often use common words in a very special context (e.g. amount of substance). They use symbols for the elements and combine these with numbers to write formulae to represent different substances. They represent chemical reactions (the interaction of substances to give new substances) by writing chemical equations. To show the structure of various species they expand basic formulae to show the necessary detail. In short, chemists have developed a distinct vocabulary and language. As with other languages it has evolved and grown as new needs arise, and consequently is not always logical, or unambiguous. WHAT IS THIS MANUAL ABOUT? This manual is an introduction to the basic concepts, vocabulary and language of chemistry. As in studying a foreign language a sound grasp of this vocabulary and language is essential to the understanding of the subject. Many student problems arise from not knowing the basic language, or from non-rigorous use of by teachers. It takes time and practice to acquire a good appreciation and command of it. WHO IS IT FOR? It is written for all students who need or wish to study chemistry for its own sake or need it as the basis of another discipline (biology, medicine, physics, engineering) at first year university level. It is of special importance for those who have not had the opportunity to study chemistry over their formative years, or who have not been fortunate enough to have experienced good teaching. However, it is also useful to those who have successfully studied chemistry over several years at school to reinforce their basic knowledge and understanding of its all important language. It is also for teachers with limited background of chemistry but who are called on to teach the subject. In teaching it is so important that basic terms and definitions are precise and accurate. HOW IS IT USED? There are several ways this manual could be used.

Page 2: Lets Talk Chemistry

If you are a complete beginner and are having to work on your own, the sequence of sections listed in the Contents is a logical one and you could work your way through the whole manual. However, it is more likely that you will be using this in conjunction with a course. In this case if you wish to review a traditional topic select from the contents the section which seems to be appropriate. However, it is envisaged that this manual will be most commonly used to look up definitions of words or concepts of concern. To do this find the word in the INDEX and then go to the section indicated. There you will find the definition or explanation you are seeking. But instead of finding just a definition, as in a normal alphabetical glossary, you will see the definition or explanation in a broader context. By reading material both before and after the entry your knowledge will be enhanced. It is this aspect of this manual which makes it different from other publications. At the end of each section there are a small number of EXERCISES on the most basic topics. Answers to these are given immediately before the index. WHAT DOES THIS MANUAL COVER? In general students who have completed the most senior level of their school’s chemistry curriculum should have covered most of the material in this manual. Notable exceptions to this are likely to be quantum numbers of section 3, spectroscopic analysis of section 12 and most of section 16 on rates of reaction and reaction mechanisms. All these topics are met early in tertiary chemistry courses, and the difficulties many students have with reaction mechanism appear to stem from a lack of understanding of the language used to express the concepts. A knowledge of elementary algebra is assumed, this being essential for quantitative work. ACKNOWLEDGEMENTS To the University of Auckland for providing one of its 1994 Teaching Improvement Grant of $1000.

To many colleagues for reading, criticising, making suggestions and proof reading at different stages of the development of this manual: Dr Robyn Dormer, Drs Paul and Sheila Woodgate, Dr Graham Wright, Heather Wansbrough (first year student), all of the University of Auckland; Gavin Peckham (University of Zululand, South Africa), Dr Siow Heng Loke (University of Malaya), May Croucher (New Zealand Forest Research Institute), Dr Bryce Williamson (University of Canterbury), and members of the University of York Science Education Group. In particular to Marten J. ten Hoor of the Netherlands, for his rigorous reading of and comments on the manual. Without his criticisms and corrections a significant number of errors or inaccuracies would have been present. To SI Chemical Data by Gordon Aylward and Tristan Findlay, 3rd edition, Wiley, Brisbane, from which the basic layout of the periodic table was taken. To General Chemistry by P.W. Atkins and J.A. Beran, 2nd edition, Scientific American Books, New York, for their excellent glossary from which many of the definitions were taken.

Page 3: Lets Talk Chemistry

CDROM - LET’S TALK CHEMISTRY

A CDrom - Let’s Talk Chemistry is also available. It contains this manual on the basic language and vocabulary of chemistry that may be printed out in full or in individual sections (chapters). As this manual is of necessity brief, this CD-ROM also contains a very much more friendly and helpful tutorial type electronic form in HTML, with fuller explanations, many figures and diagrams, and more examples and exercises all with detailed answers. The price is $25 (Australian and NZ) + $7 postage and packaging. It may be ordered by mail sending an Australian or NZ cheque to: A and B Scott, 62 Orchard Grove, Blackburn South, 3130 Vic., Australia. E-mail [email protected] or [email protected] for further information.

Page 4: Lets Talk Chemistry

1-1

SECTION 1 BASIC DEFINITIONS AND VOCABULARY ON STRUCTURE OF MATTER Chemistry: The study of matter, its composition and properties, and the changes it undergoes. Matter: Anything that has rest mass. To develop this topic we need to define the atom, the basic unit of common matter. Atom: A neutral particle consisting of a nucleus containing most of its mass, and electrons occupying most of its volume. Electron: A subatomic particle with a charge of -1. Nucleus: Consists of two types of subatomic particles; protons, each with an electric charge of +1, and neutrons which have no charge. Atomic number: Symbol Z, the number of protons in the atom. Mass number: Symbol A, the sum of the number of protons and neutrons in the atom. (Nucleon is a term for either a proton or neutron. Thus A is the number of nucleons in an atom and the number of neutrons is A-Z.) Element: A substance composed of atoms all of which have the same atomic number, i.e. the same number of protons in the nucleus, and thus the same number of electrons. The atomic number defines the element. Each element has a name (of varying historic origin) and a shorthand symbol. The periodic table on the previous page lists the elements in order of increasing atomic number. The atomic number is at the top of each entry, then the full name and then the symbol. The symbol has one or two letters. The first letter is always in capitals and the second in lower case. The origin of some symbols which may appear to bear no relationship to the full name come from Latin, Greek or German [e.g. 26, iron, Fe]. Metallic element: Elemental substance which has a "metallic" lustre, conducts electricity and heat well, and is malleable and ductile [e.g. copper]. All the elements to the left of the bold line on the periodic table are metals. Non-metallic element: Elemental substance which does not have metallic properties. [e.g. nitrogen] All the elements to the right of the dashed line on the periodic table are non-metals. Metalloid or Semimetal: An element with the physical appearance and some of the properties of a metal, but which has the chemical properties of a non-metal [e.g. arsenic].

This section introduces terms and definitions which make up the vocabulary of the language of chemistry. Starting with atoms it progresses to elements, molecules and compounds. It shows how a substance can be simply represented by a chemical formula made up of both letters and numbers. Many of the terms may be familiar to the layperson, but not necessarily their precise definitions in the context of chemistry.

Page 5: Lets Talk Chemistry

1-2

These are the elements between the bold and dashed lines on the periodic table. Isotopes: Atoms of the same element (same atomic number) but with different numbers of neutrons and hence a different mass number [e.g. all atoms of carbon have six protons but may have six (12C), seven (13C), or eight (14C) neutrons. 13C is an isotope of carbon.]. Some isotopes are unstable and emit particles or radiation over a period of time and are called radioactive. Radioactivity: The spontaneous breakdown of one type of atomic nucleus into another. Relative atomic mass: Symbol, Ar, the mean mass of an atom of an element allowing for the relative abundance of its naturally occurring isotopes relative to (divided by) one twelfth of the mass of one atom of the isotope 12C. (This topic is covered in section 8.) The value of Ar is given as the fourth item in each entry of the preceding periodic table. The chemical properties (behaviour) of the different elements can be understood in terms of the number and arrangement of the electrons in their atoms. The arrangement of the elements in the periodic table is based on this electronic structure. This topic is covered in section 3. Molecule: A discrete neutral particle consisting of a definite number of atoms bonded (held) together by chemical bonds. (Chemical bonds is the subject of section 4.) Substance: A single, pure type of matter [e.g. water, sucrose (common sugar), sodium chloride (common salt), argon (gas in many light bulbs)]. A pure substance may be elemental [e.g. oxygen, aluminium] or a compound [e.g. water, sodium chloride]. A pure elemental substance may consist of individual atoms, [e.g. He, Ne, Ar]; molecules, [e.g. nitrogen - N2, oxygen - O2, sulfur - S8, where the number of atoms in the molecule is given by a subscript number following the symbol for the element]; or large aggregates of atoms [e.g. all metals, carbon in the form of diamond or graphite]. Allotrope: Alternative form of an element differing in the way the atoms are linked [e.g. carbon - diamond and graphite; oxygen - O2 (dioxygen) and O3 (ozone)]. Each form is an allotrope. Compound: A pure substance consisting of atoms, or ions of two or more elements in a definite number ratio. It may consist of (i) discrete molecules [e.g. water, H2O; ammonia, NH3; sucrose, C12H22O11 where the number of atoms of each element in the molecule is given by the subscripted number immediately following it; or (ii) large aggregates of atoms [e.g. silica, (SiO2)x where every silicon atom is bound (joined) to four oxygen atoms, and every O atom is joined to two silicon atoms]; or (iii) ions. Binary compound: A compound containing only two elements [e.g. hydrogen chloride, HCl; water, H2O]. Ion: A charged particle. A cation is a positively charged particle, one containing fewer electrons than protons; an anion is a negatively charged particle, one containing more electrons than protons. An ion may be (i) monatomic, i.e. just one atom, [e.g. sodium ion, Na+; magnesium ion,

Page 6: Lets Talk Chemistry

1-3

Mg2+; chloride ion, Cl–; oxide ion, O2–]; (ii) polyatomic, i.e. a discrete number of atoms jointed together, [e.g. sulfate, SO4

2–; nitrate, NO3–; ammonium, NH4

+] or (iii) polymeric, i.e. large aggregate of atoms joined together, [e.g. silicate (SiO3

2–)x]. The overall charge is given as a superscript after the formula as shown. Ionic compound: A compound consisting of ions. The total charge on the cations equals the total charge on the anions, as substances are neutral. An ionic compound is an aggregate of those ions. However the charge on the ions is not shown in its formula [e.g. sodium chloride, NaCl, is made up of monatomic sodium ions, (Na+), and chloride ions, (Cl–); sodium sulfate, Na2SO4, is made up of sodium ions, (Na+ ) and sulphate ions, (SO4

2–); ammonium sulfate, (NH4)2SO4, is made up of ammonium ions, (NH4

+), and sulfate ions, (SO42–); sodium silicate,

(Na2SiO3), is made up of Na+ cations and polymeric (SiO32–)x anions.].

Matter exists normally in three states - gas (g), liquid (l) or solid (s). Phase: A particular state of matter [e.g. gas, liquid, solid]. Phase change: The change which occurs when a substance changes from one state to another, or undergoes a change in structure in the solid state. The common names for phase changes are: g → l condensation l → g evaporation l → s freezing s → l melting or fusion g → s deposition s → g sublimation Crystalline solid: A solid in which there is an orderly array of its constituents (atoms, molecules or ions). A crystalline solid may exist as a single crystal or as numerous crystals [e.g. common salt, NaCl; common sugar, sucrose, C12H22O11; diamond]. Mixture: A type of matter consisting of more than one substance and which can be separated into its components by making use of different physical properties of the substances present. Mixtures may be homogeneous or heterogeneous. Homogeneous mixture: A mixture in which any two samples have the same composition and properties [e.g. air (a gas); seawater (a liquid)]. Heterogeneous mixture: A mixture containing entities of sufficient size such that all samples are not identical [e.g. concrete; smoke]. Solution: A homogeneous mixture (usually with one component in larger amount); normally a liquid [e.g. seawater; hydrochloric acid ], but the term may be also apply to a solid, but not to a gas. Solvent: The component of a solution in largest amount [e.g. the water of seawater]. Solute: The components of a solution in lesser amounts [e.g. the salts in seawater]. Aqueous solution: A solution with water as the solvent. Crystallization: The process of a solute coming out of solution as crystals. Colloid: A mixture consisting of a finely divided phase (the dispersed phase) distributed

Page 7: Lets Talk Chemistry

1-4

uniformally in a continuous phase (the dispersion medium). Aerosols, liquid in gas [e.g. mist, fog] or solid in gas [e.g. smoke]. Emulsions, liquid in liquid [e.g. cream, fat in water, and butter, water in fat]. Sol or colloidal suspension, solid in liquid; slurry or paste when concentrated. Foam, gas in liquid or gas in solid. Gel, a polymeric dispersed phase in a liquid. The dispersed phase has a huge surface area which gives colloids their distinctive properties. Colloids are heterogeneous mixtures. Chemical formulae: Part of the language used by chemists to represent substances and compounds [e.g. H2O is the formula for water]. Molecular formula: The formula for a discrete molecule. It gives the symbol of each different element and the number of atoms of that element as a subscript immediately after the symbol for the element [e.g. ammonia, NH3; nitrogen, N2; glucose,C6H12O6; sulfuric acid, H2SO4]. There are now rules for the order in which the elements are listed based on chemical language which has evolved over time. However the order is best learned with experience (i.e. seeing it written). However, for indexes (listings) of organic compounds the order is C H O and then the remaining elements listed alphabetically. The formula may represent one discrete molecule of the substance or a macroscopic amount of it. Which is meant must be deduced from the context in which the formula is used. Empirical formula: The formula for a substance giving the simplest ratio of the atoms [e.g. glucose, CH2O]. Compounds which consist of large aggregates of atoms joined together are usually represented by their empirical formula [e.g. silica is normally represented by just SiO2 not (SiO2)x]. Elements which consist of aggregates of atoms bound together are represented simply by the symbol for the element [e.g. aluminium metal, Al]. Ionic compounds are represented by the formula for their cations and anions in their simplest whole number ratio, without showing the charge on either the cations or anions [e.g. NaCl represents sodium chloride which consists of Na+ and Cl– ions in a 1:1 ratio; Ca3(PO4)2 represents calcium phosphate which consists of Ca2+ and PO4

3– ions in a ratio of 3:2; Na2SiO3 represents a sodium silicate which consists of discrete Na+ ions and polymeric anions (SiO3

2–)x]. Note the use of brackets around polyatomic ions when a subscript is needed. A certain knowledge and experience of chemistry is required to understand fully what is meant by a given formula in a given context. (Further kinds of formulae giving more information on the structure of molecules and ions are discussed in sections 4-6.) Mineral: A naturally occurring substance of a characteristic composition and a crystalline structure. Rocks are composed of a mixture of minerals. The term is also used for organic substances such as coal and oil which are obtained by mining, but strictly these are not minerals as they are mixtures without definite chemical formulae. Chemical bonds: The forces holding atoms together in molecules or in aggregates. Commonly they are just called bonds. This topic is covered in section 4. Valence: The number of bonds an atom of a particular element can form in a molecule. [e.g. In water, H2O, one oxygen atom forms bonds to two hydrogen atoms, and each hydrogen atom forms one bond to an oxygen atom. The hydrogen atoms are not bonded to each other. The valency of O is two and of H is one. In silica, SiO2, each Si atom is bound to four O atoms and each O atom to two Si atoms. The valency of Si is 4 and of O is 2.] In simple ionic compounds containing monoatomic cations and anions the valency is simply the charge on

Page 8: Lets Talk Chemistry

1-5

the ions. [e.g. In NaCl made up of Na+ and Cl– the valency of both Na and Cl is one.] Polymer: A substance having large molecules consisting of repeated units (the monomers). Polymers do not have a definite formula as they consist of molecules of different chain length. [e.g. polyethylene, H-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2- or H-(CH2-CH2)x-H where x is a large number and the monomer is ethylene, C2H4; the polymeric silicates above.] Free-radical: An atomic, molecular, or ionic particle with an unpaired valence electron [e.g. chlorine atom, Cl· ]. Usually chemically reactive. See sections 3 - 4. To understand why pure substances have particular compositions and properties we need to know about the "electronic structure" of the atoms (i.e. the way the electrons are arranged about the nucleus of the different elements). Then we can rationalise the ratio in which atoms combine and whether they are molecular, polymeric or ionic, [e.g. why gaseous nitrogen consists of discrete N2 molecules; methane, ammonia, water and hydrogen fluoride consist of discrete molecules of CH4, NH3, H2O, HF respectively (H4C, H3N, OH2, FH are in principle equally valid formulae, but not normally used); why sodium chloride consists of Na+ and Cl– ions; why metals exist in nature mainly as cations, Mx+; why free-radicals are reactive]. The topic of electronic structure is covered in section 3. Organic chemistry: The chemistry of the compounds of carbon. Originally meant chemistry of products formed in living systems. Inorganic chemistry: The chemistry of compounds other than carbon, but includes CO, CO2 and carbonates. Nomenclature: The naming of substances. This is covered in section 6 (organic) and in section 13 (inorganic). EXERCISES From the periodic table give the name and symbol of the element, and the number of neutrons for the isotope, given the atomic number and the mass number. 1. Example: Z = 8 A = 16

Answer: oxygen, O, A-Z = 8 2. Z = 19 A = 39 3. 3. Z = 29 A = 65 4. Z = 86 A = 222 Write the formula of the following compounds from the given compositions: 5. Example: Ethane, molecular, two carbons and six hydrogens

Answer: C2H6 6. Example: Sodium carbonate, composed of Na+ and CO3

2– ions Answer: Na2CO3

7. Acetic acid, molecular, two carbons, four hydrogens and two oxygens 8. Glycine, molecular, two carbons, five hydrogens, two oxygens and one nitrogen

Page 9: Lets Talk Chemistry

1-6

9. Cystine, molecular, two nitrogens, two sulfurs, twelve hydrogens, four oxygens and six carbons

10. Potassium bromide composed of K+ and Br– ions 11. Ammonium nitrate composed of NH4

+ and NO3– ions

12. Magnesium nitrate composed of Mg2+ and NO3

– ions 13. Aluminium sulfate composed of Al3+ and SO4

2– ions

Page 10: Lets Talk Chemistry

2-1

SECTION 2 CHEMICAL REACTIONS AND EQUATIONS Chemical reaction: A chemical reaction occurs when a substance or a mixture of substances undergoes a change which produces new substances. Reactants: The initial substances of a reaction. Products: The new substances produced in a chemical reaction. Chemical equation: A symbolic representation of a chemical reaction using chemical formulae. [e.g. When gaseous dihydrogen, H2, and dioxygen, O2, are mixed at room temperature nothing happens, but when the mixture is sparked a large amount of heat is given out and water is formed. Equation: H2 + O2 → H2O

reactants product ] An arrow, normally from left to right shows the direction of the reaction. In a chemical reaction the atoms of the reactants rearrange to give new substances. Chemical bonds are broken and new chemical bonds made. Chemical bonds: The forces holding atoms together in matter. (see section 4) Balanced chemical equation: A chemical equation in which the numbers of atoms of each element is the same on both sides

[e.g. 2H2 + O2 → 2H2O]. Stoichiometry: The relative proportions in which elements form compounds or in which substances react. (From the Greek word for element, stoikheion.) Stoichiometric coefficient: The number immediately in front of the formula of the chemical substances in the balanced equation. The coefficients may be interpreted as the relative number of particles of reactants and products [e.g. For the above equation two molecules of dihydrogen react with one molecule of dioxygen to form two molecules of water]. Balancing chemical equations: Deriving the stoichiometric coefficients of reactants and products such that the equation is balanced. Although this can always be done by the mathematical method of linear equations, most equations can be balanced by inspection using the following hints. 1. If the number of atoms of one kind happen to be balanced when the formulae of

reactants and products are written down keep this so by placing 1 in front of these formulae.

2 Leave formulae containing only one kind of atom for the last. 3 Start with a kind of atom appearing in only two formulae. Two examples: Ostwald combustion of ammonia. (Combustion is a reaction with dioxygen.) The products are nitric oxide and water. Write the formulae for reactants and products,

This section introduces the language of chemical equations, the method of representing the changes which occur in a chemical reaction in which substances are consumed and new ones formed.

Page 11: Lets Talk Chemistry

2-2

NH3 + O2 → NO + H2O Hint 1 1NH3 + O2 → 1NO + H2O Hint 2 Leave O until last Hint 3 This is H. Balance H by writing 3/2 in front of H2O. 3H on left, therefore (3/2) x 2 on right 1NH3 + O2 → NO + 3/2H2O

Now O 1NH3 + 5/4O2 → 1NO + 3/2H2O One cannot have ½ or ¼ molecules. Multiply all coefficents by 4 to have only whole numbers 4NH3 + 5O2 → 4NO + 6H2O The combustion of ethane: The products are carbon dioxide and water. Write the formulae for reactants and products, C2H6 + O2 → CO2 + H2O Hint 1 Not applicable. Hint 2 Leave O until last Hint 3 Could balance C or H next. Balance carbon, C, by writing 2 in front of CO2. C2H6 + O2 → 2CO2 + H2O (2 C on both sides; 2 atoms in one molecule of ethane, C2H6, and one atom in each of 2 molecules of carbon dioxide, CO2.) Now balance H. There are 6 H atoms on the left, so write 3 in front of H2O. C2H6 + O2 → 2CO2 + 3H2O (6 H atoms in ethane and 2 H atoms in each of 3 water molecules.) Balance O. There are 2 O atoms on left and 7 O atoms on right. Write 7/2 in front of O2. C2H6 + 7/2O2 → 2CO2 + 3H2O This would be interpreted as one molecule of ethane reacting with 3½ molecules of dioxygen to give 2 molecules of CO2 and 3 molecules of water. But one cannot have half a molecule. This problem is overcome by multiplying all coefficients by two. 2C2H6 + 7O2 → 4CO2 + 6H2O 2 molecules of ethane react with 7 molecules of dioxygen to form 4 molecules of carbon dioxide and 6 molecules of water. (In section 8 another interpretation of the stoichiometric coefficient is given.) Summary: A balanced chemical equation is a concise expression for conveying much information. [e.g. the equation S8 + 8O2 → 8SO2

implies (i) a molecule of sulfur contains 8 atoms of sulfur (ii) a molecule of oxygen contains 2 atoms of oxygen (iii) sulfur and oxygen react to form sulfur dioxide, each molecule of which contains one atom of S and 2 atoms of O (as the name suggests) (iv) one molecule of sulfur requires 8 molecules of dioxygen for a complete reaction.]

Nett ionic equation: A chemical equation for a reaction occurring in aqueous solution between ionic compounds, but showing only the reacting ions. [e.g. When solutions of silver nitrate, AgNO3, and sodium chloride, NaCl, are mixed a precipitate of silver chloride forms: Ag+(aq) + Cl–(aq) → AgCl(s) The Na+ and NO3

– ions present in solution are not shown because they are not reacting. Such ions are sometimes called spectator ions. The (aq) and (s) in the equation stand for aqueous and solid respectively and make the equation more meaningful. ] Nuclear equation: An equation for a nuclear change or reaction. [e.g. the production and decay of cobalt-60:

Page 12: Lets Talk Chemistry

2-3

Co5927 + n1

0 → Co6027 → Ni60

28 + e01− + γ-photon

Normal cobalt, 59Co, absorbs a neutron (n) in a nuclear reactor to give 60Co which decays with elimination of an electron from the nucleus to give 60Ni with excess energy and this is lost as a gamma-photon.] Note that the mass numbers (superscripts) balance and the atomic numbers or charge (subscripts) balance. EXERCISES Balance the following chemical equations: 1. Example: N2 + H2 → NH3

ammonia Balance N by writing 2 in front of NH3

N2 + H2 → 2NH3 Balance H by writing 3 in front of H2

N2 + 3H2 → 2NH3 2. C6H6 + O2 → CO2 + H2O 3. H2 + Cl2 → HCl 4. CH4 + H2O → CO + H2 (benzene) 5. Fe3O4 + C → Fe + CO 6. NO + O2 → NO2 7. NaNO3 → NaNO2 + O2 8. NH3 + O2 → N2 + H2O 9. NH3 + O2 → N2O + H2O 10. CO2 + H2 → CO + H2O Write nett ionic equations for the following reactions on mixing aqueous solutions of the following soluble ionic compounds (salts). 11. Solutions of thallium nitrate, TlNO3 and potassium fluoride, KF, to give a precipitate

of thallium fluoride. 12. Solutions of copper sulfate, CuSO4, and sodium carbonate, Na2CO3, to give a

precipitate of copper carbonate. 13. Solutions of calcium nitrate, Ca(NO3)2, and sodium phosphate, Na3PO4, to give a

precipitate of calcium phosphate. 14. Solutions of magnesium sulfate, MgSO4, and barium hydroxide, Ba(OH)2, to give a

precipitates of barium sulfate and magnesium hydroxide.

Page 13: Lets Talk Chemistry

3-1

SECTION 3 THE ELECTRONIC STRUCTURE OF ATOMS OF THE ELEMENTS Electronic structure of atoms: The arrangement of electrons around the nucleus of the atom. The properties of atoms can be understood in terms of Quantum Theory, which involves the Heisenberg Uncertainty Principle and the Schrödinger Wave Equation. Quantum Theory: A theory that states that the energy of an object can only change by discrete steps. A change involves a packet of energy called a quantum. Heisenberg Uncertainty Principle: The position and momentum of a particle cannot both be known simultaneously. This implies that in an atom the position and momentum of an electron cannot both be known simultaneously. (Thus a model of an atom containing electrons in fixed orbits around the nucleus is untenable.) Schrödinger Wave Equation: A mathematical expression ascribing wave-like properties to matter. When applied to atoms it describes the properties of electrons in atoms. This equation gives rise to the concepts of energy levels, atomic orbitals and quantum numbers. Electronic energy levels: Allowed energies of electrons in atoms. Atomic orbital: A mathematical expression from the Schrödinger Wave Equation from which, for each energy level, the probability of finding the electron at different positions from the nucleus can be calculated. The atomic orbital can be depicted as an "electron-cloud" with the nucleus at the centre, the denser the cloud the greater the probability of the electron being there. Only two electrons can occupy the same orbital. Quantum numbers: Numbers which label the orbital and spin of an electron. Electron pair: Two electrons in the same orbital. They must have opposite spins.

To understand why pure substances have particular compositions and properties we need to know about the "electronic structure" of the atoms (i.e. the way the electrons are arranged about the nucleus of the atoms of different elements). Then we can rationalise the ratio in which atoms combine and whether they are molecular, polymeric or ionic, [e.g. why gaseous nitrogen consists of discrete N2 molecules; methane, ammonia, water and hydrogen fluoride consist of discrete molecules of CH4, NH3, H2O, HF respectively; why sodium chloride consists of Na+ and Cl– ions; why metals exist in nature mainly as cations, Mx+; why free-radicals are reactive. Early last century a model of the atom (Bohr model) in which electrons circulated around the nucleus in orbits, just as planets do around the sun, was developed. This model proved inadequate to explain many phenomena and was replaced in the 1920's, when it was postulated that the behaviour of electrons in atoms could be described by mathematical equations similar to those used to describe the motion of standing waves in a string. From this model the electrons can be visualised as electron clouds of various shapes with the nucleus of the atom at their centre.

Page 14: Lets Talk Chemistry

3-2

Spin of an electron: The intrinsic angular momentum of an electron. Occurs in only two senses denoted ↑ and ↓. Electron shells: The electrons in an atom exist in shells, each shell being made up of atomic orbitals or subshells. Principal quantum number: Symbol n, an integer, 1,2 3... which defines the shell. The smaller n is, the lower the energy of the electron (more energy required to remove the electron from the atom), and the closer on average it is to the nucleus. First character in designation of an orbital. Azimuthal quantum number: Symbol l, defines the subshell or kind of orbital, and can have the values 0,1,...,n-1. An orbital with l = 0 is called an s orbital; with l = 1 is called a p orbital; with l = 2 is called a d orbital; with l = 3 is called an f orbital. Second character in designation of an orbital. Magnetic quantum number: Symbol m1, specifies the particular orbital of a subshell and can have values -l, -l+1,...0,...,l-1,l. Spin quantum number: Symbol ms, specifies the spin of an electron and can have values of +½ (↑) or -½ (↓). Occupancy of shells: The first shell, n = 1, can hold 2 electrons in one orbital, labelled 1s. (l = 0 for an s orbital) The second shell, n = 2, can hold 8 electrons in four orbitals, one labelled 2s and three labelled 2p. (l = 1 for a p orbital). The third shell can hold 18 electrons in nine orbitals, one 3s, three 3p and five 3d. (l = 2 for a d orbital) The fourth shell, n = 4, can hold 32 electrons in 16 orbitals, one 4s, three 4p, five 4d and seven 4f. (l = 3 for an f orbital) Within a shell the energy levels of the orbitals (subshells) is s < p < d < f. The bottom line in each entry of the preceding periodic table gives the number of electrons in the shells in the ground state of that element [e.g. potassium, K, 2.8.8.1, has 2 in the first shell, 8 in the second, 8 in the third and one in the fourth.]. Ground state: The state of an atom when all the electrons are in the lowest allowed energy levels. Electron configuration: A statement of the arrangement of electrons in the orbitals [e.g. Cl, 1s22s2p63s2p5]. Each principal quantum number is shown only once and the number of electrons in each subshell is shown as a superscript following the symbol for the orbital. The ground state electron configuration of any element can be written down by filling the orbitals in order using the energy levels: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f [e.g. cobalt, Co, Z = 27, thus 27 electrons. 1s22s2p63s2p64s23d7. This may be written as 1s22s2p63s2p6d74s2.]

Page 15: Lets Talk Chemistry

3-3

Valence electrons: Those electrons in the outermost shell and in unfilled subshells [e.g. Cl has 7 valence electrons (3s2p5) and Co has 9 valence electrons (3d74s2)]. Valence electrons are involved in chemical bonds - section 4. The Periodic Table: A table showing the elements in rows and columns in a manner which shows up relationships between the properties of the elements. Periods: Rows of the periodic table. Elements in the same row are in the same period [e.g. calcium, Ca, and copper, Cu, are both in the 4th period]. The number of the period (row) is equal to the principal quantum number of the outermost valence shell of the atoms. Groups: Columns of the periodic table. Elements in the same column are in the same group and have the same number of valence electrons (which accounts for their similarities) [e.g. carbon, C, and tin, Sn, are both in group 14 and both have four valence electrons]. This numbering replaces a previous system, shown as Roman numbers on the table, still used by some older chemists. Blocks: Groups having the same valence orbitals. Groups 1-2 are s-block because their elements have only s valence electrons; groups 3-12 are d-block because their elements have only s and d valence electrons; groups 13-18 are p-block because their elements have s and p valence electrons. Alkali metals: The metals (elements) of group 1. Alkaline earth metals: The metals (elements) of group 2. Halogens: Elements of group 17 [e.g. chlorine]. Halide: A binary compound of a halogen and another element [e.g. HCl, CaCl2, PCl3], or with a group [e.g. CH3Cl, chloromethane but also called methyl chloride; see section 6-2]. Halide ion: Monoatomic anion of a halogen [e.g. chloride ion, Cl–]. Transition metals: The metals (elements) of the d-block. Ionisation energy: The first ionisation energy is the minimum energy required to remove an electron from a neutral atom in the gas phase: E(g) → E+(g) + e–(g) The second ionisation energy is the minimum energy to remove an electron from this gaseous ion: E+(g) → E2+(g) + e– (g) Similarly for successive ionisation energies (I.E.). The variation of I.E with position in the periodic table is important in understanding the chemical properties of the elements. In general the ionisation energy increases from left to right in a period as the number of protons in the nucleus is increasing and therefore the attractive force between it and the electron is increasing. The first I.E. of the first element of a period is much lower than that of the last element in the previous period as the electron lost is from a shell of higher principal quantum number and hence energy. Excited state: The state of an atom when an electron is in an orbital of energy greater than that in the ground state. When an electron changes energy level (orbital) a quantum of energy is emitted or absorbed as a photon.

Page 16: Lets Talk Chemistry

3-4

Photon: A particle-like package of electromagnetic radiation. The energy, E, of the photon is related to the frequency, v, of the radiation by the expression E = hv where h is the Planck constant. EXERCISES Write the electron configuration of the ground states of the following elements: 1. Example: selenium, Se

Answer: From the periodic table Z = 34; there are 34 electrons to be placed in the orbital energy series.

1s22s2p63s2p64s23d104p4 2, 10, 18, 20, 30, 34 accumulative number of electrons or 1s22s2p63s2p6d104s2p4

2. carbon 3. fluorine 4. iron 5. arsenic 6. silver 7-12. Give a possible value for the principal quantum number and for the azimuthal quantum

number for a valence electron of the elements in questions 1-6 above. 7. Example: selenium, Se

Answer: For Se the valence electrons are 4s and 4p. For 4s n = 4, l = 0 For 4p n = 4, l = 1

Give the orbital of an electron with each of the following quantum numbers: 13. Example: n = 3, l = 2

Answer: 3d 14. n = 2, l = 1 15. n = 5, l =0 16. n = 6, l =3 In which period, group and block of the periodic table are the following elements? 17. Example: Strontium

Answer: group 2, 5th period, s-block 18. xenon 19. gold 20. silicon 21. lithium 22. The second ionisation energy of sodium is much greater than that of magnesium.

Explain in terms of their electron configurations. (Hint: write down the electron configurations and see which electrons are being lost.)

Page 17: Lets Talk Chemistry

4-1

SECTION 4 CHEMICAL BONDS Chemical bonds: The forces holding atoms together in matter. In most matter atoms exist close together in aggregates. The inner electrons are held tightly by the nucleus, (i.e. they have very high ionisation energies), but the valence electrons can be attracted to the nuclei of two or more atoms simultaneously. It is this electrostatic attraction of the valence electrons to two or more nuclei that provides the forces which hold atoms together and is the basis of the chemical bond. Covalent bond: A pair of electrons shared between two atoms (nuclei) in a molecule or polyatomic ion. Single bond: One pair of shared bonding electrons; represented by a line [e.g. H-H in H2]. Double bond: Two pairs of shared bonding electrons; represented by a double line [e.g. O=C=O in CO2]. Triple bond: Three pairs of shared bonding electrons; represented by a triple line [e.g. N≡N in N2]. The chemical behaviour of different elements can be understood by considering the number and arrangement of valence electrons in their atoms. With the exception of metals, almost all atoms in stable substances have 2 (H, He), 8, 18, or an even number between 8 or 18 electrons in their outermost shell. (High temperatures and low pressures change this.) An octet (8) of electrons in the third shell (which can accommodate 18) is common. Now we can understand the formulae and properties of some simple compounds that we have met in previous sections. Methane, CH4. The carbon nucleus (atom) is surrounded by four hydrogen nuclei. Number of valence electrons = 4(C) + 4 x 1(H) = 8. These 8 are in four pairs around C and each H has

one pair. We represent methane by the Lewis structure

C HH

H

H

In most matter atoms exist close together in aggregates. The inner electrons are held tightly by the nucleus, (i.e. they have very high ionisation energies), but the valence electrons can be attracted to the nuclei of two or more atoms simultaneously. It is this electrostatic attraction of the valence electrons to two or more nuclei that provides the forces which hold atoms together and is the basis of the chemical bond. This section defines different classes of bonds and the language of depicting them on paper.

Page 18: Lets Talk Chemistry

4-2

Lewis structure: A chemical formula (diagram) which shows the arrangement of the atoms and valence electrons in the species. Water, H2O. The oxygen nucleus has two H nuclei attached. Again there are 8 valence electrons, 6(O) + 2 x 1 (H). Lewis structure:

OHH

OHH

The two pairs of dots represent the two non-bonding electron pairs in the valence shell of the O atom. However, these are not always shown as their existence can be inferred. The inner shell electrons which we can consider belonging exclusively to the one nucleus are never shown in Lewis structures. Dinitrogen, N2. 10 valence electron, 2 x 5 (N). The Lewis structure: :N≡N: Each N nucleus is surrounded by 8 valence electrons, 3 pairs of electrons being attracted simultaneously to two N nuclei. Dinitrogen contains a triple bond . Carbon dioxide, CO2, and silicon dioxide, SiO2. In CO2 and an SiO2 unit there are 16 valence electrons 4 (C) or (Si) + 2 x 6 (O). Carbon dioxide exists as discrete CO2 molecules. We can rationalise this by the Lewis structure O=C=O where each atom has 8 valence electrons and the molecule has two double bonds. Silicon dioxide (quartz) is a solid. We can rationalise this by writing the Lewis structure for a small part of the compound:

SiO

O

O

O

Si

Si

Si O

O

O

Si

Si

Si

O

O

O

Si

Si

O

Again each atom is surrounded (tetrahedrally) by 8 valence electrons, each Si with 4 bonding pairs, and each O with two single bonds and two non-bonding electron pairs. CO2 and SiO2 have different structures. The explanation for this is beyond the scope of this document. The existence of discrete molecules of CO2 implies the energy of CO2 in this form is lower than that of a polymeric structure. Sodium chloride, NaCl, is an ionic solid composed of Na+ ions surrounded by Cl– ions and vice-versa. Each Na+ and Cl– ion has eight electrons in its outermost shell. (The valence electrons are all located on the Cl– ion.) Ionic bonds hold the structure together. Ionic bonds: The electrostatic attraction between cations and anions. The bonding electrons are localised on the anions.

Page 19: Lets Talk Chemistry

4-3

Calcium phosphate, Ca3(PO4)2 consists of Ca2+ cations surrounded by PO43– anions and vice-

versa and is therefore also an ionic compound, the cations and anions being held together by electrostatic attraction. However the bonds holding the P and O nuclei together in the phosphate anion are covalent. The same is true of the bonds in a sulfate anion or a carbonate anion, etc. Metallic bonds: The bonds holding atoms of a metal together. The simplest picture is that of cations in a sea of valence electrons spread out or "delocalised" over the whole structure. Polar bond: A covalent bond in which the bonding electrons are not evenly shared, the electrons being attracted more to the more electronegative atom [e.g. H-Cl, which has a polar bond with H partially positive and Cl partially negative]. Electronegativity: A measure of the ability of an atom to attract electrons to itself in a compound. The order of electronegativity is:

F > O > Cl > N > Br > I > S > C > H > P > B > Si Polar molecule: A molecule with a positive side and a negative side [e.g. H-Cl, (δ+)H-Cl(δ-) where δ means slightly; H2O with O slightly negative and the H's slightly positive]. Bond length: The average distance between the nuclei of the two atoms bonded together. (Average because the nuclei are vibrating). Bond strength: The energy needed to break a chemical bond with the bonding electrons being equally divided between the two fragments. (see section 11) Intramolecular bonds: Bonds between atoms in a molecule. [e.g. H-O bonds of water] Intermolecular bonds: Bonds or attractive forces between different molecules. [e.g. hydrogen and Van der Waals bonds] Hydrogen bond: A weak bond between fluorine, oxygen or nitrogen in one molecule or ion and a hydrogen atom bonded to a fluorine, oxygen or nitrogen atom in another molecule or ion or in the same molecule. Properties of water and the structure of many biological substances [e.g .proteins and DNA] are explained in terms of hydrogen bonds. Van der Waals bonds (forces): The intermolecular forces which hold molecules together in liquids and solids. In vaporisation and sublimation these bonds are broken, and in condensation and deposition they are made.

Page 20: Lets Talk Chemistry

4-4

EXERCISES Show the polarity of the following bonds: 1. Example: C-Cl

Answer: (δ+)C-Cl(δ-) Chlorine more electronegative than carbon 2. C-O 3. Si-F 4. Cl-P 5. H-C 6. H-N State whether the following molecules are polar and if so show the polarity: 7. Hydrogen fluoride, HF 8. Boron trifluoride, BF3, a planar triangular molecule 9. Carbon dioxide, CO2, a linear molecule 10. Sulfur dioxide, SO2, a bent molecule 11. Ammonia, NH3, a pyramidal molecule 12. Methane, CH4, the carbon atom is tetrahedrally surrounded by the 4 hydrogens

Page 21: Lets Talk Chemistry

5-1

SECTION 5 STRUCTURAL FORMULAE Structural formula: A chemical formula which shows the groupings of atoms in a compound. Lewis structures (defined in section 4) of discrete molecules and ions are an essential part of the chemical language. They are used to show details of structure and of chemical reactions. The ability to write Lewis structures for basic molecules and ions is an essential skill for chemists. The following rules require knowledge only of the number of valence electrons of an atom of each element, and the arrangement of the nuclei, i.e. which atoms are joined or bonded together. Rules for Drawing Lewis Structures (a) Determine the total number of valence electrons in the molecule or ion by adding

together the numbers of valence electrons of each atom, and if an anion, by adding the overall charge of the ion, and if a cation, by subtracting the overall charge of the ion.

(b) Place the atoms in their relative positions. (c) Draw a line representing a single bond containing two electrons between joined

atoms. (d) Distribute the remaining electrons evenly in pairs on the outer atoms so these have up

to eight electrons (except for hydrogen which has two). Any still not used after this should be placed on the central atom.

(e) If the central atom is now surrounded by fewer than eight electrons, move sufficient

non-bonding pairs from outer atoms other than halogens to between joined atoms, thus making them bonding, to bring the number on the central atom up to a maximum of eight.

(f) Count the number of electrons "owned" by each atom assuming bonding electrons are

shared evenly. To evaluate the formal charge at that atom compare the result with the number of valence electrons of the neutral atom. Show only non-zero charges.

(g) For central atoms from the third row or later rows of the periodic table, move further non-bonding pairs to bonding positions to lower the positive formal charge on the central atom to one or zero.

Structural formulae of molecules and ions give more information than simple chemical formulae. Lewis structures show which atoms are bonded to which using the letter symbols for the atoms, and the arrangement of the valence electrons forming the bonds. A line represents a two-electron bond. Lewis structures are like words of chemical language and are found in all chemical literature. It is essential that you can write Lewis structures for molecules and ions assuming knowledge of which atoms are bonded to which. This section gives simple rules which allow you to do this. It also introduces you to the concept of the shape of molecules.

Page 22: Lets Talk Chemistry

5-2

Formal Charge: The electric charge of an atom in a molecule or ion assuming perfect covalent bonding. [e.g. in NH4

+, N "owns" four electrons and has the formal charge +1 as an N atom has five valence electrons.] Examples:

Rule CO3

2-

NO2

SO2

a

C O 4 + 3 x 6 + 2 = 24

N O

5 + 2 x 6 = 17

S O

6 + 2 x 6 = 18

B, c

CO

OO

N OO

S OO

D, e

CO

OO

N OO

S OO

E, f

4e-

6e-

7e-

CO

OO

4e-

6e-7e-

N OO

5e-

6e-7e-

S OO

f

CO

OO

N OO

S OO

structure 1

g

S OO

6e-

6e-

S OO

S OO

structure 2

Examples of the application of the rules for drawing Lewis structures

In the case of CO3

2– and NO2 above each C-O and both N-O bonds are identical although a single Lewis structure might suggest otherwise. The problem can be overcome by writing the three possible Lewis structures for CO3

2–, or the two possible Lewis structures for NO2 and drawing a double headed arrow, ↔, between them. Each individual Lewis structure for a

Page 23: Lets Talk Chemistry

5-3

molecule or ion is called a resonance structure, and the true structure of the molecule or ion is a composite of these called the resonance hybrid. (See benzene, page 6-5) Of course, many molecules or ions do not have a "central atom". Organic molecules with more than one C are obvious examples. Provided the structure is known, (i.e. which atoms are bonded to which), extending the rules with common sense will work. [e.g. C2H2, H-C-C-H, put pairs on C atoms evenly, and move both pairs to between C atoms to make a triple bond.

H C C HC CH H ]

Condensed structures. The information provided by a Lewis structure can frequently also be provided by a modified or condensed form which is easily understood once some basic ideas are mastered. Often the non-bonding electron pairs are not shown. In many organic compounds (compounds of carbon), chains of carbon atoms joined together are present. Thus butan-1-ol, for which the complete Lewis structure is can be represented as CH3CH2CH2CH2OH, where atoms joined to a carbon are shown on its right before the adjoining atom is given. Butan-2-ol is CH3CH2CH(OH)CH3. Ethanoic acid: is written as CH3COOH or CH3CO2H. A degree of condensation appropriate for the required information can be chosen. See section 6 for further condensation of the carbon skeleton of organic compounds. The structure of sulfuric acid: may be written as (HO)2SO2, which is much more meaningful than the more usual representation (H2SO4) in that it shows that the H's are joined to oxygen. Familiarity with these types of formulae comes with practice and experience based on the electron structure and valency of the atoms of the elements. Molecular shape. The shape of a simple molecule or ion describes the relative positions of the nuclei of the atoms in the molecule. H2O is described as bent because the H-O-H atoms do lie on a straight line. NH3 is described as trigonal pyramid because the N atom is above the plane of the three H's which are at the corners of an equilateral triangle. CH4 is described as tetrahedral because the C atom is in the middle and surrounded tetrahedrally by the four H

C CH

H

H

C

H

H

C

H

H

O

H

H

H

SOH O H

O

O

.......

..

..

.. ..

.

C CH

H

H

O H

O

Page 24: Lets Talk Chemistry

5-4

atoms. BF3 is trigonal planar, because the boron atom is in the centre of an equilateral triangle with fluorines at each corner. CO2 and C2H2 are linear the O-C-O atoms and H-C-C-H atoms lying in straight lines. It is difficult to give a simple description of shape of more complex molecules such as butanol, ethanoic acid and sulfuric acid above. However we can describe the arrangement of the atoms around a single atom. Thus in butanol above, each C atom is surrounded tetrahedrally by four other atoms. In ethanoic acid the carbon atom bonded to two O's and a C is in the centre in an equilateral triangle, the other carbon atom being tetrahedrally surrounded by the three H's and one C. Valence-shell electron pair repulsion (VSEPR) theory: A theory which gives a simple rationale of shape (reason for the particular arrangement or shape). The "groups" of electrons (i.e. a non-bonding electron pair, a bonding pair, the four electrons of a double bond, the six electrons of a triple bond) around the central atom (as shown in the Lewis structure) keep as far away from each other as possible. [e.g. NH3 is a trigonal pyramid - there are four pairs of electrons tetrahedrally around the N, but the shape description only involves the atoms; H2O is bent - there are four pairs of electrons tetrahedrally around the O; the carbonate ion, CO3

2–, is a planar triangle; sulfur dioxide, SO2 is bent. In sulfuric acid the S is tetrahedrally surrounded by four O's, and the H-O-S atoms do not lie on a straight line.] EXERCISES Draw Lewis structures for the following species. Unless otherwise stated the first element in the formula is central and bonded individually to the other atoms. Remember to show non-zero formal charge. 1. Tetrafluoromethane, CF4 2. Ammonia, NH3 3. Sulfur trioxide, SO3 4. sulfite anion, SO3

2– 5. Nitrate anion, NO3– 6. Nitrite anion, NO2

– 7. Ammonium cation, NH4

+ 8. Nitrous oxide, N2O, (one N central, linear) Draw the full Lewis structures of the compounds of condensed structures: 9. HCHO 10. CH3CH2CH2Cl 11. CH3COCH3 12. CH3CHCH2 13. CH3CH2OOCH3 14. CH3COOCH3 15. CH3CHBrCH2CH3 16. HOCH2CH2CHO 17. FCH2CH2CCl(OH)CH2OH 18. CH2ICH2CH2COCCCH(NH2)CH3

Page 25: Lets Talk Chemistry

6-1

SECTION 6 NOMENCLATURE AND STRUCTURE OF ORGANIC COMPOUNDS Greek and Latin prefixes play an important role in nomenclature:

Greek Latin ½ hemi semi 1 mono uni 1½ sesqui 2 di bi 3 tri ter 4 tetra quadri 5 penta quinque 6 hexa sexi 7 hepta septi 8 octa octo 9 ennea nona 10 deca deci Organic compounds: Compounds containing the element carbon [e.g. methane, butanol]. (CO, CO2 and carbonates are classified as inorganic.) See page 1-4. Special characteristics of many organic compounds are chains or rings of carbon atoms bonded together, which provides the basis for naming, and the presence of many carbon-hydrogen bonds. The valency of carbon in organic compounds is 4. Hydrocarbons: Compounds containing only the elements C and H. Straight chain hydrocarbons are named according to the number of carbon atoms: CH4, methane; C2H6 or H3C-CH3, ethane; C3H8 or H3C-CH2-CH3, propane; C4H10 or H3C-CH2-CH2-CH3, butane; C5H12 or CH3CH2CH2CH2CH3, pentane; C6H14 or CH3(CH2)4CH3, hexane; C7H16, heptane; C8H18, octane; C9H20, nonane; C10H22, CH3(CH2)8CH3, decane. Saturated compound: One having only single bonds [e.g. ethane, C2H6]. Alkane: A saturated hydrocarbon [e.g. all the above compounds]. Systematic nomenclature is largely based on the above series of saturated alkanes with the number of carbon atoms associated with the stems meth-, (1); eth-, (2); prop-, (3); but-, (4);

Many organic compounds have common names which have arisen historically, or have been given to them when the compound has been isolated from a natural product or first synthesised. As there are so many organic compounds chemists have developed rules for naming a compound systematically, so that it structure can be deduced from its name. This section introduces this systematic nomenclature, and the ways the structure of organic compounds can be depicted more simply than by full Lewis structures. The language is based on Latin, Greek and German in addition to English, so a classical education is beneficial for chemists!

Page 26: Lets Talk Chemistry

6-2

pent-, (5); hex-, (6); hept-, (7): oct-, (8); non-, (9); dec-, (10); alk-, general. The ending -ane means no unsaturation (no double or triple bonds). Alkanes may be non-cyclic (acyclic) or cyclic (contain rings). The general formula for an acyclic alkane is CnH2n+2 and for one containing one ring CnH2n. In cyclic alkanes the stem gives the number of carbon atoms in the ring. [e.g. c-C6H12 is cyclohexane, where c- means cyclic] Unsaturated compound: A compound with one or more multiple (double or triple) bonds [e.g. ethene (ethylene), CH2=CH2 ]. Alkene: A hydrocarbon containing a double bond [e.g. C3H6, CH3-CH=CH2, propene]. Alkyne: A hydrocarbon containing a triple bond [e.g. C4H6 or CH3CH2C≡CH, but-1-yne]. The endings -ene and -yne are for the double or triple bond respectively. The general formula CnH2n+2 loses two H's for each ring or each double bond and four H's for each triple bond. The position of the multiple bond is shown by a number in the name, numbering from the end of the chain to give the smallest number [e.g CH3CH2CH2CH=CH2CH2CH3 is hept-3-ene (formerly 3-heptene) not hept-4-ene]. Alkyl group: In general, an alkane minus one hydrogen atom and represented by R [e.g. CH3- is methyl (sometimes shown as Me); CH3CH2- is ethyl (sometimes shown as Et); CH3CH2CH2- is propyl (sometimes shown as Pr); CH3CH2CH2CH2- is butyl (sometimes shown as Bu)]. In straight chain alkanes the non-terminal carbon atoms are bonded to two other carbon atoms. In a branched alkane one or more carbons are bonded to three or four other carbon atoms. Primary carbon atom: A carbon atom bonded to only one other C atom. Secondary carbon atom: One bonded to two other C atoms. Tertiary carbon atom: One bonded to three other C atoms. Quaternary carbon atom: One bonded to four other C atoms. Branched hydrocarbons are named after the longest chain (saturated) or the longest chain containing the double or triple bond (unsaturated) with the branched group given by its alkyl name. [e.g. CH3C(CH3)2CH2CH3 is 2,2-dimethylbutane. ] Isomers: Compounds with the same molecular formula but with their atoms arranged differently [e.g. hexane and 2,2-dimethylbutane, both C6H14]. Constitutional (structural) isomers: Isomers having their atoms joined together in a different sequence. (Some chemists restrict this term for isomers which have different functional groups [e.g. hexene and cyclohexane]. They would classify isomers containing the same functional groups as positional isomers [e.g. 2-methylpentane, CH3CH(CH3)CH2CH2CH3 and 3-methylpentane, CH3CH2CH(CH3)CH2CH3]). Organic compounds are classified by the functional groups they contain.

Page 27: Lets Talk Chemistry

6-3

Functional group: An atom or group of atoms which give the compound distinctive chemical properties [e.g. -Cl, -OH, >C=C<, -CO2H]. Thus all organic compounds except saturated hydrocarbons have one or more functional groups. The functional group determines the class of compound. In nomenclature the functional group may be identified by a prefix, a suffix, or by the class of compound. (See below) Common functional groups and classes of compounds are: -F, fluoro-; -Cl, chloro-; -Br, bromo-; -I, iodo-; generally called haloalkanes (prefix) or alkyl halides (class of compound). [e.g. CH3CH2Cl is chloroethane or ethyl chloride. CH3CHFCH2CH3 is 2-fluorobutane or secondary butyl fluoride.] -OH, hydroxy-, giving rise to alcohols. The -OH group can be named as the prefix hydroxy-, as the suffix -ol replacing the -e of the alkane or as an alcohol. [e.g. CH3CH(OH)CH2CH3 is butan-2-ol or secondary butyl alcohol; (CH3)3COH is 2-methylpropan-2-ol or tertiary butyl alcohol]. -NH2, amino-, giving rise to amines. The -NH2 group can be named as the prefix amino-, with the suffix -amine replacing the -e of the alkane or as an amine. [e.g. CH3CH2NH2 is aminoethane, ethanamine or ethylamine.] Amines can be considered as ammonia with hydrogens replaced by alkyl groups. If just one H is replaced the amine is a primary amine, as above. If two hydrogens are replaced the amine is a secondary amine [e.g. CH3CH2NHCH3, N-methylethanamine (the N shows the methyl substituent is bonded to nitrogen) or ethyl methyl amine]. If all three hydrogens are replaced the amine is a tertiary amine [e.g. (CH3)3N is trimethylamine]. (Note that for alkyl halides and alcohols 1o, 2o or 3o refers to the carbon atom to which the halo or hydroxy group is attached. For amines it refers to the number of alkyl groups on the nitrogen. Thus (CH3)3COH is a tertiary alcohol, but (CH3)3CNH2 is a primary amine.) >C=O, carbonyl. If this is at the end of a chain the compound is an aldehyde, and named with the suffix -al. [e.g.

CH3CH2 C

O

H

is propanal or propyl aldehyde. It is more simply written as CH3CH2CHO, but never as CH3CH2COH. You need to understand that the oxygen is bonded to the carbon. This -CHO representation is used because the other would look too much like an alcohol. ] If the CO group is not at the end of a chain the compound is a ketone, and named with the suffix –one. [e.g.

CH3CH2 C

O

CH2CH3

is pentan-3-one, (or diethyl ketone). Propanone (dimethyl ketone) is commonly called acetone. It is more simply written as CH3COCH3].

C OH

O , carboxyl, giving rise to carboxylic acids. The suffix -oic and the word acid

Page 28: Lets Talk Chemistry

6-4

are used in naming. [e.g. CH3CH2CH2CO2H is called butanoic acid. Methanoic acid, HCO2H, is commonly called formic acid, and ethanoic acid, CH3CO2H, is commonly called acetic acid.] The group is also often written as -COOH instead of -CO2H. The CH3CO- group is commonly called the acetyl group. RCO- is an acyl group. The product of the reaction of a carboxylic acid with a base is a carboxylic acid salt, an ionic compound. The name of the cation is given first followed by the acid with the suffix -oate replacing -oic. [e.g. CH3CH2CH2CO2

–Na+ is sodium butanoate; CH3CO2–NH4

+ is ammonium ethanoate or ammonium acetate.] The general name for the anion is carboxylate. When the OH of the carboxyl group is replaced by another group the compound is a carboxylic acid derivative. If the OH is replaced by OR of an alcohol the compound is called an ester. The R group is given first followed by the acid with the suffix -oate replacing -oic [e.g. CH3CH2CO2CH2CH3 is ethyl propanoate]. When the OH group is replaced by NH2 the compound is a primary amide. The suffix -amide replaces -oic. [e.g. CH3CONH2 is ethanamide, more commonly called acetamide]. If the OH has been replaced by an RNH the compound is a secondary amide, or by an RR'N group a tertiary amide, and the alkyl group of the amine named as such with the prefix N to show it is bonded to the nitrogen atom [e.g. CH3CH2CONHCH3 is N-methylpropanamide]. If the OH has been replaced by a halo group the compound is an acyl halide, -oic becoming -oyl [e.g. CH3CH2COCl is propanoyl chloride]. If the OH has been replaced by a carboxylate group, OCOR, the compound is an acid anhydride. [e.g. CH3COOCOCH3 is ethanoyl anhydride or acetic anhydride. An anhydride in general is a substance formed by removing the elements of water from the compound. [e.g. 2CH3CO2H → CH3COOCOCH3 + H2O ] Multifunctional compound: A compound with more than one functional group. Nomenclature of multifunctional compounds: The longest chain containing the suffix is chosen, the priority for choosing the suffix being carboxylic acid, -CO2H, > carboxylic acid derivative, -COX > aldehyde, -CHO > ketone, -CO-, > alcohol, -OH > amine, -NH2. The second and other groups are labelled as substituents. [e.g. CH3CH(OH)CH2CO2H is 3-hydroxybutanoic acid; HOCH2CH2CH2COCH3 is 5-hydroxypentan-2-one; CH3CH(OH)CH2C(CH3)(NH2)CH3 is 4-amino-4-methylpentan-2-ol; CH3COCO2H is 2-oxopropanoic acid, (the =O of an aldehyde or ketone is called oxo when it has to be named as a substituent).] The carbon-carbon double and triple bonds are always incorporated in the chain, with lower priority than the other groups. [e.g. CH2=CHCH(OH)CH3 is but-3-en-2-ol; CH3C≡CCH2CO2H is pent-3-yn-oic acid.] For compounds with larger carbon skeletons a further condensation of structural may be used.

represents propylcyclohexane. Each line represents two carbon atoms joined by a single bond, and hydrogens which are present are not shown. The number of H's is such to satisfy the valency of carbon, 4.

Page 29: Lets Talk Chemistry

6-5

is butylbenzene. Benzene is C6H6 and is the parent of aromatic compounds. Each carbon in the benzene ring has one hydrogen attached. As a second resonance structure with the double bonds in the other three positions can be drawn, the resonance hybrid of benzene is often represented as a hexagon with a circle inside: i.e.

Aromatic compound: An organic compound containing one or more benzene rings. Phenyl group: Benzene minus one hydrogen, C6H5- (sometimes shown as Ph). [e.g. C6H5CH2CH2CO2H is 3-phenylpropanoic acid.] Benzyl group: C6H5CH2- [e.g. C6H5CH2Cl is benzyl chloride; C6H5CO2H is benzoic acid.] Many substituted benzenes have common names. PhOH is phenol, PhCH3 is toluene; PhNH2 is aniline; PhCH=CH2 is styrene. Vinyl Group: Ethene minus a hydrogen, CH2=CH-. [e.g. CH2=CHCl is vinyl chloride, CH3CO2CH=CH2 is vinyl acetate. ] Vinyl compounds are the monomers of vinyl polymers. [e.g. PVC, polyvinyl chloride; PVA, polyvinyl acetate. ] EXERCISES Write condensed formulae, but showing all double and triple bonds, for the compounds: 1. Example: 2-chlorohex-3-ene

Answer: CH3CHClCH=CHCH2CH3 2. pent-2-ene 3. oct-4-yne 4. 5-bromohept-2-one 5. 4-aminobutanal 6. pentyl propanoate 7. 2-propyl-4-hydroxyhexanoic acid 8. nonanamide 9. N-ethylmethanamide 10-18. Give the systematic name of the compounds in questions 9-17 of section 5.

or

Page 30: Lets Talk Chemistry

7-1

SECTION 7 MEASUREMENT: QUANTITIES, NUMBERS AND UNITS Quantity: A property that is measured [e.g. mass, length, time, volume, pressure]. Unit: A standard quantity against which a quantity is measured [e.g. gram, metre, second, litre, pascal; which are units of the above quantities]. Chemists measure various quantities. If the mass of a substance was found to be 6.0 grams this can be expressed as an equation m = 6.0 g Here m is the symbol for the quantity mass, 6.0 is a number and g is the symbol for the gram, a standard quantity of mass. The equation is shorthand for mass = 6.0 x (1 gram) In general an equation of measurement takes the form quantity = (number) x (unit) In print, the symbols for quantities are shown in italics and for units in regular type as is done above. The equation may be manipulated by the normal rules of algebra. Thus the heading of a table or the axis of a graph could be labelled m/g (i.e. mass in grams) and numbers only tabulated or shown on the axis. It is important to realise that the magnitude of a quantity expressed as a number without units for the quantity is meaningless unless the quantity is in fact a ratio and is dimensionless (i.e. has no units since the units in the numerator and denominator cancel). Many quantities are defined in terms of other quantities. Thus density, symbol ρ, is defined as mass per unit volume. Thus if the mass and volume of a sample of a substance are known, the density is calculated by dividing the mass by the volume, density = mass/volume or ρ = m/V. [e.g. a 5 cubic centimetre sample has a mass of 15 grams. Calculate its density. ρ = m/V = 15 g/5 cm3 = 3 g cm-3. ] For many quantities there are different systems of units. It is essential that one can convert from one set of units to another. This is simply done by replacing the units in the equation by their values in terms of the desired unit [e.g. If the velocity of a particle is 200 miles per hour

In this section the language of measurement, which is central to chemistry, will be introduced. Chemists might measure a number of variables such as pressure, temperature, mass, and volume and use the experimental values to calculate some other property. The purity of a compound or the concentration of a compound in a commercial product are possible examples. The chemist needs to be able to record data and do calculations in a precise and unambiguous format. This requires a universally adopted language involving concepts, symbols and rules. While the average lay person knows little of these, knowledge of them is essential to the practising of chemistry. This section follows the recommendations of the International Union of Pure and Applied Chemistry (IUPAC), the international governing body for chemistry.

Page 31: Lets Talk Chemistry

7-2

what is it in metres per second? 1 mile = 1609 m. 1 hour = 3600 s The velocity is 200 miles per hour v = 200 miles/hour = 200 miles (hour)-1 = 200 x (1609 m) x (3600 s)-1 = 89.4 m s-1].

Units International System of Units (SI units): The internationally adopted system which defines or expresses all quantities in terms of seven basic units, the six used by chemists being: length metre m mass kilogram kg time second s temperature kelvin K amount of substance mole mol electric current ampere A Other quantities commonly used in chemistry, and which have special names for the units derived from these basic units are: frequency hertz Hz s-1 energy joule J kg m2 s-2 force newton N kg m s-2 = J m-1 pressure pascal Pa kg m-1 s-2 = N m-2 power watt W kg m2 s-3 = J s-1 electric charge coulomb C A s electric potential difference volt V kg m2 s-3 A-1 = J A-1 s-1 Further quantities used in chemistry but without special names for the derived units are: area, m2; volume, m3; density, kg m-3; absorbance, dimensionless (therefore no units). Coherent SI units: The base units, and those derived from them. Thus all the units shown above are coherent SI units. Note that the base unit of mass is the kilogram. This is the only base unit which has a multiple prefix (see below). If, in a calculation of a quantity involving several other quantities, only coherent SI units are used, the units of the required quantity will be coherent (i.e. the appropriate base unit). The sizes of these units are often unsuitable for some measurements and the decimal multiples, shown below with the name and symbol of the prefix, are used: 10-18 atto a 10 deca da 10-15 femto f 102 hecto h 10-12 pico p 103 kilo k 10-9 nano n 106 mega M 10-6 micro µ 109 giga G 10-3 milli m 1012 tera T 10-2 centi c 1015 peta P 10-1 deci d

Page 32: Lets Talk Chemistry

7-3

Some multiple units have their own name, the three relevant for chemists being: mass tonne t 1 t = 103 kg = 1 Mg volume litre L 1 L = 10-3 m3 = 1 dm3 = 103 cm3 length ångström Å 1 Å = 10-8 cm = 10-10 m = 100 pm Units may be written out in full or the symbol used. A mixture of full word and symbol is not permitted [e.g. kgram is not permitted]. The letter s is never added to the symbol to indicate a plural. A full stop is not written after symbols except at the end of a sentence. Those symbols named after a person have a capital first letter, but when the name of the unit is written out in full a lower case first letter is used [e.g. J, joule]. When two or more symbols are combined to indicate a derived unit, a space is left between them. A space is also left between the number and the symbol for the unit, but no space is left between the prefix indicating powers of ten and the symbol to which it applies. When symbols are combined as a quotient, [e.g. metres per second], either power to the minus one or the solidus may be used [e.g. m s-1 or m/s]. But the solidus may only be used once in a derived unit to avoid ambiguity [e.g. writing kg/m/s2 for pascal could be interpreted as kg m-1 s-2 or as kg m-1 s2]. Quantity calculus: The manipulation of the mathematical equations relating quantities and their measured values using the rules of algebra. All calculations in this manual follow the rules of quantity calculus. When one quantity is multiplied by another, no space is left between their symbols, [e.g. m = ρV]. In calculations prefixes for multiples can be replaced by their numerical values. [eg. Convert the density of 3 g cm-3 into coherent SI units. The coherent units for density will be kg m-3. 1 g = 10-3 kg , and 1 cm = 10-2 m. ρ = 3 g cm-3 = (3 x 10-3 kg) x (10-2 m)-3 = 3 x 103 kg m-3. ] Some non-SI units are commonly used by chemists: temperature degrees Celsius (the temperature intervals 1 oC and 1 K are identical)

0 oC = 273.15 K 0 K = -273.15 oC

time minute min = 60 s

hour h = 60 min = 3600 s day d = 24 h

pressure atmosphere atm = 1.013 x 105 Pa = 760 torr

bar bar = 105 Pa torr (mmHg) torr = 133.3 Pa

energy calorie cal = 0.2390 J or J = 4.184 cal

electron volt eV = 1.602 x 10-19 J kilowatt hour kWh = 3.6 x 106 J

The calorie, still widely used in USA, is the energy required to raise the temperature of 1 g of liquid water 1 oC. Electron volts are useful at the atomic or molecular level, and kilowatt hours are the common unit for power consumption.

Page 33: Lets Talk Chemistry

7-4

EXERCISES Write the equations for the following statements using powers of 10 in place of prefixes. 1. Example: The density, ρ is 3 micrograms per cubic millimetre

Answer: ρ = 3 x µg mm-3 (replacing micrograms with µg) = 3 x 10-6 g (10-3 m)-3 (replacing µ with 10-6 and m with 10-3) = 3 x 103 g m-3

2. The velocity, v, is 50 millimetres per second 3. The pressure, p, is 64 meganewtons per square centimetre 4. The concentration, c, is 2.8 nanograms per microlitre Express the following numbers under the given heading in a table as a normal equation for the quantity. 5. Example: 4.2 under the heading 105 m/g Answer: 105 m/g = 4.2 m = 4.2 x 10-5 g 6. 23 under the heading 10-2 t/s 7. 7. 8 under the heading 105 p/Pa Convert the following quantities to coherent SI units. 8. Example: q = 230 cal Answer: q = 230 (4.184 J) = 962 J 9. T = 27 oC 10. t = 32 h 11. p = 61 mmHg (mmHg = 133.3 Pa) 12. ρ = 34 lb ft-3 (lb = 454 g, ft = 12 in, in = 2.54 cm)

Page 34: Lets Talk Chemistry

8-1

SECTION 8 AMOUNT OF SUBSTANCE AND ITS UNIT, THE MOLE

Amount of substance: symbol n, a quantity fundamental to chemistry. Atoms and molecules are much too small or light to be counted or weighed individually in the laboratory. The chemist therefore needs a unit to specify the quantity amount of substance of an appropriate magnitude (size) for laboratory or industrial scale work. The chosen unit is the mole. Mole: symbol mol, the unit of the quantity amount of substance. The mole is defined as the amount of substance of a system which contains as many elementary entities as there are atoms in 12 grams of carbon-12 (i.e. carbon consisting only of the isotope 12C). 12 g is an easily measurable mass. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles or specified groups of such particles. It follows from this definition of the mole that x moles of dihydrogen (H2,) will contain exactly the same number of dihydrogen molecules as there are dioxygen molecules (O2) in x moles of dioxygen or water molecules (H2O) in x moles of water. Thus it follows from the chemical equation

2H2 + O2 → 2H2O that 2 moles of dihydrogen react with 1 mole of dioxygen to give 2 moles of water, i.e. the chemical equation is also a simple way of expressing the reaction of measurable amounts of substances as well as of individual molecules. Thus the chemical equation (page 2-2)

C2H6 + 7/2O2 → 2CO2 + 3H2O is quite satisfactory when the reaction coefficients refer to amounts measured in moles. Molar mass: symbol M, the mass per mole of substance (the substance being defined by its chemical formula). (Molar means per mole in this context.) The mass of any substance is proportional to the amount of that substance and the proportionality constant is its molar mass; m = Mn. [e.g. for water m = 18 g when n = 1 mol; 18 g = M(H2O) x 1 mol;

M(H2O) = 18 g/(1 mol) = 18 g mol-1.] This is usually written as m = nM. (i.e. mass of substance is the amount in moles multiplied by the mass per one mole). Commonly used units are grams per mole [e.g. M(12C) = 12 g mol-1]. The molar mass of the atoms of each element in units of grams per mole (g mol-1) is given in the periodic table immediately under the symbol for the element. The molar mass of any substance defined by its chemical formula is the sum of the molar masses of all of its constituent atoms.

[ e.g. M(H2O) = 2M(H) + M(O) = (2 x 1.01 + 16.00) g mol-1 = 18.02 g mol-1 ] The amount of substance is normally obtained by weighing, (i.e. by measuring its mass using

Suppose a chemist wishes to carry out the chemical reaction of adding bromine to hexene to give dibromohexane, C6H12 + Br2 → C6H12Br2, starting with a known amount of hexene. How does the chemist know how much bromine is needed? The chemical equation tells you that one molecule of dibromine is needed for each molecule of hexene. But the very large numbers of molecules required for reactions on a practicable scale cannot be counted. If the mass of hexene is known how could the mass of bromine required be calculated? This section tells how this type of question is answered. It requires the introduction of a new quantity specific to chemistry , amount of substance, and it it th l

Page 35: Lets Talk Chemistry

8-2

a balance), and converting the mass of the sample to the amount of the sample in moles by

rearranging the equation m = nM to give n = Mm .

[e.g. What is the amount of copper of 20.0 g of the metal?

n(Cu) = (Cu)(Cu)

Mm = 1-mol g 63.5

g 0.20 = 0.3150 mol].

For a liquid the volume might be measured and this converted to amount in moles by using both the density and the molar mass of the substance. [ e.g. What is the amount of tetrachloromethane in a 20.0 cm3 sample? ρ(CCl4) = 1.584 g cm-3; M(CCl4) = 153.8 g mol-1. Convert from volume to mass using the density,

m(CCl4) = ρ(CCl4)V(CCl4), and then to amount using the molar mass,

n(CCl4) = )(CCl)(CCl

4

4

Mm =

)(CCl)(CCl)(CCl

4

44

MVρ {by replacing m(CCl4) by ρ (CCl4)V(CCl4)}

= 1-

3-3

mol g 153.8cm 20.0 cm g 584.1 × = 0.206 mol ]

Stoichiometry: The quantitative relationship between the amounts of reactants consumed and products formed in a chemical reaction as expressed by its balanced chemical equation. The general chemical equation

aA + bB → cC + dD implies that a moles of substance A react with b moles of substance B to produce c moles of substance C and d moles of substance D. [e.g. What amount of copper oxide could be formed from 20.0 g of copper in the reaction 4Cu + O2 → 2Cu2O

From the stoichiometry of the equations, (Cu)

O)(Cu 2

nn = 2/4 = 0.5

Therefore n(Cu2O) = 0.5n(Cu) = 0.5 x 0.3150 mol = 0.1575 mol What is the mass of the Cu2O formed? m(Cu2O) = n(Cu2O)M(Cu2O) = 0.1575 mol x (2 x 63.5 + 16.0) g mol-1

= 0.1575 mol x 143 g mol-1 = 22.5 g ] The most useful expression for the stoichiometry of the above general chemical equation is

dn

cn

bn

an (D) (C) (B) (A) ===

This equation and n = Mm are two of the most important equations used in practical

quantitative chemistry. Avogadro Constant: Symbol NA or L, the number (of entities) per mole. From many varied measurements its value has been determined as 6.022 x 1023 mol-1. Atomic mass constant: Symbol mu, One twelfth of the mass of one atom of 12C. Also sometimes called unified atomic mass unit, symbol u, previously amu.

Thus mu = 1-23

-1

A

-112

1

mol 10 6.022mol g 1 mol g 12 /

×=

×N

= 1.667 x 10-24 g

Relative atomic mass: Symbol Ar, mean mass of one atom of an element (i.e. taking into account the relative natural abundance of the isotopes) relative to (i.e. divided by) mµ. With

Page 36: Lets Talk Chemistry

8-3

the exception of the heaviest elements which have been formed from different radioactive isotopes, in general the relative amounts of the different isotopes of an element is independent of its source [e.g. chlorine, Ar(Cl) = 35.45, consists naturally of 75.5% 35Cl and 24.5% 37Cl]. Note that the relative atomic mass has no units because it is the ratio of two masses. Most reference books and periodic tables simply give Ar values for the elements. From a practical viewpoint it is most important to realise that the numerical value of the molar mass M(E) is equal to Ar(E) when the units of molar mass are g mol-1 . Atomic weight: An older term for mean molar mass of an element, M(E), but sometimes used for relative atomic mass, Ar(E). Relative molecular mass: Symbol Mr, the mass of one molecule of the substance relative to mµ. It is simply the sum of all the Ar values of the atoms in the molecule. Instead of stating the relative molecular mass of a large molecule is, for example, 20,000, biochemists often say the molecular mass is 20 000 daltons or 20 000 Da or 20 kDa. The dalton is just another (friendlier) name for the unified atomic mass constant, mµ, and is not a unit in the sense of the gram (or the second, or the mole, or the metre) because its value depends on the Avogadro constant, an experimentally derived quantity. The dalton is never used in chemical calculations. Molecular weight: An older term for molar mass, but sometimes used for relative molecular mass. The term "relative molecular mass" is also used for an ionic substance or for substances which do not exist as discrete molecules, such as MgO or SiO2 [e.g. Mr(MgO) = 40]. The formula given with the symbol Mr makes it quite clear what is meant. Some authors try and avoid this problem by using the term "relative formula mass". But this is somewhat unsatisfactory because formulae are concepts and do not have mass! EXERCISES Determine the amount of substance in the given masses of the following compounds. (Molar masses for the elements in grams per mole are given in the periodic table.) 1. Example: 25.2 g of NaCl

Answer: n = m/M M(NaCl) = (23.0 + 35.5) g mol-1 = 58.5 g mol-1

n(NaCl) = 1-mol g 58.5g 2.25 = 0.431 mol

In words this equation says the amount of sodium chloride is 0.431 moles. 2. 23 g of KNO3 3. 75 kg of C6H6 4. 33 µg of C3H7CO2C5H11 (odour of banana) Determine the masses of the given amounts of the following compounds. 5. 45 mol of graphite (C) 6. 8.2 mol of Al2O3 7. 5.3 mmol of PtCl4 Determine the mass of the product that would be formed from the given mass of reactant in the following reactions.

Page 37: Lets Talk Chemistry

8-4

8. Carbon monoxide from 24 g of methane in the steam reforming of natural gas CH4 + H2O → CO + 3H2 9. Iron from 24 kg of magnetite, Fe3O4, in steel making Fe3O4 + 4C → 3Fe + 4CO 10. Sulfuric acid from 100 t of sulfur in the manufacture of sulfuric acid S8 + 12O2 + 8H2O → 8H2SO4 11. Acetic acid from 500 L of ethanol (ρ = 0.785 g cm-3), in making vinegar CH3CH2OH + O2 → CH3CO2H + H2O

Page 38: Lets Talk Chemistry

9-1

SECTION 9 GASES AND THE GAS LAWS Ideal gas: A gas that obeys Boyle's law and Charles's law, and hence the Ideal gas equation. Boyle's law: The pressure, p, of a fixed mass of gas is inversely proportional to its volume, V, at constant temperature (i.e. p ∝ 1/V). Charles's law: The volume, V, of a fixed mass of gas is proportional to its absolute temperature, T, at constant pressure (i.e. V ∝ T). It is found experimentally that at constant pressure and temperature the volume of a gas is directly proportional to the amount of gas, n, and at constant volume and temperature the pressure is directly proportional to the amount of gas, n. Avogadro's hypothesis: Equal volumes of different gases at the same temperature and pressure contain the same number of molecules (or atoms for monoatomic gases). All this is expressed by the ideal gas equation. Ideal gas equation: pV = nRT R is called the universal gas constant, and has the value 8.314 J K-1 mol-1. This equation is particularly useful in determining the amount of a gas from its volume, pressure and temperature. [e.g. Calculate the mass of dinitrogen in 500 litres at a pressure of 200 kPa and a temperature of 100oC. n(N2) is obtained from the ideal gas equation, and the mass from the molar mass:

n(N2) = RTpV =

K373x molK J 8.314m 0.500 x Pa 10 x 200

1-1-

33

= 32.2 mol

m(N2) = nM = 32.2 mol x (2 x 14.0 g mol-1) = 903 g ]

Ideal gas: a gas for which the individual molecules or atoms occupy negligible volume, and for which there are no attractive or repulsive forces between the molecules (or atoms). Dalton's law of partial pressures: The total pressure of a mixture of gases is the sum of the partial pressures of the individual gases in the mixture. All real gases deviate from ideal behaviour, the deviation increasing with increasing p and decreasing T. Partial pressure: The pressure that a particular gas in a mixture of gases would exert if it alone occupied the container.

The mass of a gas cannot easily be measured by weighing as is done for a liquid or solid. However the volume of the gas’s container, and the pressure and temperature of the gas can be measured. This section tells you about experimental laws and a famous hypothesis which allow the determination of the amount of a gas from its volume, pressure and temperature. Then it briefly gives the simple theory that rationalises these laws.

Page 39: Lets Talk Chemistry

9-2

The kinetic theory of gases makes the following assumptions for an ideal gas: 1. Gases are made up of molecules whose sizes are negligible compared with the

distance between them. 2. There are no forces between the molecules (except in a collision). 3. Between collisions the molecules are constantly moving in straight lines and their

motion is completely random. 4. The molecules are constantly colliding with one another and with the walls of the

container, all collisions being elastic (i.e. no loss of kinetic energy). 5. The collisions with the walls of the container give rise to the measured gas pressure. Vapour: An alternative term for gas. It is usually used when it is in contact with the liquid form, solid form or solution of the same substance, or is at a temperature at which it could be made to condense by increasing its pressure. The term evaporation illustrates this. EXERCISES Assuming the ideal gas equation is obeyed determine: 1. The volume of 50 g of dioxygen at 500 K and 200 kPa 2. The mass of 5 L of HBr at 1000 oC and 4 mPa 3. The pressure of 5 kg of argon in a 5 litre cylinder at 25 oC 4. The change in pressure if a given volume of xenon at 100 kPa is heated from 298 K to

800 K 5. The total pressure in a 2 litre cylinder containing 2 g of N2O and 3 g of O2 at 298 K

Page 40: Lets Talk Chemistry

10-1

SECTION 10 CHEMICAL ANALYSIS Analysis: The determination of the composition of a substance. Composition: The composition of a substance or of a mixture can be expressed in many ways. In elemental analysis it is usually expresses as the percentage by mass of each element in the substance. For solutions or mixtures common terms are concentration, mole fraction, molality, percentage composition, and parts per million. Concentration: Symbol c, the "amount" of solute per unit volume of the solution. The "amount" may be given as mass, and c = m/V, where m is the mass of solute and V the volume of the solution (common unit g L-1), or using the formal definition of amount of substance as c = n/V, (common unit mol L-1). For a given solute, concentration in one set of units may be

converted to the other using m = nM or n = Mm [e.g. 5.50 g of sodium chloride was dissolved

in water to give 200 mL of solution. Determine the concentration.

c(NaCl) = )NaCl()NaCl

Vm( =

L 10 x 200g50.5

3-

= 27.5 g L-1

or c(NaCl) = )NaCl()NaCl

Vn( =

)NaCl()NaCl()NaCl

VMm( =

L 10 200 mol g 35.5) (23.0g 50.5

3-1- ××+

= 0.470 mol L-1

This example may be written in the following one line format: c(NaCl) = = m(NaCl)/V(NaCl) = 5.50 g/(200 x 10-3 L) = 27.5 g L-1 n(NaCl)/V(NaCl) = m(NaCl)/M(NaCl)V(NaCl)

= 5.50 g/{(23.0 + 35.5) g mol-1 x 200 x10-3 L} = 0.470 mol L-1 ] This format is only used for simple equations in the remainder of this section. The term molarity, symbol M, is an old term no longer approved by the International Union of Pure and Applied Chemistry (the body which determines the rules on the language of chemistry), but unfortunately often still used by analytical chemists for concentration in units

A large number of professional chemists are analysts. They determine things like "how much of a particular compound is in a sample of a substance", or "the amount of a particular element in a compound or mixture", or "the concentration of compound or ion in a water sample". When a chemist makes what is believed to be a new compound, an elemental analysis is carried out, i.e. the percentage by mass of each element in the compound is determined and compared with the theoretical figure for the compound. Chemical analysis is central to the practice of chemistry. Three main quantities that are measured in analysis are mass, volume, and intensity of a beam of electromagnetic radiation, giving rise respectively to gravimetric, volumetric and spectrophotometric methods of analysis. In this section, after you have been given the definitions of ways the composition of a system can be expressed, you are introduced to these three classes of analysis.

Page 41: Lets Talk Chemistry

10-2

of mol L-1. [e.g. The molarity of the above NaCl solution is 0.470; it is a 0.470 M solution of NaCl; in speech it may be described as a 0.470 molar solution. Here molar means mol L-1, not per mole. This is an example of chemists having two different meanings for the same word. The meaning is made clear by the context in which the word is used.] Mole fraction: Symbol x, the amount of a substance or entity (molecules, atoms or ions) in a mixture as a fraction of the total amounts of substances or entities in the mixture. [e.g. 1.0 g of sodium chloride is dissolved in 10 g of water. What is the mole fraction of NaCl in the solution?

x = O)H( + (NaCl)

(NaCl)2nn

n

n(NaCl) = (NaCl)(NaCl)

Mm = 1-mol g 58.5

g 0.1 = 0.017 mol

n(H2O) = O)(HO)(H

2

2

Mm = 1-mol g 18

g 10 = 0.555 mol

x = mol 0.555) + (0.017

mol017.0 = 0.030 ]

Molality: The amount of solute per kilogram of solvent. (units: mol kg-1) Percentage composition: There are various forms of this. %w/w is percentage by mass of the substance relative to the total mass. %w/V is the mass of substance in grams divided by the volume of the solution in millilitres and multiplied by 100. (Chemists here use w for mass, originating from the loose use of the word weight.) %V/V is the percentage by volume of the substance relative to the total volume. Parts per million: Symbol ppm: a ratio. For a gaseous mixture, the mole ratio, (mol/mol). For solutions, normally calculated from mg of substance per kg of solution, but for dilute aqueous solutions as mg of substance per litre of solution. For solids, the mass ratio. This section will cover three classes of chemical analysis, gravimetric, volumetric, and spectrophotometric. Gravimetric analysis: Analysis by measurement of mass. [e.g. The silver in a 100 g ore sample was recovered as 4.55 g of pure silver chloride. Determine the percentage mass of silver in the sample.

Ag(ore) → AgCl(s) Silver chloride is AgCl. Identify the quantities known and to be determined: m(AgCl) and m(Ag) respectively. From the stoichiometry determine the relationship between them: n(Ag) = n(AgCl).

n(Ag) = n(AgCl) = (AgCl)(AgCl)

Mm = 1-mol g 35.5) + (107.9

g55.4 = 0.317 mol

m(Ag) = n(Ag)M(Ag) = 0.0317 mol x 107.9 g mol-1 = 3.42 g

Page 42: Lets Talk Chemistry

10-3

% mass = 100 g 100g 42.3 100

(sample))Ag ×=×

mm( = 3.42 ]

It can be useful to derive the overall expression between the required unknown and known quantities and put all the numerical values into this expression at the end. [e.g. In the above example:

m(Ag) = n(Ag)M(Ag) = n(AgCl)M(Ag) = (AgCl)

(Ag)(AgCl)M

Mm = 1-

-1

mol g 35.5) + (107.9 mol g 107.9 g55.4 ×

= 3.42 g] Volumetric analysis: Analysis by measurement of volume. Volumetric analysis involves titration. In titration one reactant, A (the analyte), is in solution in a flask (usually a conical flask) and a second reactant, B (the titrant), in solution is added from a burette (a graduated glass tube with a tap at the bottom). Equivalence point or stoichiometric point: The stage of a titration where the volume added from the burette (the titre) contains the exact amount of B required to react with all of A according to the chemical equation for the reaction

For aA + bB → cC + dD it is where b

na

n (B) (A) = .

This stoichiometric relationship is at the centre of all volumetric analysis calculations. In titration there are commonly two scenarios. I ct(B) is known, (i.e. B is a standard solution), and nt(A) is to be determined. Here the

subscript t has been used to emphasise these are quantities in the actual titration step. The amount of A in the conical flask is determined from the titre, the volume of solution B, Vt(B), delivered from the burette:

nt(A) = bVac

ban (B)(B)

(B) ttt = .

From nt(A), the amount of A in the flask, information such as the concentration of a solution of A, the amount of A in a mixture, or the percentage purity of a substance may be determined. [ e.g. A 25.0 mL aliquot (a small known volume of solution) of hydrochloric acid was transferred with a pipette ( device which delivers a fixed known volume of solution) to a conical flask and titrated with 0.150 mol L-1 sodium hydroxide solution, the reaction being

HCl + NaOH → NaCl + H2O The titre was 23.4 mL. Determine the concentration of the HCl solution. Stoichiometry: nt(HCl) = nt(NaOH) = = ct(NaOH)Vt(NaOH)

ct(HCl) = mL 25.0

mL 23.4 L mol 0.150 (HCl)

(NaOH)(NaOH)

(HCl)(HCl) -1

t

tt

t

t ×==V

VcVn

= 0.140 mol L-1

]

II nt(A) is known, and c(B) is determined. This is called standardising the solution of

Page 43: Lets Talk Chemistry

10-4

B. The amount of A may be obtained (a) from the mass of a very pure substance A (known as a primary standard) determined by weighing and calculating n(A) from n(A) = m(A)/M(A), or, (b) if A does not have the properties of necessary for a primary standard, weighing out some other primary standard substance, X, that gives A in a stoichiometric

reaction, xX → aA so that n(A) = (X)(X) (X)

xMam

xan = .

[e.g. Potassium iodate, KIO3, a primary standard reacts with excess iodide to produce a known amount of iodine which is not a primary standard. The chemical equation is

IO3– + 5I– + 6H+ → 3I2 + 3H2O

Here X is IO3– and A is I2 and n(I2) = 3n(IO3

–) = )(KIO)(KIO3

3

3

Mm

]

(a) By weighing a primary standard A directly into the conical flask:

an

bn (A)

(B) tt = and c(B) =

(B)(B)

t

t

Vn

nt(B) = (A)(A)

(A) tt

aMbm

abn

= (because n = m/M)

c(B) = (B)(A)

(A)

t

t

VaMbm

mt(A) and Vt(B) are the quantities measured in the experiment.

[e.g. A solution of hydrochloric acid was standardised by titration against 0.200 g of the primary standard sodium carbonate, the reaction being:

2HCl + Na2CO3 → 2NaCl + CO2 + H2O The titre was 19.7 mL. Determine the concentration of HCl.

Stoichiometry: nt(HCl) = 2nt(Na2CO3) = )CO(Na)CO(Na2

32

32t

Mm

c(HCl) = (HCl))CO(Na

)CO(Na2

(HCl)(HCl)

t32

32t

t

t

VMm

Vn

=

= L} 10 19.7 mol g 16.0) 3 12.0 23.0 2{(

g 0.200 23-1- ×××++×

× = 0.192 mol L-1

(b) By weighing a larger mass of primary standard A into a volumetric flask (a

stoppered flask with a graduation mark on the neck) of fixed volume Vf, dissolving it in the solvent (usually water), then adding more solvent until the level of the solution is on the graduation mark, and then shaking thoroughly. An aliquot of the solution is transferred from the volumetric flask to the conical (titration) flask using a pipette of volume Vp. In this case the concentration of A in the volumetric flask is

Page 44: Lets Talk Chemistry

10-5

cf(A) = (A)(A)

(A) (A)(A)

ff

f

VMm

Vn

=

where the f refers to the volumetric flask. The amount of A used in the titration is

nt(A) = cf(A)Vp(A) (because c = n/V)

= (A)(A)(A)(A)

f

p

VMVm

As in (a), for the titration nt(B)/b = nt(A)/a and c(B) = nt(B)/Vt(B)

and nt(B) = a

bn (A)t = aVbc (A)(A) pf =

(A)(A)(A)A)

f

p

VaMVbm(

and c(B) = nt(B)/Vt(B) = (B)(A)(A)

(A)A)

tf

p

VVaMVbm(

Again m(A) and Vt(B) are measured in the experiment, the molar mass of A and the volumes of the flask and pipette being known.

[e.g. A second solution of hydrochloric acid was standardised against sodium carbonate, but this time 23.7 g of sodium carbonate was dissolved in water and made up to the mark in a 250 mL volumetric flask. A 10 mL aliquot was titrated with the hydrochloric acid solution. The titre of HCl was 12.6 mL. Determine the concentration of this HCl solution.

cf(Na2CO3) = f32

32f

f

32f

)CO(Na)CO(Na

)CO(Na

VMm

Vn

=

= L 0.250 mol g 106.0

g 7.231- ×

= 0.894 mol L-1

nt(Na2CO3) = cf(Na2CO3)Vp

c(HCl) = mL 12.6

mL 10 L mol 0.894 2 (HCl)

)CO(Na2

(HCl))CO(Na2

(HCl)(HCl) -1

t

p32f

t

32t

t

t ××===V

VcV

nVn

= 1.42 mol L-1 ]

(Note: The subscripts t, f and p, are introduced to help understanding - they would not normally be used in doing a calculation.) Spectroscopic analysis: Analysis involving the absorption (or emission) of light. In ultraviolet-visible spectroscopy the absorbance of a solution in a spectrophotometric cell (a container with flat transparent walls) is measured.

Page 45: Lets Talk Chemistry

10-6

Absorbance: Symbol A, log10(Io/I), where Io in the intensity of the light beam entering the solution and I the intensity leaving it. [e.g. If 90% of the incident light is absorbed by the solution, A= 1.] Optical density is an older term for absorbance. Beer's law (Beer-Lambert law): A = εcb. This states that the absorbance, A, of a solution is proportional to the concentration of the absorbing solute, c, and the pathlength, b, of the light. ε is the proportionality constant. (l is sometimes used in place of b.) Molar absorptivity: The value of ε when the concentration of the solute is measured in mol L-1. (Extinction coefficient is an older term for this.) The value of ε varies with the wavelength, λ, of the light. [e.g. Calculate the concentration of a solute, with ε = 1200 L mol cm-1, for which the absorbance is 0.450 when measured in a 1.0 cm cell.

c = cm 1.0 cm mol L 1200

0.045 1-1- ×=

bAε

= 3.75 x 10-4 mol L-1 = 375 µmol L-1 ]

EXERCISES From the masses of the compounds dissolved in the given volumes in questions 1-3 calculate the concentrations of the solutions in units of (a) g L-1 and (b) mol L-1. 1. 1 kg of sucrose, C12H22O11, in 5 L 2. 48.6 mg of KF in 100 mL 3. 5.234 g of the primary standard oxalic acid, H2C2O4.2H2O, made up to the mark in a

100 mL volumetric flask 4. A sodium hydroxide solution was standardised by titrating against the above oxalic

solution (question 3). A 10 mL aliquot of the oxalic acid took 18.52 mL of NaOH to reach the equivalence point for the reaction

H2C2O4 + 2NaOH → Na2C2O4 + 2H2O Determine the concentration of the sodium hydroxide solution.

5. The standardised NaOH solution of question 4 was used to determine the percentage

purity of a sample of ammonium sulfate, (NH4)2SO4. 0.8548 g of the ammonium sulfate was weighed into a conical flask and titrated with the NaOH solution according to the equation below, the titre being 26.75 mL. Determine the percentage purity.

(NH4)2SO4 + 2NaOH → Na2SO4 + 2NH3 + 2H2O 6. Household chlorine bleaches are aqueous solutions of sodium hypochlorite and

sodium chloride. When acidified chlorine is produced: ClO– + Cl– + 2H+ → Cl2 + H2O

The concentration of the bleach is often given as grams of "available chlorine" per litre. In measuring this an aliquot of the bleach solution is pipetted into a conical flask containing a solution of potassium iodide and some acid added slowly. The chlorine formed reacts with the iodide to give iodine:

Cl2 + 2I– → 2Cl– + I2 The amount of iodine formed is measured by titration against a standard solution of

Page 46: Lets Talk Chemistry

10-7

sodium thiosulfate, Na2S2O3: I2 + 2S2O3

2– → 2I– + S4O62–

Sodium thiosulfate is not a primary standard, and its solution is standardised by titration with a known amount of iodine formed by the reaction of a known amount of potassium iodate, KIO3, with excess iodide in an acidified solution:

IO3– + 5I– + 6H+ → 3I2 + 3H2O

In such an analysis 1.187 g of KIO3 was transferred to a 250 mL volumetric flask, dissolved in water and made up to the mark. 25.0 mL aliquots of this solution were transferred to conical flasks, excess KI solution and acid added, and the iodine formed titrated against the sodium thiosulfate solution to standardise it. The average titre was 23.48 mL. Then 10.0 mL aliquots of the bleach solution were pipetted into conical flasks containing aqueous excess KI and acid slowly added. The released iodine was titrated against the standard thiosulfate solution. The average titre was 34.70 mL. Determine the concentration of the bleach solution in terms of grams of available chlorine per litre. M/g mol-1: KIO3 = 214.0 Cl2 = 70.90 (Hint: You will have to calculate the concentration of the standard potassium iodate solution from the weighing, then that of the standard thiosulfate solution from the first titrations, and then use this figure to determine the mass of chlorine from the second titrations.)

7. A solution of an azo dye with a molar absorptivity of 17 500 L mol -1 cm-1 at 560 nm

wavelength had an absorbance of 0.528 when measured in a 0.50 cm cell at this wavelength. Determine the concentration of the dye. If the relative molecular mass of the dye was 468 calculate the mass of dye in a 20 litre vat.

Page 47: Lets Talk Chemistry

11-1

SECTION 11 THERMOCHEMISTRY Thermochemistry: Study of the heat released or absorbed by chemical reactions. Rearrangements of atoms that occur during chemical reactions involve both bond breaking and bond formation. Bond breaking results in absorption of heat from the surroundings and bond formation in release of heat to the surroundings. Heat: Energy transferred as a result of a temperature difference between a system and its surroundings. The quantity of energy transferred from the surroundings to the system is given the symbol q. The system means the substance, or reactants and products of a reaction, and the surroundings everything else. When energy is transferred from the system to the surroundings q is negative. Chemists frequently make measurements at constant (atmospheric) pressure. Energy transferred at constant pressure is given the symbol qp. A chemical equation can be made more meaningful and informative by showing the states of the reactants and products and the quantity of heat released or absorbed. [e.g. 2H2(g) + O2(g) → 2H2O(l) ∆rH = -570 kJ mol-1 implies that when 2 moles of gaseous dihydrogen reacts with one mole of gaseous dioxygen to give two moles of liquid water 570 kilojoules of energy is released from the reacting system to its surroundings.] The ∆ means change, r stands for reaction and H is the symbol for enthalpy. ∆rH is the enthalpy change of the reaction. Enthalpy change: Symbol ∆H, the change in energy of a system which undergoes a change at constant pressure. ∆H = qp Exothermic reaction: A reaction that releases energy to the surroundings. ∆rH < 0 Endothermic reaction: A reaction that absorbs energy from the surroundings. ∆rH > 0 ∆rH is negative for an exothermic reaction (the energy of the system, the reactants and products, is less after the reaction than before) and positive for an endothermic reaction (the energy of the system is greater than before). Note that the per mole (mol-1) in the units of ∆rH refers to the stoichiometric coefficients in the equation as amounts in moles. Thus if the above equation were written as

H2(g) + ½O2(g) → H2O(l), then ∆rH = -285 kJ mol-1. It is most important to understand the difference between the enthalpy change of a system, and that of a reaction. The magnitude of the former depends on the amount present in the system and has units of energy, e.g. kJ, while the amount of the latter is defined by the chemical equation and has common units of kJ mol-1.

Most chemical reactions are accompanied by the release of energy to the surroundings or absorption of energy from the surroundings, or put more simply the reacting system gets hotter or cooler as the reaction proceeds. The most common form of energy transferred is heat. This section introduces the language used in measuring and representing the heat changes that occur, and how the amount of heat released or absorbed in a chemical reaction can be calculated from tabulated data for the reactants and products of the

ti

Page 48: Lets Talk Chemistry

11-2

Latent heat: An old term for the enthalpy change of a change in state. [e.g. H2O(s) → H2O(l) ∆H = 6 kJ mol-1 . Ice has to be heated for it to melt, so ∆H is positive. This is called the enthalpy of fusion (or melting) and given the symbol ∆fusH(H2O). The symbols for enthalpy of sublimation and vaporisation are ∆subH and ∆vapH respectively.] Note: The convention used here is to place the explanatory subscript between the ∆ and H, as recommended by IUPAC, the International Union of Pure and Applied Chemistry. Until recently the subscript was placed after the H and r was omitted for enthalpy changes of chemical reactions. Calorimetry: The measurement of enthalpy change using a calorimeter. Calorimeter: An apparatus for measuring the heat absorbed or released in a chemical reaction. Heat capacity: symbol C, the ratio of the heat supplied to the temperature rise produced. i.e. the energy required to raise the temperature of a substance or system by one degree (oC or K)

when no chemical or phase changes occur. (i.e. C = T

q p

∆; SI units, J K-1)

Heat capacity at constant pressure: symbol Cp, the heat capacity when the change occurs at constant pressure, the most common condition. Specific heat capacity: Symbol cp, the heat capacity of unit mass of a substance; common units, J K-1 g-1. Molar heat capacity: Symbol Cp,m, the heat capacity of one mole of a substance; common units, J K-1 mol-1. [e.g. The absorption of 390 J of heat by a 100 g block of copper caused its temperature to increase by 10 oC. Calculate the heat capacity of the block and the specific and molar heats of copper.

Cp(block) = T

q p

∆ =

K10J 390 = 39.0 J K-1

cp(Cu) = (Cu)

(block)m

C p = g 100K J 0.39 -1

= 0.390 J K-1 g-1

Cp,m(Cu) = (Cu)(block)

nC p =

(Cu)(Cu)(block)

mMC p (as n =

Mm )

= g 100

mol g 63.5 x K J 0.39 -1-1

= 24.8 J K-1 mol-1 ]

If the reaction of a known amount of substance X, n(X), is carried out in a calorimeter and the temperature change measured, the enthalpy change for that reaction (written with 1 mole of X in the equation) can be calculated as the actual energy transferred is given by q = -Cp∆T

and (X)

- (X)

r nTC

nq

H pp ∆==∆ . Note the negative sign because if the temperature of the

calorimeter rises (∆T positive) energy has been transferred from the reaction system to the calorimeter (the surroundings) and the reaction is exothermic. [e.g. When 16 g of sulfur is combusted to sulfur dioxide 149 kJ of energy is released. Determine ∆rH for the reaction S(s) + O2(g) → SO2(g)

Page 49: Lets Talk Chemistry

11-3

mol 0.50 mol g 32g 16

(S)(S) (S) 1- ===

Mmn

1-r mol kJ 298-

mol0.50kJ 149- ===∆

nq

H p ]

Heat of formation of a substance: Symbol ∆fH, the heat (enthalpy) change when one mole of that substance is formed from its elements in their standard states. By convention an element in its most stable form at 100 kPa pressure and 25 oC (298 K) is said to be in its standard state and to have an enthalpy of formation, ∆fH, of zero [e.g. from the reaction of hydrogen and oxygen above, it follows that ∆fH(H2O) = -285 kJ mol-1 ]. Heat of combustion of a substance: Symbol ∆cH, the heat given out when one mole of that substance reacts with dioxygen to give the most oxidised products. [e.g. ∆cH(CH4) is the enthalpy change for the reaction CH4(g ) + 2O2(g) → CO2(g) + 2H2O(l) From the appropriate equation above it can be seen that ∆cH(H2) = ∆fH(H2O). ] Hess's law (of constant heat summation): The enthalpy change for a reaction is independent of the way the reaction is carried out (i.e. it depends only on the initial conditions of the reactants and the final conditions of the products). Put another way if a reaction is carried out in a number of steps the enthalpy change is the sum of the enthalpy changes for each individual step. [e.g.

(1) CO(g) + ½O2(g) → CO2(g) ∆rH1 = ∆cH(CO) = -282 kJ mol-1 (2) C(s) + O2(g) → O2(g) ∆rH2 = ∆fH(CO2) = ∆cH(C) = -393 kJ mol-1

Imagine that we burn graphite (the most stable form of carbon) and then decompose the CO2 to CO and O2:

(2) C(s) + O2(g) → CO2(g) ∆rH2 (3) CO2(g) → CO(g) + ½O2(g) ∆rH3 = -∆rH1 Note that the sign of ∆rH

changes when the reaction direction is reversed. Adding (2) and (3):

(4) C(s) + ½O2(g) → CO(g) ∆rH4 = ∆rH(CO) = ∆rH2 + ∆rH3 = ∆rH2 - ∆rH1 = -111 kJ mol-1 ]

It follows from Hess's law that for the general reaction aA + bB → cC + dD ∆rH = c∆fH(C) + d∆fH(D) - a∆fH(A) - b∆fH(B) or ∆rH = ∑∆fH(products) -∑∆fH(reactants) where ∑ means "the sum of". This is a very useful expression because the enthalpy change for any reaction can be calculated from tables of heats of formation of the compounds involved. [e.g. For the reaction F3O4(s) + 4H2(g) → 3Fe(s) + 4H2O(l) ∆rH = 3∆fH{Fe(s)} + 4∆fH{(H2O(s)} - ∆fH{Fe3O4(s)} - 4∆fH{H2(g)} = 4∆fH{H2O(l)} - ∆fH{Fe3O4(s)} because ∆fH for an element is zero. ] Bond enthalpy (strength): The energy needed to break a chemical bond, with the bonding electrons being evenly divided between the fragments [e.g. The oxygen-oxygen bond strength of dimethyl peroxide, CH3O-OCH3, is ∆rH for the reaction CH3OOCH3 → 2CH3O· ].

Page 50: Lets Talk Chemistry

11-4

EXERCISES 1. The absorption of 62.76 kJ of heat by 500 g of liquid water caused its temperature to

rise by 30 oC. Calculate the specific heat capacity of water in (i) J g-1 K-1 and (ii) in cal g-1 K-1and the molar heat capacity of water in J mol-1 K-1. (1 cal = 4.184 J)

2. It required 36.61 MJ of heat to distil 50 L of ethanol, CH3CH2OH, at its boiling point.

Calculate ∆vapH(ethanol). (ρ(ethanol) = 0.785 g cm-3) 3. The combustion of exactly 1 kg of elemental sulfur to sulfur dioxide released 9.28 MJ

of heat. Calculate ∆fH(SO2) in kJ mol-1. 4. Carbon disulfide, CS2, has a boiling point of 46 oC. The products of combustion are

CO2 and SO2. (i) Write the equation for its combustion. (ii) Calculate ∆fH(CS2) given the following ∆cH/kJ mol-1: CS2 = -1077; C = -393; S = -297

5. Calculate the approximate ∆rH for the reaction CH4 + Cl2 → CH3Cl + HCl

from the following average bond strengths / kJ mol-1: C-H = 415, C-Cl = 327, H-Cl = 431, Cl-Cl = 242 (Consider what bonds are broken and what bonds are made.)

Page 51: Lets Talk Chemistry

12-1

SECTION 12 CLASSIFICATIONS OF CHEMICAL REACTIONS Classification of reactions (the type of reaction) is based on a variety of observations or concepts. A. Classification based on observation of a phase change (a) Gas evolution: A gas is evolved from a solid or a solution.

[e.g. (i) CaCO3(s) + 2H3O+(aq) → Ca2+(aq) + CO2(g) + 3H2O(l) Calcium carbonate (limestone) dissolves in dilute aqueous acid.

(ii) 2H2O2(l) → O2(g) + 2H2O(l) Decomposition of hydrogen peroxide. ]

(b) Precipitation: A solid (the precipitate) is produced from solution.

[e.g. Na+(aq) + Cl– (aq) + Ag+(aq) + NO3– (aq) → AgCl(s) + Na+(aq) + NO3

–(aq) Insoluble silver chloride forms when aqueous solutions of two soluble salts, sodium chloride and silver nitrate, are mixed. CO2(g) + Ca2+(aq) + 2OH– (aq) → CaCO3(s) + H2O(l)

A precipitate forms when carbon dioxide is bubbled through "limewater". ] (c) Dissolution: A solid dissolves on addition of a reactant or a solvent.

[e.g. (i) a(i) above is also an example of this.

(ii) Ca3(PO4)2(s) + 6H3O+(aq) → 3Ca2+(aq) + 2H3PO4– (aq) + 6H2O(l)

Calcium phosphate dissolves in aqueous acid.

(iii) NaCl(s) + water → Na+(aq) + Cl– (aq) Common salt dissolves in water. ]

B. Classification based on change in structure of a reactant (a) Decomposition: A substance breaks down to smaller species.

[e.g. CaCO3(s) → CaO(s) + CO2(g) Decomposition of limestone to calcium oxide (quicklime) and carbon dioxide on heating.]

Many reactions may be thought to be between a substrate and a reagent. The classification is based on the change in structure of the substrate.

As the number of known chemical reactions is enormous it is useful to be able to describe a particular reaction as belonging to a certain class or classes. Just as a particular individual person could be classified by gender, age, weight, height, colour of eyes, annual income etc., and could fall into several classes, so it is with chemical reaction classification. This section gives the basis of a number of classes. One very important class, that of redox reactions, has an extensive language of its own, and this is introduced. The important classification based on how a reaction occurs is left to section 16.

Page 52: Lets Talk Chemistry

12-2

(b) Addition: The reagent adds to the substrate, often across a double or triple bond. [e.g.

(i) H2C=CH2 + Br2 → Br-CH2-CH2-Br substrate reagent

Bromine adds across the double bond of ethene (ethylene). (ii) HCH=O + H2O → H-CH(OH)2 Water adds across the carbonyl double bond of methanal (formaldhyde).

(iii) xCH2=CH2 → -CH2-CH2-CH2-CH2-CH2- Molecules add to each other to give a molecule of large molar mass called a polymer. The process is called polymerisation. ]

(c) Elimination: Atoms or groups are lost from neighbouring atoms of the substrate with

the formation of a double or triple bond. Effectively a small molecule is eliminated.

[e.g. (i)

CH3 C C

H

H

CH3

H

Br

KOH

alcoholCH3 CH CH CH3 + ( HBr )

H and Br are eliminated from adjacent carbon atoms.

(ii)

C CH3CH3CH2

H

OHC CH3CH3CH2

O

K2Cr2O7+ ( 2e- + 2H+)

Hydrogen is eliminated from adjacent carbon and oxygen atoms. This loss of two H atoms is more commonly called oxidation. See D, redox reactions. ]

(d) Substitution: One group on the substrate is replaced by (substituted for) another.

[e.g. (i) CH3-CH2-CH2-Br + OH– → CH3-CH2-CH2-OH + Br–

Br replaced by OH

(ii) H3C-CH3 + Cl2 → H3C-CH2Cl + HCl H replaced by Cl photochemically (i.e. with light). ]

(e) Condensation: Two molecules are joined and a small molecule is eliminated. Condensation can be thought of as particular type of substitution.

[e.g. CH3-CO-OH + CH3-CH2-OH → CH3-CO-O-CH2-CH3 + H2O OH is replaced by OCH2CH3 and water is eliminated. ]

C. Acid-base Certain hydrogen-containing molecules or ions react by transferring a hydrogen ion, H+, to another molecule or ion. The reactant which is the source of the H+ is called an acid, the reactant to which the H+ becomes attached is called a base, and the reaction is called acid-

Page 53: Lets Talk Chemistry

12-3

base. [e.g. HCl(g) + NH3(g) → NH4Cl(s)

The acid hydrogen chloride transfers H+ to the base ammonia to give the ionic solid ammonium chloride (composed of NH4

+ and Cl– ions). HCl(g) + H2O(l) → H3O+(aq) + Cl– (aq)

The gas hydrogen chloride acts as an acid as it transfers H+ to a water molecule which is acting as a base. ]

A hydrogen ion, H+, is just a proton, and these acid-base reactions are called proton transfer reactions. An acid is defined as a proton donor and a base is defined as a proton acceptor. This is the Brønsted-Lowry definition. A base must have a non-bonding pair of electrons which can be used to form a bond with a proton. Brønsted acid: a proton donor, i.e. a species which donates a proton in a chemical reaction [e.g. HCl in the above examples]. Brønsted base: a proton acceptor, i.e a species which accepts a proton in a chemical reaction [e.g. NH3 and H2O in the above two examples]. Oxonium (hydronium) ion: the cation H3O+, often written as H+(aq) or just H+ in equations in water (water, aqua, aq). It is important to realise the proton, H+, does not exist as such in condensed matter, i.e. liquid or solid state. The term hydronium is still widely used Hydroxide ion: the anion OH–. (As it is the O which has the formal charge of -1, HO– would be a more correct formula, but OH– is usually written). Conjugate acids and bases: When an acid loses a proton, its product is called the conjugate base of that acid. When a base accepts a proton its conjugate acid is formed [e.g. HCl/Cl–, NH4

+/NH3, H3O+/H2O, and H2O/OH– are conjugate acid-base pairs]. Neutralisation: The reaction between an acid and a base [e.g. reaction of aqueous HCl and NaOH to give NaCl and H2O ]. Salt: the product (other than water) of a neutralisation reaction between an acid and a base. Salts are ionic compounds [e.g. NaCl in the above neutralisation reaction]. Alkali: Aqueous solution of a base [e.g. aqueous NaOH or aqueous ammonia ]. D Oxidation-reduction (redox) reactions Redox reaction: A reaction in which one reactant is oxidised and another is reduced. The concept of oxidation and reduction plays a very important role in chemistry. Historically a compound was said to be oxidised if it gained oxygen and reduced if it lost oxygen. The term oxidation was later extended to include the loss of hydrogen, and reduction to include the gain of hydrogen. However with the development of the electronic structure of the elements the definitions were changed in order to encompass a wider range of reactions than just those involving loss or gain of oxygen or hydrogen: Oxidation: a process in which a species loses electrons [e.g. Cu → Cu2+ ]. Reduction: a process in which a species gains electrons [e.g. Cu2+ → Cu ].

Page 54: Lets Talk Chemistry

12-4

In a redox reaction one reactant called the oxidant (older term oxidising agent) oxidises a second reactant called the reductant (older term reducing agent) and is itself reduced. [e.g. Zn + Cu2+ → Zn2+ + Cu Zinc is the reductant and copper ions the oxidant

Cl2 + 2Br– → 2Cl– + Br2 Chlorine is the oxidant and bromide ions the reductant.] Half-reactions: Redox reactions may be separated into two half-reactions, one involving oxidation (loss of electrons) and the other reduction (gain of electrons). [e.g for the above two examples: Zn → Zn2+ + 2e– oxidation

Cu2+ + 2e– → Cu reduction Cl2 + 2e– → 2Cl– reduction 2Br– → Br2 + 2e– oxidation ]

In a balanced half-reaction equation the number of atoms of each element and the total charge must be the same on each side of the equation. Balanced redox equation: One in which there are no free electrons on either side of the reaction equation. If two half-equations (one oxidation and one reduction) with different numbers of free electrons are added to give an overall redox equation, the equations must be multiplied by integers so that the electrons cancel on addition. [e.g. For the reaction Zn + Ag+ → Zn2+ + Ag

Zn → Zn2+ + 2e– oxidation Ag+ + e– → Ag reduction x 2 and add Zn + 2Ag+ → Zn2+ + 2Ag ]

Oxidation number (oxidation state): A concept central to redox chemistry. Electronegativity: A measure of the power of an atom to attract electrons to itself when it is part of a compound. The order of electronegativity of common elements is: F > O > Cl > N > Br > I > S > C > H >P > B > Si Rules for oxidation number (state): The oxidation number of an atom in a substance may be determined from the following rules: 1. The oxidation number of an atom in an element is zero. 2. The oxidation number of an atom in a monoatomic ion equals the charge on the ion. 3. Oxygen has the oxidation number of -2, except in peroxides (-1) and when bound to

fluorine (+2). 4. Hydrogen has an oxidation number of +1 except in metal hydrides(-1). 5. Halogens have oxidation number -1 except in oxygen-halogen species. 6. The sum of the oxidation numbers of the atoms in a polyatomic species equals the charge

on that species. When an atom in a species has an oxidation number x it is said to be in the x oxidation state.The above rules, which usually give the correct number, are based on the general concept "the oxidation number of the atom in a species is the charge it would have in the most probable ionic formulation of that species". Putting this another way, it is assumed that all the electrons of a particular bond reside on the atom with the greater electronegativity. So an overall rule to determine oxidation numbers is: write the Lewis structure of the species and count the number of valence electrons on each atom assuming the bonding electrons reside on the more electronegative atom, and compare the number with that of the neutral atom; the difference is the oxidation number. [e.g. HCl; H-Cl; Cl > H; H has 0 electrons; Cl has 8

Page 55: Lets Talk Chemistry

12-5

valence electrons: H, +1; Cl, -1 SiH4; H-SiH3; H>Si, each H has 2 valence electrons, Si has 0 valence electrons: H, -1; Si, +4]. The concept of oxidation state plays a major role in the classification of inorganic compounds, but understanding of its significance and usefulness comes only with experience. A Roman numeral is often used in the name of a species to indicate the oxidation state. [e.g. The compounds CrCl3 and Na2CrO4 are called chromium(III) chloride and sodium tetraoxochromate(VI) respectively. See section 13.] In redox reactions the oxidation number of an atom in the oxidant decreases, while that of an atom in the reductant increases. [e.g. In the examples above the oxidation state of zinc has increased from 0 to +2, that of bromine has increased from -1 to 0, that of chlorine has decreased from 0 to -1, and that of silver has decreased from +1 to 0.] Rules for balancing redox equations: The ability to write balanced redox equations is an essential skill. This can be done by following a simple set of "book-keeping" rules: 1. Write down the formula for one of the reactants on the left and the formula for its product

on the right of an arrow. 2. Balance all elements other than oxygen and hydrogen. 3. Balance oxygens by adding H2O to the appropriate side. 4. Balance hydrogens by adding H+ to the appropriate side. 5. Balance charge by adding electrons to the appropriate side.

This gives a balanced half-equation. If the electrons are on the right-hand side, the reactant is the reductant (lost electrons and therefore has been oxidised). If the electrons are on the left-hand side the reactant is the oxidant (has gained electrons and therefore has been reduced).

6. Repeat rules 1-5 for the other reactant. 7. Multiply the two half equations by integers so that the number of electrons shown in each

half-equation is the same. 8. Add the two half-equations, cancelling equal number of species that occur on both sides.

The result is the balanced redox equation. [e.g. SO2 + Cr2O7

2– → SO42– + Cr3+

1. SO2 → SO42– 6. Cr2O7

2– → Cr3+ 2. SO2 → SO4

2– Cr2O72– → 2Cr3+

3. 2H2O + SO2 → SO42– Cr2O7

2– → 2Cr3+ + 7H2O 4. 2H2O + SO2 → SO4

2– + 4H+ 14H+ + Cr2O72– → 2Cr3+ + 7H2O

5. 2H2O + SO2 → SO42– + 4H+ +2e– 6e– + 14H+ + Cr2O7

2– → 2Cr3+ + 7H2O (Therefore SO2 is the reductant) (Therefore Cr2O7

2– is the oxidant) 7. x 3 x 1

6H2O + 3SO2 → 3SO42– + 12H+ +6e– 6e– + 14H+ + Cr2O7

2– →2Cr3+ + 7H2O 8. 2H+ + 3SO2 + Cr2O7

2– → 3SO42– + 2Cr3+ + H2O ]

A useful check when balancing redox equations is to note that the number of electrons in each half-reaction is the same as the changes in oxidation numbers. E. Hydrolysis Reaction between a compound and water which results in the making or breaking of the O-H bonds of water [e.g. the reaction of hydrogen chloride and water to give hydrochloric acid,

Page 56: Lets Talk Chemistry

12-6

section 12-3]. However, the term is more commonly used for reactions resulting in the breaking of O-H bonds of water and the formation of a bond between the O of the water and the compound. [ e.g. SiCl4 + 4H2O → Si(OH)4 + 4HCl TiCl4 + 2H2O → TiO2 + 4HCl ] F. Classification based on the mechanism of a reaction This topic is covered in section 16. EXERCISES The following reactions can all be classified in more than one of the classes given in this section. Give two classifications for each. The equations are not all balanced. 1. CH3CH=CH2 + H2 → CH3CH2CH3 2. 2H2O2 → O2 + 2H2O 3. SO2 + O2 → SO3 4. (CH3)2CHCH2Br + OH– → (CH3)C=CH2 + Br– + H2O 5. MnO4

– (aq) + Mn2+(aq) → MnO2(s) 6. Cu2+(aq) + I– (aq) → CuI(s) + I2(s) Give the oxidation number (state) of the italicised atom in each of the following species. 7. NH3 8. CaCl2 9. SO3 10. H2O2 11. ClO3

– 12. Cr2O72–

Balance the following redox reactions in aqueous solution. (aq has been omitted from the formulae for the dissolved ions.) 13. Cl2 + Fe2+ → Cl– + Fe3+ 14. MnO4

– + Br– → Mn2+ + Br2 15. The reaction of question 5 above 16. Zn + Fe2(SO4)3 → ZnSO4 + FeSO4

(Identify the spectator ions (i.e. those ions which are not taking part in the redox reaction), balance the essential ionic reaction and add the spectator ions at the end. Balanced equations in this form are needed in analytical determinations.)

Page 57: Lets Talk Chemistry

13-1

SECTION 13

NOMENCLATURE OF INORGANIC COMPOUNDS Inorganic compound: Compounds of elements other than carbon, but including carbon oxides, carbonates. The Greek and Latin prefixes given in section 6 also play an important role in inorganic nomenclature. Definitive rules for naming inorganic compounds have been agreed on internationally, but as these can lead to rather clumsy names many common names for well known compounds remain in use. Ionic compounds are called salts except where the anion is O2− or OH− in which case the compounds are called oxides and hydroxides respectively. Some polyatomic ions have accepted non-systematic names: NH4

+, ammonium; H3O+,

oxonium (hydronium) (also often just called hydrogen ion in aqueous solutions); OH−, hydroxide; NH2

−, amide; CN−, cyanide. In nomenclature, polyatomic ions or molecules are often considered to be made up of a positively charged species, (the charge being indicated by an oxidation number), surrounded by ligands, (neutral or negatively charged species), which each have a pair of electrons on the atom bonded (coordinated) to the central cation. The names for the two common neutral ligands, water and ammonia, are -aqua, H2O, and -ammine, NH3, respectively. Negatively charged ligands have the ending -o. [e.g. Cl−, -chloro; OH−, -hydroxo; O2−, -oxo; CN−, -cyano]. (The International Union of Pure and Applied Chemistry is considering changing the ending of simple anion ligands such as chloride to –ido. The examples above would be chlorido, hydroxido, oxido, cyanido.) A brief outline of the definitive rules and traditional naming is given below. The examples should clarify the rules.

Systematic naming rules 1 The cation has its name unmodified (e.g. the name of the element). 2 If the anion is monatomic its name is modified to end in -ide. 3 If the anion is polyatomic its name is modified to end in -ate. 4 When oxidation states are to be indicated they are shown by Roman numerals in

brackets following the name of the element. 5 The number of atoms or ligands attached to the central atom is denoted by Greek

prefixes. Examples:

In some ways the naming of inorganic compounds is quite different from those of organic compounds. This section introduces the common everyday names of well known inorganic compounds, and to the more formal systematic rules which have been developed.

Page 58: Lets Talk Chemistry

13-2

KBr is potassium bromide rules 1,2 Ca(OH)2 is calcium hydroxide rule 1 NaClO3 is sodium trioxochlorate(V) rules 1,3,4,5 FeCl3 is iron trichloride or iron(III) chloride rules 1,2,4,5 Cu(NH3)4F2 is tetraamminecopper(II) fluoride rules 1,2,4,5 NH4BF4 is ammonium tetrafluoroborate(III) rules 3,4,5 The traditional names for the third and fourth examples are sodium chlorate and chromic chloride respectively. The following points concern this more traditional method of naming: (a) The endings -ous and -ic are used for cations of elements existing in only two oxidation states, -ous representing the lower state. (b) The ending -ite is used to indicate the oxoanion of oxidation state lower than the common one which ends in -ate. An even lower one is hypo.....ite, and one higher than -ate is per.....ate. (c) The Latin or Greek stem is used for elements whose symbols come from those

languages. Examples: FeCl2 is ferrous chloride FeCl3 is ferric chloride SnCl2 is stannous chloride SnCl4 is stannic chloride KNO2 is potassium nitrite KNO3 is potassium nitrate Na2SO3 is sodium sulfite Na2SO4 is sodium sulfate NH4ClO is ammonium hypochlorite NH4ClO2 is ammonium chlorite NH4ClO3 is ammonium chlorate NH4ClO4 is ammonium perchlorate Salts can be thought of as the product of a reaction between an acid and a base. Where the base is an oxide or a hydroxide water is also a product. [e.g. HNO3 + NH3 → NH4NO3 (NH4

+ and NO3− )

H2SO4 + CaO → CaSO4 + H2O (Ca2+ and SO42−)

H3PO4 + 2NaOH → Na2HPO4 + 2H2O (Na+ and HPO42−) ]

The oxoacid corresponding to an -ate anion is called the -ic acid; to an -ite anion the -ous acid. HClO is hypochlorous acid; HClO2 is chlorous acid; HClO3 is chloric acid; HClO4 is perchloric acid. (Per is also used in two other ways; as shorthand for the peroxo group, -O-O- and for organic compounds where all hydrogens attached to carbon have been replaced by a halogen [e.g. perfluoroethane C2F6]. Thus one must look at the context of the term to determine the meaning.) For salts of polyprotic acids in which the anions still have hydrogen atoms attached, the word hydrogen is written before the anion. Thus NaH2PO4 is sodium dihydrogenphosphate, Na2HPO4 is sodium hydrogenphosphate, and NaHCO3 is sodium hydrogencarbonate. In the past the word hydrogen in salts of diprotic acids was often replaced by bi- [e.g. NaHCO3 sodium bicarbonate], but it is no longer approved nomenclature.

Page 59: Lets Talk Chemistry

13-3

Most of the covalent oxides are named according to the relative number of atoms of the element and oxygens [e.g. SO2, sulfur dioxide; SO3, sulfur trioxide; N2O3, dinitrogen trioxide]. Oxoacids may be thought of as coming from the hydrolysis of the oxide containing the element in the same oxidation state. [e.g. SO2 + H2O → H2SO3 sulfur dioxide giving sulfurous acid SO3 + H2O → H2SO4 sulfur trioxide giving sulfuric acid N2O3 + H2O → 2HNO2 dinitrogen trioxide giving nitrous acid ] Acid anhydride: The product from removing water from an acid. (Anhydride, a substance formed by removing the elements of water from a compound.) [e.g. the oxides in the above three reactions with water.] N2O and NO, two well known oxides of nitrogen, have the common names nitrous oxide and nitric oxide respectively. However they are not the anhydrides of nitrous acid and nitric acid. This emphasises the point that language of chemistry has developed historically and, as in any language, there are pitfalls for the beginner.

EXERCISES Deduce the common names for the following compounds from the names given for related compounds. 1. Cu2SO4 CuSO4 is cupric sulfate 2. Fe2(SO4)3 FeSO4 is ferrous sulfate 3. H3PO3 H3PO4 is phosphoric acid 4. K2MnO4 KMnO4 is potassium permanganate

Page 60: Lets Talk Chemistry

14-1

SECTION 14

CHEMICAL EQUILIBRIUM Reversible reaction: A chemical reaction that can occur in both the forward and reverse direction and shown by two arrows, ⇄. [e.g. PbCl2(s) ⇄ Pb2+(aq) + 2Cl–(aq) If solid lead chloride is added to water some dissolves; if aqueous solutions of soluble lead nitrate and soluble sodium chloride are mixed some solid lead chloride precipitates. The chemical reaction of the lead-acid battery, see page 15-3.] It is represented in the general case by:

aA + bB ⇄ cC + dD Reaction Quotient: Symbol Q, a mathematical expression relating the concentrations of reactants and products at any stage of a reaction:

ba

dc

Q[B][A]D][[C] =

The expression for Q may be written down by inspection of the balanced chemical equation. It is the product of the instantaneous concentrations of the reaction products each raised to the power of its stoichiometric coefficient divided by the product of the instantaneous concentrations of the reactants each raised to the power of its stoichiometric coefficient. Here the square bracket notation, e.g. [A], is used as the symbol for concentration rather than c(A). The form c(A), usually represents the overall concentration of a species disregarding any reactions which might have occurred. (See below, equilibrium constant.) Chemical equilibrium: The state of a reversible chemical reaction when the rate of the reaction in the reverse direction equals that in the forward direction; there is no change in amounts or concentrations of reactants or products with time. As the reaction is occurring in both directions we called it dynamic equilibrium. It requires an input of energy to change the composition at equilibrium, i.e. to drive the reaction in the forward or reverse direction. A reaction at equilibrium is indicated by the use of two half arrows, ⇌ as shown for the general case:

aA + bB ⇌ cC + dD Equilibrium constant: Symbol K, a quantitative expression for the equilibrium composition of a chemical reaction. It is the value of the reaction quotient, Q, at equilibrium. Thus for the general reaction above

ba

dc

K[B][A]D][[C] =

( badcK [B][A]/]D[[C] = ) where [A] etc. are the equilibrium concentrations of the reactant and product species present. Like Q the expression for K may be written down by inspection of the balanced chemical

Many chemical reactions do not go to completion. That is to say when the reactants are mixed and the chemical reaction proceeds it only goes to a certain extent, and at the end there are measurable amounts of both reactants and products present in the system. A system is then said to be at equilibrium. A measure of how far a reaction goes is given by its "equilibrium constant" which can have very small or very large values. This section introduces the language used for these reactions, and some important classes.

Page 61: Lets Talk Chemistry

14-2

equation. Again the square bracket notation is used for concentration. [e.g. For an aqueous 0.1 mol L-1 HCl solution, c(HCl) = 0.1 mol L-1, but the concentration of hydrogen chloride molecules actually present in the solution, [HCl], is very low. For an aqueous 0.1 mol L-1 phosphoric acid solution, for which we would write c(H3PO4) = 0.1 mol L-1, the equilibrium actually existing in solution is given by the equation:

H3PO4 + H2O ⇌ H2PO4– + H3O+

In this solution [H3PO4] < 0.1 mol L-1 because some of the phosphoric acid been converted to dihydrogen phosphate. c(H3PO4) = [H3PO4] + [H2PO4

–]. ] Concentration equilibrium constant: Symbol Kc, an equilibrium constant in which the equilibrium composition is expressed in terms of concentration. [ e.g. as above.] Pressure equilibrium constant: Symbol Kp, an equilibrium constant in which the equilibrium composition is expressed in terms of partial pressures of reactants and products instead of concentrations, because the concentration of a gas is directly proportional to its pressure (partial pressure in a mixture of gases).

(From the ideal gas equation pV = nRT, p = V

nRT = cRT.)

Standard (reference) concentrations and pressures: Standard or reference concentrations and pressures are useful concepts. In solutions the common standard for solutes is 1 mol L-1, and for a solvent (liquid) or solid the concentration of the pure solvent (liquid) or solid. For gaseous systems the standard state for partial pressure is 100 kPa. Units of equilibrium constants, K: In principle, the magnitude and units of an equilibrium constant will depend on the stoichiometry of the reaction and on the chosen units of concentration. To avoid having different values for the same equilibrium constant chemists use the ratio of the concentration to a standard concentration, or ratio of the partial pressure to a standard pressure, and K is dimensionless, i.e. just a number without units. [e.g. In the above reaction of phosphoric acid with water the water is both a reactant and the solvent. As the concentration of water is almost independent of the concentration of the solute and of the position of equilibrium, its concentration is almost that of pure water, and thus the value of [H2O] to put into the expression for K is one ([H2O] = c(H2O)/c(H2O) = 1). Thus

]PO[H]POH][O[H

43

-423

+

=K

where the values of the concentrations of the solute species are just the numerical values of their concentrations in units of mol L-1. For gas phase equilibria using Kp the values put into the expression for Kp are the actual partial pressures of the gases each divided by 100 kPa.] Position of equilibrium: A qualitative expression for the relative amounts of reactants and products present at equilibrium. The position of equilibrium may lie well to the left with only very small but measurable amounts of products present, or well to the right with only small but measurable amounts of reactants present, or anywhere in between. The magnitude of K is a quantitative measure of the position of the equilibrium. If K is large, then the numerator is much larger than the denominator, the concentration of the products is much greater than that of the reactants, and the equilibrium lies well to the right. If K is small the equilibrium lies well to the left, and only a small fraction of reactants are converted into products. The

Page 62: Lets Talk Chemistry

14-3

magnitude of this expression remains constant whatever other species or reactions are involved, i.e. whatever the composition of the system. [e.g. in an equimolar solution of aqueous H3PO4 and NaOH, (i.e. one made up of equal amounts of the two reactants), almost all the phosphoric acid reacts to give the dihydrogen phosphate anion,

H3PO4 + OH– → H2PO4– + H2O

but the equilibrium concentrations of the species in this solution still obey the nett expression

]PO[H]POH][O[H

43

423−+

=K

pK pK = -log10K. Many equilibrium constants have extremely small numerical values and it is often convenient, especially in tabulation, to express them in a logarithmic form. [e.g if K = 1.0 x 10-8, pK = 8.0; if K = 5 x 10-8, pK = 7.30] Chemical yield: The fraction of the limiting reactant converted to required product. Percentage chemical yield: The chemical yield x 100; a common way of expressing chemical yield. Temperature control of equilibrium: The magnitude of K varies with temperature, decreasing with increasing temperature for exothermic reactions, and increasing with increasing temperature for endothermic reactions. The larger the numerical value of ∆rH the greater the change in K with change in temperature. Thus the position of equilibrium, and hence the chemical yield can be altered by changing the temperature. K's on reaction written in reverse direction. It follows from the definition of K that if the direction of the chemical equation is reversed Kreverse = 1/Kforward. K for overall reactions: When an overall reaction can be considered as the sum of a series of reactions, K for the overall reaction is the product of the equilibrium constants of the individual reactions, as can be seen for a general case: Reaction (1) aA + bB ⇌ cC Reaction (2) cC + dD ⇌ pP + qQ Overall aA + bB + dD ⇌ pP + qQ

(overall) [D][B][A]

[Q][P] [D][C][Q][P] x

[B][A][C] )2( x )1( KKK dba

qp

dc

qp

ba

c

===

It follows from the above discussion (and from the definition of K) that if the stoichiometry of a reaction equation is altered by multiplying by 2, the value of K is squared. SOME CLASSES OF EQUILIBRIUM (1) Solubility Saturated solution: A solution in which the dissolved and undissolved solute are in dynamic equilibrium, or a solution containing the maximum amount of solid under equilibrium conditions. Solubility: Symbol s, the amount of substance dissolved in unit volume of a saturated solution.

Page 63: Lets Talk Chemistry

14-4

(a) A gas dissolved in a solvent

X(solv) ⇌ X(g)

K = [X(solv)]

(X(g))p Rearranging, p(X(g)) = K[X(solv)] This is a form of Henry's law

which states that for dilute solutions or sparingly soluble gases the concentration in a saturated solution is proportional to the partial pressure of the gas above the solution.

(b) A molecular solid or liquid substance dissolved in a solvent.

X(s) ⇌ X(solv) or X(l) ⇌ X(solv) The concentration of X in solution is independent of the excess amount of X(s) or X(l) present, and the ratio of [X(s)] or [X(l)] to the standard states of the pure solid or liquid is one (as explained under units of K), and the expression for K is: K = [X(solv)]. In this case K is the solubility of X in the solvent, K = s.

(c) An ionic solid dissolving in water Solubility constant: Symbol Ks (older term solubility product, symbol Ksp), the equilibrium constant for the dissolution of an ionic compound in water. (i) MX(s) ⇌ M+(aq) + X–(aq)

Ks = [M+(aq)][X–(aq)]. In this case it can be seen from the stoichiometry that:

s(MX) = [M+(aq)] = [X– (aq)] and s = Ks½ because Ks= s2.

(ii) MX2 (s) ⇌ M2+(aq) + 2X–(aq) Here Ks = [M2+(aq)][X–(aq)]2, and s(MX2) = [M2+(aq)] = [X–(aq)]/2.

Thus Ks = s x (2s)2 = 4s3 and s = (Ks/4)� (d) Distribution of a solute between two immiscible liquids A and B Distribution coefficient: The equilibrium for the distribution of a substance between two immiscible liquids (i.e. liquids of low solubility in each other) A and B.

X(A) ⇌ X(B)

K = [X(A)][X(B)]

(2) Decomposition of a substance [e.g. CaCO3(s) ⇌ CaO(s) + CO2(g)

Decomposition of limestone on heating (calcining). K = p(CO2)

or N2O4(g) ⇌ 2NO2(g)

Dissociation of N2O4. )O(N

)(NO 42

22

ppK = ]

Page 64: Lets Talk Chemistry

14-5

(3) Acid-base pH: pH = -log10([H3O+]/1mol L-1). The quantity most measured in aqueous acid-base equilibrium is the oxonium (hydronium) ion concentration (but usually called the hydrogen ion concentration in this context) [H3O+(aq)]. As this is often very small it is common to take the logarithm of the hydrogen ion concentration and express it as pH. [e.g. If the pH of a solution is 6, [H3O+] = 1 x 10-6 mol L-1] The operation of taking a logarithm can only be performed on a pure number. Thus [H3O+] is divided by the standard unit of concentration, 1 mol L-1. Kw: Kw = [H3O+][OH-]. In water the very important equilibrium

H2O + H2O ⇌ H3O+(aq) + OH–(aq) always exists, with the equilibrium constant Kw = [H3O+][OH–] = 1.0 x 10-14 at 25 oC and. pKw = 14. In very pure water at 25 oC [H3O+] = [OH–] = 1 x 10-7 mol L-1. Neutral solution: An aqueous solution with [H3O+] = [OH–]; at 25 oC pH = 7. Acidic solution: An aqueous solution with pH < 7.

Basic or alkaline solution: An aqueous solution with pH > 7. Acidity constant: symbol Ka, the equilibrium constant for the reaction

HA(aq) + H2O ⇌ H3O+(aq) + A–(aq) (older term, the acid dissociation constant.):

[HA(aq)]aq)]((aq)][AO[H

(HA) 3a

−+

=K

Weak acids: Acids with small values of Ka. (Ka < 1, pKa positive, position of equilibrium well to left, only a small fraction of acid molecules reacted.) [e.g. In 1.0 mol L-1 acetic acid, pKa = 4.76, [CH3CO2

–] = [H3O+] = 4.2 x 10-3 mol L-1, pH = 2.4.] Strong acids: Acids with large values of Ka (Ka >> 1, pKa negative, equilibrium well to right, almost all acid molecules reacted). [e.g. in 1.0 mol L-1 nitric acid, pKa = -1.3, [H3O+] = 1.0 mol L-1, pH = 0 ([HNO3] very small).] Weak base: A base for which the equilibrium position for the reaction

B(aq) + H2O ⇌ BH+(aq) + OH–(aq) lies well to the left, only a small fraction of molecules have accepted a proton from water. Strong base: A base for which the equilibrium position for the above reaction lies well to the right, only a small fraction of molecules remain in the solution. Basicity constant: symbol Kb the equilibrium constant for the reaction

B(aq) + H2O ⇌ BH+(aq) + OH–(aq)

[B(aq)])]aq((aq)][OH[BH b

−+

=K

Page 65: Lets Talk Chemistry

14-6

It is easily shown that Kb = acid) conjugate(a

w

KK

. Qualitatively, this expression says that the

stronger an acid the weaker its conjugate base. Strong acids have very weak conjugate bases and very weak acids have strong conjugate bases. [e.g. HNO3 is a strong acid, (pKa –1.3), NO3

– is a very very weak base; HS– is a very weak acid (pKa 19), S2– is a strong base.] Tables of Kb are unnecessary, tables of Ka or pKa of the corresponding conjugate acids being sufficient to state base strength. For an acid HA, where pKa(HA) = 7, pKb(A–) is also 7, and it can be said that the strengths of the acid and its conjugate base are the same. Polyprotic acids and bases: Acids with two, three or more hydrogens which can undergo proton transfer. Polyprotic acids can be diprotic, triprotic etc. [e.g. H2SO4- diprotic; H3PO4 - triprotic] Similarly bases which can accept more than one proton are said to be polyprotic bases. [e.g. S2–- diprotic; PO4

3–- triprotic] For polyprotic acids successive acidity constants can be written [e.g. for phosphoric acid:

H3PO4 + H2O ⇌ H2PO4– + H3O+ K = K1(H3PO4)

H2PO4– + H2O ⇌ HPO4

2– + H3O+ K = K2(H3PO4) HPO4

2– + H2O ⇌ PO43– + H3O+ K = K3(H3PO4)

Buffer solution: A solution which is resistant to change in pH (when small amounts of acid or base are added or it is diluted). A buffer solution contains significant amounts of both an acid and its conjugate base. If H3O+ is added to or produced in the solution, it is removed by reaction with the base. Like wise any OH– is removed by reaction with the conjugate acid:

H3O+ + A– ⇌ HA + H2O OH– + HA ⇌ A– + H2O

It can be seen from the expression of Ka(HA) that if a buffer solution is made up with equimolar amounts of HA and A- that the pH of the solution is equal to the pKa(HA) of the acid. The pH of a buffer solution can be controlled by taking the appropriate amounts of HA and A–. Brønsted acid: A proton donor; the molecule or ion must contain H [e.g. HCl and all the acids above in this section]. (See page 12-3) Brønsted base: A proton acceptor; the molecule or ion must have a pair of non-bonding electrons to form a bond with a proton [e.g. :NH3 and all the bases above in this section]. (See page 12-3) Lewis acid: A species (molecule or ion) which can accept a pair of electrons from a base and form a bond. This is a broader definition of an acid [e.g. carbon dioxide in its reaction with the hydroxide ion to give the hydrogencarbonate ion: CO2 + OH– → HCO3

– ]. A proton is a Lewis acid by this definition. Lewis base: An electron pair donor. [e.g. the hydroxide ion in the above example] Brönsted bases are also Lewis bases as they donate a pair of electrons to the proton. (4) Complex ion formation Complex ion: An ionic species consisting of ligands (Lewis bases) (page 13-1) bonded to a metal cation (which is acting as a Lewis acid) [e.g. Cu(NH3)4

2+; CrCl63–]. The ligands may

Page 66: Lets Talk Chemistry

14-7

be neutral molecules [e.g. NH3] or anions [e.g. Cl–]. Stability constant (Formation constant): The equilibrium constant for the overall reaction between the aquated metal cation and the free ligands to form the complex ion. [e.g

Cu2+(aq) + 4NH3 ⇌ Cu(NH3)42+

43

2

243

f ](aq)][NH[Cu])[Cu(NH

+

+

=K ]

Note: The formal convention for arrows in chemical equations is that → represents a

reaction in the forward direction, ⇄ represents a reversible reaction, and ⇌ represents a reaction at equilibrium.

. EXERCISES

Write the expression for the equilibrium constant for the following reactions. 1. 2SO2(g) + O2(g) ⇌ 2SO3(g) 2. N2(g) + 3H2(g) ⇌ 2 NH3(g) 3. BaSO4(s) ⇌ Ba2+(aq) + SO4

2–(aq) 4. 2HgO(s) ⇌ 2Hg(l) + O2(g) 5. HF(aq) + H2O ⇌ H3O+(aq) + F–(aq) 6. Ch3NH2(aq) + H2O ⇌ CH3NH3

+(aq) + OH–(aq) 7. HCO3

–(aq) + H2O ⇌ CO32–(aq) + H3O+(aq)

8. CO3

2–(aq) + H2O ⇌ HCO3–(aq) + OH–(aq)

Calculate the solubility in pure water of the following ionic solids. 9. AgBr, pKs = 12.3 10. CaCO3, pKs = 8.48 11. PbI2, pKs = 8.10 Calculate the pH of a solution if 12. [H3O+] = 1.0 x 10-3 mol L-1 13. [H3O+] = 2.4 x 10-4 mol L-1 14. [OH–] = 1.0 x 10-3 mol L-1 15. [OH–] = 2.4 x 10-4 mol L-1 16. A buffer solution was made up with H2PO4

– and HPO42–. Calculate the pH of the

solution (i) when equimolar amounts 0f each ion are used, and (ii) when the amount of H2PO4

– is twice that of HPO42–. pK2(H3PO4) = 7.20

Calculate pKa for weak acids with

Page 67: Lets Talk Chemistry

14-8

17. Ka = 1.0 x 10-6 18. Ka = 5.6 x 10-8

Page 68: Lets Talk Chemistry

15-1

SECTION 15 ELECTROCHEMISTRY Electrochemistry: The branch of chemistry that covers the relative strengths of oxidants and reductants, the production of electric current from chemical reactions, and the use of electricity to produce chemical change. Electrochemical cell: A system made up of two electrodes in contact with an electrolyte. Electrode: A conductor of electricity, commonly a metal or graphite in contact with an electrolyte in an electrochemical cell. Electrolyte: A medium (phase) which conducts electricity by the movement of ions [e.g. a molten salt] or a substance which dissolves in a solvent to give a conducting solution [e.g. aqueous sodium chloride, NaCl or any other soluble ionic compound]. Electrode reaction: A chemical reaction occurring at an electrode involving gain or loss of electrons. It is called a half-reaction. [e.g. Cu2++ 2e– → Cu; Zn → Zn2+ + 2e– ] (See page 12-4.) Redox couple: The two species of a half- reaction involving oxidation or reduction. (See page 12-3.) Represented as oxidised species/reduced species [e.g. Cu2+/Cu; Cl2/Cl–; Fe3+/Fe2+]. Half-cell: An electrode and the couple it is in contact with [e.g. Zn(s) |Zn2+(aq), a zinc rod in contact with an aqueous solution containing zinc cations; Fe3+(aq),Fe2+(aq)|Pt, a platinum rod in contact with an aqueous solution containing ferric and ferrous ions ]. The electrode may be one of the species of the couple or an inert species [e.g. zinc and platinum respectively above]. The vertical line represents a phase boundary. Galvanic cell: an electrochemical cell that produces electricity from a chemical reaction. It consists of two half-cells connected by a junction [e.g. a membrane (diaphragm) or salt bridge] allowing ions to be transferred between the electrolytes of the two half-cells. When the electrodes are connected externally by a conductor electrons flow through the conductor from the negative electrode to the positive electrode. It is normally represented by two half-cells separated by a double vertical line representing the junction allowing transfer of ions. [e.g. Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) or more simply as Zn|Zn2+||Cu2+|Cu if it is clear from the context that aqueous solutions are involved. ] Cell potential or electromotive force: symbol E, the electric potential difference between the electrodes of a galvanic cell when no current is flowing. In the above representation of a galvanic cell it is the electrode potential of the right hand electrode minus the electrode

Some systems involving redox reactions can be designed so that the reactants (and products) are partially separated from each other, and the reaction leads to an electric current being produced in an external circuit, and which can be used for many useful purposes. Batteries and their many uses are the obvious examples. Electrical energy can also be used to drive non-spontaneous chemical reactions to produce desired products in processes known as electrolysis. This section introduces the language and concepts of these processes collectively known as electrochemistry.

Page 69: Lets Talk Chemistry

15-2

potential of the left hand electrode. It is normal to show the negative electrode on the left and the positive electrode on the right and thus express the cell potential as a positive quantity. The SI unit of E is the volt, symbol V. Standard hydrogen electrode: symbol she, a reference electrode defined as having zero electrode potential. It consists of a platinum electrode in contact with gaseous dihydrogen and aqueous hydronium ions under defined conditions. Electrode potential (of a couple): symbol E(oxidised form/reduced form), also called reduction potential or redox potential is the electrode potential of that couple relative to the standard hydrogen electrode. More precisely it is the cell potential of a galvanic cell in which one half-cell is the standard hydrogen electrode, and is negative if the electrode of the couple is the negative electrode of that cell and positive if the electrode is the positive electrode of that cell [e.g. E(Cu2+/Cu = 0.34 V; E(Zn2+/Zn) = -0.76 V; E(Cl2/Cl–) = 1.40 V ]. The magnitude of a redox potential is a measure of the relative strength of an oxidant or reductant. The more positive the redox potential the stronger the oxidant (i.e. the greater its potential to oxidise some other species) and the weaker the reductant. Conversely the more negative the redox potential the stronger the reductant and the weaker the oxidant. Strong oxidant: An oxidant of a redox couple with a large positive redox potential [e.g. F2, E(F2/F–) = 2.9 V; Cr2O7

2–, E(Cr2O72–/Cr3+) 1.4 V. The half-reaction corresponding to this

couple in aqueous solution is that given on page 12-5, 14H+ + Cr2O7

2– + 6e– → 2Cr3+ + 7H2O ]. Strong reductant: a reductant of a redox couple with a large negative redox potential [e.g. Na, E(Na+/Na) = -2.7 V]. Table of electrode potentials: A table listing the electrode potentials of redox couples. As electrons flow spontaneously from the negative to the positive electrode the electrode potentials can be used to deduce the direction of a redox reaction. [e.g. Can Cl2 oxidise copper, Cu? Since this reaction would involve flow of electrons from Cu to Cl2 it can be seen from the electrode potentials above that the redox reaction can occur as E(Cu2+/Cu) is negative with respect to E(Cl2/Cl–). A table in which the electrode potentials increase down the table (most negative at the top and most positive at the bottom) can be regarded as an energy diagram. A spontaneous reaction involving two couples will be that in which electrons lose energy by moving from a reductant of one couple to the oxidant of another couple below it in the table. [e.g. Couple E/V

Zn2+/Zn -0.76 Cu2+/Cu 0.34 Cl2/Cl– 1.40

Electrons can flow from Zn to Cu2+ but not from Cu to Zn2+. Thus zinc metal can reduce copper ions to copper metal but copper metal cannot reduce zinc ions to zinc metal. ] Cell reaction: The overall chemical redox reaction occurring in the cell. What distinguishes a cell reaction from a general redox reaction is that the reactants are physically separated in the cell and the processes of oxidation and reduction are accompanied by electron flow in an external conductor [e.g. for the cell Zn2+|Zn||Cu2+|Cu the half-reactions will be: Zn → Zn2+ + 2e! (as the zinc electrode is negative) and Cu2+ + 2e– → Cu (as copper is the positive electrode)

Page 70: Lets Talk Chemistry

15-3

and the overall reaction the sum of these, Zn + Cu2+ → Zn2+ + Cu ]. Electrolysis: the use of electric current to drive chemical reactions. Electrical energy is converted into potential chemical energy, and the non-spontaneous reaction is forced to occur. Electrolytic cell: an electrochemical cell for electrolysis. Consists of two electrodes in contact with electrolyte(s), one connected to the positive side and one connected to the negative side of a direct current power supply. A species is oxidised at the electrode connected to the positive side (electrons are "sucked" out of the electrode) and a species is reduced at the electrode connected to the negative side (electrons are "pushed" in). The design of cell varies. The electrodes may both be in the same electrolyte [e.g. in decomposition of water from aqueous sulfuric acid, or copper-plating a metal from a copper sulfate solution and a copper electrode ] or in different compartments separated by a device allowing migration of ions between the two electrolytes [e.g. the production of H2, Cl2 and NaOH from the electrolysis of aqueous NaCl]. As for a galvanic cell, half-reaction equations and an overall chemical equation can be written for the cell reactions. [e.g. for the above three electrolytic reactions the equations are as follows: Decomposition of water: 2H+(aq) + 2e– → H2 electrode connected to -ve side

2H2O → O2 + 4H+(aq) + 4e– electrode connected to +ve side overall 2H2O → 2H2 + O2

Copper-plating: Cu2+(aq) + 2e– → Cu(metal plated) electrode connected to -ve side

Cu(electrode) → Cu2+(aq) +2e– electrode connected to +ve side

overall Cu(electrode) → Cu(metal plated) Aqueous NaCl: 2H2O + 2e– → H2 + 2OH– electrode connected to -ve side

2Na+ + 2Cl– → 2Na+ + Cl2 + 2e– electrode connected to +ve side overall 2Na+ + 2H2O + 2Cl– → 2Na+ + 2OH– + H2 + Cl2

Here Na+ ions migrate through a diaphragm from the compartment where Cl2 is being produced to the compartment where H2 is being produced.] Anode: the electrode at which a species is oxidised. Thus for a galvanic cell it is the negative electrode, and for an electrolysis cell it is the electrode connected to the positive side of the power supply. Cathode: the electrode at which a species is reduced. Thus for a galvanic cell it is the positive electrode, and for an electrolysis cell it is the electrode connected to the negative side of the power supply. Reversible galvanic cell: a cell in which the direction of the cell reaction can be reversed by electrolysis [e.g. the common car battery (lead-acid battery)

-ve Pb,Sb | Pb | PbSO4(s) |H2SO4(aq) | PbSO4(s) | PbO2 | Pb,Sb +ve Pb,Sb represents the electrodes, a grid made of an alloy which supports the reductant, Pb, and the oxidant, PbO2. To recharge the battery electrons are pushed into the left-hand electrode and sucked out of the right-hand electrode by applying an external power supply]. Faraday's laws of electrochemistry: The 1st law states that the mass of the substance being oxidised or reduced is proportional to the charge passed. The 2nd law states that this mass is directly proportional to the molar mass of the substance and inversely proportional to its

Page 71: Lets Talk Chemistry

15-4

change in oxidation state, z. Combining these laws the charge passed, Q ∝ zm/M (Q ∝ zn), i.e. Q = znF where Q is the charge passed, n is the amount of substance oxidised or reduced and F is the proportionality constant, called the Faraday constant. When z = 1 and n = 1 mol, Q = 1 mol x F = F mol, i.e. the Faraday constant is the charge per mole of electrons. Faraday constant: symbol F, the molar electric charge, i.e. the charge per mole of electrons, 96 485 C mol-1. EXERCISES Couple

Half-reaction

E/V

Zn2+/Zn

Zn2+ + 2e– ⇄ Zn

-0.76

Sn2+/Sn

Sn2+ + 2e– ⇄ Sn

-0.14

Cu2+/Cu

Cu2+ + 2e– ⇄ Cu

0.34

I2/I–

I2 + 2e– ⇄ 2I–

0.54

Fe3+/Fe2+

Fe3+ + e– ⇄ Fe2+

0.77

Ag+/Ag

Ag+ + e– ⇄ Ag

0.80

Br2/Br–

Br2 + 2e– ⇄ 2Br–

1.10

O2/H2O

O2 + 4H+ + 4e– ⇄ 2H2O

1.23

Cl2/Cl–

Cl2 + 2e– ⇄ 2Cl–

1.40

From the above table 1. Deduce which of the following species is the strongest oxidant: I2, Ag+, Sn2+, O2 2. Deduce which of the following species is the strongest reductant: Cl–, Sn, Ag, Fe2+ Write the cell, the chemical equation for the overall cell reaction and determine the cell potential for a galvanic cell made from the redox couples: 3. Zn2+/Zn and Fe3+/Fe2+ 4. Br2/Br– and I2/I– 5. Ag+/Ag and Cl2/Cl– Predict whether 6. Sn can reduce Br2 7. O2 can oxidise Cl– 8. Ag can reduce Cu2+ 9. Fe3+ can oxidise I–

Page 72: Lets Talk Chemistry

16-1

SECTION 16 RATES OF REACTION AND REACTION MECHANISM Chemical kinetics (kinetics): The study of the rates of chemical reactions and the steps by which they occur. Rate: Rapidity of change, a property involving time. Some chemical reactions are very fast and appear instantaneous to the eye on mixing the reactants, while others are very slow and the products only form over minutes, hours or even longer. Reaction rate: The change in concentration of a reactant or product divided by the time it takes for the change to occur. Common units are mol L-1 s -1. [e.g. for the reaction between potassium permanganate and oxalic acid, 6H+ + 2MnO4

– + 5H2C2O4 → 2Mn2+ + 10CO2 + 8H2O the decrease in concentration of permanganate over 40 s was found to be 0.12 mol L-1 Thus the average change in concentration over this period was ∆[MnO4

–]/t = -0.12 mol L-1/40 s and the reaction rate could be expressed as the rate of loss of MnO4

– = 3.0 x 10-3 mol L-1 s-1. The rate of this reaction could also have been expressed as the rate of loss of H2C2O4 (7.5 x 10-3 mol L-1 s-1) or as the rate of formation of CO2 (1.5 x 10-2 mol L-1 s -1).] In this example the numerical value of the rate depends on which reactant or product concentration is being measured. To overcome this it is possible to express the rate of reaction defined by its chemical equation. Rate of reaction: Symbol, v, for the reaction aA + bB → cC + dD the rate is equal to (rate of formation of C)/c or (rate of formation of D)/d or (rate of loss of A)/a or (rate of loss of B)/b [e.g. for the previous example v = 1.5 x 10-3 mol L-1 s -1]. Most chemical reactions take place in a series of elementary reactions (steps). Elementary reaction: A one step reaction. There are three classes of elementary reaction. Unimolecular reaction: An elementary reaction involving only one particle [e.g. N2O4(aq) → 2NO2(aq) ]. Bimolecular reaction: An elementary reaction involving only two particles (the most common type). The particles may be the same or different [e.g. 2Br·(aq) → Br2(aq) or HF(aq) + NH3(aq) → F–(aq ) + NH4

+(aq)]

Some reactions appear to occur instantaneously on mixing the reactants; that is to say too fast to follow by eye. Others are much slower and they can be followed over seconds, minutes or hours. The study of how fast reactions occur, i.e. the study of reaction rate, is known as kinetics, and the specific language and concepts which have been developed are introduced in this section. Studies of how the rate of a reaction depends on the concentrations of reactants, and sometimes products, have provided evidence for how an overall reaction proceeds, i.e. what we call the mechanism of a reaction. This section also introduces some basic classes of reaction mechanism which are widely used in describing and understanding chemical reactions.

Page 73: Lets Talk Chemistry

16-2

Termolecular reaction: An elementary reaction involving only three particles. (Common in gas phase reactions where a third body is required to disperse the energy released from bond formation when two particles combine to give one [e.g. 2Cl·(g) + Cl2(g) → 2Cl2(g) The molecule of chlorine initially present gains energy released by the formation of the new Cl-Cl bond].) Molecularity: The number of particles (molecules, ions or radicals) taking part in an elementary reaction - 1 for unimolecular, 2 for bimolecular and 3 for termolecular. Reaction mechanism: The sequential elementary steps of a reaction. A + B → X X + C → P + Q The overall reaction is A + B + C → P + Q and the species X is called a reaction intermediate. Reaction intermediate: A species produced in one step of a chemical reaction but which is consumed in a later step. It is not a product of the overall reaction, and hence does not appear in the overall equation. Rate-determining step: The slowest elementary step which effectively determines the overall rate of a reaction. Rate equation (law): An equation relating the instantaneous rate of the reaction to the concentrations of reactants, products, or any other permanent species (such as catalysts) taking part in the reaction at that instance [e.g. for the reaction H2O2(aq) + 2H+(aq) + 2I– (aq) → I2(aq) + 2H2O at pH of 7 the rate law is found by experiment to be: v = k[H2O2][I–] Rate coefficient (constant): Constant of proportionality in a rate law [e.g. k in the above rate law]. Reaction order: (a) With respect to a single species - the power to which the concentration of a single species is raised in a rate law [e.g. the above reaction is first-order in H2O2 and first-order in I–]. (b) Overall - the sum of the powers of the species in the rate law [e.g. the above reaction is second-order overall. k is a second-order rate constant]. For a reaction between two species to occur they must collide, and most reactions take place in a series of bimolecular collisions, i.e by a series of bimolecular elementary steps. If the step involves two different species, A and B, the rate of collision, and therefore the rate of reaction, is proportional to the concentrations of A and B. Thus for the elementary bimolecular reaction A + B → C + D rate of reaction = k[A][B] where k is called a bimolecular rate constant. A bimolecular rate constant is clearly 2nd order rate constant. For elementary reactions the order of the reactions are the same as the molecularity. As the units for rate of reaction are mol L-1 s-1 it follows from the rate laws that the units of 1st order, 2nd order and 3rd order rate constants are respectively s-1, L mol-1 s -1 and L2 mol-2 s-1. As bonds are often broken in elementary reactions not all collisions between reactant species lead to reaction, only those which collide with sufficient energy to overcome the activation energy barrier.

Page 74: Lets Talk Chemistry

16-3

Activation energy: Symbol Ea. The minimum energy needed for colliding species to react, or cross the barrier. The magnitude of k depends on the height of the barrier, i.e. on the activation energy. The greater the Ea, the smaller the k, and the slower the reaction. The activation energy, Ea, is determined form an Arrhenius plot. Arrhenius plot: A plot of ln k or ln rate against 1/T where T is the absolute temperature (temperature in kelvin). The slope of an Arrhenius plot equals -Ea/R where R is the universal gas constant. This follows from the Arrhenius equation. Arrhenius equation: k = Aexp(-Ea/RT) ( RTEAek /a −= ) where A is called the Arrhenius pre-exponential factor. The kinetic energy of molecules, and hence the fraction of collisions with energy in excess of Ea increases with increasing temperature. Therefore the rate of a reaction, and the magnitude of k, increase with temperature. The greater the Ea, the more sensitive the rate and rate constant to changes in temperature. Unlike equilibrium constant expressions, the rate law for a reaction cannot be derived from inspection of the overall chemical equation but must be determined experimentally. Only for elementary reactions (i.e. one step reactions) can the rate law be deduced directly from the equation (as explained above). Information on reaction mechanisms can be obtained from experimentally obtained rate laws. While a rate law for a reaction can be derived for a postulated mechanism (because the rate law for each elementary step can be written), agreement between the experimental rate law and postulated mechanism does not prove the mechanism. Different mechanisms can lead to the same rate law. Integrated rate equation: An equation which relates the concentration of a species at a given time to its initial concentration and the time, and which involves the rate constant [e.g. for the 1st order reaction A → Products [A]t = [A]oe-kt where [A]t and[A]o are the concentrations of reactant A at time t and at the beginning of the reaction respectively]. Half-life: Symbol t½, the time for the amount of a reactant in limiting amount (i.e. not in excess) to fall to half its original value. Commonly applied to radioactive isotopes [e.g. the half-life of cobalt-60, 60Co, is 5.3 years]. Mean life-time: Symbol τ, the time for the amount of a reactant in limiting amount (i.e. not in excess) to fall to 0.368 (=1/e) of its original value. This is the average time a molecule must wait before undergoing reaction. Chain reaction: A reaction in which an intermediate formed in an initiation step produces further reactive intermediates in the propagation steps until removed in a termination step. [e.g. For the photochemically initiated reaction between toluene and bromine

C6H5CH3 + Br2 → C6H5CH2Br + HBr Br2 → 2Br· initiation

Br· + C6H5CH3 → HBr + C6H5CH2· propagation steps making

C6H5CH2· + Br2 → C6H5CH2Br + Br· one propagation cycle

2Br· → Br2 termination

Page 75: Lets Talk Chemistry

16-4

Chain length: The average number of propagating cycles between initiation and termination. Catalyst: A substance which increases the rate of a reaction without being consumed in the overall reaction. It provides an alternative pathway or mechanism which increases the rate of the overall reaction. The catalyst is involved in the new pathway but if it is consumed in one step it is regenerated in a later step. Homogeneous catalysis: The catalyst is in the same phase as the reactants [e.g. sulfuric acid in the esterification of an organic acid and alcohol ]. Heterogeneous catalysis: The catalyst is in a different phase from the reactants. A solid catalyst adsorbs reacting species onto its surface where the reaction takes place [e.g solid V2O5 in the reaction of gaseous SO2 and O2 to form SO3 ]. Classification and representation of reaction mechanisms A detailed reaction mechanism uses structural formulae (Lewis or condensed as appropriate), and shows the details of the movement of electrons in bond breaking and bond making in each elementary step. For addition, substitution and some elimination reactions each step is considered a reaction between a substrate (the main reactant compound) and a reagent (a second reactant) which converts the substrate into a new species. The reagents are defined as nucleophiles, electrophiles or free-radicals. Nucleophile: A chemical species seeking a nucleus with which to form a bond. To do this the species must have a pair of non-bonding electrons available to form a bond. A nucleophile is also a Lewis base. Lewis base: An electron pair donor in a chemical reaction (see page 14-6). Electrophile: A chemical species seeking a pair of electrons on a second species with which to form a bond. An electrophile is also a Lewis acid. Lewis acid: An electron pair acceptor in a chemical reaction (see page 14-6). Free-radical: A chemical species with an unpaired valence electron seeking an electron with which to form a bond. (see page 1-4) Adduct: The product of an addition reaction. Carbocation: A trivalent species with a positive charge, an electrophile and a Lewis acid [e.g. (CH3)3C+, the positive carbon having only six valence electrons]. The mechanistic classification is combined with the structural classification and may include the reaction order (1st or 2nd) of the rate-determining elementary step and further information on the substrate. [e.g. CH3Cl + I– → CH3I + Cl– nucleophilic substitution second order, SN2, (nucleophilic attack of iodide ion on carbon replacing chlorine in one step, synchronous making and breaking of bonds)]. In representing details of bond breaking and making electron movement is shown using a "curly arrow", for electrons moving as pairs, and for electrons moving singly. The tail of the arrow is close to the electron(s) moving and the head shows where they are moving

Page 76: Lets Talk Chemistry

16-5

to. Bond breaking and making is described as heterolytic or homolytic. Heterolytic bond breaking or making: The pair of electrons of a bond move together [e.g. breaking (CH3)3C-Br → (CH3)3C+ + Br–

making and breaking I: – + CH3-Cl → I-CH3 + :Cl– ]. Homolytic bond breaking: Electrons of a bond divide evenly to give two free-radicals [e.g. CH3O-OC2H5 → CH3O· + ·OC2H5 ]. Homolytic bond making: Pairing of an unpaired electron with a second electron from a bond of the substrate to form a new bond and a new radical; or dimerisation of two radicals to give a molecule [e.g. Cl· + CH2=CH2 → Cl-CH2-CH2· or Br· + ·Br → Br2 ] Full reaction schemes showing electron movement are given in the examples of mechanisms below. Note how it is not necessary to show non-bonding electron pairs. The tail of the curly arrow implies the presence of an electron pair. (i) Nucleophilic substitution second order, SN2 [e.g CH3Cl + I– → CH3I + Cl– ]

(Nucleophilic attack of iodide ion on carbon replacing chlorine in one step, synchronous making and breaking of bonds.) (ii)

Nucleophilic substitution first order, SN1 [e.g. (CH3)3CBr + H2O → (CH3)3COH + HBr ] (The slow dissociation of the

substrate to (CH3)3C+ and Br– followed by fast addition of water to the carbocation and loss of proton.)

H

H

O(CH3)3C + :OH2 (CH3)3COH + H3O+

+H

HO(CH3)3C(CH3)3C+ :OH2

(CH3)3C+ + Br -Br(CH3)3C

(iii) Electrophilic addition [e.g. CH3CH=CH2 + HCl → CH3CHCl-CH3 ]

(Addition of a proton from HCl to the carbon utilising an electron pair of the double bond followed by fast addition of chloride to the carbocation intermediate.)

I- C

H

HH

ClH

H

C

I H

+ Cl -

Page 77: Lets Talk Chemistry

16-6

+ + Cl -

CH3

HH

H

H

CC

ClH

CH3

H H

H

C C

CH3

C

H

CH3+

Cl-

CH

CH3

Cl

CH3

(iv) Elimination second order, E2

[e.g. (CH3)2CH-CH2Br + OH– → (CH3)2C=CH2 + H2O + Br– ] (Transfer of a proton from the secondary carbon atom to the hydroxide ion and loss of bromide ion in one step.)

+ H2O + Br -

CH3

CH3H

H

CC

-OHH

H

Br

C

H

CH3

CH3

C

(v) Electrophilic aromatic substitution [e.g. C6H6 + HNO3 → C6H5NO2 + H2O ]

(The first and slow step involves the addition of NO2+, an intermediate species from

nitric acid, to a carbon atom of the benzene ring utilising a pair of electrons from the ring. This is followed by a fast proton transfer from the intermediate adduct.)

2HNO3 ⇌ H2O + NO2+ + NO3

+ HNO 3

NO2- NO2O+

H

NO2

+

+

O:-O

N

H+

O

O

N

Note that the hydrogens are not shown on benzene itself, but when addition to a carbon atom occurs, the H atom on that carbon is shown.

Page 78: Lets Talk Chemistry

16-7

(vi) Free-radical addition [e.g. Cl· + CH2=CH2 → CH2Cl-CH2· ]

CH2CH2ClCH2CH2Cl..

(vii) Hydrogen abstraction [e.g. C6H5CH3 + Cl· → C6H5CH2· + HCl ] (This could be classified as attack of a chlorine atom on hydrogen replacing a benzyl group, i.e. a free-radical substitution reaction, but the simpler term hydrogen abstraction is always used for the removal of a hydrogen atom from a substrate by a free-radical.)

Cl+ HCl. .CH2HCH2

Note: (1) that electron pairs and H atoms are not necessarily shown. They can be implied from the symbolism, or from the number of valence electrons in the structure shown. (2) how important it is to do "electron book-keeping", and show formal charge. This is the language used in research publications involving discussion of reaction mechanism - using the condensed structures consistent with the level of detail required. EXERCISES 1. Calculate the average rate of the following reactions in units of mol L-1 s-1 from the

given concentration changes. (a) CH3CH=CH2 + Br2 → CH3CHBrCH2Br when [Br2] changed from 0.34 to

0.25 mol L-1 in 90 seconds. (b) 6H+ + 2MnO4

– + 5H2C2O4 → 2Mn2+ + 10CO2 + 8H2O when [MnO4

–] changed from 0.40 to 0.27 mol L-1 in 48 minutes. 2. The bimolecular reaction H3O+ + OH– → 2H2O is one of the fastest known with

a bimolecular rate constant of 1.0 x 1011 L mol-1 s-1. Calculate the rate of the reaction when equal volumes of 0.10 mol L-1 solutions of hydrochloric acid and sodium hydroxide are mixed. Remember that on mixing the two solutions the concentration of each reactant is changed.

3. The reaction in question 1(a) is an electrophilic addition and occurs in two steps. In

the first slow step a bromine atom becomes attached to C1 and a bromide ion is formed. In the second step the bromide ion adds to the bromo-carbocation intermediate. Show the mechanism using curly arrows, and give the rate law.

4. Give a structural and mechanistic classification of the reaction between toluene and

bromine given as an example of a chain reaction on page 16-3.

Page 79: Lets Talk Chemistry

Ans-1

ANSWERS TO EXERCISES 1-2. potassium, K, 20 1-3. copper, Cu, 36 1-4. radon, Rn, 136 1-7. C2H4O2 1-8. C2H5O2N 1-9. C6H12O4N2S2 1-10. KBr 1-11. NH4NO3 1-12. Mg(NO3)2 1-13. Al2(SO4)3 2-2. 2C6H6 + 15O2 → 12CO2 + 6H2O 2-3. H2 + Cl2 → 2HCl 2-4. CH4 + H2O → CO + 3H2 2-5. Fe3O4 + 4C → 3Fe + 4CO 2-6. 2NO + O2 → 2NO2 2-7. 2NaNO3 → 2NaNO2 + O2 2-8. 4NH3 + 3O2 → 2N2 + 6H2O 2-9. 2NH3 + 2O2 → N2O + 3H2O 2-10. CO2 + H2 → CO + H2O 2-11. Tl+(aq) + F−(aq) → TlF(s) 2-12. Cu2+(aq) + CO3

2−(aq) → CuCO3(s) 2-13. 3Ca2+(aq) + 2PO4

3−(aq) → Ca3(PO4)2(s) 2-14. Mg2+(aq) + SO4

2−(aq) + Ba2+(aq) + 2OH−(aq) → Mg(OH)2(s) + BaSO4(s) 3-2. 1s22s2p2 3-3. 1s22s2p5 3-4. 1s22s2p63s2p6d64s2 3-5. 1s22s2p63s2p6d104s2p3 3-6. 1s22s2p63s2p6d104s2p6d95s2 (on rules given, but 1s22s2p63s2p6d104s2p6d105s on chemical evidence) 3-(7-12). C, n = 2 l = 0 (s) or 1 (p); F, n = 2 l = 0 (s) or 1.(p); Fe, n = 4 l = 0 (4s) or n = 3 l = 2 (3d) As, n = 4 l = 0(s) or 1(p) Ag, n = 5 l = 0 (5s) or n = 4 l = 2 (4d) 3-14. 2p 3-15. 5s 3-16. 6f 3-18. group 18, 5th period, p-block 3-19 11, 6, d 3-20. 14, 3, p 3-21. 1, 2, s 3-22. Mg would be losing its second 3s electron, but when Na, with only one 3s electron loses a 2nd electron it is a 2p electron which is of much higher energy. 4-2. (δ+)C-O(δ-) 4-3. (δ+)Si-F(δ -) 4-4. (δ-)Cl-P(δ+) 4-5. (δ +)H-C(δ-) 4-6. (δ+)H-N(δ-) 4-7. Yes, (δ+)H-F(δ-) 4-8. No, a symmetrical molecule 4-9. No, a symmetrical molecule 4-10. Yes, S(δ+) and O's (δ-) 4.11. Yes, N(δ-) and H's(δ+) 4-12. No, a symmetrical molecule 5-1. 5-2. 5-3. 5-4.

5-5. 5-6. 5-7.

(Note: In all answers the non-bonding electrons on halogens are not shown.)

C FF

F

F N HH

H

SO

O

OS

OO

O

NOON

O

O

O

N H

H

H

H

Page 80: Lets Talk Chemistry

Ans-2

5-8. 5-9. 5-10.

5-11. 5-12. 5-13. 5-14

5-15.

The non-bonding electrons on oxygen in answers to 5-16 - 5-18 are not shown. 5-16. 5-17.

5-18.

6-2. CH3CH=CHCH2CH3 6-3. CH3CH2CH2C/CCH2CH2CH3 In answers 6-4 - 6-9 below (=O) represents an oxygen atom bound by a double bond to the carbon atom immediately to its left. The structure of 6-4 showing all non C-C andC-H bonds is shown in the box for clarity. 6-4. CH3C(=O)CH2CH2CHBrCH2CH3 6-5. NH2CH2CH2CH2C(=O)H 6-6. CH3CH2C(=O)OCH2CH2CH2CH2CH3

N ON N N OC

HH

OC CH

H

H

H

H

C

H

H

Cl

CCC

O

H

H

HH

H

H

C C

H

H

H

CH H

H C C

H

H

H

O

H

H

O C H

H

H

C

H

H

CH O

O

C

H

H

H

C CH

H

H

Br

H

C

H

H

C H

H

H

C CO

H

H

H

H

H C

O

H

C CF

H

H

H

H

C

Cl

O

H

C O

H

H

H

C CI

H

H

H

H

C

H

H

C C

O

C C C

H

N

H

H

H

H

H

CH3CCH2CH2CHCH2CH3

O Br

Page 81: Lets Talk Chemistry

Ans-3

6-7. CH3CH2CH(OH)CH2CH(CH2CH2CH3)C(=O)OH 6-8. CH3CH2CH2CH2CH2CH2CH2CH2C(=O)NH2 6-9. HC(=O)NHCH2CH3 6-10. (5-9) methanal 6-11. 1-chloropropane 6-12. propanone 6.13. propene 6.14. ethyl methyl peroxide 6-15. methyl ethanoate 6-16. 2-bromobutane 6-17. 3-hydroxypropanal 6-18. 2-chloro-4-fluorobutan-1,2-diol 7-2. 5.0 x 10-2 m s-1 7-3. 6.4 x 1011 N m-2 7-4. 2.8 x 10-3 g L-1 7-6. t = 2.3 x 103 s 7-7. p = 7.8 x 10-5 Pa 7-9. 300 K 7-10. 1.15 x 105 s 7-11. 8.13 x 103 Pa 7-12. 5.45 x 102 kg m-3 8-2. 0.227 mol 8-3. 960 mol 8-4. 209 nmol 8-5. 540 g 8-6. 836 g 8-7. 1.79 g 8-8. 42 g 8-9. 17.4 kg 8-10. 306 t 8-11. 512 g 9-1. 32.5 L 9-2. 153 ng 9-3. 62 MPa 9-4. 168 kPa 9-5. 172 kPa 10-1. 200 g L-1 0.584 mol L-1 10-2. 0. 486 g L-1 8.36 x 10-3 mol L-1 10-3. 52.34 g L-1 0.4151 mol L-1 10-4. 0.4483 mol L-1 10-5. 92.7 10.6 cf(KIO3) = 0.02219 mol L-1 c(Na2S2O3) = 0.1417 mol L-1 c(Cl2) = 17.4 g L-1 10-7. 6.03 x 10-5 mol L-1 0.565 g 11-1. 4.184 J g-1 K-1 1.00 cal g-1 K-1 75.4 J mol-1 K-1 11-2. 43.0 kJ mol-1 11-3. -298 kJ mol-1 11-4.(i) CS2(l) + 3O2(g) → CO2(g) + 2SO2(g) (ii) 90.0 kJ mol-1 11-5. -101 kJ mol-1 12-1. addition, reduction 12-2. decomposition, redox 12-3. addition, oxidation 12-4. elimination, acid-base 12-5. precipitation, redox 12-6. precipitation, redox 12-7. -3 12-8. +2 12-9. +6 12-10. H, +1 O, -1 12-11. +5 12-12. +6 12-13. Cl2 + 2Fe2+ → 2Cl− + 2Fe3+ 12-14. 16H+ + 2MnO4

− + 10Br− → 2Mn2+ + 5Br2 + 8H2O 12-15. 2H2O + 2MnO4

− + 3Mn2+ → 5MnO2 + 4H+ 12-16. Zn + Fe2(SO4)3 → ZnSO4 + 2FeSO4 13-1. cuprous sulfate 13-2. ferric sulfate 13-3. phosphorous acid 13-4. potassium manganate

14-1. (g)][O(g)][SO

(g)][SO

22

2

23 14-2. 3

22

23

(g)](g)][H[N(g)][NH

14-3. [Ba2+(aq)][SO42-(aq)] 14-4. [O2(g)] 14-5.

[HF(aq)](aq)]F)][aq(O[H3

−+

14-6. (aq)]NH[CH

aq)]((aq)][OHNH[CH

23

33−+

14-7. (aq)][HCO

aq)](O(aq)][H[CO

3

32

3−

+−

14-8. (aq)][CO

(aq)](aq)][OH[HCO2

3

3−

−−

14-9. 7.08 x 10-8 mol L-1

14-10. 5.75 x 10-5 mol L-1 14-11. 1.26 x 10-3 mol L-1 14-12. 3.00 14-13. 3.62 14-14. 11.00 14-15. 10.38 14-16. (i) 7.20 (ii) 6.90 14-17. 6.00 14-18. 7.25

Page 82: Lets Talk Chemistry

Ans-4

15-1. O2 15-2. Sn 15-3. Zn|Zn2+||Fe3+,Fe2+|Pt, 1.53 V 15-4. Pt|Br2,Br−||I2,I−|Pt, 0.56 V 15-5. Pt|Cl2,Cl−||Ag+|Ag, 0.60 V 15-6. Yes 15-7. No 15-8. No 15-9. Yes 16-1. (a) 1.0 x 10-3 mol L-1 s-1 (b) 2.3 x 10-5 mol L-1 s-1 16-2. 2.5 x 108 mol L-1 s-1 16-3.

Rate law: rate = k[CH3CH=CH2][Br2] 16-4. Substitution, free radical

C CCH3 H

HH

Br Br

C CCH3

HH

H

Br

Br

C C

Br

H

CH3

H

H

Br