Chemistry for CAPE® Unit 1 Chemistry for Roger Leroy CAPE® Norris Barrett Annette Maynard-Alleyne Jennifer Murray Unit 1 3 Great Clarendon Oxford It University furthers and Oxford © Roger The First This in means, Press, as Enquiries should must impose British Data by © the No the sent the and Barrett, of in in permitted by law, scholarship, registered other trade mark Jennifer 2015 Press may in any writing by of licence rights outside Department, in the 2015 be or Oxford by any University under terms organization. scope Oxford reproduced, form or circulate same Library this work condition Cataloguing on in in any any other of the University form above Press, at and you must acquirer Publication Data available 978-1-4085-1668-3 10 9 8 7 Printed 6 5 in Great Britain by CPI Group (UK) Ltd., Croydon CR04YY Acknowledgements Cover: Mark Lyndersay, Lyndersay Digital, Trinidad www.lyndersaydigital.com Illustrations: Page Wearset make-up: Index: Indexing Ltd, Wearset Boldon, Tyne and Wear Ltd Specialists (UK) Ltd Photographs Page 2 iStockphoto Although we copyright cases. the the this notied, earliest Links and If to for have holders made the effort publication publisher will to trace this has rectify and not any contact been errors all possible or in omissions opportunity. third party information materials work. every before websites only. contained are provided Oxford in any by disclaims third party Oxford any in good faith responsibility website of countries 2012 reprographics reproduction Oxford. asserted publication in a of research, Maynard-Alleyne, Press transmitted, permission in is certain been Ltd University Oxford University this or the Annette have Oxford Rights of Kingdom excellence University Thornes appropriate to UK United above. not this the part of worldwide. authors prior 6DP, department Oxford by OX2 objective system, expressly the in Nelson concerning be address You of retrieval with a Leroy published without or agreed Press reserved. a is publishing Norris, published rights the by rights edition stored Press illustrations moral Oxford, University’s University Original All the education Text Street, referenced for in all at Murray 2012 Contents Introduction v Chapter 4. 1 Module 4 Redox Redox reactions reactions 48 and oxidation 1 state Chapter 1 Atomic 1. 1 The 1.2 Isotopes 1.3 Energy structure structure of the 2 atom 4.2 Redox radioactivity levels and emission 6 Sub-shells 1.5 Ionisation and atomic orbitals energies 8 10 Revision questions 2. 1 2 matter and bonding Covalent The gas laws 52 5.2 The kinetic theory of 5.3 Using the ideal 5.4 Changing state gas gases 54 equation 56 58 6 Energetics 60 14 6. 1 Enthalpy 6.2 Bond 6.3 Enthalpy 6.4 Hess’s 6.5 More 6.6 Lattice changes 60 energies 62 14 bonding – dot and changes by experiment 64 cross diagrams law 66 16 2.3 More dot and 2.4 Ionic metallic 2.5 Electronegativity and 52 and forces of attraction 2.2 Kinetic theory 12 Structure States of 5 5. 1 Chapter Chapter 50 4 spectra 1.4 equations 2 Chapter and 48 cross diagrams bonding enthalpy changes 68 18 energy: the Born–Haber 20 cycle 70 and 6.7 intermolecular forces 22 Properties of giant 24 Properties of simple More about lattice energy Exam-style questions 2.6 2.7 structures 26 2.8 Shapes of molecules (1) 28 2.9 Shapes of molecules (2) 30 Revision questions 3 The 3. 1 Equations 3.2 Calculations mole and Module 74 Module Chapter 2 7 Rates of 7 . 1 Following the 7 .2 Calculating 7 .3 Collision theory reaction course of a 76 reaction 76 32 concept moles using the 34 rate of reaction and 78 rate of reaction 80 7 .4 Rate 82 7 .5 Order of 34 equations mole concept reaction from 36 3.3 Empirical 3.4 Using Avogadro’s and molecular formulae 3.5 Solution law experimental data 84 Reaction 86 38 7 .6 mechanisms 40 Revision questions concentration titrations 42 More 44 about titrations 8. 1 Revision questions 88 and Chapter 3.6 1 molecular compounds Chapter – 72 8 Chemical Characteristics of equilibria 90 chemical 46 equilibria 8.2 The equilibrium 90 constant, K 92 c iii Contents 8.3 Equilibrium calculations involving K 94 Chapter 96 Groups c 8.4 The equilibrium constant, K 12 II, The IV chemistry of and VII 140 p 8.5 Le Chatelier’s 8.6 Solubility principle product, 98 12. 1 Properties of Group 12.2 Group II elements 140 100 K sp 8.7 Solubility product II carbonates, nitrates and and the sulphates common ion effect 12.3 Chapter 9 Acid–base equilibria Brønsted–Lowry theory of Group IV elements and their tetrachlorides 104 12.4 9. 1 142 102 Some 144 properties of Group IV acids oxides and bases 12.5 9.2 Simple pH calculations Relative IV and oxides 148 Equilibrium calculations involving weak acids 9.4 stability of Group 106 cations 9.3 146 104 Acid and 12.6 Group VII: the 12.7 Halogens 12.8 Reactions of halogens 150 108 and hydrogen halides 152 base dissociation constants halide ions 154 110 Revision questions 9.5 Changes in pH during 156 acid–base titrations 112 Chapter 9.6 Acid–base indicators 9.7 Buffer 9.8 Calculations 13 Transition elements 13. 1 solutions An introduction to transition 116 elements involving 158 buffer 13.2 solutions Physical properties of transition 118 elements Chapter 10 Electrode and equilibrium 10. 1 Introducing potentials 13.3 120 standard 160 Coloured ions and oxidation states 13.4 162 Complex ions and ligand electrode exchange potential Electrode potentials and Complex 10.4 Using and redox reactions 166 122 Chapter Redox ions cell potentials 10.3 164 120 13.5 10.2 158 114 reactions and cell diagrams 14 Qualitative tests for 124 ions 168 Ø E values to chemical 10.5 predict change Electrode 126 potentials electrochemical and cells 128 Revision questions 130 Exam-style questions – Module 2 14. 1 Identifying cations (1) 168 14.2 Identifying cations (2) 170 14.3 Identifying anions Exam-style questions Period 11. 1 11 The 3 174 sheets 176 revision questions 178 elements of 3 134 Glossary 183 Index 186 Physical properties of the Period 3 elements in 134 11.2 Patterns 11.3 Properties of chlorides iv Module 3 Answers to Chapter – 132 Data Module 172 Period 3 Period elements 3 oxides 136 and 138 Answers to questions exam-style CD Introduction This Study Guide has been developed exclusively with the Caribbean ® Examinations candidates, Council both in ) (CXC and out of to be used school, as an additional following the resource Caribbean by Advanced ® Prociency Examination It prepared (CAPE ) programme. ® has been teaching and by a team examination. with The expertise contents are in the CAPE designed to syllabus, support learning ® by providing the features and for guidance this activities On from condence T est of in Y ourself you problem to it an achieve easier syllabus. course is in an type Do answering activities to inside specically the which your includes electronic techniques: candidate answers could skill syllabus format! level short answers be and improved. and questions. designed questions study to and concepts examination-style example where examination refer key examination understanding, examination sections sample show your to the examination with Chemistry CAPE master and CD, good provide are to in remember interactive questions, build you best requirements examiner will your for developing activities multiple-choice refer This the the essay activities you make you Marks and feedback These help Guide assist Y our answer of on Study to to included requirements full Inside tools and guide so to provide helpful that experience feedback you can will revise areas. unique examination combination practice will of focused provide syllabus you with content invaluable and interactive support to help you ® reach your full potential in CAPE Chemistry. v 1 Atomic 1. 1 The structure structure Learning outcomes Dalton’s John On completion should be able of this describe Dalton’s describe the atom in relative the section, atomic structure terms of charges sub-atomic of theory the position, and the atom atomic theory Dalton thought of atoms as being hard spheres. He suggested that: you to: of masses all atoms of the atoms cannot atoms of atoms combine same be element broken different down elements to form are any have more exactly alike further different complex masses structures ( compounds). of particles. Later changes to Dalton’s atomic centuries, Dalton’s theory scientists was have first atomic theory described modified the in 1807. theory to fit Over the the past evidence two available. 1897 Thomson discovered suggested sea of the the electron ‘plum-pudding’ positive (which model of he the called atom: a ‘corpuscle’). electrons He embedded in a charge. 1909 Rutherford suggested positively-charged planetary model that nucleus with most in of the electrons the mass middle of of the the surrounding atom atom. the was This in led a tiny to the nucleus. 1913 Bohr suggested distances that electrons depending on their could only orbit the nucleus at certain energy. John Dalton (1766–1844) 1932 Did you know? Scientists now Chadwick think that matter discovered the presence of the neutron in the nucleus. is 1926 onwards made called up of sub-atomic ‘leptons’ electron is a and type of particles Schrödinger , Born terms probability lepton of and neutrons are of the Heisenberg of finding described it at any the one position point. of This an led electron to the in idea and of protons and ‘quarks’. An made atomic orbitals (Section 1.4). up quarks. electrons electron (–) cloud shells of electrons electron (–) electron proton sea of + charge nucleus Figure 1.1.1 (+) nucleus Three different models of the atom: a The plum-pudding model; b The Bohr model; c The atomic orbital model neutron nucleus Figure Figure 1.1.2 The nucleus of an atom contains protons and neutrons. The electrons are outside the nucleus in fixed energy levels. 2 1.1.2 shows a useful model of an atom based on the Bohr model. Chapter Masses and The actual very small. charges of masses and sub-atomic charges of protons, 1 Atomic structure particles neutrons and electrons are –27 For example, a proton has a mass of 1.67 × kg 10 and a –19 charge of +1.6 comparison. Relative × C. 10 These are The relative shown mass in the masses and charges give an easier table. Proton Neutron 1 1 Electron 1 _____ 1836 Relative charge +1 Sub-atomic The effect electrons of is a particles strong shown 0 and electric field –1 electric fields on beams of protons, neutrons and below. Exam tips + beam + of beam + of beam of The electrons protons direction an electron be found deflection + towards deflection plate – towards no rule from the deflect by a the movement magnetic eld of Fleming’s can as shown direction below. left-hand of the Remember (1) magnetic eld The effect of an electric field on beams of electrons, protons and neutrons is Charged in effect plate The Figure 1.1.3 of neutrons particles plates move with electrons and towards the same protons, the plates charge. the If with we electrons opposite use are the charge same deflected and voltage to a away to N → points the S (2) in that the the centre nger direction opposite to electron flow. much first greater extent. This is because electrons have a very small mass finger compared with protons. field left hand centre finger Sub-atomic particles and magnetic fields current thumb Figure 1.1.4 shows an evacuated glass tube. When heated by a low voltage motion supply, electrons accelerated they are towards collide with produced the the metal from the cylinder , fluorescent tungsten C. screen. They When filament, produce the a north F . They glow pole are when of a magnet Key points is brought moves near the downwards. electron The beam, direction the of glow the on the fluorescent movement can be screen predicted from Fleming’s left-hand rule. The direction of movement is at right angles The atomic to t the direction Remember of that conventional the the magnetic direction field of and the electron conventional flow is opposite theory has developed to the experimental evidence. current. that of the Most is current. in of the protons the mass nucleus and of the which atom contains neutrons. The electron flow electrons F are outside the nucleus C in shells. conventional magnetic field Electrons, protons and neutrons current have characteristic relative N masses fluorescent and charges. screen movement Figure 1.1.4 Beams are The effect of a magnetic field on a beam of electrons of protons deflected directions Using different opposite deflected apparatus, direction since to they the have beams of electrons. no protons Beams of will be deflected neutrons will in not the be by in and opposite electric magnetic elds. electrons and Neutrons are not deflected. charge. 3 1.2 Isotopes Learning outcomes and radioactivity Atomic Atomic On completion should be able of this section, dene to: ‘mass Atomic number’ number and mass number (Z) you (It number also number equals ( Z) the is the number number of of protons electrons in a in the neutral nucleus atom of and an the atom. position and of the element in the Periodic T able.) ‘isotopes’ dene ‘relative ‘relative atomic isotopic mass’ mass’ using and the Mass Mass number number (A) ( A) is the number of protons plus neutrons in the nucleus 12 C scale know of how to calculate an relative So atomic write and masses equations some the number of neutrons in an atom is A − Z (mass number − atomic number). involving α-, β- Isotopes γ-radiation state atom. uses of radioisotopes. Isotopes are atoms with the same atomic number but different mass numbers. We show the mass number and atomic number of an isotope like this: electron mass number → atomic number → 23 Na proton Relative hydrogen 1 atomic and ← element symbol 11 relative isotopic mass 12 atoms The mass of a single atom is too small to weigh, so we use the C proton (carbon-12) neutron to this atom standard, Relative as a standard. which atomic has mass a We mass (A compare of the exactly 12 masses of other atoms units. ) r Relative atomic mass (A ) is the weighted average mass of naturally r 12 occurring mass deuterium 1 proton, 1 of atoms exactly of an 12 element on a scale where an atom C of has a units. atoms neutron Relative Relative isotopic isotopic mass mass is the mass of a particular isotope of an element 12 on a scale where Calculating tritium 1 proton, Most naturally mass numbers atom C of accurate occurring has a mass relative elements of exactly atomic exist as a 12 units. masses mixture of isotopes. So their atoms 2 are not always exact. We have to take into account the neutrons proportion Figure 1.2.1 an of each isotope. This is called its relative isotopic abundance . The three isotopes of The relative atomic mass for neon is calculated below from the relative hydrogen isotopic masses and relative abundance. Did you know? Isotope Isotopes with 2, 8, 20, 28, 50, Relative isotopic mass Relative abundance/% 82 20 Ne and 126 neutrons extremely or protons stable. These 20 ‘magic 21 Ne numbers’ closed are thought nuclear the full shells shells of the to 90.9 are 21 0.30 22 8.8 indicate comparable noble to 22 Ne gases. (20.0 × 90.9) + (21.0 × 0.3) + (22 × 8.8) ______________________________________ A = = r 100 4 20.2 Chapter 1 Atomic structure Radioactivity Isotopes of some spontaneously. break The down, table elements These rays shows or are have particles three nuclei described types are of Name of Type of emission emitted Alpha helium as which break radioactive given out. These down (decay) isotopes . are called As the nuclei emissions. emission. particles/rays Stopped by Exam tips (α) nuclei charged (positively thin sheet of paper A particles) common α-particles Beta (β) electrons (produced by nuclear 6 mm changes) thick are not: (γ) very thick lead He be written for each of that atoms. They helium error are is nuclei to think that electrons from the the nucleus. The β-particle radiation from can suggest sheet electrons Equations to helium are . Another outside electromagnetic is 2+ aluminium foil high frequency are they β-particles Gamma error these types of decay. For the arise from electron the nucleus not shells. example: α-decay The isotope charge of 2 produced units has lower a mass than the number original 223 of → units lower and a nuclear 4 219 Ra Rn 88 4 atom: + He 2 86 β-decay The mass one. This number stays the same but the number of protons increases 1 is because a neutron by 1 n) ( is changed into a proton ( 0 p) and an 1 0 electron e): ( –1 1 ( For example carbon-14 0 1 n) → 0 ( p) + ( 1 decays to e) –1 nitrogen-14 14 → it emits a β-particle: 0 14 C N 6 when + e –1 7 Key points γ-decay γ-rays can be emitted along with α- or β-particles or in a process called ‘electron capture’. A proton is converted to a neutron, so the Mass of number stays the same but the atomic number decreases by number is the total number mass one. protons + neutrons in an For isotope. example: 37 0 Ar + 18 37 e → –1 Isotopes are same number different Uses of of with the protons numbers of but neutrons. radioisotopes atoms Cl 17 T racers for searching for faults in pipelines and for studying of certain organs in the body, I e.g. atomic are and isotopic compared using the 12 131 working Relative masses the is used to study C thyroid scale. function. In medicine, Dating for radiotherapy in the treatment of cancers. The relative element is atomic the mass weighted of an mean of 14 once the age of objects, e.g. using C to date objects which the were living. atoms of the isotopes they contain. 241 Smoke detectors often use Am. When an unstable isotope decays 235 Generating source of power , e.g. U is used in many nuclear reactors as rays a or particles are emitted. energy. α-, β-, and γ-radiation characteristic have properties. 5 1.3 Energy levels Learning outcomes completion of this section, electrons be able amount explain spectra energy explain how data from give evidence levels for the lines emission of discrete electrons of the Balmer series in We arranged when of have energy a that energy levels certain required the are electrons quantum electron is energy levels electrons The absorbs Lyman say in the them. The and atoms energy for fixed a values change is of energy. called a The smallest quantum of to: energy. in and spectra you fixed should emission Energy quanta The On and of then which in are is the in radiation to are be energy their it in or quantised. quantised. lowest then said discrete levels ground can an The move atom is possible state . up excited The to a state . most for When of electron energy an are stable each an higher When electrons level. excited emission electron falls to a lower energy level a quantum of radiation is given out. spectra The understand how energy between energy levels is the two difference electron moves to a lower energy level. in electrons’ related to the energy frequency of radiation. electrons ground Figure 1.3.1 in electrons state excited in state An electron can jump from one energy level to another by absorbing or emitting a fixed amount (quantum) of energy The difference frequency of in energy radiation between by the two energy levels is related to the relationship: –1 energy (J) → ΔE = hν ← frequency of radiation (s ) ↑ –34 Planck’s In any atom principal electron further shells. away Atomic When an there They from electrical frequencies. It is The several levels are the in or or (6.63 × possible energy principal given –1 J s 10 levels. energy numbers n = ) These levels . 1, n = 2, n thermal radiation An emission energy is is emitted spectr um passed only at differs through certain 3, also etc. from a a gaseous normal or visible light that: lines up of called going sample wavelengths separate converge (get lines. closer to each other) as their increases. lines get closer together 6 = called are spectra the made Figure 1.3.2 are They nucleus. emission element, spectrum are quantum constant Part of a simplified line emission spectrum of hydrogen frequency of Chapter Interpreting the Each line moving lines in the from seen are Lyman hydrogen higher Balmer fall (seen back series electrons to emission a lower emission spectrum energy is level. Atomic structure spectrum a result Among of the electrons several series of the: series electrons a hydrogen 1 fall in to (seen back the the in to ultraviolet n = the the 1 visible n = region), energy 2 where previously excited level. region), energy where previously excited level. n = 5 n = 4 n = 3 n = 2 n = 1 frequency emission spectrum Lyman Figure 1.3.3 series Balmer series A line emission spectrum is the result of electrons falling from higher energy levels to lower energy levels The energy associated with each line is found by using the relationship Did you know? ΔE = hν. Y ou wavelength), can the see that more the energy greater is the frequency released. The (the point smaller where the the lines There eventually come together is called the convergence limit . This are spectrum an electron falling from the highest possible energy level. If the other more energy than this, it becomes free from the pull of the nucleus atom. The The This Then Bohr energy is model the is model of levels mirrored of atom the get by an closer the hydrogen converted lines an spectral the infrared called and Pfund the Paschen, series. ion. n = 5 n = 4 towards lines shown in are atom together atom to emission of Brackett the of electron region. These has sets represents getting the outside closer opposite, a of the together . quantum of n = 3 atom. In the energy Bohr n = 2 moves n = 1 the A hydrogen further The quantum more When levels, electron the energy electron emitting of an in the energy electron loses n = will of level excite has, energy, radiation 1 it the (ground the electron higher will fall characteristic state) the down to to the energy to the the n level lower n = = it 3 2 level. level. will be in. energy frequencies. Key points electron shells (principle Electrons When occupy specic energy levels (quantum levels) in the quantum Figure 1.3.4 electrons gain specic quanta of energy they move from shells, n) atom. lower The Bohr model of the atom. to The principal energy levels get closer higher When energy excited, emitting emission The levels. They electrons radiation of become lose ‘excited’. energy, they fall together towards the outside of the atom. back characteristic frequency. This to is lower the energy origin of levels the line spectrum. energy difference between any two energy levels is given by ΔE = hν 7 1.4 Sub-shells Learning outcomes and Shells The On completion of this section, atomic and principal be able know that shell for split n into the = 2 principal and shells) know relative the quantum above sub-shells quantum can be (subsidiary Notice terms know and energy energies of how electronic and to determine the term a hold the of the after energy, by d. level first the apart are For four from named quantum element does than neutral protons. the s, first, p, d. are split Figure into 1.4.1 not the calcium always example shells in the conform scandium in terms order to the of the 4s of shows the increasing the sub-shells, pattern of sub-shell s is energy. in followed at a by p lower 3d. is The atom, the number maximum shown in the of number electrons of is electrons equal each to the shell number and of sub-shell can table. atoms ions understand shells These the sub-shells conguration for that of followed For shells quantum (sub-levels). to: sub-shells sub-shells you sub-shells should orbitals Principal Total quantum electrons number of shell quantum in Maximum principal number of electrons in sub-shell ‘atomic shell orbital’ describe the shapes of s and s p d p orbitals. n = 1 2 2 – – n = 2 8 2 6 – n = 3 18 2 6 10 4d 5s n = 4 4p ygrene 3d 4s Atomic orbitals n = 3 3p gnisaercni Each sub-shell contains one or more atomic orbitals . An orbital is a 3s region Each of space orbital where can hold there a is a high maximum probability of two of finding electrons. So an the electron. number of 2p orbitals n = 2 2s each in other each in sub-shell shape (see is: s = Figure p = 3 and d = 5. Orbitals differ from 1.4.2). 1s n = 1 z prinicipal 1, quantum z z y sub-shell z y y y shell x Figure 1.4.1 x x x The principal quantum shells, apart from n = 1, are split into sub-shells s orbital p orbital p x Figure 1.4.2 Note that: A 2s The orbital p y orbital z The shapes of s and p orbitals orbital three is larger different than p the orbitals 1s are orbital and mutually a at 3p is right larger angles than to the 2p. each other . Here of are two electrons examples present 2 in of how to write the type of orbital and the number them: electrons in orbital 5 electrons ↓ ↓ 2 First quantum s 8 shell → 1s sub-shell ↑ 5 Third quantum p shell → 3p sub-shell ↑ Chapter The electronic configuration of 1 Atomic structure atoms Exam tips The electronic electrons in a 1 sure configuration particular shows atom. The the number procedure and type of the This is: diagram remember Make you know the order of the orbitals, 1s 2s 2p 3s etc., as may the help order you of lling the well sub-shells. as the number of electrons in the atom. 1s 2 Fill up lowest and 3 a Stop the sub-shells energy, p when Example e.g. sub-shell you to an a s the maximum sub-shell maximum have the of correct can amount have a starting maximum with of 2 those of electrons number of configuration of sodium 2p 3s 3p 3d 4s 4p 4d 5s 5p electrons. 1 Electronic 2s 6. (Z = 11): 6s 2 2 1s Example 6 2s 1 2p 3s 2 Did you know? Electronic configuration 2 1s 2 6 2s of 2 2p 3s scandium 6 3p 2 4s (Z = 1 3d 21): 2 (or 2 1s 6 2s 2p 2 3s 6 1 3p 3d 2 4s ) The electronic also Note that the 4s sub-shell is filled before the 3d (see Figure be p , x For electronic configuration of ions to the ions are atom. formed In one general, or the more electrons electrons are are lost or p , removed gained from from or the be p y example, show the can separate . (see Figure 1.4.2). z the conguration can When to 1.4.1). orbitals The conguration written of electronic sodium (Example written: added 2 outer 1s 2 2s 2 2p 2 2p x 1 2 p y 3s z sub-shell. Example 1 2 Magnesium atom Example 2 Nitrogen atom 2 Note and that ions the are 2 2 3s 2+ ; Magnesium 3 2p electronic less 2p 2 2s 1s 6 2s 1s ion 3– ; Nitride ion configuration straightforward (see (N of (Mg 2 ) 1s some Section 2 2s 2 ) 2 1s 2s 6 2p 6 2p transition element atoms 13.1). Key points The principal shells Each An quantum named s, sub-shell orbital is a p, d, can shells (apart from the rst) are divided into sub- etc. hold region of a maximum space where number there is of a electrons. good chance of nding an electron. s orbitals Each The Electronic to are orbital spherical can maximum the hold in a number and maximum of congurations energy shape sub-levels of electrons of atoms placed in p have 2 in an hour-glass shape. electrons. each and ions order of sub-shell is are found increasing s by = 2, p = adding 6, d = 10. electrons energy. 9 1) 1.5 Ionisation Learning outcomes energies Ionisation The On completion of this section, first be able know energy , ΔH , is the energy needed to from each atom in one mole of atoms of an element state to form one mole of gaseous ions. For example: + explain data know can ionisation evidence for how be –1 (g) Cl + e ΔH electron energy The +1260 kJ mol of Size of charge conguration ionisation value the ionisation energy depends on: sub-shells nuclear (greater between determined from successive = i1 → energy how give one its influencing Cl(g) in to: the factors ionisation remove i1 gaseous ionisation you electron should energy to energies. the remove charge: number nucleus these In of and any one period, protons), outer electrons. So the electrons and tends ΔH the higher greater to the the the nuclear attractive more increase force energy with needed increase in i1 nuclear charge. Distance outer of the electrons between them outer are and electrons from the the from nucleus, nucleus and the the the nucleus: less The attractive lower the value further force of the there is ΔH i1 Shielding: between of full Electrons the in nucleus inner shells, full and the inner the shells outer greater the reduce electrons. shielding the The and attractive greater the force the lower number the value of ΔH i1 First ionisation Figure 1.5.1 shows energies how the across values of a ΔH period change across the first three i1 periods. first Did you know? ionisation –1 energy/kj He 2500 The p , p x and p y orbitals in mol a z particular Ne sub-shell are equal 2000 in F energy. The electrons are Ar added N 1500 Cl one by one to these separate H before they are paired P Be 1000 orbitals C Si S up. Ca B 500 This minimises repulsions between Li Na K electrons. The electrons in 0 the 0 separate p orbitals of nitrogen 10 5 are shown 20 and atomic oxygen 15 number here: Figure 1.5.1 First ionisation energies for the first 20 elements in the Periodic Table 2p We can energy use the with three ideas increasing above proton to explain the change in ionisation number . 2s Across a period there is a general increase in ΔH . The increased i1 1s nuclear charge because nitrogen 2 1s The 2 2 1s 2p extra successive of protons) electrons outweighs added are the being other placed two in factors the same oxygen 3 2s the (number electron in 2 2s 4 2p oxygen has outer electron there is the shell. same There number is of not much inner difference electron in shielding since shells. to When a new period starts, there is a sharp decrease in . ΔH The outer i1 go into an orbital which is already electron half-lled. The extra repulsion goes in the same orbital a shell further from is Both This these also factors explains outweigh why 10 the effect decreases ΔH i1 repulsion. nucleus and there of greater is more nuclear called charge. spin-pair the of shielding. electrons into down a group. Chapter The decrease in ΔH from Be → 2p sub-shell, B is due to the fact that the 1 Atomic structure ionisation energies, fifth i1 electron the in B nucleus slightly. charge. goes than These A into the two similar the 2s sub-shell. factors reason The outweigh explains which shielding the the is effect dip in slightly also of the from increases increased from ΔH further Mg → nuclear Al. i1 2 Mg The decrease = in 1s 2 2s ΔH 6 2p 2 2 3s from N Al → O is = 1s due 2 2s to 6 2p 2 3s 1 3p spin-pair repulsion. The i1 eighth in it it. electron The easier can O increased to remove Successive We in goes into repulsion the eighth ionisation remove electrons associated with ionisation energies. an are similarly already charged atom one electrons given the by are one. called symbols The the an electron particles makes , ΔH energies Exam tips successive ΔH i1 For has electron. an these They the which energies from removing of orbital , ΔH i2 , etc. i3 example: In + O 2+ (g) → O O + e ΔH = +3390 kJ mol 1 3+ (g) → O + e ΔH = +5320 kJ mol can configuration from work out the energies. electron Figure ionisation configuration 1.5.3 shows of how an this energies element is done. from In successive The it may also be possible to distinguish between s atoms or by small changes in always successive ionization ions are always 3 The charge on the ion formed is some and p same as the number of the sub- (successive) shells are ions. gaseous. the instances ions formed positive 2 ionisation The –1 (g) i3 We that: –1 (g) i2 2+ Electron considering remember ionisation energy energies. number e.g. for ionisation is a 4+ the fourth energy the ion formed ion. 5 lom 1– 2 electrons difficult to very remove Jk/ygrene Key points 4 8 electrons easy to less remove The rst ionisation energy is noitasinoi 11+ the 1 energy needed to remove a electron 3 easily mole removed of electrons from gaseous a mole of atoms. gol 01 The ionisation energies required nucleus to 2 0 1 2 3 4 5 6 7 8 9 10 remove electrons time from number of electrons a mole of at a gaseous removed atoms Figure 1.5.3 one 11 are called the successive a Successive ionisation energies for a sodium atom; b How the arrangement ionisation energies. Ionisation energy of electrons can deduced from these ionisation energies on the nuclear distance from Trends be of the in the values charge, outer nucleus and ionisation explained using depend the electrons shielding. energy these can three factors. The values ionisation atom of provide electronic successive energies of an evidence for the conguration. 11 Revision 1 State and explain later 2 the four main any the protons, of Dalton’s modifications discoveries Explain tenets concerning direction electrons and and questions to the his atomic theory in theory light 13 of neutrons of in deflection an the atoms atom. degree Write electronic and a Ca b Zn c Cl d Mn a Define b Write configurations of the following ions: of electric field. 2+ 3 What and is the rationale for relative than their charges absolute of using the relative sub-atomic masses and masses particles rather 14 charges? the an energy 4 How many protons, the following electrons and neutrons are the equation of ‘first ionisation showing the first energy’. ionisation calcium. in atoms? 15 27 a term State the three factors ionisation Al that influence the value of the energy. 13 39 b K 16 19 131 c Explain energy I the general across a trend of an increase in ionisation the first ionisation period. 53 239 d Pu 17 94 5 Calculate the relative atomic 10 that its two isotopes abundances 6 a Write of equations following of boron and and B have 81.3% showing the there energy between: a Mg and Al b P is a decrease in relative respectively. α-decay of 18 the and S The following ionisation radioisotopes: group 238 i why given 11 B 18.7% mass Explain of data energies the show the first of element X. Suggest an Periodic Table X seven belongs successive which to. U 92 222 ii Ra 88 –1 Ionisation b Write equations following showing the β-decay of energy number Enthalpy/kJ mol the radioisotopes: 1st 737 .7 234 i Th 90 2nd 1450.7 14 ii C 6 3rd 7 State the 8 Explain uses how of four an named atomic 7732.7 radioisotopes. emission spectrum is 4th 10 542.5 5th 13 630 6th 18 020 7th 21 produced. 9 10 How is the a Lyman b Balmer in Write equation the meaning State held 12 series produced difference 11 series in that all of two the maximum an s, a What is b Draw a: 12 hydrogen between of the the i 2s ii 2p p and an d orbital indicates energy symbols number orbital? spectrum? the levels energy and state the used. of sub-shell. atomic orbital emission electrons that can be 711 Chapter 1 Atomic structure – revision questions 13 2 Structure and 2. 1 States matter of Learning outcomes bonding States of Substances On completion of this section, be able understand the between forces states of with high and melting melting points attraction point their atoms (or ions). have strong Substances with forces low of attraction melting points have to: weak matter of you between should and forces forces of attraction between their molecules. relationship of attraction and a weak forces c d b matter between strong molecules attractive know that single covalent bonds forces are formed electrons by sharing between a the pair of atoms weak understand covalent bonding strong bonding within molecules in forces terms of overlap of atomic Figure 2.1.1 The strength of forces and arrangement of particles in solids, liquids and gases orbitals understand sigma and pi Giant structures with ionic or covalent bonds are solids bonding. with high break solid with the many are only within the molecular solid to are overcome a lot between of energy the to particles. and do not move. melting strong. weak so points: But it the does The forces forces not between take much these. xed arrangement only low low are fairly close together and do vibrate. melting are strong. are weak points. But so it the The forces does not forces within between take the the much energy to these. are randomly close together arranged. They but are slide more over or each less other . the and Gases The does + in have Particles are They molecules liquid have overcome molecules between forces arrangement molecules move. Liquids atom xed solids the Particles not one in molecules energy of takes vibrate. Molecular electrons strong It structure They attractive forces points: a Particles giant melting have forces not very low between take melting the much points molecules energy to and are boiling very overcome points. weak so it these. + Particles are far apart and move randomly. gas Figure 2.1.2 a The formation of a covalent bond Covalent A electron H H in pair covalent bond b Two ways of showing the covalent bond in hydrogen 14 of covalent two bond is formed neighbouring attractive electron bonds bonding forces clouds are are atoms in when usually by balance the strong. the and force pair with nuclei It a are needs a of of the a attraction electrons repulsive certain lot of between between forces distance energy to the between apart. break nuclei them. The the Covalent them. Chapter Sigma A bonds covalent bond and is pi orbital formed which when combines atomic orbitals contributes one (Section unpaired 1.4) The ‘joined’ orbital is called to of the atomic orbitals, a the molecular orbital . The greater combining stronger is the covalent (σ called its a line bonds drawn formed is (σ valency valentia covalent (atoms) an atom – (from meaning the powerful). means joining ‘the power together’. The bonds) concept Sigma of bond. of bonds power the So Sigma bonding the Latin overlap and overlap. electron is bond. Structure Did you know? bonds The Each 2 bonds) between are the symmetrical formed two about by nuclei. a line the The overlap of electron joining the atomic density two orbitals of the along bond nuclei. many of hydrogen together valency one with of is based atoms a given oxygen oxygen hydrogen nucleus valency can is in how combine atom, two combine atoms on e.g. the because with two water. + s atomic molecular orbital orbitals Figure 2.1.3 When a p modied slightly Two s-type atomic orbitals overlap to form a molecular orbital orbital to altered orbital, one similar thing a σ bond combines include of is in some shape the so ‘lobes’ happens with s and that, of the when an s orbital, some on p combining hour-glass two p the molecular character . to shape orbitals The p form a is becomes molecular becomes combine orbital orbital smaller . ‘end- on’. In A each case formed. Exam tips molecular nucleus orbital Although of + remember of modified p a s π electron bond has density, that electron these density two you areas must two areas represent only bond orbital one bond. Do not confuse it with a orbital double molecular bond, which contains one σ orbital bond and one π bond. + Key points modified p modified orbital p bond orbital Figure 2.1.4 Solids with have bonds (π bonds) Liquids bonds (π bonds) are formed by the sideways overlap of p their and melting Pi The electron density of the bond formed is not of atoms gases points points attraction or have ions. low because of the atomic weak forces orbitals. melting strong forces between Pi high The formation of sigma bonds by ‘end-on’ overlap of atomic orbitals between their symmetrical molecules. about a line overlapping joining to form the a π two nuclei. bond (see Figure also 2.1.5 Section shows 2 p atomic orbitals 2.9). A covalent the force the of nuclei pair of bond is formed attraction of two electrons by between atoms and the which forms the bond. -bond Sigma on’ p z orbital p Pi bonds overlap bonds atomic are formed by ‘end- orbitals. by ‘sideways’ orbital z overlap Figure 2.1.5 are formed of The formation of a pi bond by ‘sideways’ overlap of p of atomic orbitals. atomic orbitals z 15 2.2 Covalent Learning outcomes bonding Simple dot A On completion should be able of this section, shared pair – dot and of and cross diagrams cross diagrams electrons in a single bond is represented know that is formed a pair of a single when covalent two atoms bond of share draw a variety and of cross draw compounds bonds dot electron and is a bond + fluorine cross deficient compounds octet atoms fluorine molecule The electron pairs in a fluorine molecule diagrams for compounds with is usually of only the outer electrons which are used in covalent bonding. an Electron expanded pairs in the outer shell which are not used in bonding are called electrons. lone pairs Dot and shell cross diagrams electrons Use a cross dot to Draw pairs in to the can covalently represent represent outer and orbital Example: pair with only It and shared diagrams for Figure 2.2.1 line. electrons dot single single electrons covalent F a to: a by you the the the contains a used outer outer electrons number be bonded of in to pairs. of the arrangement The electrons electrons pairs maximum show shell shell lone to molecules. two also pairs from from emphasise It main of one another the reects of points atom and a atom. number the outer are: fact of bond that each electrons. Methane If possible, electrons are arranged so that each atom has four pairs of H electrons around conguration). H + C it (an octet Hydrogen is of electrons/ an exception the – it noble can gas only electron have two C electrons When around pairing its nucleus electrons, a when forming covalent bond is covalent formed bonds. between an H electron Figure 2.2.2 from one atom (dot) and an electron from another atom Drawing a dot and cross (cross). diagram for methane Examples of dot and cross diagrams a b + H Cl Cl H 2 H + O O H H hydrogen water, H O 2 chloride Did you know? When we draw dot and cross (the diagrams, there is no difference 3 in + H N N extra , the electrons pair. The dots book-keeping within and each electron crosses exercise to are just keep comes from H a a metal H atom when forms track electr O H ammonia, NH hydroxide an it ion) ion 3 of the electrons. The electron density Figure 2.2.3 in lone pairs of electrons is, Dot and cross diagrams for: a hydrogen chloride; b water; c ammonia; d the however, hydroxide ion greater than that of bonding pairs of – electrons. For more information see Note that we can draw dot and cross structures for ions such as OH and + Section 2.8. by NH the same rules. The square brackets 4 show 16 applying that the charge is spread evenly over the ion. around the ion Chapter Electron deficient Some molecules are molecules covalent unable bonds. to These complete molecules the are octet said of to electrons be when electron example is boron trichloride, . BCl and bonding they is a This only has six common error to try and write deficient . dot An Structure Exam tips It form 2 electrons and cross diagrams for ionic around 3 compounds the boron as if they were covalent atom. molecules. of + 3 Cl B a metal likely to ion very If you and be have a compound non-metal ionic unless then the it is metal Cl is compared To see small with how to and the highly charged non-metal write dot and ion. cross Cl diagrams for Section Figure 2.2.4 ionic compounds see 2.4. Dot and cross diagrams for boron trichloride, BCl 3 Molecules Some to with molecules more than electrons. An 8. can an expanded octet increase These example the molecules is sulphur number of electrons are to have said hexauoride, an . SF in their outer expanded This has shell octet 12 of electrons 6 around the sulphur atom. + 6 S F F F Figure 2.2.5 F Dot and cross diagrams for sulphur hexafluoride, SF 6 Key points A single When of bond atoms form is formed covalent when bonds two each atoms atom share usually a has pair a full of electrons. outer shell electrons. Electron their covalent deficient outer Atoms atoms in a compound have fewer than 8 electrons in shell. with an expanded octet have more than 8 electrons in their outer shell. Dot and cross molecule or diagrams show how the electrons pair together in a ion. 17 2.3 More dot Learning outcomes and Molecules Some On completion of this cross section, atoms be construct for The describe dot and cross with by bonds bond is sharing two shown by a pairs of double electrons. line, e.g. A for double oxygen, bond a diagrams double co-ordinate dot and bonding two bonding) and cross oxygen atoms oxygen molecule diagrams b involving co-ordinate bonds. 2 + 2 O atoms C atom carbon dioxide c C H 4 H C + H C H H C ethene 4 H Figure 2.3.1 When The atoms atoms Dot and cross diagrams for: a oxygen; b carbon dioxide; c ethene atoms triple 2 C share bond is three shown pairs by a of electrons, triple line, a e.g. triple for bond is nitrogen formed. N ≡N. + Figure 2.3.2 A dot and cross diagram for nitrogen Co-ordinate A co - ordinate provides bonding one a The the both to atom atom lone bonding bond the occur second we with with of (dative electrons atom pair a the bond) covalent is formed bond. For when 1: lone pair an of to orbital. orbital (an complete electron the outer decient shell of atom) both Ammonium ion, NH 4 + H H + N H + H N H + Figure 2.3.3 atom electrons unlled unlled electrons H one co - ordinate need: with the covalent for + Example A dot and cross diagram for an ammonium ion, NH 4 18 is O=O. + covalent construct bonds double bonds (dative form multiple to: molecules triple able with you formed. should diagrams accepts atoms. Chapter In this 2 Structure and bonding example: + The H ion has The nitrogen The lone Each space atom pair on for has the a two more lone pair nitrogen electrons of in its outer shell. electrons. provides both electrons for the bond. + H (H The atom = 2; N now = displayed co - ordinate has 2, a gas electron conguration. 8). N for mula bond noble as an (formula →. The showing head of all the atoms → and points bonds) away shows from the the atom H Figure 2.3.4 The displayed formula for an + which donates Example 2: the lone ammonium ion, NH pair . Compound 4 for med between and BF NH 3 3 F F H B F + H N B F H N H H F Figure 2.3.5 H F A dot and cross diagram for the co-ordination compound, BF NH 3 In this B example: has the uorine its 3 atoms. outer N has The Both a simple shell lone Example 3: There to of conguration still the room nearest for 2 noble 2, 6 when more gas bonded electrons to to be three added to conguration. electrons. donates and is form pair nitrogen boron electronic its nitrogen Aluminium lone pair have the chloride, to the unlled simple orbital electronic of boron. conguration 2, 8. AlCl 3 Aluminium atom gas is 2 chloride electrons is an short electron of the 8 decient electrons molecule required – the aluminium for the nearest noble Did you know? conguration. Hydrated At room temperature, aluminium chloride exists as Al Cl 2 AlCl 6H 3 This is because the lone pairs of electrons on two of the chlorine form co - ordinate bonds with the aluminium O is an chloride, ionic compound, 2 atoms but can aluminium molecules. 6 anhydrous aluminium chloride a molecule with the formula Al Cl 2 which has between Cl is atoms. co-ordinate two of the Cl 6 bonding and Al Cl Cl atoms. The ionic Al hydrated because the compound water is molecules Al 3+ stabilise the Al co-ordinate ions. They bonding do between this Cl molecules and the highly charged 3+ Al Figure 2.3.6 A dot and cross diagram for the Al Cl 2 ion. molecule 6 Key points A double three In bond pairs of the two pairs of electrons and a triple bond contains electrons. co-ordinate for contains (dative covalent) bonding one atom provides both by water electrons bond. 19 2.4 Ionic and Learning outcomes metallic The formation of On completion of this section, When be able know that are compounds are transferred from metal atoms to non-metal atoms, formed. to: electrons ionic you ions should bonding ions are formed The outer shell of both ions formed have the noble gas electron when conguration: atoms gain or lose electrons + K write dot and cross e O + 2e → O 8, 8,1 2– [2,8,8] 2,6 [2,8] structures 2– + K + 2, ionic → diagrams for describe the structure of metals. Positive formed In an ions by are gain ionic formed of by loss of electrons and negative ions are electrons. compound the number of positive and negative charges 2+ must balance, e.g. for calcium chloride the ions present are Ca and 2+ Cl Exam tips . So the simply, formula for calcium chloride is [Ca ] 2[Cl ], or more CaCl 2 1 Remember that for in Groups to I III, metal the atoms charge Dot on as the the ion formed group is the number. For non- When show metal the atoms charge number in Groups V on –8. the For ion is and cross diagrams for ionic structures same writing only dot the and outer cross diagrams electron shells for ionic because structures these are we the usually ones involved in to VII, electron transfer . diagram for Figure 2.4.1 shows how to construct a dot and cross group example, magnesium oxide. Note: S forms 2– the sulphide ion, S (6 − 8 = –2). The ions formed conguration 2 Remember ions do not that + e.g. H some contain of have the a full nearest outer shell noble gas). of electrons (electron positive metals, The charge on The square brackets the ion is placed at the top right. + and NH ions. indicate that the charge is spread throughout 4 ion. Mg Mg 2+ 2,8,2 Figure 2.4.1 2,6 2– [2,8] [2,8] Constructing a dot and cross diagram for magnesium oxide a b + Cl 2– Na 2+ + [2,8,8] [2,8] O Ca + Na 2– 2+ [2,8,8] [2,8] Cl + [2,8] [2,8,8] Figure 2.4.2 20 Dot and cross diagrams for: a calcium chloride; b sodium oxide the Chapter Ionic 2 Structure and strong bonding bonding ionic between The ionic bond is an electrostatic force of attraction between oppositely charged charged ions. formation regularly The of The a net giant arranged electrostatic attractive ionic in a forces str ucture. between In this three-dimensional attractive forces these structure lattice between ions the (see ions act results the also in ions in ions the are Section all bonds oppositely 2.6). directions Cl and the bonding is very strong. + Na Metallic bonding Figure 2.4.3 Most metals exist in a lattice of ions surrounded a ‘sea’ of Part of a giant ionic lattice delocalised of sodium chloride electrons. any Delocalised particular atom electrons or ion. are They those are free which to are move not associated between the with metal ions. outer electrons of metal become delocalised ‘sea’ Figure 2.4.4 of delocalised electrons The structure of a typical metal. (Note: the diagram shows only one layer of atoms for clarity. In reality there would be other layers on top of and beneath the layer shown.) In this structure: The number electrons The positive attraction of lost This to strong delocalised by each charges the electrons metal are held delocalised electrostatic depends on the number of atom. together by their strong electrostatic Exam tips electrons. attraction acts in all Remember directions. between So metallic bonding is usually that the metallic big and difference ionic bonding strong. is: The strength of metallic bonding increases with: Ionic: increasing decreasing positive charge on the the Metallic: size of the metal negative charges are ions. ions the negative charges are ions electrons. increasing number of delocalised electrons. Key points Ions are formed Ions generally Ionic bonding when have is a full the charged ions. Metallic bonding is atoms net a gain outer or shell lose of electrons attractive force lattice of positive electrons. between ions in a (noble gas positively ‘sea’ of structure). and negatively delocalised electrons. 21 2.5 Electronegativity Learning outcomes and Electronegativity Electronegativity On completion of this section, be able the meaning of terms ‘electronegativity’ ‘bond polarity’ describe the three types intermolecular forces: dipole–dipole, to the ability attract the of a particular bonding pair of atom involved electrons to in covalent itself. van and of Electronegativity increases across Electronegativity decreases down The a period from Group I to Group VII. the order of electronegativity is: any F group. > O > N > Cl > Br…> C > H. weak permanent Polarity in molecules: bond polarisation der Waals, hydrogen formation to: understand is you bond should intermolecular forces If the electronegativity values of the two atoms in a covalent bond are in a covalent bond are bonding. the If same, the different, a we say that the electronegativity we say that bond values the is of bond non-polar the is two atoms polar b We ± H H H + – can also apply the idea of polarity to molecules: F In a non-polar molecule the centres of positive and negative charge coincide. centre of ion – centre of centre of charge the same + charge – In a polar coincide. Figure 2.5.1 molecule the centres of positive and negative charge do not charge This means that one end of the molecule is slightly a Hydrogen is a non-polar negatively + charged, , δ and the other end is slightly positively charged, δ molecule because the centres of + and – charge coincide. b Hydrogen fluoride is The degree polar because the centres of + and – the sign of →. + polarity The → is measured points to the by a dipole partially moment. negatively This charged is shown end of by the charge do not coincide. molecule. + Examples: + δ δ δ δ → + H If we tell if know the bonds The act – the Cl Cl shape molecule so that molecule positive ← + and is of as a they then negative a – Br molecule whole oppose is each non-polar charge is and polar other , (see the the or polarity non-polar . they Figure may 2.5.2) of If each the cancel because bond, we polarities each the can of other centre the out. of same. H a Cl b c + + O C C Exam tips H We often talk about bonding + H + in Cl + Cl + Cl general terms as being strong Cl Many chemists add the term so dipoles add so dipoles polar molecule polar each other bonding, especially weak bonding we are talking is more between as forces may accurate between help to make between forces (strong molecules, 22 refer to molecules. the within bonding) molecules to The polarity in: a water; b trichloromethane; c tetrachloromethane about Weak it non-polar covalent Figure 2.5.2 bonding. When so ‘bonding’ for molecule stronger cancel however molecule reserve or dipoles weak. Cl Cl intermolecular forces them It distinction Inter molecular in neighbouring forces arise molecules. because There molecules and forces permanent van hydrogen dipole–dipole between der W aals forces (weak). bonding. of are forces the three attraction types of between the dipoles intermolecular force: Chapter Intermolecular metallic forces dipole forces bonding. are > generally van der are The in weak compared comparative the order: with strength hydrogen covalent, of these bonding ionic 2 Structure and and intermolecular > permanent dipole– W aals. CH CH 3 Permanent dipole–dipole forces 3 + C Per manent bonding dipole–dipole forces are the weak attractive forces + O CH between C O CH 3 3 + the of δ the dipole neighbouring of one molecule molecule (see Figure and the δ of the dipole of a Figure 2.5.3 2.5.3). Permanent dipole–dipole forces between two propanone molecules van der Waals forces van der W aals molecules, Electrons So the than forces including in of atoms electron attraction noble are density gas are atoms, always may in be not permanent. have van der All atoms W aals and forces. motion. greater in one part of the molecule another . Did you know? So an This instantaneous dipole can dipole induce the is formed. formation of a dipole in a neighbouring molecule. The two The e.g. neighbouring molecules attract each other because of their peculiar lower and high more dipoles. than electron here density and temporarily surface extensive in many molecules greater + + + two water + of ice tension, are hydrogen can form an bonds water due to bonding molecules. Water two pairs water, than hydrogen of average per atoms electrons. of So two molecule. + temporary single of other have lone hydrogen + properties density attraction atom + between Figure 2.5.4 and van der Waals forces are only temporary because the electrons in the molecules of atoms are in constant movement Hydrogen bonding Key points Hydrogen bonding is a special form of permanent dipole bonding. It requires: one molecule atom. These with are an the H atom most covalently electronegative bonded to an F , O or Electronegativity an N atom attract atoms in a is the covalent electrons in ability bond the of to bond to itself. a second molecule having a F , O or N atom with a lone pair of electrons. Electronegativity together a H-band 3 shown dashes or 3 of molecules + H + determine + polar whether used a to molecule is + or non-polar. H H + + H van der Waals forces are based on H temporarily molecules structure be H H + Figure 2.5.5 can dots + lone differences the b by + with induced dipoles. pair Hydrogen bonding: a in pure water; b between water and ammonia Hydrogen between pair a of bonding electrons hydrogen covalent occurs molecules on atom bond to having F, O or a lone N and attached F, O or by a N. 23 2.6 Properties of giant structures Lattices Learning outcomes A On completion should able describe of be their this section, you giant lattice with to: The structures in terms ionic covalent how the properties structures of related to their and a regular bonding types ionic, of three-dimensional between giant e.g. the structures sodium arrangement particles are called of particles. giant Lattices str uctures . are: chloride, magnesium oxide (see Section 2.4 diagram) giant giant covalent silicon (sometimes called giant molecular), e.g. diamond, dioxide metals are three for structures, is strong giant lattices understand giant of metallic, e.g. copper , iron (see Section 2.4 for diagram). structure. Properties of Melting the (see and large boiling number Section Soluble giant in of ionic structures points are strong bonds high: it takes between a lot the of energy oppositely to break charged ions 2.4). water: the ions can form ion–dipole bonds with water molecules. + + H H O + H O + + + + H H H H + O O H H O H + + Figure 2.6.1 Conduct can in only the + Water molecules forming ion–dipole bonds around + and – ions electricity move only when when molten molten or or dissolved dissolved in water . in water: They the cannot ions move solid. solid a cathode Figure 2.6.2 liquid b anode cathode anode a The ions cannot move in the solid state. b The ions are free to move in the liquid state. Exam tips Brittle: when ions the of repulsive A common suggest error that in exams electrons are is substances sodium chloride such as conduct force forces. is applied, charge So the are the next lattice to layers each breaks of ions other , slide. there When are many many easily. to involved a when a same b molten electricity. force First is ask ionic metal yourself if (compound with remember non the the of structure metal). ‘ion’ in applied reactive If it ionic. is ionic, Figure 2.6.3 a A small force is applied. b The ions move out of position and ions with the same charge are next to each other so the lattice breaks along this plane. 24 Chapter Properties of Melting the and large giant boiling number of covalent 2 Structure and bonding structures points are strong covalent high: it takes bonds a in lot the of energy network silicon to of break atoms. atom oxygen atom weak forces between layers Figure 2.6.4 The structures of: a diamond; b graphite; c silicon dioxide Did you know? Insoluble with in water: the atoms are too strongly bonded to form bonds Each carbon bonded Apart from electrons some the layers Apart soft the ions when in because layers Melting that a and do are free they there are slide is to conduct move. These hard: Graphite electrons there are conducts move no ‘spare’ because anywhere it along weak van each huge is three-dimensional too der difcult W aals other to forces when a network break. between force is of Graphite the layers. π to three electron carbon the the directions over electricity: applied. are different can not electrons. voltage graphite bonds Properties of they delocalised from strong So graphite or has is atom in graphite is water . a p atom. These delocalised rings in bonding. along the others. This in the the So the in around not a all structure graphite but a each electrons form system graphite layers, leaves orbital by conducts between layers. applied. metals boiling points are strong forces high: it takes a lot of energy to break Key points the large number the delocalised Insoluble in attraction of electrons water (see (although between the ions of attraction Section some and the ions and 2.4). react the between with water): delocalised the force electrons is too form bonds with water . Some metals react with water ionic because Giant structures and Conduct electricity when solid or molten: the delocalised electrons to move when a voltage is have boiling high points Malleable (can be beaten into they in shape) and ductile (can be drawn Giant when a force is applied, the layers of metal ions can slide other . The attractive ionic forces between the metal ions and conduct can still act in all directions. So when the layers slide, or can easily form. This leaves the metal in a different in are water. They electricity dissolved in when water new because bonds structures soluble the molten electrons strong over only each many directions. into generally wires): have many applied. ions. are bonds free a of electrons. because have lattice they melting lose structures repeating strong to Giant regularly of shape the ions are free to but move. still strong. a b Metals conduct because are free some to electricity of move the electrons throughout the force structure. applied Giant not they Figure 2.6.5 a A small force is applied to a metal. b The metal ions slide out of position, covalent conduct do not electrons structures electricity have free or do because moving ions. but there are still strong forces between the ions and the delocalised electrons. Metallic and structures ductile atoms can are because slide malleable the over layers each of other. 25 2.7 Properties Learning outcomes of simple Molecular Substances On completion of this section, be able know solids how the molecular related to that with This a simple reects a molecular regular structure packing of the such as iodine, molecules in a I , can lattice. form The to: understand simple lattices 2 forces compounds you crystals. should molecular their compounds of are structures simple often properties have between the simple molecular brittle. They delocalised do molecules structures not conduct electrons nor are the which weak are van solids electricity der at because W aals room they forces. So, temperature have are neither ions. molecular a lattice structure understand bonding of how hydrogen influences the properties molecules describe the molecular and solubility of compounds non-polar in polar solvents. Figure 2.7.1 The lattice structure of iodine (only shown in two dimensions) Melting and boiling points of simple molecular structures Many simple melting and overcome with CH CH 3 CH 3 the weak molecular down a molecular boiling when because It such does as not van der W aals forces keeping van der W aals forces increase are it intermolecular lattices, heated. substances points forces iodine take the liquids does gases. take the sulphur , are much energy They much between or lattice or not to have energy molecules. also easily overcome low to Solids broken the weak together . with: CH 3 3 increasing number of electrons increasing number of contact in the points molecule in the molecule (contact points CH 3 3 3 3 are So b CH places the melting CH and boiling come points close of together). down each group as the the noble number of gases and electrons the (and the 3 relative H increase CH 2 mass) increases. CH 2 2 Pentane CH and 2,2-dimethylpropane have the same number of because there electrons, CH 2 3 but Figure 2.7.2 points molecules 2 halogens H the CH 2 CH where the boiling point of pentane is higher . This is are more 2,2-dimethylpropane, a has contact points for van der W aals forces to act over . The total van der a lower boiling point than pentane; b because there are more contact points for van der Waals forces to act over W aals forces Large molecule molecules, temperature number W aals 26 per are therefore greater . So the boiling point is higher . of such because electrons). forces to act as straight they have Also over . chained very there are high polymers, relative many are solids molecular contact points at room masses for the (large van der Chapter The anomalous W ater has a much properties of higher boiling point water Group VI hydrides. The boiling than expected points of Structure and bonding Exam tips by comparison with It other 2 the hydrides from S H is not always safe to assume to 2 that T e H increase gradually due to increased van der W aals forces. W ater compounds with hydrogen has 2 bonding a much higher boiling point than the other hydrides because it always extensively compared vaporise hydrogen-bonded. with water other than Since hydrogen-bonding intermolecular the other forces, Group VI it takes a is lot or dipole–dipole bonding is have a higher boiling point stronger more energy to than those with van der Waals intermolecular forces. hydrides. the boiling point of For SbH example, (not 3 H 400 O H-bonded) 2 NH is higher than (H-bonded). This is that of because we 3 )K( H 300 Te are 2 tniop H S molecules masses, SbH Se with being very more 3 2 2 than 200 10× the mass of NH gniliob 3 100 Figure 2.7.3 0 0 2 3 4 5 Most less solids dense arranged The are in an also has are a hydrides order to is Most and are molecules non-polar formed the are the form of than ethanol, C hydrogen water . to and the OH and to if the in ice is are bonding. water . bonding. are are such as iodine, and however , so simple than . This sulphur is in Did you know? are between soluble because in they water . You can float of the strong bonding in the water. a piece paper H + H some filter needle hydrogen Place paper. on water the a Put needle the surface very paper still sinks, water. When the needle the should + + H C remain H H 2 Figure 2.7.5 of H + + and of filter on + + H H needle + H a because on + The structure of ice forces 3 with Figure 2.7.4 water those molecules NH solute dissolve intermolecular stronger ammonia, between molecules non-polar will, new forces solute 5 bonds hydrogen however , substances They Some by hydrogen molecules the Ice, molecules liquid attractive Many solute themselves. H due in between them. hexane solvent 2 can in as as liquids. the stabilised molecular attracted molecules e.g. tension strength such because together covalently-bonded not between is structure greater insoluble solvents solute water , the corresponding This close surface usually themselves. butane, as simple dissolve, solvent their water . lattice not high Solubility of and than (liquid) ‘ open’ molecules W ater denser than The boiling points of the Group VI period In comparing different H on the surface of the water. 5 Hydrogen bonding between: a ammonia and water; b ethanol and water Key points Simple molecular solids can form lattices solids have melting with weak forces between the molecules. Simple are between Hydrogen-bonded points molecular weak forces than total van of simple a and low have many der Waals forces molecular and boiling points because there molecules. molecules expected The the relatively are soluble between compound at higher in molecules room melting and boiling water. may determine the state temperature. 27 2.8 Shapes of Learning outcomes molecules VSEPR theory The On completion of this section, be able predict the to shapes and in out electron the pair shapes of repulsion theory molecules. It uses (VSEPR the theory ) following can be rules: Pairs simple of electrons in the outer shells of the atoms in a molecule repel bond each angles shell work to: valence you used should (1) molecules other and move as far apart as possible. This minimises and repulsive forces in the molecule. ions. Repulsion than the Repulsion than the of various a below and lone-pairs lone-pairs lone-pairs between molecules information shapes lone-pairs between between repulsion Shapes of The between repulsion and and bond-pairs bond-pairs with only shows how (of electrons) bond-pairs and (of theory is greater electrons. electrons) bond-pairs single VSEPR of of is greater electrons. bonds is used to work out the molecules. b H Four bond pairs. No lone pairs around C. H 109.5° Equal repulsion. So bond angles all 109.5°. C H Shape: H H H tetrahedral + ion (NH is also this shape) 4 H d c Three bond pairs. No lone pairs around B. F 120° F Equal So bond angles all 120°. F B B repulsion. Shape: trigonal planar F f e T wo bond pairs. No lone So bond pairs around Be. 180° Equal repulsion. angles both 180°. Be Shape: h g Six F F bond Equal S 90° S F linear pairs. No repulsion. Shape: lone So pairs bond around angles all S. 90°. octahedral F F F Exam tips j i Three greater Remember that ‘tetra-’ bond pairs. One lone pair around N. repulsion epulsion means four. Greater repulsion between lone and H So a tetrahedron has four faces. N bonding pairs. So bond angles close up to H H H 107°. H less Shape: pyramidal 107° repulsion + l k Figure 2.8.1 A tetrahedron Three bond Greater H means eight. So pairs. an has One lone between pair lone around and O. bonding So bond angles close up to 107°. H Shape: octahedron repulsion H H ‘Octa-’ pairs. O H pyramidal. (The this n m greatest repulsion O also has 3 107° eight faces. ion CH T wo shape.) bond Greatest pairs. T wo repulsion lone pairs between around lone pairs, O. less O H H repulsion H between lone pairs and bond medium and least repulsion between bonding 104.5° repulsion Figure 2.8.2 An octahedron So bond angles close up to 104.5°. least Shape: repulsion 28 pairs H non-linear, V-shaped pairs. Chapter Shapes of multiple molecules and compound ions 2 Structure bonding Did you know? with bonds Experimental The and information below shows how VSEPR theory is used to work out the the data shows carbon–oxygen bond that all lengths in 2– shapes of another carbon dioxide as well as oxo ions (those containing oxygen and non-metal). CO ions are the same. Similarly all 3 the sulphur–oxygen the SO bond lengths in 2– a ion T wo 180° groups of two bonding C No O lone pairs. Equal to bond angles both hybridisation same. This is 2– Three orbitals. bond has that a of bond a length double and single linear bond d the 180°. between Shape: of repulsion. Each So 2– the pairs. due c are 4 b groups of bond pairs. between the relevant atoms. No O lone O pairs. Equal repulsion. So 120° C bond C O angles 120°. O Shape: trigonal planar 120° (The nitrate ion is also trigonal planar .) 2– e 2– f Four O groups of bond pairs. No O lone O S O pairs. Equal repulsion. So S bond O angles 109.5°. O Shape: tetrahedral 109.5° O Key points The shapes theory which electrons and In VSEPR repulsion bond depends around theory, > angles a on in the particular lone-pair bond-pair : : molecules number of can lone be predicted pairs and using VSEPR bonding pairs of atom. lone-pair bond-pair repulsion > lone-pair : bond-pair repulsion. 29 2.9 Shapes of Learning outcomes molecules Hybridisation of orbitals Methane On completion of this (2) section, has four C–H bonds of equal length. The four unlled C atomic you 1 orbitals should be able can be thought of as being mixed so that each s has character 4 to: 3 p and character . This process of mixing atomic orbitals is called 4 predict angles the in shapes simple and bond organic allows compounds understand molecules 3 hybridisation . a formed the shapes using hybridisation of the with of idea atomic These greater and equal mixed overlap also allows repulsion of orbitals atomic each called orbitals bond between are to be when the orbitals. sp a same. They use ideas describe of the of orbital all σ is bonds of orbitals hybridisation shape are them. a Hybridisation molecular b to benzene. + 3 sp orbital Is from C orbital from C-H H molecular orbital 3 Figure 2.9.1 a An sp orbital from carbon combines with a 1s orbital from H to form a molecular orbital making up a C–H bond; b Methane has 4 directional hybrid orbitals Other hydrocarbons with single bonds also form molecular orbitals from 3 hybridised sp orbitals. -bonds 3 Figure 2.9.2 The structure of ethane. The molecular orbitals formed from sp hybrids allow each bond to be a σ-bond. The The structure of dot and cross ethene diagram and shape of a ethene are shown below. b H H 117 .3° C C H Figure 2.9.3 In ethene, H a Dot and cross diagram for ethene; b Shape of an ethene molecule one singly occupied 2s orbital and two of the three singly 2 occupied 2p orbitals in each carbon atom hybridise to make three 3 orbitals. form σ These bonds have which approximately 120° similar are shapes arranged with each to in other a orbitals. sp plane since sp 2 These making there is a sp bond equal orbitals angle repulsion of of the electrons. The remaining form The -bond 30 2p orbitals from each carbon atom overlap sideways to bond. H–C–H expected, Ethene has three sp bond because angle of the in ethene is 117.3°, rather inuence of the π-bond the of the carbon than whose the 120° electron density orbitals in one plane and a π-bond above and below this plane π -bonds 2 Figure 2.9.4 a at right This angles bond to angle that of allows the plane maximum overlap and of the hydrogen orbitals. atoms. is Chapter 2 Structure and bonding Resonance In methane particular and ethene positions. over three or over these atoms. Benzene, more H C 6 shows a , In atoms, These has six is structures. electrons are localised, substances, allowing electrons carbon the some are atoms of said i.e. they molecular the to electrons be arranged are in orbitals free extend movement delocalised in a ring. Figure 2.9.5(a) 6 representation structure the some a single The of benzene. (composite) bonds between The form the ↔ means which carbon lies that the between atoms are actual these neither two double nor Exam tips single is bonds. called a They are resonance somewhere in-between. The composite structure hybrid It is a common resonance a b two or They are single Figure 2.9.5 a Two possible ways of representing benzene; b A modern representation of many lines the of think that mixtures a structures. We by Figure the of structure. ‘in-between’ cases (see to are more forms represent in error hybrids use can structure of dashed 2.9.5(b)) benzene 2 In benzene, hybrid the orbitals six carbon (one to atoms each form hydrogen a hexagon atom and with two to three localised other carbon sp 2 atoms). 120°. This These The The 3 leaves orbitals six orbitals sp a single overlap electrons are p arranged orbital sideways involved H can to on in each form move a a plane. of the So six the carbon delocalised freely around bond system the angles are atoms. of π bonds. ring. H 2 Figure 2.9.6 The structure of benzene: a The p orbitals before overlap (the sp hybrids are shown by the straight lines); b The delocalised system of π bonds 2 In graphite, the other form move an each carbon carbon atoms. extensive along conducts the atom The system layers of when forms three remaining delocalised a voltage is p localised orbitals electrons. applied. sp hybrid overlap These This is orbitals sideways electrons why on to to can graphite electricity. Key points Hybridisation orbital The shapes theory with and Benzene is of s and mixed of simple the a p atomic orbitals results in the formation can predicted of an character. idea organic of molecules be using VSEPR hybridisation. resonance hybrid with a planar ring. 31 Revision 1 a Using the concept of atomic questions orbitals, explain 8 a how: b the three types of intermolecular forces of attraction. i sigma ii pi and bonds Which your State of b two the attraction are formed. these Identify bonds is stronger? i Explain answer. type in of intermolecular force the following of molecules: CO ii NH iii HF iv ICl 3 2 Draw dot and cross diagrams for the following molecules: a S H 9 2 b PCl Magnesium all oxide, solids. Using diamond structure and and aluminium bonding are explain the 3 c differences, CO if any, among these three solids in terms 2 of 3 a How is a co-ordinate the following properties: bond formed? a melting point b electrical conductivity c solubility in + b O H has a co-ordinate bond. Draw a dot and 3 cross 4 Use dot diagram and cross to show the diagrams bonding to in illustrate this the ion. 10 in a 5 the following potassium ionic Using aluminium a An element has has filled Would in the relatively bonding a is b does one electron in its outer explain Place inner shells. this c iodine vaporises very not conduct electricity Describe the easily. type Explain why water has a much or point low? the following melting of this Explain element your be 12 answer. the other Group VI Which of higher melting the following Period 3 metals in order sodium point, molecules or BCl PCl b What Using VSEPR a ‘electronegativity’. is the trend in O H 3 electronegativity as you go down Group VII? b H S 2 c BCl d CCl e PH 3 c Explain your answer to part (b). 4 7 Identify the following polar: a BF b CH c CO d NH 3 Cl 3 2 3 32 would ? have Explain a your 3 theory, state and explain magnesium + Define point answer. of the following a boiling hydrides. point: 13 aluminium 6 higher of element. melting high increasing iodine: brittle 3 c why shell than bonding and oxide 11 b structure compounds: sulphide b and water. bonding molecules as polar or non- 3 molecules and ions: the shapes of Chapter 2 Structure and bonding – revision questions 33 3 The 3. 1 Equations mole Learning outcomes On completon should be able of ths concept and Writing secton, The four The atom dene ‘molar the chemical steps below equations show how counting is shown by: terms = , a H simple = , O chemical = equation. x. 1: W rite down the formulae of all reactants and products: ‘mole’ and mass’ construct balance C to: CH (g) + O 4 to you Step moles molecular and (g) → CO 2 (g) + H 2 O(g) 2 onc Step 2: Count the number of atoms in each reactant and product: equatons. CH + O 4 Step 3: Balance one of the → CO 2 atoms + H 2 xx e.g. O 2 xx x hydrogen: Exam tips CH + O 4 When balancng Neer alter the formula of a Step 4: Keep compound. Balance by → The rst be the balancing, one type puttng numbers at + CH a formula. atom balance n Oxygen s balanced whch s should easest combuston Y ou of to can the often instead equation easest ratio of atom xx at a time until all atoms 2O just if state you → CO 2 the + 2H 2 xxxx number of O 2 xx xx each type of atom on either side prefer . of reactants of the and products reaction. In in the a balanced equation equation for the is called combustion the of last. methane Numbers way O 2 to stoichiometry 2H xx reactons. The balance balance: of one + 2 xx 4 the front CO 2 equatons: n front through formula. the multply atoms n all the dioxide): above, 2 the stoichiometry is 1 (methane): 2 (oxygen): 1 (carbon (water). the Ionic Ionic equations compounds carbonate as water , ions the well include as salts acids separate. and For such as alkalis. ammonium When ionic sulphate 4 SO 2 (aq) → 2NH 4 SO 2 in 2– (aq) + SO 4 (aq) → 2H (aq) 4 + H sodium dissolve example: + ) (NH and compounds 2– (aq) + SO 4 (aq) 4 + NaOH(aq) When ionic reaction. ions. Step T o 1: compounds The ions write W rite an that ionic the full react, play Na only no (aq) some part in + of the the 2: W rite the + CuSO charges on chemical (aq) those 2+ Mg(s) + Cu (aq) ions reaction → equation, MgSO 4 Step OH 3: Cancel the + SO spectator 4: The (aq) → Mg (aq) are ionic: 2– SO (aq) + Cu(s) (aq) + Cu(s) 2+ (aq) equation Mg(s) Cu(s) 4 → Mg is 4 that which remains: 2+ 34 the ions: Cu ionic + which 4 Step in spectator e.g. 2+ 2– (aq) (aq) substances 2+ Mg(s) part called 4 4 Step take are equation: balanced Mg(s) → + Cu (aq) 2+ → Mg (aq) + Cu(s) Chapter 3 Exam tips For precptaton the ons s makng added to reactons up the aqueous t s often precptate. sodum easer For chlorde, to wrte example, lead an onc when chlorde s equaton from aqueous lead ntrate precptated. The ons 2+ whch take part n the reacton are Pb and Cl , so the equaton s: 2+ Pb (aq) + 2Cl (aq) → PbCl (s) 2 State State the symbols symbols reactants reactant (s) (l) mole relative grams is often added products. to equations These are to placed show after the the physical state of formula of each solution in water) product: solid The The and are and liquid and molar atomic called a (g) mass mole of gas (aq) aqueous (a mass or relative that molecular substance mass of (abbreviation a substance mol). We use in the Exam tips 12 C scale One as mole specific a is standard the in amount particles (atoms, comparing of masses substance ions or which molecules) accurately, has as the there so: same are number atoms in of exactly Accurate relate are n gen the atomc masses Perodc Table to 12 12 g of C the isotope. four sgncant gures, e.g. A [Na] r = The molar mass, M, is the mass of one mole of specied substance a grams. So the term ‘molar mass’ can apply to ionic compounds as 22.99. You wll well For molecules we use the term ‘relative molecular mass’. rounded-up of M are molar g mol mass of a compound such as sodium sulphate, Na SO 2 ths e.g. alue by adding the relative atomic masses together , taking the number of each type of values: Na = = 23. In dong dong mole smple at ths leel n chemstry take these ‘rounded-up’ alues as atom: 23, S = 32; O =16 accurate problems r Secton 2Na A [Na] when is beng A n into we account use 4 calculatons calculated gen r Use calculatons. The alue for The –1 units be as calculatons, molecules. often in 1S wth enough to gnore sgncant gures any (see 3.5). 4O –1 M = (2 × 23) + 32 + (4 × 16) = 142 g mol Key points A mole s the amount of substance whch has the same number of 12 speced partcles Molar Chemcal equatons reactants and In mass onc s the as there mass of show are 1 atoms mole equal of a n exactly 12 g substance numbers of n each of the C sotope. grams. type of atom n the products. equatons the spectator ons are not shown. 35 3.2 Calculations Learning outcomes Simple We On completon should be able of ths using secton, can the mole nd the mole concept calculations number of moles by using the relationship: you to: mass of substance (g) ____________________ number of moles (mol) = –1 perform the calculatons mole based molar on mass (g mol ) concept. Worked How example many moles 1 of magnesium chloride, MgCl are present in 19.1 g of 2 magnesium chloride? values: (A Mg = 24.3, Cl = 35.5) r Exam tips Step 1: Calculate the molar mass of MgCl = 24.3 = 95.3 g mol + (2 × 35.5) 2 When wrtng mportant to about make moles, clear t –1 s what type mass (g) 19.1 ___________ Step of partcles we are referrng to. 2: Use the relationship: moles = molar we just say moles of chlorne, t clear whether ts Cl atoms = mass 0.200 mol 95.3 s Worked not _____ = If example 2 or Cl 2 molecules. For example n 35.5 g of What mass of calcium hydroxide, Ca(OH) , is present in 0.025 mol of 2 Cl there are 0.5 mol of Cl 2 but molecules 2 1.0 mol of Cl atoms. Ca(OH) ? 2 values: (A Ca = 40.0, O = 16.0, H = 1.0) r Step 1: Calculate the molar mass of Ca(OH) = 40.0 = 74.0 g mol + 2 × (16.0 + 1.0) 2 –1 Step 2: Rearrange mass Step 3: nd the mass the mass the molar the balanced Worked Fe = Step of the O 2 O the a of × in terms molar values: products specic mass of of mass: mass mass (g) , = 0.025 × 74.0 = 1.85 g in a reaction we use: reactant this reactant equation. 3 maximum is formed (stoichiometric) example Calculate equation moles masses oxide, the = Substitute Reacting T o (g) reduced mass by of iron excess formed carbon when monoxide 399 g (A 3 of 1: Calculate the relevant O 2 2: W rite the = formula 3: Multiply (2 × 55.8) stoichiometric O 2 masses (Fe and 55.8, O Fe each Fe (s) + 3CO(g) simple × 16.0) = 159.6 → 2Fe(s) + 3CO (g) 2 formula mass in grams by the relevant number: O 2 Use (3 ): 3 equation: + 3CO → 2Fe → 2 + 3CO 3 2 159.6 g 4: + 3 stoichiometric proportion to × 55.8 g calculate the (111.6 g) amount 111.6 ______ 399 g → × 159.6 36 = 3 Fe Step iii) r Fe Step Fe 16.0). 2 Step iron( values: 399 = 279 g Fe of iron produced: Chapter Worked We can used be to do similar obtain used, but Calculate water example a the when a 3 The mole concept 4 calculations given slightly nd of different minimum methane to amount mass the method of undergoes maximum product. The is shown methane of reactants shown above can here. needed complete mass method to produce combustion 9.0 g of values: ( A r C = 12.0, Step 1: H = 1.0, Calculate O the = 16.0). relevant formula masses and (CH H 4 CH = 12.0 + (4 × 1.0) = 16.0; H 4 Step 2: O = Calculate moles of mass (g) Use the (g) + 2O they (g) (g) + 2H 2 moles 18.0 the mass of can = moles since also example Calculate the the O(g) be × × molar 16 = mass 4.0 g calculations applied from to the electrons masses mass have of of the hardly ions. The atoms any mass from of ions which mass. 5 maximum aqueous 0.25 mole different methane: methane: (g) and of 2 0.50 mol method excess CO the → signicantly derived calculate 0.25 mol equations are → to 2 mol Worked with = 0.50 mol → Calculate mole not 16.0) 18 2 = are × 1 mol mass The = mass equation 4 Ionic (1 9.0 = stoichiometric CH 4: + ____ = molar Step 1.0) water: ___________ 3: × 2 moles Step (2 O): 2 mass silver of ions silver formed values: (A Zn when = 3.27 g 65.4, Ag of = zinc reacts 108). r + Step 1: Calculate the relevant formula masses (Zn and Ag ): + Zn Step 2: W rite the = 65.4; stoichiometric Ag Step 3: Multiply each + 2Ag atomic stoichiometric 2+ (aq) mass → Zn 65.4 Step 4: Use + 2Ag in grams + by 2Ag(s) the relevant 2+ (aq) g simple (aq) number: + Zn(s) 108 equation: + Zn(s) = → Zn (aq) + → proportion to 2Ag(s) 2 calculate the × 108 g amount (216 g) of silver produced: 3.27 _____ 3.27 g × → 216 = 10.8 g Ag 65.4 Key points The mole requred The concept to form mole a concept formed from a can be used specc can gen be used mass to mass of calculate of to the masses of reactants product(s). calculate the mass of specc products reactants. 37 3.3 Empirical Learning outcomes and Empirical formula An On completon of ths secton, empirical be able for mula and of a molecular formula compound shows the simplest whole number you ratio should molecular formulae of the elements present in a molecule or formula unit of the to: compound. deduce emprcal formulae usng A absolute or molecular for mula in compound The formula deduce Many of shows the total number of each type of atom relate present masses masses elements n a molecule. a molecular formulae. simple molecular Many of an ionic inorganic compound molecules is always have the its empirical same formula. empirical and formulae. organic molecules have different empirical and molecular formulae. Some empirical and molecular formulae are shown below. Compound Emprcal formula Molecular formula hydrogen HO H peroxde O 2 butane C H 2 benzene C 5 2 H 4 CH C 10 H 6 sulphur doxde SO SO 2 cyclohexane 2 CH C 2 Deducing an example Calculate the 12 1 empirical 5.95 g H 6 empirical formula Worked contains 6 tin and formula 7.1 g of a compound chlorine of values: (A tin Sn = and chlorine 119, Cl = which 35.5). r Step 1: Calculate the number of moles of each Sn Cl 5.95 7.1 _____ _____ = 0.05 mol = 119 Step 2: Divide each by the lower number 0.05 W rite the formula of moles 0.2 ____ = 0.1 = 0.5 3: 0.2 mol 35.5 _____ Step element. 0.4 0.5 showing the simplest ratio: SnCl 4 Worked In this example example, we 2 are given information about % by mass rather than mass. Calculate iodine by the which mass values: (A r 38 empirical contains C formula 8.45% = of a compound carbon, 12.0, H = 2.11% 1.0, I = of carbon, hydrogen 127.0). and hydrogen 89.44% and iodine Chapter Step 1: Divide % by 3 A r C H 8.45 = 2: Divide = by lowest = 2.11 0.704 ______ ______ 1 = 0.704 the 0.704 127.0 number = W rite 2.11 1.0 0.704 3: ______ 0.704 ______ Step 89.4 _____ 12.0 Step I 2.11 _____ 3 = 0.704 formula showing 1 0.704 the simplest ratio: CH I 3 Exam tips 1 After Step e.g. P (n 1 to ths 2 you 2.5 O. case by may In 2). be ths So left wth gures case you must the formula s P The emprcal formulae farly smple 1.46 O. It s of ratos. After therefore emprcal formula of The molecular example, the Step 2 adsable P you type of may round be left up the formula as formula the empirical the molar the empirical Worked A is a simple formula the we number wth 1.46 a to norganc rato such 1.5. Ths ones, as wll 1 are P to ge an multiple of butane of the H C empirical formula, need to empirical has twice formula. the For number of 10 H C . T o deduce the 5 know: formula mass of the formula example compound numbers, whole 3 2 molecular a molecular formula molecular atom whole get especally 4 each to 5 compounds, to not up O 2 Deducing the many are O 2 2 that multply has an compound mass. 3 empirical formula CH and a molar mass of 84. 2 Deduce its empirical formula values: (A C = 12.0, H = 1.0). r Step 1: Find the Step 2: Divide empirical the molar formula mass of mass: the 12.0 + compound (2 by × 1.0) the = 14.0 empirical formula Key points mass: An emprcal formula shows the 84 ___ = 6 smplest whole number rato of 14 atoms Step 3: Multiply each atom in the empirical formula by the in Step a compound. number deduced n Emprcal formulae are deduced 2: usng × CH 2 6 = C masses or relate masses elements present n H 6 12 of the a compound. Molecular formulae total number a formula A of unt show atoms of a molecular formula can the formula relate of f the the n compound. deduced from mass the present be emprcal molecular compound s known. 39 3.4 Using Avogadro’s Learning outcomes Avogadro’s Avogadro’s On completon of ths secton, be able law states and that pressure equal have volumes equal of all numbers gases of at the same molecules to: At law you temperature should law apply Aogadro’s law to room temperature and pressure (r .t.p.) 1 mole of any gas occupies deduce 3 . 24.0 dm the stochometry of a At standard temperature and pressure (s.t.p.) it occupies reacton 3 22.4 dm nolng deduce gases molecular formulae from Applying Avogadro’s law to the synthesis of water from hydrogen and oxygen: combuston data. 2H (g) + O 2 2 Did you know? 2 (g) 1 volumes number of atoms n a 1 atoms s ery large: 6.02 × We 10 –1 atoms called mol . Although the Aogadro number was rst Loschmdt. That for s the Aogadro ths alue constant, deduced why volume 2 mol volumes 3 3 24 dm 48 dm (at r .t.p.) can use Avogadro’s numbers of molecules numbers of moles. law in in the mole same calculations volume of because gas, there if there are also are equal equal s the by Johann the constant O(g) 2 2 mole 23 of 2H mol 3 48 dm The → 2 mol Worked example Calculate the 1 symbol s –1 L mass of ethane (M = 30 g mol 3 ) in 240 cm of ethane gas at r .t.p. 3 Step 1: Change cm Step 2: Calculate 3 to dm 3 : 240 cm 3 ÷ 1000 = 0.240 dm 3 volume (dm ) _____________ the number of moles using: moles = 24 0.240 ______ = = 0.01 mol 24 Step 3: Calculate Worked Calculate ethane is mass: example the mass (g) = moles × M = 0.01 × 30 = formed at r .t.p. when 0.3 g 2 volume completely of carbon burnt in dioxide excess oxygen values: (A C = 7.50 g of 12.0, r H = Step 1.0). 1: Calculate the number of moles of ethane: Exam tips 7.50 _____________________ = (2 You may nd example of CO 2 by t easer to 12.0) + (6 × 0.25 mol 1.0) do Worked calculatng formed from × the 7 .50 g of mass Step 2: W rite the relevant ethane, stoichiometric mole equation for the reaction and identify the ratios: 2 .e. 2C H 2 30 g 2 × ethane 44 g CO produces = (g) + 7O 6 (g) → 4CO 2 (g) + 6H 2 2 mol O(l) 2 4 mol 88 g CO 2 2 4 __ Step So 0.25 g ethane produces 88 × 3: Calculate the number of moles: 0.25 mol ethane × → 0.25 mol 2 0.25 CO 2 = 22 g CO 2 CO formed from 0.25 mol ethane = 0.5 mol 2 22 g CO = 2 22/44 = CO 2 0.5 mol CO 2 3 Step 4: Calculate the volume of CO at 2 40 r .t.p: 0.5 × 24 = 12 dm Chapter Deducing the The worked deduce the Worked stoichiometry of example below stoichiometry example shows of a how a we mixture At to the end deduce Step of 1: the the of hydrogen reaction the reaction can use Avogadro’s there of unbalanced and is 20 cm only the of water oxygen present. is law reacted Use together . Apply Avogadro’s equation: (g) H law: + O (g) → 2 3: W rite the 3 volumes 2H 1 1 (g) volume mole(cule) + O 2 Deducing The worked deduce the Worked a O(l) 2 20 cm mole(cule)s equation: H 2 40 cm 2 Step law reaction. 3 2: to Avogadro’s 2 Step concept 3 stoichiometry W rite mole 3 40 cm of The reaction. 3 A 3 (g) → 2H 2 O(l) 2 molecular formula example below molecular example shows formula of how a we can compound use Avogadro’s using law combustion to data. 4 3 Propane contains carbon and hydrogen only. 3 reacts with Deduce for the Step 1: exactly the oxygen, 125 cm molecular formula W rite the information H x (g) + of propane Find the 1 (g) carbon and write Deduce → H xCO 4: propane is formed. balanced equation ratio 5 (g) (g) + of 5O of C O(l) 2 3 gases and 3 (g) → use Avogadro’s 3CO (g) + yH 2 atoms: law: volumes 2 number yH 2 volumes y + equation: 75 cm 1 mol C O(l) 2 H x be Step of dioxide a unbalanced 3 volume x the 125 cm simplest C 3: of 2 3 Step 75 cm below O y 25 cm 2: 25 cm reaction. C Step When 3 → 3 mol CO y (so x must 2 3) Deduce the number of H atoms: C H 3 (g) + 5O y (g) → 3CO 2 (g) + 2 O(l) yH 2 6 of the so 4 oxygen so 4 moles which Step 5: W rite the 10 oxygen atoms of come atoms must water from balanced are the react react with with formed carbon hydrogen containing 8 to form water hydrogen atoms propane. equation: H C 3 (g) + 5O 8 (g) → 3CO 2 (g) + 4H 2 O(l) 2 Key points Aogadro’s law temperature states and that pressure The stochometry of The molecular formula from combuston a data equal reacton of by olumes contan a the can smple of same all gases number be deduced molecular by at of the applyng Aogadro’s compound applyng Aogadro’s same molecules. can be law. deduced law. 41 3.5 Solution Learning outcomes concentration The This On completon of ths secton, concentration of refers be able to the amount a of titrations solution solute dissolved in a solvent to make a you solution. should and In chemistry the units of solution concentration are usually to: –3 expressed understand concept how can be the mole used in moles per cubic decimetre ). (mol dm the is the molar to number determne This concentration concentraton concentration of solution ) (mol dm of moles of solute (mol) _____________________________ –3 of = 3 volume of solution (dm ) solutons –3 In know how ttratons to carry and how out to calculations involving process the to change mass results gures in grams 3 understand n the use chemcal of solution concentration in mol dm remember: acd–base to change that moles that volume to moles 3 cm to dm 3 by dividing volume in cm by 1000 sgncant solute = concentration × volume calculatons. moles of solute ______________ of solution = concentration Simple calculations Worked example Calculate the involving concentration 1 concentration of a solution of potassium hydroxide ( M = 3 56.0) containing 3.50 g KOH in of 125 cm solution. 3.50 _____ Step 1: Convert grams to moles: = 0.0625 mol KOH 56.0 3 Step 2: Change cm Step 3: Calculate 3 to dm : 3 125 cm 3 = 0.125 dm 0.0625 _______ concentration: –3 = KOH 0.500 mol dm 0.125 Worked example Calculate the 2 3 mass of magnesium chloride, MgCl (M = 95.3) in 50 cm 2 –3 of a 0.20 mol dm solution 3 Step 1: Change cm Step 2: Calculate of magnesium 3 to dm : chloride. 3 50 cm 3 = 0.050 dm Exam tips –3 When dealng wth number sgncant gures of moles, which equals: concentration (mol dm 3 × volume (dm ) = 0.20 × 0.050 = 0.010 mol MgCl 2 (s.f.): Step zeros e.g. before 0.036 numbers has 2 are not s.f., Zeros e.g. after When dong seeral The nal data 42 s.f., s.f. calculaton do wth round up round up for answer. answer the to the lowest proded. should same be number number n answer signicant not expressed as 3 are steps. Only your nal s.f. a steps, between has moles to grams: = 0.010 × 95.3 = 0.95 g (to two gures) s.f. a gure 0.0400 Convert signicant The 3: the of is to two gures in signicant the data is gures two. because the least number of ) Chapter Worked example 3 The mole concept 3 –3 What of volume sodium of a solution hydroxide, of NaOH concentration (M = 0.10 mol dm contains 2.0 g 40)? 2.0 ____ Step 1: Convert grams to –2 moles: = 5 × 10 mol NaOH 40 moles 2: Calculate volume in dm (mol) _______________________ 3 Step : = –3 concentration (mol dm ) –2 5 × 10 ________ = 0.10 3 = 0.50 dm Acid–base titrations A titration solution is of used concentration to determine unknown Fill a the acid). of a burette the amount concentration. solution with acid of of alkali known The of substance procedure for present in a determining the is: concentration (after washing it with burette Record Put a the initial known burette volume of reading. alkali into the ask using a volumetric acid pipette. Add an acid–base Add the indicator to the alkali in the ask. volumetric acid slowly from the burette until the indicator changes pipette colour Record the (end the point). nal burette reading (nal – initial burette reading is called titre). Repeat the process until two or three successive titres differ by no 3 more than 0.10 cm alkali white T ake the average of these titres. We usually ignore the rst titre, and tile since indicator this is the rangender titration (rough titration). For example, in the Figure 3.5.1 table below, the fourth and fth titres would be selected and The apparatus used in an averaged. acid–alkali titration Frst ttre Second ttre (rangender) (cm Thrd ttre 3 Fourth ttre 3 ) (cm 5th ttre 3 ) (cm 3 ) (cm ) 3 (cm ) 32.92 32.85 32.00 32.25 32.35 Key points –3 Molar concentraton (mol dm ) = number of moles of solute (mol) ÷ 3 olume Acd–base colour of When soluton (dm ttratons rapdly at are the processng ). carred end out usng an ndcator whch changes pont. ttraton results, the alues selected should be from two 3 or three The successe answer gures as to the a ttres whose calculaton lowest s number alues are expressed of no to more the than same sgncant gures n 0. 10 cm number the data of apart. sgncant proded. 43 3.6 More about Learning outcomes Calculating In On completon of ths secton, be able know how to calculate concentration solution concentration by titration from titration results we need know: to: order solution you to should titrations to calculate the volume and concentration of the solution in the burette, e.g. the soluton acid concentraton from ttraton know a the results of experments how ttraton to to use the results from calculate the volume the balanced of a the solution equation for in the the ask reaction. the Worked stochometry of example 1 reacton 3 know some examples of redox 25.0 cm of a of aqueous solution of sodium hydroxide is exactly neutralised by 3 ttratons. –3 12.2 cm sulphuric acid of concentration 0.100 mol dm . –3 Calculate the concentration in of mol dm the sodium hydroxide solution. Step 1: Calculate the moles of acid (moles = concentration × volume in 3 ) dm 12.2 _____ 0.100 × –3 = 1.22 × 10 mol H SO 2 4 1000 Step 2: Use the stoichiometry of the equation to calculate moles of NaOH: H SO 2 (aq) + 2NaOH(aq) Na × SO 2 mol 2 + 2H O(l) 2 –3 mol 10 (aq) 4 mol –3 1.22 → 4 1 2.44 × 10 mol 3 Step 3: Calculate the concentration of NaOH (moles ÷ volume in dm ): –3 2.44 × 10 ___________ –3 = = 0.0976 mol dm 0.0250 Using titration data to deduce T o nd the volumes of Worked stoichiometry both we to know the concentrations example 2 –3 of a 0.0250 mol dm solution of a metal hydroxide –3 with 1: the hydrochloric hydroxide. Calculate was titrated 3 0.10 mol dm neutralise Step and reactants. 3 25 cm need stoichiometry the acid. Deduce number of the It required 12.5 cm stoichiometry moles of each of of this acid to reaction. reactant: 25 _____ moles hydroxide = 0.0250 × –4 = 6.25 × 10 mol 1000 12.5 _____ moles of hydrochloric acid = 0.10 × –3 = 1.25 × mol 10 1000 Step 2: Deduce the simplest mole ratio of hydroxide to hydrochloric –4 hydroxide 6.25 OH Step 3: W rite the M(OH) 44 10 2HCl mol 1 → : hydrochloric : stoichiometric + 2 × acid: –3 HCl equation: MCl + 2 2H O 2 acid 1.25 × 10 2 mol Chapter 3 The mole concept Redox titrations Redox The titrations end change point of one examples are Potassum involve in of these the given oxidation titrations reactants or reduction is detected by use of reactions either a (see using special Section the redox 4.1). colour indicator . Some below: manganate(vii) ttratons 2+ This can be used ammonium to iron( ii) nd the concentration sulphate. The of, reaction is for example, carried out Fe in ions the in + of acid When ions): (H Did you know? presence usng solutons manganate(vii) for 2+ MnO (aq) + 5Fe + (aq) + 8H 2+ (aq) → Mn potassum you 3+ (aq) + 5Fe (aq) + 4H 4 O(l) 2 cannot purple very light very green pale very pink keep the MnO ions in the potassium manganate( vii) are soluton for ery pale long yellow because brown manganese(iv) The of ttratons, deep purple deposts oxde of are formed. It in 4 s also not easy to see the menscus 2+ colour . When the potassium manganate( vii) is added to the Fe ions, the n the burette because potassum 2+ ions MnO are converted to Mn ions which are almost colourless. After 4 manganate(vii) 2+ all the has Fe reacted, the purple colour of the MnO ions are s one of the most visible. 4 ntensely-coloured 3+ So the colour change at the end point is yellowish (because of the known. ions) to compounds Fe For ttratons nolng ths purple. compound burette t s permssble readngs from the to top take of the Iodne–thosulphate ttratons menscus. These are used concentration in of reactions a solution where of iodine hydrogen is liberated, peroxide by e.g. to adding nd the iodide ions: + 2I (aq) + H O 2 The iodine with a liberated solution of in (aq) + 2H (aq) → I 2 this sodium reaction as an thiosulphate, (aq) + 2Na 2 The end iodine The the point of present end point titration S 2 brown O 2 the sodium used is sharpened ask as up. the blue-black in the blue-black to colourless (aq) → S O 2 2NaI(aq) O(l) solution, + Na by end of S adding point is iodine change a few and gives a is The of starch to the colourless. solution solution when end when brown starch colourless sharper (aq) 6 occurs from drops neared. O 4 colourless titration change titrated 3 2 colour is : colourless thiosulphate The presence 2H 2 3 colourless is + aqueous Na 2 I (aq) 2 not. to is This point. Key points Dchromate(vi) ttratons The reactions acid. For of potassium dichromate are carried out in the presence of The can example: the 2– Cr O 2 2+ (aq) + 6Fe + (aq) + 14H 3+ (aq) → 2Cr concentraton be found relatonshp + 6Fe (aq) + 7H soluton usng concentraton –3 (mol dm light a O(aq) 2 very of ttraton 3+ (aq) 7 orange by green very green ) = number of moles pale of yellow solute (mol) ÷ olume of 3 soluton (dm ) together wth the 2– The Cr O 2 ions in the potassium dichromate( vi) are orange in colour . 7 stochometrc 2+ When the potassium dichromate is added to the the 2– ions, Fe equaton from the Cr O 2 7 reacton. 3+ ions are converted titration is sulphonate point is not + then to ions Cr obvious so phosphoric from green a acid) to which redox is are green. indicator added. The (barium The colour end point of the diphenylamine change at the The can end violet-blue. stochometry be found by concentratons are of a reacton ttraton of both f the solutons known. Redox ttratons change n often colour of nole one of a the reactants. 45 Revision 1 Balance the following molecular questions equations, 9 then Pentane contains carbon and hydrogen 3 write the ionic equation for 20 cm each: only. When 3 of pentane reacts with 160 cm of oxygen, 3 a (aq) FeCl + NaOH(aq) → Fe(OH) 3 b (s) + HCl(aq) → MgCl (aq) + H 2 Na S 2 S(s) NaCl(aq) a of product is carbon dioxide is produced, the other O 2 (aq) + HCl(aq) → steam. (g) 2 NaCl(aq) + H 3 O(l) Deduce + the molecular formula of pentane. 2 + SO (g) 2 2 100 cm 3 Mg(s) c + 10 Calculate the molar concentrations of the following solutions: Define: 3 a a mole 3.33 g KNO in 250 cm of solution 3 3 b molar Calculate b mass the 2.65 g CO Na 2 molar mass of in 75 cm of solution 3 the following 11 Calculate the mass of solute in the following compounds: solutions: ) Pb(NO 3 Ca (PO 3 3 2 a –3 of 25 cm ) 4 a 0.01 mol dm solution 3 2 b 750 cm of NaCl –3 of a 0.27 mol dm solution of NH NO 4 CuSO ·5H 4 12 3 a Calculate 7 . 1 g the Na number of moles in A student titrated a standard solution of 3 the following: SO 2 3 O 2 hydrochloric acid solution obtained and against 25 cm of sodium hydroxide the following volumes of acid: 4 20 g CaCO 3 3 b Calculate the mass 0.025 mol 2.0 of Burette the following: readngs/cm 1 2 3 4 25.80 37 .90 NaHCO 3 Fnal olume –3 × 10 mol HCl Intal olume 4 Calculate burn the 5.5 g of mass of propane oxygen (C H 3 5 Write the following molecular and required to 0.50 Actual olume ttre empirical formula of 26.65 25.65 25.50 the a Complete b From the table above. compounds: a dinitrogen b disulphur the results compound in the table, determine the tetroxide of acid required to neutralise the dichloride sodium A 1.00 ). 8 average volume 6 12.50 completely contains 0.48 g of carbon and hydroxide. 0.08 g –3 13 A student titrated a 0.2 mol dm solution of –1 of hydrogen and has a molar mass of 56 g mol . 3 potassium Find the empirical and molecular formulae of manganate() against 25.0 cm of the iron() solution. The average volume of potassium compound. 3 manganate() 7 Calculate the masses of the following volumes at r.t.p. solution. 3 a 6 dm of O 2 3 b 120 cm of SO 2 8 When butane (C H 4 steam and carbon ) is burnt completely in oxygen, 10 dioxide are produced. Calculate 3 the volume of 46 butane is of steam produced completely burnt. at s.t.p. when in this titration was 27 .50 cm of Calculate gases used 56 cm the molar concentration of the iron() . Chapter 3 The mole concept – reson questons 47 4 Redox 4. 1 Redox reactions reactions Learning outcomes What A On completon of ths secton, be able are simple definition reduction understand terms of state reactions? is the of oxidation loss of is oxygen. the gain But if of we oxygen. look at redox electron reactons transfer n are oxide taking with place at hydrogen, the same we can see that oxidation oxdaton (oxdaton know deduce an the number) oxdaton an atom Copper( ii) oxdaton number oxdaton n a + H (g) → Cu(s) + H phrase OIL remember of electron redox usng Is oxide is oxidised. reduction are There two been rules. wll help reactons n you losing Loss both are oxidised and reactions taking other or gets are reduced. reactions place at the same Is Gan (of gains oxidation oxygen and time. ways of finding out whether electron transfer changes in oxidation or not a substance has Oxidation is loss Reduction is gain of state. in redox reactions electrons. of electrons. terms reacts with chlorine electrons). Mg to + form Cl → magnesium chloride: MgCl 2 Reducton Hydrogen where reduced: Magnesium (of oxygen Redox to transfer: Oxdaton gets Electron transfer RIG O(l) 2 of Exam tips The reduction rules number compound number and state and definition of and 2 n simple reaction time: CuO(aq) change A the to: copper( ii) redox oxidation you of should and 2 electrons). Each magnesium atom loses two electrons from its outer shell. This is oxidation. 2+ Mg Each chlorine complete its atom outer in the shell. → chlorine This is + Mg 2e molecule gains one electron to reduction. – Cl + 2e → 2Cl 2 Equations like separately are preferable, to these called write showing half these the oxidation equations . half It is equations or also reduction reactions acceptable, though not as as: 2+ Mg − → 2e Mg and Cl → 2Cl − 2e 2 Oxidation We can extend reactions or in a discussion 48 (oxidation definition covalent of redox numbers) to compounds include by oxidation using and oxidation reduction states numbers). oxidation ion the involving (oxidation An states state (oxidation compound oxidation to show number number ) the has is degree been a of number given oxidation. abbreviated as In to the each atom following ‘OxNo’. Chapter Oxidation number 1 OxNo 2 The OxNo of each 3 The OxNo of an refers to a on the single atom atom ion ion, in or an arising e.g. 4 The OxNo of uorine 5 The OxNo of an is 6 Redox reactons rules ion in from a oxygen compound. is 0, single e.g. atom Mg is = the 0, H = 0. same as the 3+ Mg in a element 2+ charge 4 = +2, compounds atom in a 2– Fe = is +3, Br always compound = –1, S = –2. –1. is –2 (but in peroxides it –1). The OxNo hydrides it of is a hydrogen –1), e.g. atom : CH H in = a compound +1, HCl: H is = +1 +1, (but in NaH: metal H = –1. 4 7 The e.g. sum in of all the OxNo’s of atoms/ions 2 3+ The a compound is zero, 3 2– 2Al 8 in O Al sum of = the 2 × (+3) OxNos = in a +6 3O compound = ion 3 × (–2) equals = the –6 charge on 3– the 3 9 ion, O e.g. atoms The most negative in = the 3 × atoms = –6 electronegative OxNo +5 − element of 6 N = in = –1 a +5 (the and OxNo charge compound is on of the given ion) the OxNo. Applying oxidation The ion, NO (–2) of many number elements in Did you know? rules Groups V to VII have variable oxidation The numbers. We work these out using the oxidation number rules. of example, what is the oxidation number of S in SO H 2 Roman numbers partcular Applying rules Applying rule 5 and 2H 6: + H = 4O +1 + S and = 0 O so = of –2 2(+1) + 4(–2) + S = that So OxNo of S = reactions Increase in show partcular called names n the the oxdaton atom ‘Stock (or state on). Ths chlorde nomenclature’. shows that ron +6 has Redox the ons) 0 Iron(ii) (or ? s 7: atoms 4 compounds after For and oxidation oxidation number of an OxNo of +2. Potassum manganate(vii) shows that manganese an OxNo the state an atom or ion in a reaction is Chlorne(i) has oxde shows of +7 . that chlorne oxidation. has Decrease in oxidation number of an atom or ion in a reaction an OxNo of +1. is reduction. When copper( ii) nitrogen change oxide as reacts with ammonia, each atom of copper and shown. Key points N 3CuO + oxidised 2NH 3Cu + 3 +2 Cu N + 3H 2 –3 0 O 2 0 reduced Oxdaton s loss Reducton s gan Redox nto The copper in the copper oxide is reduced to copper (OxNo change equatons two to electrons. of electrons. can be dded equatons, one from showng +2 half of oxdaton and the other 0). reducton. The nitrogen in the ammonia is oxidised to nitrogen gas (OxNo change from –3 to Oxdaton s ncrease n oxdaton 0). state. Reducton oxdaton Some atoms other can oxdaton atoms be found number decrease hae xed numbers. The of s n state. n a usng oxdaton number compound oxdaton rules. 49 4.2 Redox equations Learning outcomes Balancing We On completon should be able deduce of ths half balance equatons from balancing then understand agent and know how terms ablty of the terms reducng to order by: the number the two of electrons half lost equations and gained together . agent Construct dsplacement , IO 1 a balanced in acidic ionic equation solution using for the the two reaction half of zinc equations with shown iodate(v) below: 3 n 2+ reducng reference example oxdsng elements oxdsng/ by equations equatons ions, half putting Worked two equations you to: balanced releant secton, half Equation 1: Oxidation of zinc to zinc( ii) Equation 2: Reduction of iodate( V) ions: Zn → Zn + 2e to to iodine: reactons. 1 + IO + 6H + 5e → I 3 Each IO Zn ion atom gains loses 5 2 electrons electrons when when + 3H 2 2 O 2 oxidised. Each I atom in the reduced. 3 T o balance Multiply the electrons: equation 1 by 5 (10 electrons in total): 2+ 5Zn Multiply equation 2 by 2 → (10 5Zn + electrons 10e in total): + 2IO + 12H +10e → I 3 Put the two equations together and – 5Zn + + 6H 2 2IO cancel + + 12H the electrons: 2+ → 5Zn + I 3 Balancing Worked + 6H 2 equations example O 2 using oxidation O 2 numbers 2 Exam tips W rite a balanced equation for the reaction of ClO ions in acidic solution 3 2+ When balancng numbers, are per the oxdaton oxdaton atom – you with Fe Step 1: 3+ ions to the same Cl ions, Fe W rite down in the unbalanced oxidation on each If a of present, states identify the that acd + Fe + + H atoms which 3+ → Cl + Fe + H 3 the Step queston equation, number sde equaton. water . number . 2+ atoms and numbers ClO of ions must change compare form 2: Deduce the O 2 changes in OxNo s oxidation change number . –6 wrte 2+ ClO + Fe + + 3+ H Cl + Fe + H 3 O 2 + H When on the balancng numbers, left. usng remember +5 oxdaton you to select two OxNo reactants products whose 3: Balance the is change. If +1 oxidation number changes so that a total oxidation oxdaton the reacton ClO + 6Fe + + H 3+ → Cl + 6Fe + 3 s the 0. 2+ numbers change +3 and number two –1 usually Step need +2 disproportionation H O 2 reaction + Step (see Secton 12.8) you wll 4: Balance present) need to select three the charges by balancing and then other molecules H such ions as (or OH water . speces. 2+ ClO + 3 50 the only 6Fe + + 6H 3+ → Cl + 6Fe + 3H O 2 ions if Chapter Reducing During A An The a redox reducing ability order of reactions. another an of on agent a its A electrons gains substance electrode to oxidising or chemical solution act as a reaction For copper( ii) gets is a better + reducing Cu gets reduced. agent Section ability is one gaining Silver where one zinc a electrons). not can out atom displaces better So the displace agent reagents displacement replaces copper from 2+ → than Zn (aq) copper oxidising reaction copper put sulphate: + (Zn is Cu(s) better at 2+ is Cu does oxidising We carrying 2+ electrons). or 10.1). by example, (aq) agent oxidised. reducing 2+ Zn(s) Zinc reactons agents and (see reducing reaction. of and electrons potential displacement a aqueous loses agent their in and oxidising Redox reaction: oxidising depends in agents 4 agent goes from to than the copper( losing 2+ Zn (Cu is better at right. ii) sulphate solution because 2+ silver is a better oxidising agent than copper and the ion Cu is a better + reducing If we agent react salts, we ←––– than different can build metal is a Mg –––→ metals up a with metal better metal to ion this is aqueous reactivity reductant Zn According silver Ag (better Fe a better reactivity solutions series , at different releasing Sn oxidant series, (better metal electrons) Cu at accepting magnesium will ←––– Ag electrons) displace silver –––→ from nitrate. Mg(s) + 2AgNO (aq) → Mg(NO 3 But of e.g. silver will Displacement halogen halide will ions not displace reactions displace (see a Section magnesium can less ) 3 also from involve reactive one (aq) + aqueous non-metals. from 2Ag(s) 2 an magnesium A more aqueous nitrate. reactive solution of 12.6). Key points Redox lost equatons and Redox A more reacte In of ts can n equatons soluton ganed be deduced releant can be f a reacton, the The reducng agent t s by wll loses by the balancng dsplace better oxdsng balancng numbers of electrons equatons. balanced element salt half at agent a less losng gans electrons oxdaton reacte changes. element from a electrons. electrons and number becomes and becomes reduced. oxdsed. 51 5 Kinetic 5. 1 The theory gas laws Learning outcomes On completion should be able of this Compressing section, We can are continuously picture gases gases as a collection of randomly colliding with each other and with the particles walls of which the to: container state Boyle’s draw graphical law and Charles’ law representations because in which they are they are constantly placed. hitting The the gas particles walls of force Boyle’s law the exert a pressure container: to pressure and Charles’ (Nm (N) _________ –2 illustrate moving you ) = 2 area (m ) law When calculate volumes and we together . of gases under calculate using different Charles’ Boyle’s volumes temperatures The the volume molecules hit of the a gas, walls the of molecules the container get squashed more often closer and so different the conditions decrease pressures of gas pressure increases. law and gases conditions under using law. increased pressure Did you know? Figure 5.1.1 When the volume of the container is decreased, the gas molecules are squashed closer together and hit the walls of the container more often The chemical high industry often pressures involving in gases, industrial e.g. in the uses very processes –2 We measure pressure in newtons per square metre (Nm ) or in pascals Haber –2 1 Nm process for cases it is making convenient to ‘atmospheres’ the is ammonia. unit of pressure. This with Standard is atm) atmospheric larger: 1 atm = pascal (Pa). pressure is 101 325 Pa (often rounded down to it numbers Boyle’s than 1 101 000 Pa). as because smaller = such use (abbreviated easier to deal In law 101 000 Pa. Boyle’s law states: temperature, Boyle’s law pressure Boyle’s is too the For a fixed volume obeyed high. as of a long Figure number gas as 5.1.2 is the of moles inversely temperature shows two of a gas at proportional is not graphical too a fixed to its low pressure . or the representations of law. a b emulov emulov 0 0 0 pressure 0 1 pressure Figure 5.1.2 Boyle’s law shown graphically: a As the volume of gas decreases the pressure increases; b A plot of volume against 1/pressure shows proportionality 52 Chapter Under different conditions we can represent Boyle’s law 5 Kinetic theory mathematically as: P V 1 initial pressure = P 1 × Worked example A a V 2 initial volume = nal pressure 2 × nal volume 1 3 gas has volume of volume this gas of 1.2 dm when the . Its pressure pressure is is 100 kPa. increased to What 250 kPa is the at constant temperature? Step 1: Substitute the values into V P 1 = P 1 V 2 : 100 × 1.2 = 250 100 2: Calculate the new volume: V V 2 × 1.2 __________ Step × 2 3 = = 0.48 dm 2 250 Heating As the move gases temperature faster and If the further apart. increases. reach a If point temperature In pressure So we chemistry gas the is to with the the gas the xed of volume pressure), force the increases temperature absolute use a constant, theoretical the often of (at increased remain volume called we increases wall decrease when is a the as a because molecules the gas, of the they the molecules have must more get temperature we gas must is emulov energy. of hit eventually zero. This temperature. absolute temperature scale (Kelvin 0 temperature scale). The units of this scale are kelvins (K). 0 Kelvin So temperature minus 15 °C in = °C kelvin + Figure 5.1.3 273. is –15 temperature + 273 = (K) As the temperature of a gas increases, its volume increases in direct 258 K. proportion (Charles’ law) Charles’ law Exam tips Charles’ law pressure, the (kelvin) states: For volume of temperature. Charles’ a a fixed gas Figure number is directly 5.1.3 of moles of a proportional shows a gas to graphical at its constant absolute representation You of can Charles’ law into Boyle’s one law and equation if law. pressure Under combine different conditions we can represent Charles’ law mathematically and temperature both change: as: P V 1 V P 1 2 ____ 2 ____ = initial nal volume _____________________ volume V ___________________ ___ 1 nal T = initial temperature (K) V example T 1 ___ 2 2 = temperature (K) T 1 Worked T 2 Key points 2 3 At 27 °C occupy a at gas occupies 227 °C a volume assuming that of 600 cm the . pressure What volume remains does the gas Standard pressure temperature 273 K Step 1: Convert °C to 27 °C = 27 + 227 °C = 227 273 + = and 273 Kelvin temperature Boyle’s the values into the = °C + 273 300 K = 500 K V V Substitute 101 325 Pa K: ___ 1 2: are respectively. law: At temperature, Step and constant? 600 ___ 2 = equation T T 1 volume of a gas V ____ : constant the ____ 2 = 300 is inversely proportional to its 500 2 pressure. 600 ____ Step 3: Calculate the new volume: = V 3 × 2 500 = 1000 cm Charles’ law: At constant 300 pressure, is the directly absolute volume of proportional a to gas its temperature. 53 5.2 The kinetic Learning outcomes theory General Most On completion of this section, be able know the gaseous know the know real characteristics the the and of gases know basic either exist as small isolated molecules, atoms, e.g. e.g. ammonia helium and and sulphur argon. in a gas: assumptions of are arranged are far move apart randomly from (are each disordered) other theory differences ideal the or particles rapidly and randomly. between gases The are gases the state kinetic properties of to: The gases you dioxide, should of ideal gas kinetic theory and ideal gases equation. The kinetic theory of Gas particles are Gas particles do Gas particles have Collisions they A gas the which obey Real at many moving not between has these energy of Boyle’s have each other . volume. gas particles are characteristics the law made different gases differences do particles and is Charles’ are obey elastic, is a i.e. called an measure law no energy is lost when ideal of gas. their In such a temperature. gas, Ideal exactly. We ignore The very attraction high and the law theory and and the very at is volumes between of temperatures. noticeable kinetic pressure measurements Boyle’s especially The cannot accurate pressures not temperatures. At randomly. attract no that: gases Scientists gases states collide. kinetic gases gases of low high always the particles The Charles’ very not the volumes results law pressures many show exactly. obeyed of that These and very low because: particles. is not zero. temperature: nitrogen 1.5 emulov × The So particles the are volume volume of the closer of the together . particles container they is not are negligible placed compared with the in. 1.0 erusserp the gas is ideal The forces Attractive The of attraction forces between between the the particles particles pull cannot them be ignored. towards each other . 0.5 carbon pressure volume 0 0 200 400 is lower than expected for an ideal gas and the effective dioxide 600 At of the extremely gas high is lower than pressures expected. and extremely low temperatures the 800 particles may be so close to one another that repulsive forces between 5 pressure (kPa × 10 ) the Figure 5.2.1 clouds cause deviations from ideal gas behaviour . Hydrogen, nitrogen and carbon dioxide show deviations from Boyle’s law over a range of pressures 54 electrons Gases and which do pressures not are obey called Boyle’s real law gases and Charles’ law at all temperatures Chapter The For us ideal an the ideal gas gas ideal gas 5 Kinetic theory equation we can combine Boyle’s law and Charles’ law. This gives equation : PV = nRT Where: P is the pressure V is the volume in pascals, n is the number R is the gas T is the temperature Pa. 3 in of cubic metres moles of (1 m ). –1 mol (8.31 J K in 1000 dm gas. –1 constant 3 = kelvin ). (K). mass in grams (m) _________________ Since number of moles = we molar equation mass can rewrite the ideal gas (M) as: mRT _____ PV = M If we PV = know nRT, four we of can the ve quantities calculate the in the ideal gas equation, fth. Key points The kinetic motion An at ideal gas laws Gases The theory a gas states variety is one under all of that gas particles are always in constant random speeds. which obeys pressures do not obey the ideal gas equation gas is and laws PV the = gas laws. Real gases do not obey the temperatures. at high pressures and low temperatures. nRT. 55 5.3 Using the Learning outcomes On completion should be able perform of this ideal Calculations section, you calculations gas equation using the Worked example Calculate the ideal gas equation 1 to: using volume of gas in a weather balloon which contains the 0.24 kg ideal gas of helium at a temperature –1 equation He (A = 4.0; R = 8.31 J K of –25 °C and a pressure of 60 kPa –1 mol ) r know how to determine the Step relative molecular mass of 1: Change pressure calculate with low boiling points or the inserting relevant data temperature number of into the correct units and moles: gases 60 kPa by and liquids = 60 000 Pa –25 °C = –25 + 273 = 248 K into 240 ____ the ideal gas equation. 0.24 kg = 240 g moles of He = = 60 mol 4 Step 2: Rearrange the ideal gas equation into Exam tips the form you need: nRT ____ PV = nRT → V = P 1 When doing gas law calculations, 60 think of the units that you Step 3: Substitute the gures: V (in ) m × 8.31 × 248 ________________ 3 always 3 = = 2.06 m 60 000 are likely to have to change to as ‘molPaK’ moles mol pressure Deducing relative We simple molecular masses Pa temperature can use a weighing method to deduce the relative molecular K mass of a gas. 3 2 Make sure that you change cm A 3 or dm 3 to m 3 (1 m large flask of known volume is lled with gas and the mass of the gas is 3 = 1000 dm ) found. the If values method used, of the is its temperature in the ideal suitable accuracy in is and gas the pressure equation school limited to are known, nd the laboratory because of we value when problems can of M. dense substitute Although gases involving are the all this being buoyancy air . Worked example A volume 2 3 flask flask is of 100 kPa 2.00 dm and the contains temperature 3.61 g is of 20 °C. –1 molecular Step 1: mass Change of the gas pressure, (R = and gas. The pressure the = 100 000 Pa 2.0 dm 20 °C Step 2: Rearrange and molar the ideal mass, = gas M) to PV = 20 + temperature to their correct 2.00/1000 273 M = (in the the 3.61 the × × 10 3 m form showing mass, mRT _____ → M = M Substitute 2.00 subject: mRT 3: = 293 K _____ Step units: –3 = equation make PV values: 8.31 × 293 _____________________ M = = –3 100 56 000 × 2.00 × 10 the ). 3 100 kPa in relative –1 mol 8.31 J K volume a Calculate 44 (to two signicant gures) m, Chapter Deducing the A similar point The relative method can be molecular used to mass of deduce the a volatile molar mass of 5 Kinetic theory liquid a low boiling liquid. apparatus is shown below: thermometer hypodermic syringe gas syringe syringe Figure 5.3.1 The the bulb Apparatus used for nding the relative molecular mass of a volatile liquid procedure Let light oven gas is: syringe Record this Record volume Draw Inject reach a particular temperature in the syringe oven. temperature. some of some of the the of the air liquid in the into liquid gas the from syringe. hypodermic the syringe hypodermic and syringe weigh into the it. gas syringe. Allow liquid in Record the The the vapour the mass calculation V olume of to gas is vaporise and record the nal volume of air + syringe. of the carried vapour = hypodermic out nal in gas the syringe same syringe way and as volume the for – a pressure. gas initial but: gas syringe volume Mass of liquid hypodermic = initial mass of hypodermic syringe – nal mass of syringe Key points The ideal equation, temperature or equation known. The The by is are relative volume gas of relative (at the (at of of a can if be the gas mass of of liquids vapour temperature used other can temperature volume known moles mass known molecular measuring vaporised number molecular gas PV = nRT, to physical be found and by pressure, quantities weighing a volume, in the known pressure). with low produced and deduce boiling when a points known can be found mass of liquid pressure). 57 5.4 Changing Learning outcomes state The The On completion should be able of this section, liquid molecules describe to: the liquid state in are not motion, proximity explain the of liquid: distances arranged, due to although the effect of there weak may be some order intermolecular over forces of terms which are constantly being broken up as they gain kinetic and energy arrangement a regularly attraction of in you short state from neighbouring molecules particles terms ‘melting’ are have close more together kinetic each other in and energy than in a solid and so are able to slide over ‘vaporisation’ explain of changes energy in changes proximity, state and in changes arrangement in Most of fairly random way. substances that are liquid at room temperature are either: and motion a terms covalently bonded molecules with considerable van der W aals forces of particles. attraction, e.g. molecules bromine with dipole–dipole signicant bonding e.g. hydrogen bonding, e.g. water or propanone. Did you know? Water deal in of the up hydrogen smaller exist in to groups the 18 ice-like state structure. This extensive or liquid of water Changing state When is its a solid molecules The with energy bonds broken as is a flickering water. increases solid the vibration of the particles. The increases. The forces The solid of attraction between the particles weaken. more are melts slide when over enough each energy is transferred to make the other . continuously within The sometimes crystal melting point is the temperature when the solid is in equilibrium the with to the molecules. and formed structure. This of an being transferred temperature particles Hydrogen heated: Larger with between arranged great to perhaps molecules structure randomly due a bonding. structure, water is has the liquid. referred structure of For an ionic attractive molecular overcome When a The the The less is energy of energy lattice. energy of required the to melting required intermolecular transferred from is is So for forces break point melting holding the is – strong very just the ionic high. For enough molecules a to together . heated: The forces lot the weak liquid escape a in solid, particles. solid forces increases temperature attraction the liquid of between and the the movement liquid the become particles a and energy of the increases. vapour . weaken This until process they is called vaporisation The liquid boils equilibrium when between enough the energy liquid and is the transferred gas to allow an phase. Exam tips It is the important boiling liquid. This is equilibrium 58 to point distinguish some called with between particles have evaporation. the vapour, evaporation enough Boiling which occurs appears and energy as boiling. to when bubbles Even escape from the whole within below the liquid the is liquid. in Chapter These changes temperature because the molecules of state stays energy (or can be constant ions) absorbed rather shown at the goes than on a heating melting to break raising liquid point the the curve. and forces at 5 Kinetic theory The the boiling between point the temperature. vaporising ) C°( erutarepmet gas heating up solid liquid heating up melting 0 solid 0 heating 5 up 10 15 time Figure 5.4.1 20 (min) The change in temperature when water is heated at a constant rate. Note how the temperature stays constant as the solid melts and the liquid turns to vapour When a vapour attracted It to is each cooled, other . the particles Energy is lose released kinetic and the energy vapour and begin turns to to be liquid. condenses When each a liquid other is more sufciently low cooled, and for Summarising the more the particles slowly. liquid to changes of lose kinetic Eventually solidify. The the energy and slide temperature liquid past is freezes. state energy absorbed vaporisation melting condensing freezing released Figure 5.4.2 Energy is required to melt and vaporise a substance. Energy is released when a substance condenses and freezes Key points Particles over in each Most liquid substances molecular a Particles are not ordered, are close together and move by sliding other. which structure in a solid are with are liquid weak at room temperature attractive forces generally arranged in a have between lattice. The the a simple molecules. particles only vibrate. When a solid particles When They a melts, the together. The liquid energy particles vaporises, the input overcomes begin particles to slide gain the forces over energy one and keeping the another. move further apart. move freely. 59 6 Energetcs 6. 1 Enthalpy Learning outcomes changes Enthalpy Particles On completion should be able of this section, have know that to: energy enthalpy two main types change of energy: you Potential closer and energy two is the attracting energy due particles are to to the position each other , of the the particles. lower the The potential changes energy. accompany chemical reactions understand the meaning of Kinetic energy is the energy associated with the movement of the particles. term ‘enthalpy change’ Energy describe the terms has content and is be the absorbed total to amount separate of energy particles from (potential each and other . kinetic) in The heat chemicals. ‘endothermic’ The to ‘exothermic’ draw energy exothermic prole and heat content is also called the enthalpy (symbol H). diagrams for We endothermic in cannot their measure normal enthalpy physical by states itself. at So it was 101.325 kPa decided pressure that and all elements 298 K have reactions. zero enthalpy. These conditions are called standard conditions . The Ø is symbol We can, used to however , show standard measure conditions. enthalpy changes . These occur when heat Exam tips energy is exchanged with the surroundings in a chemical reaction. The Ø symbol In Chemistry it is useful to for mean. You letters have and already enthalpy change is . ΔH what between the enthalpy of the products sigma and enthalpy change come pi. The Greek and the is the reactants. they Ø across Ø ΔH = Ø H − H products two: The know difference some Greek standard reactants letter –1 The capital change delta in enthalpy change Δ is something. change in used and to So denote ΔH units of enthalpy change are kJ mol a means ΔT means Exothermic and endothermic reactions temperature. T wo important – terms system sur roundings reaction and An the is chemical – these carried anything decrease any into in common include reaction Ø are reactants out, dipping endother mic surroundings which and the the the not reaction absorbs chemistry energy container taking − from the in which in e.g. a the a reaction thermometer . surroundings. The temperature. Ø H products part mixture, Ø H are: products air , solvents in is positive. So ΔH reactants is positive. reaction Ø For example: CaCO (s) → CaO(s) + CO 3 All e.g. reactions thermal requiring An a continuous decompositions. endothermically, exother mic surroundings (g) e.g. Some ammonium reaction increase Ø releases in of heat dissolve are in endothermic, water energy to the surroundings. example: C(s) + O (g) → 2 All combustion acids 60 are reactions exothermic. The Ø is negative. So ΔH reactants is negative reaction Ø For 572 kJ mol temperature. H products + chloride. Ø – H = reaction input salts –1 ΔH 2 CO (g) 2 are –1 ΔH = −394 kJ mol reaction exothermic. Many reactions of metals with Chapter Energy Energy the 6 Energetics prole diagrams prole relative usually diagrams enthalpy includes the reaction the enthalpy the (enthalpy of change, an reactants formulae pathway (exothermic), the on a prole the of and show: products reactants and on the y-axis, this products x-axis downward upward diagrams) arrow, arrow, ↑, ↓, indicates indicates energy energy released absorbed (endothermic). H (g) + I 2 (g) 2 ygrene –1 H = +151 kJ mol 2HI reaction Figure 6.1.1 pathway Energy prole diagram for the thermal decomposition of hydrogen iodide (endothermic) The diagram The So for reaction the the endothermic mixture enthalpy of gains the reaction energy products is so shows the higher that: surroundings than that of cool the down. reactants. Ø So ΔH is positive. reaction Mg(s) + 2HCl(aq) ygrene –1 H = –456.7 kJ mol MgCl (aq) + 2 H (g) 2 Key points reaction pathway Figure 6.1.2 Energy changes in chemical in heat Energy prole diagram for the reaction of magnesium with hydrochloric acid reactions result loss or (exothermic) gain: The diagram for the exothermic reaction shows that: In an an enthalpy exothermic transferred The reaction mixture loses energy so the surroundings warm So the enthalpy of the reactants is higher than that of the ΔH. reaction, the heat is surroundings. up. In to change, an endothermic reaction products. heat is absorbed from the Ø So ΔH is negative. reaction surroundings. Exothermic negative reactions ΔH reactions values, have have endothermic positive ΔH values. Energy prole relative and diagrams energies products of and the the show the reactants enthalpy change. 61 6.2 Bond energes Learning outcomes Bond On completion of this section, making Breaking the and bond bond between breaking two atoms requires energy. It is you endothermic. should be able to: know that chemical Making new bonds between two atoms releases energy. It is reactions exothermic. take place through energy changes involved in If the energy released breaking and know that bond breaking endothermic and bond making in new bond bonds, breaking the is greater reaction is than the energy endothermic. If the energy We how bond in absorbed making in new bond bonds, breaking the is less reaction is than the energy exothermic. is exothermic understand making is released in making absorbed bond can draw energy prole diagrams to show these changes. energies a can be used enthalpy to calculate change of the 2H + 2Cl reaction bond making bond describe energy energy can and affecting how affect H ygrene bond the factors bond chemical + Cl 2 2 H 2H–Cl reactivity. reaction pathway Exam tips Do not make the mistake in b thinking 4H that in during the In many bonds reaction. in Section are break particular 7 .5 for broken bonds into the break sequence details). showing all + 2O bonds only during more diagrams the reactions Particular a all ygrene certain level reaction compound atoms. form a bond bond breaking making 2H–H +O=O H 2H and Energy the atoms reaction breaking then reforming accounting’ only. pathway are for Figure 6.2.1 ‘atom O 2 (see Energy prole diagrams for: a an endothermic reaction – the thermal decomposition of hydrogen chloride; b an exothermic reaction – the synthesis of water from hydrogen and oxygen Bond Bond energies energy particular is bond the in bond dissociation bond broken is amount one energy. put of mole after of The this energy needed gaseous symbol to break molecules. for bond one This energy is is mole also E. of The type to units the of bond of is is to energies When the bond energy required broken. released type bond energy V alues being of same are a So C=C always bonds –1 . kJ mol break are new E(C=C) double positive are the amount of broken. For example: = the energy +610 kJ mol they amount absorbed 2O(g) ΔH = + 496 kJ mol –1 2O(g) 62 → O=O(g) ΔH = – 496 kJ mol refer of to bonds energy when –1 → refers bond. because formed, as O=O(g) of symbol. –1 The a called the same Chapter Factors For affecting atoms of the bond same energy bonds. This is type, double because bonds there is a have greater higher bond attractive energies force than compounds bonding pairs of electrons and the nuclei of the atoms forming compared with only one electron if = + ethene the pi in the C=C sigma bond bond. You is weaker can see than this pair . –1 E(C–C) as the the bond such between bond two Energetics Did you know? In single 6 you compare the bond energies –1 E(C=C) 350 kJ mol = + 610 kJ mol of the carbon–carbon single and –1 E(C≡C) The bond energies of the = halogens +840 kJ mol double are: exposed –1 E(Cl–Cl) = + bond –1 244 kJ mol E (Br–Br) = + = + electron and the reects density smaller the of more the amount pi of 193 kJ mol overlap –1 E(I–I) bonds. This of the p atomic orbitals z 151 kJ mol 2 compared The bond energies decrease down the group as the distance between nuclei increases. This is because there is less force of the lengthens. bonding The pair bond of electrons energy for and uorine, the nuclei however , as is the much sp bonding in sigma molecular orbitals. (see attraction Section between the the the atomic with 2.9). bond lower than –1 expected close each The to (E(F–F) each other bond atom = other + ). 158 kJ mol so the electron This is clouds because in the the F atoms neighbouring are very atoms repel considerably. energies increases in of carbon–halogen bonds also decrease as the halogen size. –1 E(C–Cl) = –1 E(C–Br) +340 kJ mol = +280 kJ mol –1 E(C–I) The bond atoms in energy a of a molecule. particular For = +240 kJ mol type of bond can be affected by other example: –1 E(C–F) in CH F = + –1 452 kJ mol E(C–F) in CF 3 For this reason Calculating The ‘balance energies is = + 485 kJ mol 4 we often use enthalpy sheet shown average changes method’ below for for bond energies, using calculating the bond E energies enthalpy changes using bond reaction: H H–C–H + 2 O=O → O=C=O + 2 H–O–H Key points H Bonds broken (endothermic +) Bonds formed (exothermic –1 −) /kJ mol E(C–H) = 4 × 410 = 1640 2 × E(C=O) = 2 × 740 = × E(O=O) = 2 × Bond energy 496 = 992 4 × E(O–H) = 4 × 460 = to gaseous = +2632 kJ Total = enthalpy change = +2632 – 3320 = Bond energy and is is the break energy one bond in mole one of mole a of molecules. energies calculate –688 kJ mol a Bond endothermic making –3320 kJ –1 Overall an –1840 particular Total is bond –1480 needed 2 and exothermic. × breaking process –1 /kJ mol 4 Bond the can be used enthalpy to change in reaction. reactivity The strength of the bond –1 The bond Nitrogen break the energy only of reacts bonds to decomposes the very N ≡N under get the readily triple very bond harsh reaction because is high conditions started. the very O–O (994 kJ mol because Hydrogen bond it is ). difcult between to peroxide energy is . ease of decomposition of hydrogen different halides is to the strength of the hydrogen–halogen bond (see is different A high value of bond energy may also contribute related in atoms compounds. The given only –1 150 kJ mol slightly two Section to a molecule being 12.7). relatively unreactive. 63 6.3 Enthalpy changes Four types of Learning outcomes On completion of this section, Standard be able experment enthalpy enthalpy change change of reaction: The enthalpy change when you the should by amounts of reactants shown in the to: equation react to give products Ø under standard conditions. Symbol ΔH r describe enthalpy changes 1 of Example: 2Fe(s) + 1 Ø O (g) → Fe 2 2 reaction, O 2 (s) ΔH 3 –1 = –824.2 kJ mol r combustion, neutralisation and Standard enthalpy change of neutralisation: The enthalpy change solution when 1 mole alkali under of water is made in the reaction between an acid and an Ø know how to determine these standard conditions. Symbol ΔH n enthalpy changes by experiment Example: HCl(aq) + NaOH(aq) → NaCl(aq) + H O(l) 2 Ø use experimental calculate data enthalpy –1 = ΔH to –57.1 kJ mol n changes. Standard enthalpy mole substance of a change is of solution: dissolved in a The very enthalpy large change amount of when water 1 (to Ø innite dilution) under standard conditions. Symbol ΔH sol + Example: NaCl(s) + aq → Na Ø (aq) + Cl (aq) –1 ΔH = +3.9 kJ mol sol Standard when 1 enthalpy mole of a change of substance combustion: is burnt in The excess enthalpy oxygen change under standard Ø conditions. Symbol ΔH c Example: –1 Ø (g) CH + 2O 4 Finding We can (g) → CO 2 enthalpy calculate (g) + 2H 2 changes many O(l) ΔH 2 enthalpy = –890.3 kJ mol c using changes a calorimeter from the results of experiments thermometer involving can, a a vacuum procedure plastic calorimeter. ask or A a calorimeter more can complex be piece a of apparatus. cup, The a metal general is: lid reaction React known Measure Calculate the amounts of reactants temperature the energy change transferred in of a known the using volume of solution. solution. the relationship q = mcΔT mixture Where polystyrene cup m is q the is the mass energy of transferred solution in the in joules (J), calorimeter –1 c is ΔT Figure 6.3.1 polystyrene A polystyrene drinking cup the is specic the heat capacity temperature of change water (rise or in grams (g), –1 °C (Jg fall) –1 ). V alue of c = 4.18 Jg –1 °C (°C). Ø Calculate the enthalpy change per mole of specic reactant for , ΔH sol can act as a simple calorimeter Ø per mole of water formed for or ΔH for the number of moles shown n Ø in the equation for ΔH r Ø ΔH by calorimetry: worked example r 3 0.90 g of solution was to zinc in stirred 27.9 °C. a was added to calorimeter . constantly. Calculate 75.0 cm The The the –3 of 0.25 mol dm copper( ii) sulphate temperature enthalpy of change the for is copper(ii) in excess. solution this rose reaction, sulphate The from A Zn mixture 18.0 °C = r Step 1: Substitute q Step 2: the = Calculate data 75.0 the × into the 4.18 energy per × equation: (27.9 mole of – q 18) Zn: = = mcΔT 3103.65 J 0.9 g → 3103.65 J 65.4 _____ So 65.4 g → × 3103.65 = 225 531.9 J 0.9 –1 Standard enthalpy signicant 64 gures) change of reaction = –230 kJ mol zinc (to two 65.4. , Chapter Enthalpy changes of neutralisation are found in a similar way to the 6 Energetics above Exam tips example and by adding measuring known the concentrations maximum and temperature volumes rise. of Enthalpy acid and changes alkali of In solution are found by adding a known amount of solute to a known calorimetry assume of water in a calorimeter and measuring the maximum rise in experiments we volume that: temperature. 3 Standard enthalpy A simple apparatus is shown in Figure change of used to combustion measure the enthalpy by 1 cm of water Aqueous weighs 1 g. calorimetry change of if combustion they solutions are are treated as water. 6.3.2. The solution specic When has heat a the same capacity solid is added as water. to a lid solution, mineral the mass of the solid is wool ignored. water calorimeter draught screen cap spirit burner Figure 6.3.2 Measuring the enthalpy change of combustion The procedure is: Exam tips Weigh Put the spirit burner and cap with the fuel. You a known amount of water in the calorimeter and record should know the main sources its error in calorimetry experiments: temperature. Light the spirit temperature Remove The results above. We the are need mass of of burner , the spirit in in the the and let calorimeter replace a cap the similar cap way rises and as the for fuel by burn about until the 10 °C. the the water the temperature the mass the molar fuel mass in rise the of of heat / to the solution air from and from the the burning fuel worked example calorimeter the loss water reweigh. know: of water burner , processed to remove loss of and thermometer heat to evaporation the calorimeter of fuel. water burnt of the fuel. Ø The value of is ΔH calculated per mole of fuel burnt. is enthalpy c Key points Enthalpy change of reaction, ΔH , the change when the r amounts of reactants Enthalpy change of shown in the combustion, equation ΔH , is the react to enthalpy give products. change when 1 mole c of a substance Enthalpy is change burnt of in excess oxygen. neutralisation, ΔH , relates to the enthalpy change for n + the reaction: H – + OH → H O 2 Enthalpy change of solution, ΔH , is the enthalpy change when 1 mole of a sol substance dissolves to form a very dilute solution in water. Ø The symbol Values for indicates ΔH , r using a ΔH , c the ΔH enthalpy and n ΔH , change can often under standard be found by conditions. direct experiment sol calorimeter. 65 of 6.4 Hess’s law Learning outcomes On completion should be able of this Enthalpy section, change of formation Standard enthalpy mole compound change of for mation: The enthalpy change when 1 you of a is formed from its elements under standard to: Ø conditions. Symbol ΔH f explain and use the term Ø Example: ‘enthalpy state change Hess’s C(graphite) + 2H C(graphite) law (g) → CH 2 of formation’ is used here (g) because –1 ΔH 4 [CH f graphite is the (g)] = –74.8 kJ mol 4 most stable form of carbon. Ø The value of of ΔH an element in its normal state at room temperature is f use Hess’s law to calculate zero. enthalpy reaction changes and of formation, combustion. Hess’s Hess’s Did you know? law is an example of the to of Thermodynamics chemical states that states as reactions. The 6.4.1 of that the the route illustrates total by enthalpy which Hess’s law. the It change reaction shows for a chemical reaction is proceeds. that the enthalpy change is First the law law independent Figure Hess’s law same, whether you go by the direct route or the indirect route. applied First law Energy cannot be created We can found use Hess’s directly by law to calculate experiment, e.g. enthalpy the changes enthalpy which change of cannot be formation of propane. or destroyed Ø Deducing enthalpy change of reaction using ΔH values f A + B C direct + D An reactants enthalpy cycle (Hess cycle) is constructed as shown in Figure 6.4.2. products route H H r r reactants products NH + HCl NH 3 X + Y H H f shows the route via H f (NH f ) H 3 +H products Cl) 4 (HCl) f intermediates 1 elements in 1 N + 2H 2 2 + Cl 2 2 2 and Y standard Figure 6.4.1 (NH f indirect reactants X Cl 4 states The enthalpy change is the Figure 6.4.2 same whichever route is followed Ø Using Hess’s law we can see that Ø ΔH + f ΔH example Calculate ΔH ΔH f Ø = – f products Ø ΔH r Worked Ø = r Ø So ΔH reactants ΔH products f reactants 1 Ø for the reaction: SO r (g) 2H S(g) → 3S(s) + values: SO f (g) = 2H 2 –296.8 kJ mol O(l) 2 –1 Ø ΔH + 2 –1 , H 2 S(g) = –20.6 kJ mol , 2 –1 H O(l) = –285.8 kJ mol 2 The stages W rite Draw Apply are: the the balanced enthalpy Hess’s equation. cycle with the elements at the law. SO (g) + 2H 2 H O(l) 2 [SO f 2 ] 2 2 × H f +2 × H f [H S] 2 3S(s) + O (g) 2 Figure 6.4.3 66 bottom + 2H (g) 2 [H O] 2 as shown. Chapter According to Hess’s Energetics law: Ø Ø ΔH 6 [SO f ] + 2ΔH 2 Ø [H f S] + ΔH 2 Ø = 2ΔH r [H f O] 2 Ø –296.8 + 2 × (–20.6) + = 2 ΔH × (–285.8) r Ø so ΔH = 2 × (–285.8) + 296.8 + (2 × 20.6) r Ø so –1 = ΔH –233.6 kJ mol r Ø Deducing enthalpy change of reaction using ΔH values c The enthalpy cycle is constructed as shown below and a worked example shown. H r reactants H product(s) +O c H +O 2 c 2 (reactants) (products) combustion product(s) Figure 6.4.4 Worked example Calculate ΔH 2 Ø for the reaction: 3C(graphite) + 4H r (g) → C 2 H 3 (g) 8 –1 Ø ΔH values: C(graphite) = –393.5 kJ mol , c –1 (g) H = –285.8 kJ mol , 2 –1 H C 3 (g) = –2219.2 kJ mol 8 H r 3C(graphite) + 4H C (g) 3 × H H 3 2 (g) 8 [C(gr)] c H + 4 × H [C c (g)] 8 (g)] [H c H 3 2 3CO (g) + 4H 2 O(l) 2 Figure 6.4.5 According to Hess’s Ø ΔH law: Ø + Ø ΔH r [C c H 3 ] = 3ΔH 8 Ø [C(graphite)] + 4 ΔH c [H c (g)] 2 Ø + ΔH (–2219.2) = 3 × (–393.5) + 4 × (–285.8) r Ø so ΔH so ΔH = 3 = –104.5 kJ mol × (– 393.5) + 4 × (–285.8) + 2219.2 r Ø –1 r Ø Note: In this reaction, ΔH is also the enthalpy change of formation of Key points r propane. Enthalpy refers Exam tips of a to of formation the formation compound from under Remember change standard of its 1 mole elements conditions. that: Hess’s law states that the total Ø ΔH of an element is 0. enthalpy change in a reaction is f independent of the route. In enthalpy correct cycles, make sure that you follow the arrows round in the direction. Hess’s law calculate Look carefully at the state symbols Ø e.g. ΔH [H f when looking –1 O(g)] 2 = –241.8 kJ mol up enthalpy ΔH [H f –1 O(l)] = be used enthalpy to changes which changes, Ø but can cannot be determined directly by –285.8 kJ mol 2 experiment, e.g. ΔH from r ΔH of reactants and ΔH or f products. c 67 6.5 More enthalpy Learning outcomes An We On completion of this section, enthalpy can use be able construct using enthalpy cycle bond in Figure energies using bond energies. It is often 6.5.1 to easier , calculate however , to an enthalpy use the ‘balance to: sheet’ the cycle you change should changes Hess cycles using method (see Section 6.2). bond energies H r reactants explain of the term hydration, enthalpy ΔH bond hyd explain the term lattice products change of energy, energy bond reactants of energy products ΔH latt gaseous use Hess ΔH or cycles to atoms determine Figure 6.5.1 Δ H sol hyd Worked example Calculate ΔH 1 Ø for the complete combustion of ethene to carbon dioxide r and water . E(C=C) +610, E(C–H) E(C=O) H +740, +410, E(O=O) E(H–O) +496, +460 H H r H H E(C=C) + 4 × E(C–H) + 3 × E(O=O) 2 C(g) + 6O(g) + 4 × E(C=O) + 4 × E(H-O) 4H(g) Figure 6.5.2 Ø According to Hess’s law: ΔH = E(reactants) + E(products) are because r Note: The values of E(products) negative bond formation is exothermic. E(C=C) + 610 + 4 × E(C–H) + 3 × E(O=O) + 4 × + 4 × 410 3 × 496 + 4 × + Ø Δ (–740) + + 4 4 × × E(H–O) (–460) –1 = H E(C=O) –1062 kJ mol r Did you know? We can calculate theory from the Enthalpy lattice size energy and change of Standard enthalpy mole specied hydration change of and hydration: The lattice enthalpy energy change when 1 in of a gaseous ion dissolves in enough water to form an charge Ø innitely dilute solution under standard conditions. Symbol Δ H hyd of the ions and packed. This the way they sometimes are gives slightly 2+ Examples: Mg 2+ (g) + aq → Mg Ø (aq) ΔH –1 = –1920 kJ mol hyd Ø different values from the lattice Cl (g) + aq → Cl (aq) –1 ΔH = –364 kJ mol hyd energy obtained by experiment. Lattice ‘Lattice energy’ is a commonly energy: The enthalpy change when 1 mole of an ionic compound used Ø is formed from its gaseous ions under standard conditions. Symbol ΔH latt term and is often used to mean 2+ the same as lattice enthalpy. The difference between small we the two is Example: often use the 68 to mean the same + 2Cl (g) → CaCl (s) ΔH –1 = –2258 kJ mol latt so energies are always exothermic two when terms (g) 2 Lattice that Ca Ø thing. new ‘bonds’ are made. because energy is always released Chapter Ø Enthalpy change of of can calculate hydration the using enthalpy an change enthalpy of Energetics Ø solution from ΔH and ΔH latt We 6 hyd solution or the enthalpy change cycle. H latt gaseous ions ionic solid cation(s) H hyd H sol + anion(s) H hyd ions in aqueous solution Figure 6.5.3 Here is a worked Calculate the example: enthalpy change of solution of magnesium Ø the following data: –1 [MgCl ΔH latt –1 = –2592 kJ mol chloride Ø , using 2+ ΔH 2 [Mg ] = hyd Ø , –1920 kJ mol ] –1 ΔH [Cl ] = – 364 kJ mol hyd H latt 2+ Mg (g) + 2Cl Mg Cl (g) (s) 2 2+ H [Mg ] hyd H + 2 [MgCl sol × H [Cl ] 2 ] hyd 2+ Mg (aq) + 2Cl (aq) Figure 6.5.4 According to Hess’s law: Ø Ø ΔH [MgCl latt ] + ΔH Ø [MgCl sol Ø So ΔH 2 Ø [MgCl sol ] = [Mg [Mg Ø ] + 2 × ΔH hyd Ø ] + 2 × + 2 × [Cl ] hyd Ø ΔH [Cl ] − ΔH hyd Ø ΔH 2+ ΔH hyd –1920 So = 2+ ΔH 2 ] 2 [MgCl latt (–364) − ] 2 (–2592) –1 [MgCl sol ] = –56 kJ mol 2 The value also be of the found enthalpy from this change type of of hydration enthalpy the lattice the enthalpy change of solution the enthalpy change of hydration cycle for if a we particular ion can know: energy of one of the ions. Key points ΔH can be found by using a Hess cycle involving bond energies of r reactants Enthalpy and products. change of hydration, ΔH is hyd of gaseous Lattice ions energy, dissolves ΔH is in the water the enthalpy enthalpy change when 1 mole , to form a change very dilute when 1 solution. mole of an ionic latt compound A Hess is formed from cycle can be drawn its gaseous relating ions. the enthalpy changes ΔH , sol ΔH ΔH and hyd . latt. 69 6.6 Lattce energy: Learning outcomes On completion of this section, Some more In to order be able Born–Haber enthalpy calculate lattice cycle changes energy, we need to use ionisation energies, you enthalpy should the change of formation and two other enthalpy changes. These two to: are: explain the change of terms ‘enthalpy Standard atomisation’ mole ‘electron enthalpy change of atomisation: The enthalpy change when 1 and of gaseous atoms is formed from an element in its standard state afnity’ Ø under standard conditions. Symbol ΔH at calculate lattice energy from an Ø Examples: appropriate Born–Haber Na(s) → Na(g) ΔH –1 = + 107.3 kJ mol at cycle. 1 Ø Cl (g) → Cl(g) ΔH 2 2 –1 = + 122 kJ mol at Ø V alues of ΔH are always positive (endothermic) because energy has to be at absorbed to remember Electron 1 mole break that bonds the afnity: of holding enthalpy The electrons rst are the change electron added to a atoms of together . atomisation afnity mole of is the gaseous It is is per important atom enthalpy atoms formed. change to to form when 1 mole Ø of gaseous ions . X Symbol ΔH ea1 Ø Examples: Cl(g) + e → Cl (g) –1 ΔH = −348 kJ mol = −200 kJ mol ea1 Ø S(g) + e → S (g) –1 ΔH ea1 As with anions ionisation with energies, multiple successive electron afnities are to form charges: Ø O(g) used + e → O (g) –1 ΔH = −141 kJ mol = +798 kJ mol ea1 2– O (g) + e → O Ø (g) –1 ΔH ea2 Exam tips It is important enthalpy that change you put the calculations correct to nd Ø is generally negative but ΔH ea1 be is always is being added to a negatively charged and ΔH values are always positive. at Ø ΔH are always negative. latt Ø 70 when Make you sure carry you because ion. The Ø values values f +) positive, overcome. Ø ΔH or out know these. the incoming ea2 electron ΔH (– energy. Ø ΔH to signs lattice may be positive or negative. repulsive forces have Chapter Born–Haber We can lattice of Energetics cycles construct energy 6 an an enthalpy ionic cycle and use Hess’s law to determine the solid. ions H in latt ionic gaseous solid state H H f (several steps) elements in standard their states Figure 6.6.1 Ø ΔH consists of several steps: 1 atomise metal → atomise ionise Ø So for NaBr: ΔH Ø = ΔH 1 Hess’s law: + ΔH + = can show lattice energy these cycle . of = ΔH ΔH [Na] + ΔH i1 [Br] ea1 Ø − ΔH f changes The sodium + Ø ΔH all Bor n–Haber Ø Ø [Br] → f latt a atoms ΔH latt Ø We metal Ø ΔH 1 So ionise at Ø ΔH → atoms Ø [Na] at Ø By non-metal non-metal on 1 an energy Born–Haber bromide as cycle shown level can in diagram. be used Figure to This is called determine the 6.6.2. + Na (g) + Br(g) + e H [Br] ea1 –325 + Na H (g) + Br (g) [Na] i1 +494 H [ at Na(g) + Na(g) + Br(g) ] Br 2 1 +112 H 2 Br (g) 2 Exam tips [Na] at H 1 Na(s) +107 + 2 [Na Br] latt Br (l) 2 It H [Na is easier to determine lattice Br] f energy –361 Na the Figure 6.6.2 by calculating the sum of Br(s) atomisation energies, ionisation A Born–Haber cycle to determine the lattice energy of sodium bromide. All energies and electron afnities –1 gures are in kJ mol Ø rst (ΔH Ø ). Then apply ΔH 1 Ø ΔH ΔH 1 From the enthalpy change Ø see that: Ø = ΔH values ΔH 1 shown on Ø [Na] + ΔH at the Born–Haber + ΔH we [Na] + ΔH i1 + 107 + 112 + 494 make + (–325) kJ mol ΔH Ø = ΔH latt likely to f mistakes confused by with too the signs many or enthalpy –1 = +388 kJ mol changes 1 Ø less [Br] –1 Ø = are ea1 be ΔH . You can Ø Ø [Br] at cycle = latt Ø – this way. Ø − ΔH f 1 –1 = 361 − (+388) = −749 kJ mol Key points Enthalpy change of atomisation, ΔH , is the enthalpy change when 1 mole at of gaseous Electron atoms afnity, is formed from ΔH is the the element. enthalpy change when 1 mole of electrons is ea added A to 1 mole Born–Haber changes ΔH , latt of gaseous cycle ΔH , f is an ΔH , at atoms. enthalpy ΔH and cycle which includes the enthalpy ΔH ea 71 6.7 More about Learning outcomes lattce The energy Born–Haber cycle for magnesium oxide + Magnesium On completion of this section, 2 should be able . So understand how Born–Haber with and the radius to cycles multiple explain an ion with a 2 when constructing a charge. Born–Haber The cycle oxide for ion has magnesium a charge oxide of we to: have forms you involving the take into account: of ionic the rst and the the rst and second second ionisation energies of magnesium and ions charges effect on to construct charge magnitude of The enthalpy magnesium cycle oxide electron and are afnities Born–Haber shown below. of cycle The oxygen. to calculate relevant the lattice enthalpy energy changes (all of in −1 lattice ) kJ mol energy. are: Ø ΔH Ø [MgO] = −602, ΔH f = +148, ΔH = +736, ΔH i1 [O] = +249, at Ø Ø [Mg] ΔH Ø [Mg] at Ø [Mg] = +1450, ΔH i2 [O] = −141, ie1 Ø [O] ΔH = +798 ea2 H 2+ Mg latt 2– (g) + O Mg O(s) (g) H several H steps f 1 Mg(s) + O 2 Figure 6.7.1 (g) 2 Enthalpy cycle to determine the lattice energy of magnesium oxide 2+ Mg 2– (g) + O H 2+ Mg (g) ea2 (g) + O(g) + 2e 2+ H Mg ea1 (g) + O O(g) + e (g) + e + H Mg [Mg] i2 (g) + Exam tips H Mg(g) [Mg] + O(g) i1 Take care with the signs for 1 H Mg(g) [O] + at the rst and second 2 O (g) 2 electron 1 H afnities. When added they positive may give a Mg(s) [Mg] at together value. H + 2 O (g) 2 [MgO] f MgO(s) Remember that an arrow, ↓, Figure 6.7.2 pointing energy downwards is given out means and an arrow, From ↑, pointing upwards Born–Haber cycle to determine the lattice energy of magnesium oxide that means the enthalpy Ø energy is absorbed. Take change values given above we can see that: that care to ΔH Ø = ΔH 1 Ø [Mg] + ΔH at Ø [O] + ΔH at [Mg] Ø + ΔH i1 [Mg] Ø + ΔH i2 Ø [O] + ΔH ea1 [O] ea2 Ø draw these correctly if there is a ΔH = (+148) + (+249) = +3240 kJ mol + (+736) + (+1450) 1 −1 second electron afnity involved. Ø ΔH Ø = ΔH = –602 latt Ø − f ΔH 1 −1 72 − (+3240) = –3842 kJ mol + (−141) + (+798) Chapter Born–Haber cycles for AlCl , Na 3 O and Al 2 6 Energetics O 2 3 3+ Aluminium chloride, AlCl , has an Al ion and 3 Cl ions. So in the 3 calculation: Ø ΔH Ø [Cl] and ΔH at [Cl] must be multiplied by 3. ea1 3+ The energy required to form Al Ø (g) from Al(g) is Ø ΔH + + Sodium( i) oxide, Na O, has 2 Ø ΔH i1 + ΔH i2 i3 2− Na ions and 1 O ion. So in the + ΔH 2 calculation: Ø ΔH Ø [Na] and ΔH at [Na] must be multiplied by 2. i 2− The energy required to form O Ø (g) from O(g) is Ø ΔH ea1 3+ Aluminium oxide, Al O 2 , has 2 ea2 2− Al ions and 3 O ions. So in the 3 calculation: Ø Ø [Al] ΔH must be multiplied by 2 and ΔH at [O at ] must be The energy required to form Al multiplied Ø (g) from Al(g) is ΔH 2− The energy required to form O + ΔH from O(g) is Lattice affects the value of energy oppositely The the arises charged size of the opposite smaller less from ions. ions. charge ions. So, exothermic electrostatic value Small than for as the The any the lattice ions large given size of are ions. + ΔH of attracted The = effect is e.g. = The seen charge we to on compare sized ions, charged the ions the we ions of if the LiCl ions. the = that very the is lattice to ions higher energy of on gets increases: −1 −780 kJ mol cation = Ions size is KCl kept = −711 kJ mol constant: −1 −848 kJ mol with opposite lattice see is strongly density −1 −1031 kJ mol strongly more , Cl cation −1 LiF between −1 NaCl −848 kJ mol same attraction charge −1 LiCl ea2 on: The anion the i3 energies? force depends ΔH Ø ΔH ea1 What Ø + i2 Ø (g) 3. Ø i1 by 2 3+ a high charge energies of LiF charge than and the lattice energy much higher than LiBr of LiF ions = are −803 kJ mol attracted with a low MgO, which have MgO which has which has singly more charge. If similar doubly charged ions: Ø −1 [LiF] ΔH = Ø −1049 kJ mol The of way the [MgO] = −3923 kJ mol latt the ions however , −1 ΔH latt ions also than are ar ranged affects ionic the in value charge and of the the ionic lattice. lattice energy. The It arrangement has less effect, size. Key points When constructing successive Born–Haber values for ΔH and i The value lattice of lattice energy is energy more cycles ΔH involving must be used ions with when multiple charges, appropriate. ea depends exothermic if on the the size ion is and charge smaller and of the has a ions. The higher charge. 73 Exam-style Answers to all exam-style questions can questons be found on the – Module 1 accompanying CD 3 10 Multiple-choice questions of 25.0 cm aqueous ron(ii) 3 20.5 cm 1 What s the relate atomc mass (A ) of sulphate requred –3 of 0.02 mol dm potassum manganate(vii) bromne, r for complete reacton. The onc equaton for the 79 gen that the relate abundance of Br s 50.5% reacton s: 81 and Br s 49.50%? 2+ 5Fe + – (aq) + MnO (aq) + 8H 3+ (aq) → 5Fe (aq) + 4 A (79 × 0.495) – (81 × 0.505) 2+ Mn (aq) + 4H O(l) 2 B (79 × 0.495) + (81 × 0.505) –1 What C (79 × 0.505) – D (79 × 0.505) + (81 × s concentraton (g mol ) of the soluton of 0.495) ron(ii) (81 the × sulphate? 0.495) (A alues: Fe = 56; S = 32; O = 16) r 2 Whch of the followng partcles wll be deflected to 25.0 1 _____ A the greatest degree by an 0.02 __ × × 20.5 A Alpha B Protons partcles C Electrons D Neutrons 152 × 152 5 20.5 1 _____ B 0.02 __ × × 25.0 5 20.5 52 X 3 × electrc eld? s a radosotope. Whch of _____ the followng C 26 0.02 × represents the resultng undergoes beta nuclde, Y, when × 5 × 152 × 5 × 152 25.0 52 X 26 25.0 52 D 52 Y A 48 B Y 27 4 _____ decay? C Whch of s the 20.5 D Y 24 the followng × 56 Y 28 0.02 28 electronc conguraton Structured questions for the copper(i) A [Ar] 10 1 [Ar] 9 4s 3d 9 C on? 3d B [Ar] 3d 1 2 11 4s a 4s D [Ar] Descrbe what Whch of the followng equatons represents onsaton energy of A A Na(g) → Na B Na(g) → Na + + e + 2e C Na D Na 2+ (g) → Na – (g) loss of potassum + has of an be used + 2+ (g) → Na to nd the concentraton the followng e of e peroxde. The two half equatons are – (g) + 2e – + + 8H – + 5e 2+ → Mn + 4H H successe O 2 + element to manganate(vii) 4 An gan [2] soluton MnO 6 or number – (g) hydrogen + of sodum? – (g) terms oxdaton agent. standard can + n n the b second change 3d oxdsng 5 happens electrons AND 10 O 2 → O 2 + – 2H + 2e 2 –1 onsaton energes measured n kJ mol : i Usng the 1st 736 4th 10 500 balanced 2nd 1450 5th 13 600 between 3rd 7740 group of the Perodc Table s the By to belong V B Whch hae of the IV C the followng largest 8 KF B Whch of lattce III haldes a and [2] to the changes n numbers of the elements, show that D of manganate(vii) s the oxdsng II potassum would and hydrogen peroxde s the reducng agent. [4] energy? KCl C the followng KBr descrbes D the shape KI of When aqueous soluton of the soluton was bromne was potassum added halde, to KX , an a aqueous brown obsered. the 2– carbonate wrte reacton element c A the to? agent 7 aboe, show mangangate(vii) reference potassum A to peroxde. makng oxdaton lkely equatons potassum hydrogen ii What half equaton i Wrte the formula ii Usng your of the halde on, X. [1] ? on, CO 3 A octahedral C pyramdal B trgonal D tetrahedral balanced planar aqueous 9 What s the molar mass of 0.07 g of a gas answer n part equaton for bromne and the i aboe, wrte reacton a between KX. [2] whch iii Whch of the elements, or X Br 2 , has the 2 3 occupes a olume of 120 cm at r.t.p.? greater oxdsng ablty? Explan your answer 3 (1 mole of gas occupes a olume of 24 dm at r.t.p.) usng A 74 14 B 16 C 28 D 32 an approprate half equaton. [4] Module 12 a State THREE t b relates The knetc deal i to assumptons theory howeer State the ii a does the knetc deal theory an all exst under gases n whch behaour of t 71. 11% are and aganst gas P on the Use Why b i the deal of gas moles equaton of a gas to excess of C, oxygen 2.22% of H and used? [1] ‘emprcal formula’ and ‘molecular If the molar mass ts of oxalc acd emprcal formula s , 90 g mol and molecular formula. the c The [6] concentraton ttratng the occupes [2] calculate same calculate whch s Dene ii [2] number 26.67% –1 why 2 i contan questions formula’. explan showng and CO to Exam-style of O. a gases axes. c s found [4] PV deal as [3] realty. occur. graph cures for that not condtons deatons Sketch assumes ths deate from the of gases. 1 t wth a manganate(vii) a 2 moles of of oxalc standard acd soluton. They potassum can soluton be found of react n manganate(vii) by potassum a rato to 5 of moles 3 olume of R = at 0.072 dm –1 8.31 JK –48 °C and 3.4 atm. of oxalc acd. In one such ttraton t was found –1 3 mol [5] that t requred an aerage olume of 21.5 cm of –3 ii If the mass of the gas n part aboe s 0.02 mol dm 0.94 g, potassum manganate(vii) to react 3 calculate ts relate molecular mass. wth [1] i 13 a State Hess’s law. 25.0 cm Descrbe The enthalpes and water of formaton of carbon what you acd. of the would obsere at the end reacton. [1] doxde ii –1 are oxalc [1] pont b of Calculate the number of moles of oxalc acd –1 and –393 kJ mol –286 kJ mol used n the ttraton. [3] respectely. –3 iii i Use ths nformaton to draw an Calculate of cycle dagram for the standard enthalpy of ethane H (C 2 the oxalc Calculate the standard concentraton ) (mol dm [1] ). Usng the molar mass gen n part aboe, [2] 6 calculate ii molar acd. of iv formaton the energy enthalpy the mass concentraton of the oxalc of formaton acd. of c A ethane. student was experment of soluton n of requred the lab to to carry out determne ammonum 15 an the ntrate. The enthalpy The rst 3 are gen a polystyrene cup, a onsaton energes of the elements n Perod as follows: student Symbol was [1] [2] measurng of Na Mg Al S P S Cl Ar 496 738 578 787 1012 1000 1251 1520 cylnder element and i a thermometer. Why used was n a polystyrene the cup and not a beaker experment? [1] Frst onsaton energy/ ii State one assumpton that was made when –1 kJ mol performng the calculatons for ths reacton. iii Gen the [1] that water the was temperature dagram, ntal temperature 27 .3 °C 15. 1 °C, clearly and draw showng the nal an was endothermc or Dene ii Usng an prole Usng the nformaton n exothermc. part iii term ‘rst magnesum equaton energy of onsaton (Mg) llustratng the as an energy’. example, the rst [1] wrte onsaton element. [1] i Usng the table, state what the general trend [4] n iv the the b reacton i was energy whether a of aboe, the alue we go of across the rst the onsaton perod and energy account for s as ths –1 calculate when the 16.0 g enthalpy of of soluton ammonum ntrate n kJ mol general dssoles trend. [3] n ii Study the table and account for any 3 100 cm of [4] water. rregulartes (A values: N = 14; H = 1; O =16; C = 12. r The enthalpes onsaton of n the energes general across trend the of the rst perod. [5] –1 formaton of carbon doxde and water are –393 kJ mol c i Sketch a graph to show the rst e –1 and –286 kJ mol respectely. The standard successe combuston onsaton energes of magnesum –1 of ethane s –1560 kJ mol ). aganst ii 14 Ethanedoc acd, commonly called oxalc How the do an organc acd that occurs naturally n these alues help anmals and oxygen. When s made oxalc of acd s carbon, burnt hydrogen n electrons of remoed. successe [2] to dentfy the group onsaton to whch plants magnesum and of acd energes s number excess belongs. [3] and oxygen 75 7 Rates 7 . 1 Following of Learning outcomes reaction the Rate of A On completion of this course section, study be able design reaction reaction of the rate kinetics, of we reaction carry is out called reaction experiments (and carry out) kinetics. measure In order the rate to at study which are used up or products are made. suitable change experiments for affect in concentration studying factors (or which to to: reactants a you reaction should of reaction amount) of reactants or products rate. rate of reaction = time Rate of reaction −3 is in this change concentration terms as −1 Following the method nature expressed for s mol dm The usually taken of used the continuous course of to follow products methods and and the a reaction progress reactants. sampling of The a reaction methods depends fall into on two the groups: methods. Continuous methods A particular period gas syringe of physical property of the reaction mixture is monitored over a time. Measuring volume of gas given off The volume time of intervals a gas given using a gas off in a syringe reaction or can upturned be measured measuring at various cylinder initially delivery full of water . For example in the reaction of calcium carbonate with tube rubber bung hydrochloric various acid, times as the the (s) CaCO volume reaction + of carbon 2HCl(aq) → CaCl 3 marble hydrochloric dioxide can be measured at proceeds. (aq) + CO 2 (g) + H 2 O(l) 2 acid chips Figure 7.1.1 Colorimetry The reaction of calcium This is suitable for reactions in which there is a change in the intensity carbonate with hydrochloric acid can be of colour during the reaction. For example in the reaction between iodine found by measuring the volume of carbon and dioxide given off at various times propanone in aqueous COCH CH 3 The ‘cell’ light iodine is (aq) acidic + I 3 brown solution: (aq) → CH 2 in colour COCH 3 but the I(aq) + HI(aq) 2 other reactants and products are sensitive colourless. As the reaction proceeds, the intensity of the brown colour cell fades. by A or We can use transmitted lter is chosen a colorimeter through so that the the to measure reaction correct the amount of light absorbed mixture. wavelength of light falls on the meter reaction light filter source Figure 7.1.2 76 mixture reaction mixture, mixture light-sensitive A colorimeter proceeds the in more more cell. the light As light is cell. is the The more absorbed reaction transmitted intense and less between through is the of transmitted iodine to colour the and the reaction through to propanone light-sensitive cell. the Chapter If 7 Rates of reaction Changes in electrical conductivity the total reaction, number we can conductivity . mixture. mixture We This of use uses occurs. ions follow a in the a reaction reaction by conductivity alternating Conductivity meter current is mixture changes measuring so to monitor that measured in no during changes in the reaction electrolysis siemens, a electrical of the S. Did you know? dipping conductivity Hydrogen electrode ions have ions very conductivities ions. This measured meter present hydroxide electrical compared means conductivity reaction and high changes even as that if can there reactants with other electrical be are or other ions products. + Hydrogen mixture ions are really H O ions 3 Figure 7.1.3 (see A conductivity electrode connected to a meter Section appear An example of a reaction that can be followed by this method by is: to ‘chain thought 9. 1). move Hydrogen extremely conduction’ to be – ions rapidly they transmitted are rapidly + (CH ) 3 CBr(l) + H 3 O(l) → (CH 2 ) 3 COH(aq) + H (aq) + Br (aq) 3 + between H O ions and water 3 As the reaction increases as proceeds, hydrogen the ions electrical and conductivity bromide ions are of the solution molecules. formed. Other methods There may be a change in pH as some reactions proceed. This may be + ions due to H can be use Some to or OH ions monitor these reactions can reaction mixture not accurate very as be a being type followed gas unless is produced of by gas consumed. A pH meter reactions. measuring released. the or has This a the can be relatively decrease done high by in mass of weighing, molar the but is mass. The sampling method Small and samples analysed, after the are samples proceeding, removed usually for It is cooled A chemical are from using a taken, the reaction specic they mixture chemical must be at reaction. quenched to particular times Immediately stop the reaction example: rapidly so that the reaction slows down signicantly. Key points does not can affect be added the which substance to prevents be the reaction continuing but analysed. A catalyst can be The course of be followed An example is the reaction of propanone with iodine in acidic particular solution: or CH COCH 3 (aq) + I 3 (aq) → CH 2 COCH 3 sample is taken at reaction can I(aq) + a particular time and by measuring property of a a reactant product. HI(aq) 2 A a removed. added to sodium Suitable methods of following a carbonate reaction directly changes in are measuring + solution. reaction. Sodium The carbonate reaction reacts mixture with can the then be H ions titrated which with catalyse the to determine the concentration of i volume of gas, ii sodium electrical thiosulphate the conductivity, iii colour iodine. intensity Some by at or iv pH. reactions taking a various reaction can be followed sample for time analysis intervals as the proceeds. 77 7 .2 Calculating Learning outcomes rate Reaction For On completion of this section, many be able describe with rate and reactions, progress of reaction rate a changes reaction magnesium reacts with dilute as the reaction hydrochloric acid, how reaction rate varies the hydrochloric acid falls and the volume of hydrogen gas given off time + 2HCl(aq) → (aq) MgCl + H 2 know rate When concentration rises: Mg(s) proceeds. the to: of reaction you excess should of how to calculate graphically. (g) 2 reaction We use molar square brackets concentration of to indicate molar hydrochloric concentration. So [HCl] means acid. −3 [HCl] mol dm 0.8 Time/s 0.6 0 0.4 15 0.2 43 0. 1 100 180 b a H 2 f o emulov [HCl] 0 0 0 Figure 7.2.1 time 0 (s) time (s) The progress of the reaction beween Mg and HCl can be followed: a by measuring the decrease in concentration of HCl or; b the increase in volume of H 2 −3 We can see that [HCl] decreases from 0.8 to 0.6 mol dm in the rst 15 seconds. ∆ 0.8 [HCl] _______ We can write this ∆ The graph decreasing measure i.e. at point is the 70 s. on a curve with the very can curve 0.6 −3 If we gets small, 0.013 mol dm −1 s 15 shallower make calculate (see = time which time. rate We – _________ = as we time obtain this Figure the rate a by with time. interval rate at a drawing So over the particular a rate which tangent point at a 7.2.2). 0.6 md lom 3– 0.4 ]lCH[ 0.2 0 20 40 60 80 100 120 time Figure 7.2.2 78 (s) 140 160 180 in time particular 0.8 0 is we 200 Chapter T o calculate Select Draw (α) a a on straight the the Calculate of reaction the line line at at a particular corresponding this point (the to Rates of reaction time: a particular tangent) so that time. the two angles same. tangent convenient rate point look Extend the 7 line to meet the axes of the graph This is the rate (or at values). the gradient of the graph. of reaction. 0.54 _____ In the example −3 above, −3 10 mol dm The negative the rate of reaction at 70 s is = −3.4 × 157 −1 s [HCl] sign shows that the rate is decreasing with time. 0 Initial rate of reaction 0 The of initial the reaction is calculated by drawing a tangent at the start can nd by Looking at proceeds, rate rate during how reaction drawing the the at a (s) Drawing a tangent to show of in varies at with several Figure reaction particular reaction rate tangents gradients rate a hydrochloric Figure 7.2.3 time initial rate of reaction in reactant each of curve. Changes We rate 7.2.4, decreases. time the concentration different we In corresponds can points see Figure to a on that as 7.2.4(a) particular of the a the we specic graph . reaction can see that concentration of acid. a b 0.8 V final H 2 ]lCH[ f o emulov 0.6 0.4 0.2 V t t time Figure 7.2.4 time a The rate of change of [HCl] and; b rate of volume change of H both 2 decrease with time In rate experiments particular results of products example reactant, experiments to in show are interested than in how product, measuring what Figure of V we rather happens increase to the the varies in concentration with time. We concentration concentration of of can or a use the volumes reactants. of For 7.2.4(b): hydrogen – V nal of hydrogen is proportional to [HCl] at time t t Key points For To most reactions, calculate graph, rate tangents the specic the tangent. of are rate of reaction reaction drawn times). The at at rate decreases specic specic of the times from points reaction as is on given a the by reaction proceeds. concentration–time graph the (corresponding gradient (slope) to of 79 7 .3 Collision Learning outcomes On completion should be able of this theory explain rate of reaction Collision theory section, In order to break to react, the reactant particles must collide with enough energy you specic bonds in their molecules. They must also collide with to: the and the effect concentration, of correct react particle size on rate of together come so into that the parts of each molecule that are going to contact. and The temperature orientation minimum energy that colliding particles must have in order to react reaction is called the activation energy , symbol . E At this minimum energy, the a explain the effect of catalysts, molecules including enzymes, on the rate have a particular conguration called the transition state of reactions The describe the role of enzymes and concentration, pressure and surface area in on industrial effect of rate of reaction biological Concentration processes. The more concentrated (molecules or ions) in a a solution, given the volume. greater is the number There are more increases. So, rate of chances particles of collisions a occurring. transition state collision frequency of reaction increases. Pressure ygrene This E The is relevant for gaseous systems. The higher the pressure, the greater a is the number of molecules in a given volume. There are more chances of reactants collisions products reaction pathway reaction transition The collision frequency increases. So, rate of increases. Size of The b occurring. solid greater particles the total surface area of a solid, the greater is the number of state particles gas that are molecules. If exposed we to break collide a large with lump molecules of solid in into a solution smaller or pieces with we ygrene E a expose products react a larger faster surface than larger area. So, if the mass is the same, smaller particles particles. reactants Effect of temperature on The reaction curve particles in any gas or solution are moving at different speeds. The against the Activation energy for: a an exothermic reaction; b an endothermic reaction reaction pathway The Figure 7.3.1 Boltzmann distribution rate of energy of fraction a of particle depends particles distribution that on have a its speed. particular A graph energy of is energy called a Boltzmann cur ve average energy )E( fo with of particles sufficient energy to htiw noitcarf ygrene selcitrap number react E a Figure 7.3.2 Boltzmann distribution curve. The area under the graph represents the total energy The shaded energy energy. 80 to area react shows when number of particles (E) the they proportion collide, i.e. of molecules equal or that greater have than enough the activation Chapter The effect of temperature on Increasing slightly. the An temperature increase molecules. The much. the But activation in Figure the reaction rate kinetic proportion 7.3.3). of increases If there energy, changes the temperature average energy activation in rate of are there more will shape with (see the more distribution energy does energy the particles be of the however , molecules markedly Rates of reaction reaction increases energy, 7 not with to area energy all curve the increase equal shaded successful of or above under equal to collisions very the or and the graph above the increases. T > T 2 1 T 1 There are )E( a E T at a T more greater than particles energy than at T 2 1 2 fo Did you know? E a htiw noitcarf ygrene selcitrap with Most Figure 7.3.3 Boltzmann distribution curve at a lower 1 (E) enzymes they above are left become at 40 °C for inactive temperatures too long. The and a higher temperature, T energy if weak attractive forces holding the temperature T 2 protein are chain in overcome its and correct the position protein Catalysis changes Catalysts path (a speed up different activation energy successfully the rate of mechanism) and leads to a therefore reaction with by lower greater providing activation proportion of a different energy. the reaction particles longer to Lower colliding reacting. b enzymes are = uncatalysed a cat = catalysed a ygrene a fo ygrene E E cat a reaction E however, tangled E pathway energy up which is Increasing size) are protein substrate they catalysts. They have and bound catalyse. They an catalyses to the This catalyse active the site most part reaction. enzyme). is on because (A Enzymes the of of the reactions their surface substrate are very substrate ts is a rather in the in the like a key ts a lock. This allows the bonds which are and formed to be in their in correct reaction the frequency of particle temperature rate because particles have than activation the rate of increases. Increasing increases more energies greater energy. active to The Boltzmann distribution be curve broken chain others. decrease increases reaction reactant specic into in which site with rapidly protein concentration (or collisions loosely reactions protein the a A catalyst lowers the activation energy of a reaction; b A greater number organisms. the about E Enzymes bonds the many at Key points a because living the to function because no cat of particles have energies greater than or equal to the activation energy Enzymes In substrate, temperatures. Above reactant Figure 7.3.4 able is conformation reaction. excess site a htiw noitcnarf E gets active correct the of denatures E E energy the presence 60 °C, selcitrap activation in catalyse higher a shape. The shows the relative number positions. of particles having particular energies. active site protein Catalysts increase the rate substrate of Figure 7.3.5 reaction by providing an The substrate ts the active site alternative reaction with activation pathway of an enzyme similar to a key tting a lock Many the enzymes are manufacture citric of important drugs and in processes in the such as manufacture baking, of brewing chemicals and such as lower Enzymes catalyse are proteins specic energy. which reactions. acid. 81 7 .4 Rate equations Learning outcomes Introducing In On completion of this section, Section be able understand reaction’, ‘rate This saw how corresponded to to a calculate particular the rate of reaction concentration of at a particular reactant. If we to: plot we equations you time. should 7.2, rate the ‘rate terms ‘order equation’ of and a graph of rate of reaction of Mg with acid against [HCl], we get a 2 curve. line But if we showing a plot the rate proportional of reaction against we [HCl] get a straight relationship. constant’ interpret concentration–time b md lom( 3– reactions. s 1– order md lom( 3– second ) graphs for s and a 1– zero, rst and ) concentration–rate etar etar 2 [HCl] [HCl] 2 Figure 7.4.1 Rate of reaction of HCl with Mg plotted: a against [HCl]; b against [HCl] 2 W e can write this proportionality The proportionality The overall The power the rate expression to case, hydrochloric How do The which equation particular order equation. example constant (in we is the called called the case that 2) of reaction can for rate the is only the cannot be of a conduct By be found (g) is order second reactant of the order rate k[HCl] is raised reaction. with In respect to determined by at [H ] from conducting a the stoichiometric series of experiments. For + 2NO(g) → 2H a O(g) + N varying the (g) 2 concentration of each of the time. keeping [NO] constant, experiments show that constant, experiments show that 2 rate = ]. k[H 2 Did you know? By varying [NO] keeping [H ] 2 2 rate The rate equations for reactions or ions may not include present in = k[NO] some compounds the Combining these two rate equations we get the overall rate equation: chemical 2 rate = k[H ] [NO] 2 equation. order of In some reaction instances, can the be fractional. We say 82 that the reaction is rst order with respect to H , 2 with respect in this equation? 2 experiments one varying = constant particular the 2 reactants reaction reaction: 2H We of equation called reaction rate rate acid. we deduce the It the concentration this say is mathematically: to NO and third order overall. second order Chapter How We can can we deduce the order of deduce the order of reaction plotting a graph of reaction plotting a graph of concentration The shape of these graphs is rate with Rates of reaction reaction? respect against of a 7 to a single concentration reactant against reactant of by: reactant time. characteristic. a b 2nd order k[A] rate = order rate = k[A] order lom( etar lom( 1st k zero md order 3– zero ) md 3– fo A fo noitartnecnoc = s 1– ) noitcaer 2 rate 2nd order (deep curve levelling out) –3 concentration Figure 7.4.2 of A (mol dm ) time (s) a How concentration of reactant A affects reaction rate; b How concentration of reactant A changes with time Units of The For units the k of k are reaction: different NO(g) + for different CO(g) + O orders (g) → of NO 2 reaction. (g) + CO 2 For example: (g) 2 2 the rate Step 1: equation is: Rearrange rate the = rate Exam tips k[NO] equation in terms of k: is easy to confuse units in section. The following facts rate ______ k It 2 important: –3 2: Substitute the are = [NO] Step this units of rate and concentration: Units of rate are mol dm –1 s –1 –3 mol dm (don’t forget –1 the s ). s ___________________ = –3 mol dm –3 × mol dm Units of k for zero –3 are Step 3: Cancel the mol dm order reactions –1 s units: –1 Units of k for rst order reactions s ___________________ = –1 –3 are –3 × s mol dm 1 Step 4: Rearrange with the positive index rst: _________ 3 is the same as dm –1 mol –3 mol dm 3 = dm –1 mol –1 s Key points Rate of reaction is related to concentration of reactants by the rate equation. m The general form constant, orders Order [A] with of of and [B] respect reaction concentration or the rate are to A can be equation is concentrations and rate of = k[A] n [B] reactants where and m k and is n the are rate the B. determined from concentration against graphs of reaction rate against time. 83 7 .5 Order of Learning outcomes reaction from Order of For On completion of this section, a zero be able deduce order appropriate interpret graphs of reaction from to So second data a reaction, given the a reactant initial order method rate–concentration from obtained the when decomposes rate reactions, concentrations concentration–time and order initial graph of rates shows a concentration constant rate against of time decline with (see Section to: 7.4). reaction from data you respect should experimental in of the results of a of given hydrogen the reaction initial is rate several reagent. of an can by the be calculated experiments Figure peroxide presence given at using 7.5.1(a) four gradient. by and tangent different starting the initial catalyst. rst the shows different enzyme For The results concentrations initial rates are graphs then perform calculations to nd k plotted against the initial concentrations of (Figure 7.5.1(b)). This shows that rate = O k[H 2 rate of reaction by in the understand rate the ]. So the peroxide reaction is rst 2 substituting order values hydrogen or with respect to hydrogen peroxide. equation meaning of –3 mol dm s 1– 30 ) 0.40 ‘half-life’. 0.6 dm –3 0. 12 mol dm 10 3 mol 0.3 0.2 etar 0.06 0.4 fo O fo emulov 2 noitcaer mol mc( 3 0.20 O 2 mc ( 3 ) –3 20 0.5 dm 0. 1 0 0 0 50 100 150 time 200 0 0.2 [H O 2 Figure 7.5.1 0.4 –3 (s) a The initial rate of decomposition of H O 2 O b A plot of initial rate of decomposition of H 2 ] mol dm 2 at different concentrations; 2 against [H 2 O 2 ] shows the reaction to be 2 rst order Deducing We can a value for the deduce concentrations from the acid the and overall initial catalysed rate of rates. rate constant reaction The hydrolysis from table of limited below methyl data shows data showing obtained methanoate: + H HCO CH 2 + H 3 O –3 [HCO CH 2 → HCO 2 + ]/mol dm [H H + CH 2 OH 3 –3 –3 ]/mol dm Initial rate/mol dm –1 s 3 –3 0.50 1.00 0.54 × 1.00 1.00 1. 10 2.00 1.00 2.25 2.00 2.00 4.48 10 –3 × 10 –3 × 10 –3 × 10 + Doubling the concentration of HCO CH 2 doubles So rate the is , keeping [H ] constant, 3 rate. rst order with respect to CH [HCO 2 ]. 3 + Doubling the concentration of H keeping [HCO CH 2 doubles the rate. + So 84 rate is rst order with respect to [H ]. ] 3 constant also Chapter 7 Rates of reaction + The overall rate equation is: rate = k [H ][HCO CH 2 The rate constant experimental from the can values table be into deduced the rate by substituting equation. ]. 3 any T aking of the × Half-life Half-life half its is sets line of of values above: –3 0.54 the rst –3 = 10 k (0.50) × and order of the time original taken value. (1.00) So k = 1.08 × 10 3 –1 dm mol –1 s reaction for the Symbol: concentration of a reactant to fall to t 1/2 Figure 7.5.2 shows decomposition half-lives are of several successive hydrogen constant. half-lives peroxide. This is Y ou can characteristic for see of a the enzyme-catalysed that rst successive order reaction. 0.4 0.3 H[ 2 O 2 ] 0.2 Did you know? 0. 1 The half-life for may 22s 23s used a rst to nd order the rst reaction order 21s rate 0 0 be 20 40 constant using the relationship: 60 0.693 time ______ t = 1/2 Figure 7.5.2 The successive half-lives for the decomposition of H O 2 are constant (about 2 k 22 s). So the reaction is rst order with respect to hydrogen peroxide For zero order order reactions reactions the the successive successive half-lives half-lives decrease. second ]A[ ]A[ t t time Figure 7.5.3 For increase. time a Successive half-lives for a zero order reaction decrease; b Successive half-lives for a second order reaction increase Key points Order of reaction experiments the reaction The rate rate ii can be found determining the by i measuring change in initial rate concentration in of several a reactant as proceeds. constant can be found by substituting relevant values into the equation. Half-life In a rst is the order time taken for reaction, the half-life concentration is independent of of a reactant the to halve. concentration of reactant. 85 7 .6 Reaction Learning outcomes mechanisms The rate determining Chemical On completion of this section, be able We understand the term ‘rate determining step’ deduce call this the in reaction organic chemistry mechanism . Each occur of in these a number steps of may to: involve especially you steps. should reactions, step possible mechanism determining reaction mechanisms from intermediates reaction step which is are called controls highly the the rate overall reactive. The deter mining rate of slowest step . reaction. So step The in in the rate the reaction sequence, appropriate Step 1 Step 2 Step 3 data. Reactant A + → Reactant Intermediate X → slow The overall reactants The rate to in Example Aqueous the of reaction that rate 1: The sodium reaction order is in the suggests step. So hydroxide ) on Step rate 1, the conversion 3 Chemists First CBr + OH with to → respect OH are usually (CH ) ions. to So COH + Br 3 2-bromo -2-methylpropane rate = k[(CH ) ion not is not react involved directly in with the OH and CBr]. 3 rate determining ions. 3 think step: OH does those 2-bromo -2-methylpropane: 3 order the of 2-bromo-2-methylpropane hydrolyses CBr equation 3 respect that ) (CH C fast 3 This Product step. hydrolysis of rst with depends appear 3 zero → X. determining (CH The Y fast intermediate substances involved Intermediate B that the ionisation of reaction occurs in two steps: 2-bromo -2-methylpropane slow + (CH ) 3 CBr → (CH 3 ) 3 C + Br 3 + Second step: reaction of (CH ) 3 C with OH 3 fast + (CH ) 3 The on slow the step rate at determines which (CH rate with 86 equation the because intermediate. + OH → (CH 3 the ) 3 the C ) 3 overall CBr rate ionises. of COH 3 reaction. Hydroxide So ions the do rate not depends appear 3 they are involved in the rapid ionic reaction in Chapter Example 2: The Propanone reacts reaction of with iodine propanone in the with presence of an 7 Rates of reaction iodine acid catalyst. The + catalyst provides ions H for the reaction. + H CH COCH 3 + I 3 → CH 2 COCH 3 I + HI 2 + The rate equation for this reaction is: rate = k[CH COCH 3 A proposed mechanism ][H ] 3 is: + OH O Exam tips fast + step 1: CH C CH 3 H + CH 3 3 3 intermediate A You need learning + OH not any worry of about these OH mechanisms at present. In rate slow + step CH 2: CH 3 3 C CH 3 + H of 2 reaction questions you will be intermediate A given about OH fast steps 3 and CH 4: C CH 3 + the can see that this mechanism information mechanism. Remember that we cannot 2 deduce We relevant fast I 2 any is consistent with the rate the reaction directly from equation. the mechanism rate equation. + Intermediate A is derived from the CH COCH 3 and H ions which react 3 The best we can do is to see if the + together to form it. This is why both COCH CH 3 rate in equation. the rate Iodine is we by are given reference propanone contains in a fast H appear in the 3 step later on, so does not appear reaction mechanism with rate the is consistent equation. equation. Predicting order of If involved and a to reaction the with rate mechanism, determining bromine ions. OH reaction from The in given we step. alkaline proposed a can predict For the example, solution. reaction reaction The order the alkaline mechanism mechanism of reaction reaction of solution is: slow Step 1: CH COCH 3 + OH → CH 3 COCH 3 + H 2 O 2 fast Step 2: CH COCH 3 The slow + Br 2 step → CH 2 involves one COCH 3 Br + Br 2 molecule of CH COCH 3 the rate equation should be: rate = COCH k[CH 3 and one OH ion. So 3 ][OH ] 3 Key points The rate The rate of order number The determining step in a reaction with respect mechanism determines the overall reaction. rate of of reaction molecules equation involved provides in to the evidence particular rate to reactants determining support a shows the step. particular reaction mechanism. 87 Revision 1 a Benzenediazonium chloride questions decomposes above 3 5 °C to produce phenol, hydrochloric acid For the below nitrogen as shown in the following reaction X + Y → Z, the tabulated results and were obtained. equation: + C H 6 N 5 Cl (aq) + H 2 O(l) → C 2 H 6 N OH(aq) + Experiment 5 (g) + [X]/ used b The describe to study equation for propanone I a (aq) suitable the rate the of method this that could Initial rate/ –3 –3 mol dm mol dm mol dm 1 0.030 0.015 0. 1242 2 0.010 0.015 0.0138 3 0.010 0.045 0.0414 2 Briefly [Y]/ –3 HCl(aq). –1 s be reaction. reaction of iodine with is: + CH 2 COCH 3 (aq) → CH 3 COCH 3 I(aq) + 2 a What is the order with respect to X? b What is the order with respect to Y? c What is the overall d Write the e Determine + H Explain to 2 The why monitor questions shown a colorimetric the rate of which follow this are (aq) + I method (aq). is suitable order of the reaction? reaction. based on the graphs rate equation for the value of the the reaction. rate constant for this reaction. below: I 4 The following etar graph shown questions below, experimental data. concentration of a are which It based was a student’s generated from represents reactant on [Y], the change over a in period of 75 [reactant] seconds. ]tnatcaer[ II 0.6 0.5 md 3 time lom III 0.4 0.3 ]Y[ ]tnatcaer[ 0.2 0. 1 time 0.0 IV 20 40 etar time a 60 80 (seconds) i What is the initial ii What is the instantaneous rate? rate at [reactant] a Which of the graphs represent a zero 20 seconds, 60 seconds? order reaction? b b How would represents you determine a rst or second whether Graph order reaction i What is ii What are to the reactant How would Graph IV this change, if it were labelled 88 a second axes? term ‘half-life’? the values of the rst two half-lives reaction? Giving your order reaction, with reasoning, state the order to reaction represent the shown? c c by with for respect meant II the same with respect to reactant Y. of the Chapter 5 The sketch below distribution for a represents sample of the gas Boltzmann at a 7 temperature a of 7 Consider Step 1: Rates the reaction reaction (g) NO of + SO 2 500 K. Step 2: – revision mechanism: (g) → NO(g) ii Which iii b a Copy the diagram and label the A What is + O (g) is the is → 2NO the same axes, sketch the graph to show distribution would change = ][F k[NO to if the Giving i What is by the rate 2 your reasoning, is state consistent the effect of a decrease on the rate of a Step 1: Step 2: + NO NO F → FNO 2 + F graphs, along with and notations, to explain this Single step half-life relate to the concept ii of + What F 2 mechanism: → 2FNO 2 would reaction ). (t if 2 happen the A sample tested to the rate concentration of of F the were 2 1/2 a slow fast effect. 2 questions F FNO appropriate 2NO The following + 2 → 2 6 these reaction? II symbols of observed equation? 2 your the in I Use which with 390 K. temperature ii equation, temperature rate c equation? ] mechanisms decreased step? how i the fast rate-determining described 2 On (g) equation? axes. rate b the rate slow 2 overall step is reaction the (g) 3 2 What + SO 2 2NO(g) i questions of a after nuclide 16 What fraction with a half-life of 4 days is while the sample has the concentration of NO 2 remained days. of doubled, the same? decayed? 1 __ A 4 3 __ B 4 7 __ C 8 15 __ D 16 b Radioactive newly decay prepared is a rst order radioactive process. nuclide has –6 constant (rate approximate c A 1 hour B 1 day C 1 week D 1 month The half-life much of a constant) half-life of of 1 the sample 10 If a decay –1 s , what is the nuclide? of Thorium-234 12 g × a would is 24 hours. remain How after 96 hours? A B 3 g C 0.75 g D d 1.5 g A 6 g sample after its of a radioactive nuclide is tested 6 days preparation. 7 __ It is found of that the sample has decayed. 8 What is the A 2 days B 4 days C 6 days D 7 half-life of this nuclide? days 89 8 Chemical 8. 1 Characteristics Learning outcomes equilibria Reversible Many On completion of this section, be able understand chemical equilibria reactions reactions then continue stops. We until say one the of the reaction reactants goes to runs out. completion, The e.g. reacting to: magnesium chemical you reaction should of the and principles of hydrogen. with This hydrochloric reaction is acid to form magnesium chloride and irreversible. physical–chemical Some reactions are reversible e.g. equilibria forward state the characteristics of CuSO 5H 4 system in dynamic reaction a O CuSO + 2 5H 4 O 2 equilibrium. backward Heating blue copper( II) turns it back Reversible place react at to show hydrated in time reactants at reactions do equilibrium reaction iodide the called the the the go can we white copper( (backward backward the sulphate reaction). reactions reactions . as anhydrous II) The reactants take products form completion. use gas to the and equilibrium iodine sign vapour to Y. For produce gas: approach alone. and time to it sulphate forward (g) + I 2 products turns anhydrous II) same hydrogen H We the equilibrium not reaction of to copper( which are sulphate water These an hydrogen blue same give example, to copper( II) Adding reactions the products. T o hydrated sulphate. reaction equilibrium Figure (g) Y 2HI(g) 2 by 8.1.1(a) starting shows off how with the either reactants concentrations of alone H or and 2 when a mixture of hydrogen (0.02 mol dm ) and iodine tube When –3 ) (0.02 mol dm reached, and there products. is is heated no The in further same a sealed change in equilibrium at the is 500 K. concentration reached when equilibrium of we the start is reactants with –3 hydrogen iodide (0.04 mol dm ) alone a (see Figure 8.1.1(b)). b and [I ] noitartnecnoc noitartnecnoc equilibrium reached [HI] [HI] falls md lom( 3 2 falls ) ] 2 md lom( 3– ) [H equilibrium reached [H ] and 2 [I ] 2 rises rises time Figure 8.1.1 time The change in concentrations of reactants and products as equilibrium is –3 approached: a starting with 0.02 mol dm –3 H (g) and 0.02 mol dm 2 –3 0.04 mol dm 90 HI(g) I (g); b starting with 2 I 2 –3 change Chapter Characteristics of the It is dynamic: to products to reactants. At molecules and equilibrium equilibrium of molecules the rate of reactants are products are the forward and Chemical equilibria state continually of 8 being continually backward converted being converted reactions are equal. The concentration change. and They backward Equilibrium where none mixture. the are only Many simple example equilibrium If we put energy reactions, molecules increases. reached the we a where rate as is write the from this a can solution liquid equilibrium rates A of closed escapes take and in no from place gases the to the the the do not forward in are system the is one reaction open beakers if involved. of moving to the pressure system equilibrium is become from liquid. its with the A by a the container . the most vapour . molecules condense. is closed a increasing exerted called a in the As vapour attractive Eventually liquid to more the forces a point vapour is at liquid–vapour vapour vapour in equilibrium pressure . For water , as: O(l) H Y 2 a and to in molecules water vapor chemistry vapour They other vapour The closed are its water surface. each cause physical and container , molecules from a system . products concentration closer reached. in closed or at the same. equilibrium closed molecules liquid can an the get products because however , in between a escape, the same in place escape They equilibrium with to a is equilibrium of up water start between set in the reactants takes and This are occurs the Liquid–vapour A reactants reactions of reaction of constant. H O(g) 2 b c vapour liquid Figure 8.1.2 liquid a A liquid is placed in a sealed container; b Liquid molecules move from liquid to vapour at a greater rate than they return from vapour to liquid; c At equilibrium the molecules move from liquid to vapour at the some rate as they move from vapour to liquid Key points Chemical occur at equilibrium the Equilibrium are the is is dynamic: the backward and forward reactions time. reached when the rate of forward and backward reactions same. Equilibrium In a same closed only occurs system, a in a liquid closed is in system. equilibrium with its vapour. 91 8.2 The equilibrium constant, K c Learning outcomes The equilibrium constant, K c If On completion of this section, we high should be able put temperature construct and both iodine equilibrium with the product, terms deduce a sealed and hydrogen tube iodine and are heat present it in at a the constant gas phase, of (g) units of + I 2 concentration the iodide: expressions H in in hydrogen to: together hydrogen you When K we measure the (g) Y 2HI(g) 2 concentration of each reactant and product at c equilibrium, describe an experiment a value for nd there is a relationship between these concentrations. to For determine we this reaction the relationship is: K c 2 [HI] _______ = constant (K ) c [H ][I 2 The constant is called ] 2 the equilibrium constant , K c −3 [HI] in means K c refers whole value writing start equilibrium take care to solids that appear of K is the is same called in the (at concentrations however the of in mol dm terms of concentrations. the equilibrium same expression temperature hydrogen, iodine and and pressure) hydrogen iodide we any equilibrium reaction we can write an equilibrium expression the is based on the stoichiometric equation. For the equation: equation. Their mA concentration constant in with. which stoichiometric equilibrium iodide ignore For any hydrogen c whatever expressions, the relationship Exam tips When to of c The The concentration remains much there + nB Y pC + qD constant is. So p the q [D] [C] ________ The equilibrium expression is: = K c equilibrium expression for m the [A] n [B] reaction: 2 ] [NH _________ 3 Example 1: N (g) + 3H 2 Fe O 2 (s) + 3CO(g) Y 2Fe(s) + 3CO 3 (g) Y 2NH 2 (g) K 3 = c 3 (g) [N 2 ][H 2 ] 2 2 is [SO ] __________ 3 3 [CO] Example 2: 2SO (g) + O 2 (g) Y 2SO 2 (g) K 3 = c 2 ______ K [SO = ] [O 2 c ] 2 3 [CO ] 2 Units of K c These vary with the form of the equilibrium expression. They have to be −3 worked in the out by substituting equilibrium in mol dm expression. For each the ‘concentration boxes’ example: 2 −3 ] [SO (mol dm __________ 3 = K of −3 ) × (mol dm ) ___________________________________ units = . c 2 [SO ] −3 [O 2 ] (mol dm −3 ) × (mol dm ) −3 × (mol dm ) 2 ) ___________________________________ So: units = = −3 ) × (mol dm ) 1 _________ 3 = dm −1 mol −3 mol dm If the no 92 top units. and bottom of the expression cancel completely then there are Chapter Determining a value of K by 8 Chemical equilibria experiment c The value of K for the reaction c CH COOH + C 3 can be The found using procedure Make up H 2 ethanoic acid a OH Y CH 5 titration COOC 3 ethanol method. The reaction of known concentration and catalyst, e.g. 3 and Use a Shake of ethyl burette the H O 2 water catalysed by an acid. to 2 mol dm HCl. ethanoate, water deliver mixture temperature volumes of the 3 water , 2 cm 3 2 cm ethanol, 1 cm ethanoic –3 of 5 cm amounts is and 3 acid + 5 ethanoate is: mixtures reactants H 2 ethyl to well reach each and of (Y ou and the allow could to start with known acid.) liquids it also into stand a for stoppered 1 week at bottle. room equilibrium. −3 After 1 week, hydroxide the reacts Find the added of a Calculate acid in the number concentrations of reaction the moles acid ethanoic reaction start the as mixture an with indicator . mixture sodium 1 mol dm During decreases as this the time, ethanoic ethanol. of and the the of the hydrochloric acid catalyst you concentrations: hydrochloric of of titration. and Moles whole concentration similar of the phenolphthalein with exact by Calculation using amount acid titrate at of the ethanoic start of acid, the a, ethanol, experiment b, from water , the c, volumes used. acid mixture at by experiment. equilibrium titration Call – this = moles moles of of acid calculated hydrochloric acid at in the x. Exam tips Moles of ethyl experiment because ethanoate Moles – for of is moles every is – is = moles ethanoic of of acid ethanoic ethanoic at acid at equilibrium acid reacted, a the (a – mole start x). of of This the is It ethyl at for to equilibrium moles is of ethyl every = moles ethanoate mole of of at ethanol at equilibrium ethanol reacted, a the (b – mole start (a of – important equilibrium formed. because ethanoate of mole ethanol experiment This ethanoate of the the equation given. The way that value you example, formed. you of write you K have depends for the if write you reaction the write Moles + of moles for every water of at ethyl mole of equilibrium ethanoate water at = moles of water equilibrium formed, a mole of (c + at (a ethyl start – of x)). experiment This ethanoate is is because been on equation. the HI as 2HI Y H between + 2 expression I the For equation H , 2 the according c the x)). ethyl that expression I and 2 equilibrium 2 is: formed. [H ][I 2 ] 2 _______ kc The concentrations expression to nd a are then value of substituted into the = 2 equilibrium [HI] K c Key points For an equilibrium concentration of reaction reactants there and is a relationship products and the between the equilibrium constant K . c p [C] q [D] _______ The general form of the equilibrium expression is K = c m [A] [A] and [B] [C] and [D] p, q, m, n = reactant = product concentrations are number of stoichiometric K can be n [B] concentrations moles of relevant reactants/products in the equation. calculated from the results of appropriate experiments. c 93 8.3 Equilibrium calculations involving K c Learning outcomes Calculating K from equilibrium concentrations c On completion should be able of this section, In some or moles, know how to calculations equilibrium present at calculations equilibrium. we may When be given doing the these amount, in calculations grams we rst to: need equilibrium you to calculate the concentration of each of these substances. perform involving constant, the K Worked example 1 . c Propanone reacts with CH hydrogen COCH 3 At equilibrium present as as HCN Y to were form CH 3 there well + cyanide an addition C(OH)(CN)CH 3 0.013 mol 0.011 mol of the of product: 3 propanone addition and product, 0.013 mol of HCN C(OH)(CN) CH 3 3 CH . The volume of the reaction mixture was 500 cm . Calculate K 3 c −3 Step 1: Calculate CH the concentration COCH 3 + of each HCN Y CH 3 0.013 in mol dm : C(OH)(CN)CH 3 1000 3 1000 _____ 1000 _____ × 0.013 _____ × 0.011 500 × 500 −3 = substance 500 −3 0.026 mol dm −3 0.026 mol dm 0.022 mol dm [CH C(OH)(CN)CH ] ____________________ 3 3 Step 2: W rite the equilibrium expression: K = c [CH COCH 3 ][HCN] 3 0.022 ______________ Step 3: Substitute the values: K = 3 = 32.5 dm −1 mol c 0.026 Calculating K using initial × 0.026 concentrations and c concentration of Worked 0.29 mol allowed are example of to ethanoic reach present. product 2 acid and equilibrium. Calculate 0.25 mol At of ethanol equilibrium are mixed 0.18 mol of together ethyl and ethanoate K c Step 1: W rite the below balanced equation with CH COOH + C 3 Step 2: the necessary information it. At the start: At equilibrium: Calculate the (According OH Y CH 5 COOC 3 H 2 0.25 mol the 1 mol O 2 0 of moles equation, for at equilibrium: every 1 mol of COOC CH 3 formed, + H 5 0 0.18 mol number to H 2 0.29 mol of H O is H 2 5 formed.) 2 CH COOH = 0.29 – 0.18 mol = 0.11 mol (since for every 0.18 mol 3 of COOC CH 3 H C 2 CH OH = 3: formed, 0.18 mol COOC 0.25 H – 0.18 mol 2 W rite the formed, = 0.07 mol equilibrium H ][H expression O] 94 is consumed) (since H and OH for is every 0.18 of consumed) 5 substitute 0.18 × 0.18 0.11 × 0.07 the values. = = 4.2 (no units) c that the COOH ___________ K COOH][C 3 of C 2 = [CH same of 5 COOC [CH 0.18 mol c Note CH 3 _____________________ 3 2 5 2 K of 5 5 3 Step H 2 we do number of H 2 not have OH] 5 to know concentration equilibrium expression. the terms concentrations in the since numerator and there are the denominator Chapter Worked example mixture were put of in hydrogen a sealed Chemical equilibria 3 −2 A 8 (2.40 tube × and 10 left −2 mol) to and come to iodine (1.38 equilibrium. × At 10 mol) equilibrium −3 1.20 × mol 10 of iodine was present. Calculate K c Step 1: W rite the below balanced equation with the necessary information it. H (g) + I 2 (g) the start: 2.40 At equilibrium: × Y 2HI(g) 2 −2 At −2 mol 10 1.38 × 10 1.20 × 10 mol 0 −3 Step 2: Calculate Amount the of number iodine of moles (1.38 so, mol × at H −2 ) 10 equilibrium: consumed −2 = at mol − (0.12 × 10 −2 ) = 1.26 × 10 = 1.14 × 10 equilibrium 2 −2 = (2.40 × −2 ) 10 − (1.26 × 10 −2 ) mol −2 (1.26 × mol 10 I were consumed so the same amount of H 2 consumed because is 2 these have the same mole ratio in the equation.) so, mol HI at equilibrium −2 = 2 × 1.26 × stoichiometric produces Step 3: W rite equation 2 mol the −2 mol 10 of = 2.52 every × 10 mole of expression and 2 iodine in which substitute −2 [HI] (2.52 _______ (since the reacts HI.) equilibrium ×10 the values: 2 ) ___________________________ = K mol K c = = 46.4 (no units) c −2 [H ][I 2 ] (1.14 × 10 −3 ) × (1.20 × 10 ) 2 Key points Values of K can be deduced from the concentration of reactants and c products present Concentrations calculated from at of equilibrium specic the together reactants equilibrium or with the products expression, equilibrium at expression. equilibrium knowing the value can of be K c together and at with relevant concentration data at the start of the experiment equilibrium. 95 8.4 The equilibrium constant, K p Learning outcomes Mole fraction For On completion of this section, reactions than should be able to understand term measure mixtures of concentrations. gases, When it we is easier are to dealing measure with pressures mixtures of to: gases involving you the ‘partial meaning of the we mixture need in to know terms of the fraction moles. This is of a particular called the gas mole present in the fraction pressure’ number of moles of a number of moles of gases particular gas _________________________________________ construct equilibrium mole expressions fraction . = total in terms of partial a mixture of 0.2 mol H and 0.3 mol O 2 deduce units of the mixture pressures In in the mole fraction of H 2 is: 2 K p 0.2 _________ perform calculations involving K = p 0.2 Partial The is exerted called the partial pressure, by the partial partial The 0.4 0.3 pressures pressure gases + molecules pressure pressure pressures of all = the of total of a that particular gas. pressure gases in the gas Symbol, × mole mixture in a mixture of p fraction add up to the total P T P = p T + p 1 + p 2 … 3 Example: A mixture of gases contains 1.2 mol N , 0.5 mol H 2 and 0.3 mol of O 2 . 2 5 The total pressure is 8.0 × kPa. 10 Calculate the partial pressure of hydrogen. 0.5 _______________ 5 partial pressure of = H (8.0 × 10 ) × 5 = 2 × 10 kPa 2 1.2 Using the equilibrium constant, + 0.5 + 0.3 K p Equilibrium constants in terms of partial pressures, K , may be calculated p in a similar way to those for except K that no square brackets are used. c For example, the equilibrium N (g) expression + 3H 2 (g) for Y the 2NH 2 reaction: (g) 3 Exam tips is: 2 Although the standard pressure p is approximately 1.01 × 10 NH ___________ 3 K = p 5 3 Pa, pN × p H 2 industrial chemists often use 2 the Units of K have to be worked out in a similar way as those for p atmosphere (atm) as a unit K . The c of units are calculated using pascals, Pa. 5 pressure, 1 atm = 1.01 × 10 Pa. For the reaction N (g) 2 You should be prepared to from calculations using + 3H (g) Y 2NH 2 (g) , the units are worked out 3 do atmospheres the equilibrium expression as follows: as 2 the base unit as well as pascals. ) (Pa) ___________ 1 ______________________ cancelling _____ = 96 × (Pa) 2 ) × (Pa) × (Pa) –2 = 3 (Pa) (Pa) Pa Chapter Worked Sulphur example dioxide reacts with oxygen (g) + to O 2 equilibrium, the Chemical equilibria 1 2SO At 8 mixture form (g) Y sulphur 2SO 2 trioxide: (g) 3 contains 12.5 mol SO , 87.5 mol O 2 and 2 7 100 mol . SO The total pressure of the mixture is 1.6 × 10 Pa. 3 Calculate K p Step 1: Calculate partial pressures: 12.5 _____ p 7 = × 1.6 × 6 10 = 1.0 × 10 Pa SO 2 200 87.5 _____ 7 = p × 1.6 × 10 × 1.6 × 10 6 = 7.0 × 10 = 8.0 × 10 Pa O 2 200 100 ____ 7 = p 6 Pa SO 3 200 Step 2: W rite the equilibrium expression: 2 p SO ____________ 3 K = p 2 p SO × pO 2 Step 3: Substitute the partial 2 pressures: 6 (8.0 × 2 ) 10 ________________________ K = –6 = 9.1 × 10 –1 Pa p 6 (1.0 Worked 2 mol of example nitrogen × 10 2 ) 6 × (7.0 × 10 ) 2 and 6 mol of hydrogen are mixed and allowed to reach 7 equilibrium mixture at 680 °C contains and 3 mol 2 of Pa ×10 pressure. ammonia. At Calculate equilibrium, the K p N (g) + 3H 2 Step 1: Find the number (g) Y 2NH 2 of moles (g) 3 of nitrogen and hydrogen at equilibrium: Moles N = 2 – 1.5 = 0.5 mol 2 1 (For each mol NH formed mol 3 of N has been used, 2 2 1 __ i.e. 3 × mol) 2 Moles H = 6 – 4.5 = 1.5 mol 2 (For each 2 mol of NH formed 3 mol H 3 are consumed Key points 2 3 __ i.e. 3 × mol) Partial pressure = total pressure × 2 mole fraction Step 2: Calculate partial of = 2 × 10 = 2 × 10 0.1 = 0.2 × 10 × 0.3 = 0.6 × 10 Pa 2 7 7 = 2 × 10 the equilibrium of partial 0.6 = 1.2 × 10 reactants gures: 2 p 7 NH = constant pressures in can be Pa deduced from (1.2 ___________ 3 K The 7 × 3 Substitute gases. terms NH 3: mixture Pa 2 Step a 7 H p in 7 × N p gas pressures: 7 p of ×10 2 In a and reaction partial pressures of products. involving gases, ) _______________________ = p 3 pN × p 7 H 2 (0.2 ×10 ) 7 × (0.6 ×10 3 the quantities of reactants or ) 2 products –13 K = 3.3 × 10 at equilibrium can be –2 Pa p deduced from expression the and K equilibrium together with p relevant partial pressure data. 97 8.5 Le Chatelier’s Learning outcomes principle Position of Position On completion should be apply able of this section, effect products in principle concentration If the of change pressure temperature If on an that the Le changes concentration, temperature relative amounts of reactants and of position products of increases equilibrium relative moves to to the the reactants, we products, we right. concentration the of position reactants of increases equilibrium relative moves to the to the left. Chatelier ’s changed, pressure affect principle: the If position of one or more equilibrium factors shifts that in affect the an equilibrium direction which in the the change or values of Le Chatelier ’s principle relates to changes in concentration, pressure and temperature. K c the that opposes and the equilibrium are K to mixture. in reaction how equilibrium or say know an to concentration, refers principle you say the equilibrium Le Chatelier’s to: Le Chatelier’s explain of equilibrium: p apply the Le Chatelier’s Haber Process principle and to the Note: of . K Catalysts They have only no speed effect up the on the rate of position of equilibrium or the value reaction. c Contact Process. Position of The effect of equilibrium: changing changing concentrations can be concentration seen by referring to the reaction: COOH CH + C 3 Increasing position More the of Increasing position acid concentration ethanoate the of OH Y CH 5 to ethanoic right to of the (to are + H 5 acid (or oppose O 2 ethanoate water ethanol) the added shifts the reactant(s)). formed. ethyl left. H 2 ethyl water concentration equilibrium of the and COOC 3 ethanol equilibrium ethyl H 2 ethanoic ethanoate More (or ethanoic water) acid shifts and the ethanol are formed. Removing the right water Note: water (to are from oppose the the reaction removal shifts of the water). position More of ethyl equilibrium ethanoate formed. Changing the concentration has no effect on the value of or K K c Position of Change of gas (g) in + I 2 Exam tips is a common error to a change in the (g) on only position Y affects each side of 2HI(g). reactions of the equilibrium This of gas change on in the left pressure and concentration reaction such because does, 2 moles the the of however , the are different example, there numbers is no reaction: stoichiometric gas on change the equation shows 2 the right. position of equilibrium in or (g) + 3H 2 affects on there For as: N pressure where equation. pressure suggest a that pressure changing the p 2 moles A It equilibrium: molecules change H in to and value of K (g) Y 2NH 2 (g) 3 or c K . If we add more reactants, more Increase in pressure shifts the position of equilibrium to the right p where products are formed until the K is restored. Remember it are fewer gas molecules (2 moles on the right and 4 moles value on of there the left). This opposes the increase in the number of gas molecules is c per only a change in temperature the equilibrium volume Decrease K pressure increases. in pressure makes decreases). the The molecules position of move further equilibrium apart shifts (their to (Temperature Only where Changes the constant. concentration Remember TOCK when that changes unit the gas molecules are more concentrated. ). c Note: Changing the pressure has no effect on the value of or K c 98 K p the left Chapter Position of A change in equilibrium: temperature the value of K or For an endothermic value of K or K More products c . reaction, So the increase position of in p We temperature equilibrium increases shifts to the the right. formed. So for the Y (g) H + I 2 increasing the and (g) ΔH 2 temperature products an value exothermic of or K K c reactants more +9.6 kJ mol effect of temperature of the position the hydrogen position iodide . So the increase position of in formed. So for the increase the r shifts reaction, of equilibrium in favour So energy + O Y the of dioxide temperature equilibrium 2SO and decreases shifts to the reaction the the left. i.e. More in the (g) + in 3H 2 yield can shifts in = Increasing the (g) the in the increase direction of in direction energy, energy absorbed. is absorbed in endothermic reactions. position Haber of equilibrium in favour processes Process: Y 2NH (g) –1 ΔH 3 increased the goes −197 kJ mol r industrial 2 be the –1 ΔH Ø N increases particles. oxygen. Maximising the yield synthesised (g) 3 temperature temperature the reaction: 2 increasing sulphur (g) in of opposing Energy The equilibrium. decomposes. Ø (g) 2 is = −92 kJ mol r by: pressure: more product (NH ) is formed because the 3 position of equilibrium Decreasing the shifts temperature: in favour increases of fewer the molecules. yield of NH because the 3 reaction is Removing shifts to Using the a exothermic. ammonia the right catalyst. equilibrium in by condensing favour This of speeds it: fewer up the position of equilibrium molecules. the reaction rate but has no effect on yield. Did you know? We use compromise although rate of conditions reaction for increases the as synthesis of temperature ammonia increases, because the yield of In ammonia decreases. So a temperature of about 420–50 °C is used to the Contact reasonable yield at a fast enough Process, a high give pressure a is not used because: rate. The yield of SO produced for 3 The key acid is: reaction in the Contact Process for the manufacture of sulphuric SO is fairly high at fairly low 2 temperatures. Ø (g) 2SO + O 2 (g) Y 2SO 2 (g) –1 ΔH 3 = −197 kJ mol r We can increase the yield of SO by decreasing the the pressure and Highly are using a catalyst. The corrosive theory behind this is less as for the Haber easily contained oxides in the the reactor same sulphur temperature, 3 increasing vessel at high pressure. Process. Key points Le Chatelier’s principle states: When the conditions in a chemical eq uilibrium change, the position of eq uilibrium shifts to oppose the change. Increase the in right. concentration Decrease in of reactants concentration shifts of the position reactants, shifts of the equilibrium position to to the left. A change in temperature affects the value of K or K c concentration The to or conditions get the to –1 = p are 2SO Ammonia principle on of For Le Chatelier’s understand the An use reaction: Ø the can p are 2HI(g) of equilibria K c Chemical Did you know? changing temperature changes 8 best pressure used in yield of have the no Haber product effect on Process at a but in values. and Contact reasonable changes p these rate of Process are chosen reaction. 99 8.6 Solubility product, K sp Learning outcomes Solubility Many On completion of this section, ionic soluble should be able construct for equilibrium solubility expressions deduce units for perform calculations solubility to be dissolve in insoluble. water But but even some are ‘insoluble’ only ionic slightly compounds is some ions saturated in when solution no more to a very solute small will extent. dissolve in We it. say that Solubility a solution is measured in (or mol dm sometimes in g/100 g water). K sp appear −3 product, K sp or to: form compounds you product. Potassium involving chloride 35.9 g/100 g Calcium hydroxide solution Silver is soluble – is described a saturated solution contains water . contains chloride is 0.113 g/100 g regarded as as sparingly soluble – a saturated water . ‘insoluble’. It is in fact sparingly soluble −4 – a saturated Solubility When a solid. they is The move contains 1.93 × g/100 g 10 water . product sparingly equilibrium the solution soluble or established ions from move solution ‘insoluble’ between from to the the salt is ions solid to added in to water , solution solution at and the an the same ions in rate as solid. water molecules – I not ions shown + ions saturated solution + of Ag – ions and I ions solid Figure 8.6.1 At equilibrium ions move from saturated silver iodide solution to silver iodide solid at the same rate as they move from solid to solution We can write the equilibrium in AgI(s) the Y diagram above + − Ag (aq) + I as: (aq) + [Ag − (aq)][I (aq)] _______________ The related equilibrium expression is: K = [AgI(s)] But the concentration of solid is constant, so we can write the expression as: + = K [Ag − ][I ] sp K is called the solubility product . The square brackets show solubility sp −3 in mol dm Solubility saturated of their . product solution relative is the of a product sparingly of the concentrations soluble salt at 298 K 1: ion Co(OH) ratio Co 2OH Al O 2 100 ion the in a power 2+ K ion 3 = [Co = [Al − ][OH ] sp 3+ 2: to − and 2 Example each concentrations. 2+ Example of raised ratio 2Al 2− and 3O 3+ K sp ] 2 [O 2− ] 3 2 Chapter Units of We solubility calculate the 8 Chemical equilibria product units for K in the same way as for general equilibrium sp expressions involving K c Example: 2+ = K [Co 2 ][OH ] −3 units are (mol dm −3 ) × (mol dm 2 3 ) = mol −9 dm sp Solubility Worked A product example saturated calculations 1: Solubility solution of lead product from iodide, PbI , contains solubility 0.076 g PbI 2 solution. Calculate for K lead iodide. in 100 g 2 Molar mass of lead iodide = sp −1 461.0 g mol −3 Step 1: Calculate the concentration 0.076 the in 1.65 × 10 −3 mol dm 100 concentration of each 2+ [Pb ion in −3 ] = 1.65 × 10 solution. −3 mol dm −3 ] [I mol dm −3 = 461.0 Calculate solution _____ × 2: the 1000 ______ Step of = 2 × (1.65 × 10 −3 ) mol dm −3 ] [I = 3.30 × 10 −3 mol dm 2+ Step 3: W rite the equilibrium expression: K = [Pb 2 ][I ] sp −3 Step 4: Substitute the values: K = (1.65 × 10 −3 ) × (3.30 × 10 2 ) sp −8 = 1.80 × 10 −8 Step 5: Add the units: K = 1.80 × 10 3 −9 mol dm sp Worked example Calculate the 2: Solubility from solubility product −3 solubility in mol dm of calcium sulphate, CaSO . 4 −5 = K 2.0 × 10 2 mol −6 dm sp Step 1: W rite down the equilibrium expression: 2+ K = [Ca 2− ][SO sp Step 2: Rewrite the equilibrium expression 2+ In CaSO , [Ca 3: = [SO the so K = −5 4: Complete the × 10 calculation 2+ [Ca ] [Ca of one ion only: 2 ] sp value: 2.0 Step terms 2+ ] 4 Substitute in 2− ] 4 Step ] 4 2+ = and [Ca add 2 ] units: −5 = √ 2.0 × 10 −3 = 4.47 × 10 −3 mol dm Key points Solubility of ions in account product a is an saturated the of relative The value and concentration K can The solubility expression solution number be of of a showing each calculated the sparingly ion using in equilibrium soluble salt. It concentration takes into solution. the relevant equilibrium expression sp of K and of a of ions present sparingly the formula of at soluble the equilibrium. salt can be calculated using the value salt. sp 101 8.7 Solubility product and the common ion effect Learning outcomes Predicting Solubility On completion should be able explain the the of this section, precipitation product can be used to when two when we solutions are mixed. principles common ion underlying mix sulphate? solutions The possible of the solubility selective whether example: soluble reaction will a precipitate we get a will form precipitate salts barium chloride and sodium is: effect + Na 2 relate For to: BaCl predict you product precipitation to of SO 2 → BaSO 4 + 2NaCl 4 the ions Sodium chloride is very soluble in water but barium sulphate is sparingly soluble. know how to determine solubility 2+ product by experiment. For barium sulphate K = [Ba 2− ][SO sp 2+ If the ionic product −10 ] = 1.0 × 2 10 mol −6 dm 4 [Ba 2− ][SO −10 ] is greater than 1.0 × 10 2 mol −6 dm 4 a precipitate of barium sulphate 2+ If the ionic product ][SO [Ba will form. −10 2− ] is less than 1.0 × 10 2 mol −6 dm 4 barium Worked Will a sulphate will example precipitate × 10 1.20 × 10 in solution. It will not precipitate. 1 form −5 6.00 remain when an aqueous solution containing −3 mol dm −5 barium chloride is added to an equal sodium of BaSO sp = 1.0 × 10 2 mol −6 dm 4 2+ 1: of sulphate? −10 K Step volume −3 mol dm Calculate the concentrations: [Ba −5 ] = 3.00 × 10 2− [SO −3 mol dm −6 ] = 6.00 × 10 −3 mol dm 4 2+ Step 2: W rite the equilibrium expression: K = [Ba 2− ][SO sp Step 3: Substitute the = (3.00 10 = 1.80 gures to give the −5 × × (6.00 Step 4: Compare 2 mol 10 ionic product: −6 ) −10 × ] 4 values of × 10 ) −6 dm ionic product and K : sp −10 1.80 so Worked What silver a precipitate example from × 10 −10 10 will is greater than 1.0 × 10 form. 2 concentration chloride −4 1.0 × of a silver ions solution is needed of sodium = 1.8 in solution chloride to precipitate containing −3 mol dm ? −10 K of AgCl × 10 2 mol −6 dm sp + Step 1: W rite the equilibrium expression: K = [Ag = [Ag ][Cl ] sp −10 Step 2: Substitute the gures: (1.80 × 10 ) + −4 ] × (1.0 × 10 ) + Step 3: Rearrange to nd [Ag ]: −10 1.80 ] × 10 ____________ + [Ag = −6 = 1.8 × 10 −3 mol dm −4 1.0 × 10 + is 102 the minimum concentration of Ag needed to cause precipitation. Chapter Precipitation We can use and qualitative solubility product analysis precipitate data compounds to choose from suitable their reagents solutions. For silver nitrate to test for halide ions such as , Cl Br and I . stones are soluble in water as are most nitrates. But silver halides have low halide solubility ion product. results in a So, adding precipitate only a few drops of by precipitation of present substances silver nitrate in the urine. The a depends on many factors. to If a produced Most deposition very are we naturally halides equilibria to example the use Chemical Did you know? Kidney selectively 8 the urine is too concentrated, e.g. the solubility product of some ions + (aq) Ag These precipitates can be + Cl (aq) distinguished → by AgCl(s) their characteristic may be such as exceeded calcium solubility product by higher experiment ions The method is based on making a saturated solution of a and analysing the solution to nd the oxalate than normal concentration of equilibrium ion which calcium is present hydroxide, in the Ca(OH) solid as well. For sp and of may up. A calcium also example in favour shift of the a to of insoluble calcium nd salts for K build concentration precipitation particular may particular the substance precipitates colours. calcium Determining and phosphate such as calcium carbonate and : 2 calcium Add enough solid is calcium present as well reach equilibrium. Filter off Titrate acid of the samples Calculate the the titration Find solid known by K hydroxide as a calcium of the to a known solution. volume Shake and of leave water for 24 so that hours phosphate, conditions become especially too if the alkaline. to hydroxide. calcium hydroxide solution with hydrochloric concentration. concentration of the calcium hydroxide solution from results. substituting in the equilibrium expression sp 2+ = K [Ca 2 ][OH ] sp The The salt common common caused common by with ion ion effect effect adding the a is the reduction solution dissolved of salt. a in the solubility compound The which solubility of of has in BaSO a an dissolved ion in water is 4 −5 1.0 × −3 mol dm 10 −3 . But the solubility of BaSO in 0.10 mol dm 4 −9 sulphuric acid, SO H 2 difference by , is only 1.0 × 10 −3 mol dm . We can explain referring to equilibrium 2+ = K [Ba expression: 2− ][SO sp −10 ] = 1.0 × 10 2 mol −3 dm 4 2− The this 4 concentration of −3 SO ions in the solution = 0.10 mol dm 4 Key points 2− (ignoring the very small concentration of ions SO from barium 4 sulphate). −10 Substituting the concentration 2+ So [Ba ] −9 = 1.0 × 10 values: 1.0 × 2+ = 10 [Ba ] × The common to the of a ion = solubility of barium sulphate in H dissolved quantitative effect work salt by adding a SO 4 solution and practical which involving preparing with contains the an ion dissolved in salt. chemistry In refers solubility −3 mol dm common common effect in 0.1 2 The ion reduction and weighing solids When the ionic product in the (gravimetric equilibrium expression for K sp analysis), it is important that no solid is lost. The common ion effect can exceeds help here. For example: the concentration of sulphate in aqueous be determined precipitates is washed will be with ions with lost dilute in the sulphate gravimetrically barium in sulphate. water , the sulphuric wash some ltrate. acid liquid remains If a by barium So the rather ensures adding precipitate and that is water . the of chloride. barium sulphate precipitate than barium ions ltered The will presence maximum in dissolve then of amount water and washed sulphate of product is a precipitated. This sulphate and solubility solution salt can the barium The can the in solubility be found product by concentration solution by of a salt determining of the titration ions or other methods. precipitated. 103 9 Acid–base 9. 1 Brønsted–Lowry Learning outcomes equilibria completion should be able explain the of this section, A acid and acid is a substance acids acids which and neutralises and bases a base to form a salt and water . base is a substance base which 2HCl(aq) + neutralises acid to form a salt and water: CaO(s) → an CaCl (aq) + H 2 using Brønsted–Lowry describe the behaviour acids and theory differences of strong and bases you to: terms of Simple definitions of An On theory acid in Another denition of base an acid is: a substance O(l) 2 salt water which ionises in water to + weak form bases. hydrogen ions. A hydrogen ion, , H is sometimes referred to as a proton. + HCl(g) Another denition water form to of a + base hydroxide aq is: a → H (aq) + substance Cl that (aq) dissolves or ionises in soluble in ions. + KOH(s) Many water metal is oxides called an or + aq hydroxides → K are (aq) + bases. OH A (aq) base which is alkali Did you know? + The H O ion is called Brønsted–Lowry theory of a 3 hydroxonium frequently ion. called a t is ‘hydronium’ alone we in talk ion. H ions aqueous about and bases rarely An A acid is a proton donor . base is a proton acceptor . occur solutions. When hydrogen or + ‘oxonium’ acids less ions, we + When hydrogen donated to the chloride water . So gas reacts is HCl with a water , the H Brønsted–Lowry ion acid . of the The acid is water (g) + really mean a H O ion. t is often accepts a proton so it is a Brønsted–Lowry base. 3 convenient, however, especially in + H donated + calculations and talk to about write this hydrogen ion as H + ions. HCl(g) + H O(l) → H 2 acid When ammonia, NH O (aq) + Cl (aq) 3 base , reacts with water , it accepts a proton from the 3 + water and becomes an ion. NH So ammonia is a Brønsted–Lowry base. 4 The water donates the proton, so it is a Brønsted–Lowry acid. + H donated + NH (g) + H 3 base Y ou will notice that Brønsted–Lowry For example, following water NH (aq) + OH (aq) 4 do can not act is set as have methanoic equilibrium Y acid acids when O(l) 2 an to acid acid or involve reacts a base aqueous with – it III) up: donated + Y HCOOH + HClO 2 base 104 acid + HCOOH 2 amphoteric solutions. chloric( + H is ClO 2 acid, the Chapter Conjugate Every away acid its Every equilibria pairs has base a a conjugate base – the ion left when acid – the ion formed an acid has given has a conjugate when an acid – COOH (aq) + H 3 O (l) Y CH 2 acid Strong has proton. CH they Acid–base proton. accepted Strong 9 and acids weak ionise dissociate acids (almost) completely. + COO (aq) + H 3 base and base (aq) conjugate acid bases completely For O 3 conjugate in solution. We sometimes say example: + (l) HNO + H 3 O(l) → H 2 O (aq) + NO 3 (aq) 3 + or more simply HNO (l) + aq → H (aq) + NO 3 Hydrochloric Weak acids ionise equilibrium present weak lies than Organic acid, nitric acid and (dissociate) to the (aq) 3 left. sulphuric only There acid partially are many in are all strong solution. more The molecules acids. position of acid citric acid of ions. acids such as methanoic acid, ethanoic acid and are acids. Exam tips + CH COOH(l) + H 3 O(l) Y CH 2 COO (aq) + H 3 O (aq) 3 t Strong bases ionise completely in solution, is important to distinguish e.g. between the strength of an acid and + NaOH(s) + aq → Na (aq) + OH (aq) its Weak bases equilibrium ionise (dissociate) lies the to left. only There partially are many in solution. more The molecules position of of base concentration. degree of Strength to the or weak. Concentration ionisation refers – strong refers to 3 present than the ions. number of moles per dm . A –3 4 mol dm Ammonia and amines are examples of weak solution of ethanoic acid bases. is concentrated, but it is a weak acid. + CH NH 3 Comparing The table bases. weak shows Strong + H acids bases of same Y CH and (aq) + OH (aq) 3 bases pH much Strong NH 3 typical have concentration. O(l) 2 some acids same the (g) 2 values lower bases for pH have strong values much and than higher weak weak pH acids acids values of and the than weak concentration. Key points Concentration pH of acid acid or base strong pH weak pH acid strong pH base weak base The Brønsted–Lowry theory states that donors and acids are bases proton are proton –3 0 1.0 mol dm 2.4 14 11.6 acceptors. –3 2 0.01 mol dm Strong acids have a 3.4 higher electrical 12 conductivity and 10.6 react more Strong calcium carbonate or magnesium. This is because strong than hydrochloric weak acids (for acid) have example a higher ethanoic concentration acid). of bases are completely aqueous solution. Weak ionised in acids and rapidly acids are hydrogen only partially ionised in (for aqueous example and (almost) bases with acids solution. ions Strong equal acids and distinguished their bases concentrations aqueous by the of can pH be values of solutions. 105 9.2 Simple pH Learning outcomes calculations pH We On completion should be able of this section, can compare the relative acidity or alkalinity of substances using the you pH scale. We express ion concentration this as the logarithm to the base 10 of the hydrogen to: −3 define the terms pH and (in ): mol dm K w + pH = −log [H ] 10 perform pH, calculations hydrogen concentration and and involving hydroxide K . ion Most pH values neutrality we (neither use in acidic chemistry nor are alkaline) on a being scale pH of 7. 0 to V ery 14 with concentrated w acids pH can have values The The p p pH above values below 0 and very concentrated alkalis can have 14. notation notation chemists to numbers and (as in express pH) is small is −log a modied values shown logarithmic simply. The scale used relationship by between the below. 10 −12 Number −log The p 1 × (number) notation constants, −7 10 1 12 can also be = pH pH of Strong 1 × 7 used in −log [OH ] pK 10 The −5 10 0 10 5 relation to 4 10 1 × 0 other 10 −4 equilibrium e.g. pOH Simple × −log [K 10 ] pK a = −log w [K 10 ] w calculations strong acids = a are acids completely ionised in water . So the pH can be easily calculated. −3 1 mol dm −1 HCl = −log (1 × 10 = −log (1 × 10 −3 0.01 mol dm ) = pH 0 pH 2 −2 HCl −3 0.002 mol dm ) = −3 HCl = −log (2 × 10 ) = pH 2.7 Exam tips + You must make sure that you know how to convert pH to H concentration + and H most concentration common to method pH. This may vary from calculator to calculator. The is: + pH e.g. pH 3.6 to [H ] x 3.6 + [H ] 106 ± → (shift) → 5 10 −4 → answer (2.51 × 10 ) −5 e.g. 1.8 × 10 1.8 f → you are not to → sure – pH EXP ask your → ± → teacher. log → ± → answer (4.74) Chapter Worked example Calculate the pH of −5 3.32 × 9 Acid–base equilibria 1 a solution whose hydrogen ion concentration is −3 mol dm 10 + pH = −log [H ] 10 −5 = −log (3.32 × 10 ) = 4.48 (note that there are no units to pH) 10 Worked example Calculate the 2 hydrogen ion concentration of a solution of pH 10.5. + pH = −log [H ] 10 + ] [H The ionic −pH = −10.5 10 = product of −10 10 = water, 3.2 × 10 −3 mol dm K w W ater can act substance is molecules as either dissolved can act as an in an acid it or (see acid base depending Section by 9.1). donating In on what pure protons to type water , other a of few water water molecules. + H O + H O 2 2 acid base Y H O + OH 3 conjugate acid conjugate base + We can simplify this equation to: H O Y H + OH 2 + [H ][OH ] __________ For which the equilibrium expression is: K = [H O] 2 Since the the value concentration for K. The of water expression is very then high, we can incorporate this into becomes: + K = [H ][OH ] w −14 K is called the ionic product of water . Its value is 1.00 × 10 2 mol −6 dm w Did you know? Using K to calculate the pH of alkalis w Why We can use K to calculate the pH of strong bases. We need to know is the pH of water 7? the w We concentration of hydroxide ions in the solution and value of can see this if we use the ionic K w expression for Worked example Calculate pH water: + 3 [H −14 ][OH ] = 1.00 × ] = [OH ] 10 + of a solution of potassium hydroxide of concentration and [H −3 0.0250 mol dm + So [H 2 ] −14 = 1.00 × 10 + Step 1: W rite the ionic product expression in terms of [H ] ions. + and [H −14 ] = √ 1.00 × 10 −7 = 1.00 × 10 −7 −14 K = 1.00 × 10 2 mol −6 −log dm (1.00 × 10 ) = 7 10 w + Step 2: W rite the equilibrium expression in terms of [H ] Key points K ______ w + [H ] = [OH ] Step 3: Substitute the values into the pH is a measure ion concentration, −14 1.00 ] × = the [H hydrogen ]. + 10 ____________ + [H of + expression. −13 = 4.00 × pH = −log [H ]. 10 10 0.0250 K is the ionic product of water w −13 Step 4: Calculate pH: − log (4.00 × 10 ) −14 = pH 12.4 (value 10 The = pH 1.00 of be found a = [H 10 strong using + K × ][OH the 2 mol base −6 dm ). can expression − ]. w 107 9.3 Equilibrium calculations involving weak acids Learning outcomes The acid dissociation constant, K a On completion should be able of this section, The equilibrium The equation law can be applied to aqueous solutions of weak acids. you for the partial ionisation of ethanoic acid in water H O is: to: + COOH(aq) CH understand the dissociation term constant, K ’ a perform calculations involving + H 3 ‘acid Because that its O(l) Y CH 2 water is present concentration at is a COO (aq) + 3 very constant high and concentration, we (aq) 3 can we simplify can this assume equation to: + the use of K , pH and [H ]. + a CH COOH(aq) Y CH 3 The equilibrium COO (aq) + H (aq) 3 expression for this reaction is: + [CH COO ][H ] _______________ 3 K = a [CH COOH] 3 K is the acid dissociation constant . For a monobasic weak acid such a −3 as ethanoic acid, the units of are K mol dm a −3 A high value for K e.g. 40 mol dm indicates that the position of a equilibrium is well over to the right. The acid is almost completely ionised. −5 A low value for e.g. K 1.3 × 10 −3 mol dm indicates that the position a of equilibrium is well over to the left. The acid is only very slightly ionised. We can also use values pK to compare the strengths of acids. a = pK −log a The [K 10 general ]. a equilibrium expression for K a We can based write on the a general general equilibrium expression for all monobasic acid reaction: + HA HA A The represents represents general the the unionised conjugate equilibrium Y + H A acid. base expression of is the acid. therefore: + [H ][A ] ________ K = a [HA] + Since the concentration of H and A must be equal, we can simplify this further: + [H 2 ] _____ = K a [HA] In calculations involving K we make two assumptions. a If the K value for the weak acid is very small, we can assume that the a concentration as the of acid concentration molecules of remaining undissociated acid at equilibrium molecules in is the the same original acid. We can ignore water . For water , K , w 108 the most is hydrogen work very this small. is ions which acceptable arise from because the the ionisation ionic product of for Chapter Calculations involving 9 Acid–base equilibria K a We can calculate the value of K for a weak acid if we Exam tips know: a the concentration the pH of acid n calculations involving K , a of the solution. remember that concentration We can calculate the pH of a solution of a weak acid if we the value of the pH and pH to hydrogen ion concentration. concentration Worked ion converted K a be know to hydrogen can example of the acid. 1 −3 Deduce the pH of a solution of 0.05 mol dm −5 K for ethanoic acid = 1.74 × ethanoic acid. −3 10 mol dm a Step 1: W rite the equilibrium expression. + [H 2 + ] [H ____________ K 2 ] _____ = or K a = a [CH COOH] [HA] 3 + Step 2: Rearrange the equation + [H to make [H ] = K × Substitute Step 4: Find the subject: [CH values: [H COOH] 3 + 3: the 2 a Step 2 ] 2 −5 ] = (1.74 × 10 −7 ) × 0.05 = 8.7 × 10 + [H ] by taking the square −7 √ 8.7 × = 10 root: −4 9.33 × −3 10 mol dm + Step 5: Convert [H + ] to pH: pH = − log [H −4 ] = − log 10 pH Worked = (9.33 × 10 ) 10 3.03 example 2 −3 The pH of a 0.01 mol dm of methanoic acid is 2.9. Deduce the value of K a + Stage 1: Convert Stage 2: W rite pH the to [H + ]: [H equilibrium −2.9 ] = −3 10 = 1.26 + [H 2 + ] = [H Substitute into 2 ] = K a [HCOOH] 3: −3 mol dm _____ or a Stage 10 expression. __________ K × the [HA] equilibrium expression: −3 (1.26 × 10 2 ) ______________________ K = a 0.01 −4 K = 1.59 × 10 −3 mol dm a Key points + For a weak acid, [H ] and pH can be calculated + [H using the expression: − ][A ] _______ K = a [HA] − Where [A ] is the concentration of concentration unionised of ionised acid and [HA] is the acid. 109 9.4 Acid and Learning outcomes base Using dissociation pK constants values a We On completion of this section, can use −log a be able values to compare the acidity of weak acids. a = pK should pK you [K 10 ]. Some examples are shown in the table. a to: −3 describe pK , K a perform and pK b Acid b calculations Equilibrium involving K and pK b and solution K /mol dm pK a + H SO 2 , a aqueous a sulphuric(IV) pK in Y −2 H + HSO 3 1.5 × 10 5.6 × 10 1.82 3 pK b w + hydrofluoric HF Y methanoic HCOOH −4 H + F + acid Y 3.25 −4 H + HCOO 1.6 ×10 3.80 Exam tips + ‘carbonic acid’ CO + H 2 We can pH for calculate a dibasic the acid way as for a Y −7 H + HCO 4.5 such as H S in a The higher the pK monobasic value, the weaker the because the 6.35 acid. acid. can use pK values in calculations rather than K a is 10 a We This × 3 approximate 2 similar O 2 value of K values. a is a1 much greater than K . a2 Worked example 1 e.g. −3 Deduce the pH of a solution of 0.001 mol dm benzoic acid. pK benzoic a – H S(aq) Y HS + (aq) + H (aq) acid = 4.2. 2 –8 K = 8.9 × a1 10 –3 mol dm Step 1: Convert pK to K a – HS 2– (aq) Y S + H (aq) So = K : pK a + aq) 10 −pK a K = 1.2 × a2 10 − log [K 10 ] a −4.2 a –13 = a K = −5 10 K a = 6.3 × 10 −3 mol dm a –3 mol dm Step 2: W rite the equilibrium expression: + We can ignore the H ions arising + [H 2 ] _____ = K from the second ionisation their concentration a because [HA] is insignificant + Step compared with those 3: Rearrange the equation to + the first make [H 2 ] the subject: arising form ionisation. [H 2 ] = K × [HA] a + Step 4: Substitute Step 5: Find the values: [H 2 ] −5 = (6.3 × 10 −8 ) × 0.001 = 6.3 × 10 + [H ] by taking the square root: −8 √ 6.3 × −4 10 = 2.51 × 10 + Step 6: Convert [H −3 mol dm + ] to pH: pH = −log [H ] −4 = −log 10 pH The = (2.51 × 10 ) 10 3.6 basic dissociation constant, K b For calculations constant, . K involving For a example, weak in the alkali, we can use the basic dissociation reaction: b + NH (g) + H 3 We can write the O(l) Y NH 2 equilibrium (aq) + OH (aq) 4 expression: + [NH + ][OH ] [BH ____________ 4 K ][OH ] ___________ = or more generally K b = b [NH ] [B] 3 + Where We [B] make is the weak similar expression (see base and assumptions Section [BH to ] is those the for a conjugate weak acid acid. in applying 9.1). + Since [BH ] = [OH ] we can simplify the expression 2 [OH ] _______ = K b [B] 110 further to: this Chapter Worked example Calculate the 9 Acid–base equilibria 2 −3 pH of a 0.01 mol dm solution −4 K (ethylamine) = 6.5 × of −3 10 −14 mol dm K b Step ethylamine. = 1.0 × 10 −3 mol dm w 1: W rite the equilibrium expression: 2 [OH ] _______ K = b [B] 2 Step 2: Rearrange the equation to make [OH ] the subject: 2 [OH ] = K × [B] b 2 Step 4: Substitute Step 5: Find the [OH ] values: by taking [OH the −4 ] = (6.5 square 6.5 × 6: Use K = [H −6 ) × −3 = 10 2.55 × 10 calculate [H + Step 10 0.01 = 6.5 × 10 root: −6 √ × −3 mol dm + ][OH ] to ]: w −14 K 1.0 ______ w + [H ] = ×10 ___________ −12 = = 3.92 × 10 −3 mol dm −3 [OH ] 2.55 × 10 + Step 7: Calculate pH: pH = −log [H −12 ] = −log 10 pH = (3.92 × 10 ) Exam tips 10 11.4 A quick way question Using pK and pK a to such get as an answer ‘calculate to the a pH values b of a solution of a strong base of −3 A useful expression which relates K to K w and K a concentration is 0. 1 mol dm ’ is to b use the expression 14 –log [OH ]. 10 pK + pK a = pK b w + This works log [OH because − log [H ] − 10 This can also be written as: ] = 14 10 −14 −log K 10 Worked − log a K = 10 b example 3 14 (the 14 derives from −log (1.0 × 10 )) 10 −3 Deduce the value of K for a 0.01 mol dm solution of aqueous ammonia. a pK for ammonia = 4.7 b Step 1: Apply the relationship pK + pK a pK + 4.7 = 14 So pK a Step 2: Convert = pK b to calculate pK w = 14 – : a 4.7 = 9.3 a pK to K a : a −10 pK = −log a K 10 = −log a (9.3) K 10 = 5.0 × 10 −3 mol dm a Key points For a using weak the base the general concentration of hydroxide ions can be calculated expression: + − ][OH [BH ] __________ K = b [B] + where [BH ] is the the concentration pK = a −logK and a concentration of pK unionised = b of the protonated base (salt) and [B] is base. −logK b 111 9.5 Changes Learning outcomes On completion should be able of this in pH pH–titration section, you Measuring describe changes in pH explain titrations the titrations curves pH changes changes Figure in strong or weak acid strong or weak base. pH when reacts with a a when 9.5.1 electrode from all the the are is an graphs acid shows placed burette time. continuously manually to by the is the in at The added base curves showing how the pH of an acid or base during changes acid−base acid–base to: Titration during a pH added apparatus the acid slow is using in and a data in the the an alkali used the to or an follow flask constant recorded recording alkali to and rate. alkali these the The added to changes. alkali added solution is an A acid. pH gradually kept stirred either: logger pH attached after xed to a computer volumes of or acid have been flask. (alkali) Titration The output to: graphs curves in Figure weak monoprotic acids in when it 9.5.2 acids which only reacts. The are one show the titrated hydrogen exact shape results with ion of per the obtained alkalis. molecule curve when strong Monoprotic of depends acid on acids can be whether and are donated the acid meter and data base are strong or weak. logger chart recorder a pH sensor Figure 9.5.1 b pH PC pH 14 14 12 12 acid Monitoring the pH as a base is added to an acid equivalence 10 10 point 8 8 6 6 4 4 2 2 0 equivalence point 0 0 5 10 15 20 25 30 35 40 45 50 –3 volume of 0. 10 mol 0 5 10 15 20 25 3 base dm added (cm ) c 30 35 40 45 50 –3 volume of 0. 10 mol dm 3 base added (cm ) d pH pH 14 14 12 12 equivalence 10 10 point 8 8 6 6 4 4 2 2 0 equivalence point 0 0 5 10 15 20 25 30 35 40 45 –3 volume of 0. 10 Figure 9.5.2 mol dm 50 0 5 10 15 20 25 3 base added (cm ) 30 35 0. 10 mol dm 45 50 3 base added (cm ) pH titration curves for: a a strong acid and a strong base; b a strong acid and a weak base; c a weak acid and a strong base; d a weak acid and a weak base 112 40 –3 volume of Chapter Apart very from steep units. The called the steep the point for of acid the where at a the and weak single which equivalence portion Strong curve portion the acid drop base point . Y ou and of has can weak base exactly see base, the changes the curves pH neutralised that this by the show Acid–base equilibria a several acid corresponds 9 to is the curve. strong base −3 Figure 9.5.2(a) shows the titration of 0.10 mol dm hydrochloric acid −3 (strong acid) with sodium 0.10 mol dm HCl(aq) + NaOH(aq) → hydroxide NaCl(aq) (strong + H base). O(l) 2 We can see why the pH changes so rapidly around the equivalence point 3 by calculation. Just before the equivalence point there 3 NaOH and is 24.9 cm of 3 HCl 25 cm present. The excess 0.1 cm of acid is present in 3 solution. 49.9 cm Exam tips 0.1 _____ + So [H ] = 0.1 −3 × = 0.0002 mol dm = pH 3.7 49.9 f 3 When there is 0.1 cm + [H −3 of alkali in excess: −14 ] = 1.0 × 10 [OH ] −11 /0.0002 = 5 × 10 = 0.0002 mol dm −3 mol dm So pH = you with titrate an titration 10.3. images The equivalence point is at approximately pH 7 because the salt base (in the flask) (in the burette), curves will be of those that the are the pH mirror shown, formed e.g. for is a acid a strong acid and a strong neutral. base: Strong acid and weak base pH −3 Figure 9.5.2(b) shows the titration of 0.10 mol dm hydrochloric acid 14 −3 (strong acid) with aqueous 0.10 mol dm ammonia (weak base). 12 HCl(aq) + NH (aq) → NH 3 The ammonium base and is chloride acidic as a formed result of is Cl(aq) 10 4 the salt of a strong acid and a weak 8 hydrolysis: 6 + NH Cl(aq) + H 4 So the O(l) → NH 2 equivalence point (aq) + H 3 is at about O (aq) + Cl (aq) 3 pH 4 5. 2 0 Strong base and weak acid 0 5 10 15 20 25 30 35 40 −3 Figure 9.5.2(c) shows the titration of 0.10 mol dm ethanoic acid (weak −3 acid) with sodium 0.10 mol dm hydroxide (strong base). Key points + CH COOH(aq) + NaOH(aq) → CH 3 COO Na (aq) + H 3 O(l) 2 The and sodium is ethanoate alkaline CH as a COO formed result (aq) of + is the salt of a weak acid and a strong hydrolysis: H 3 O(l) → CH 2 COOH(aq) + OH the equivalence point is at shape and weak pH–titration acid base and the strength of curve the used. (aq) about pH For a strong acid and a strong 9. there sudden base a on base Weak of depends 3 So The base is pH a very change large and corresponding acid to the end point of the titration. −3 Figure 9.5.2(d) shows the titration of 0.10 mol dm ethanoic acid (weak For a strong acid and a weak base −3 acid) with aqueous 0.10 mol dm ammonia (weak base). or a weak acid and a strong base, + CH COOH(aq) + NH 3 (aq) → CH 3 COO NH 3 there (aq) pH The ammonium ethanoate formed is the salt Both leaving the the ethanoate equivalence and point ammonium of at around ions pH a weak acid and a COO 3 (aq) + H O(l) 2 → CH large and sudden can undergo + to the point of the titration. hydrolysis 7. COOH(aq) corresponding weak For a there CH a fairly change end base. is 4 OH weak is a acid and gradual a weak change base in pH as (aq) 3 the acid is added to the base. 113 9.6 Acid–base Learning outcomes On completion should be able of this indicators Introduction to section, An acid–base pH changes know that indicator is a indicators substance which changes colour when the you over a certain range of pH values. Indicators can be thought to: of acid–base specific indicators as being weak acids whose conjugate base has a different colour . are + HIn used to determine the end indicator of acid–base titrations Y molecule the strength of the In deprotonated indicator depending colour on + H point acid A colour B and + In more acidic (or less alkaline) conditions the concentration of H of H base ions explain the applied to explain the term ‘pH range’ is higher . as So the position of So the indicator In equilibrium moves to the left. indicators basis for shows colour A. selecting + appropriate indicators for use more ions acid–base alkaline (or less acidic) conditions the concentration in is lower . titrations. So the position So the indicator Indicator usually indicators to of be its a yellow greyish at change is That a is colour moves to the right. B. pH 6 and colour the this the point, two blue over colour at of a pH the the colour extremes. pH 7.6. range indicator At For an of of one or should the two indicator example, units. change in may be bromothymol intermediate point it For the blue appears green. colour not At between pH The shows useful, range. intermediate is equilibrium range Indicators middle of 6 why adding acid to alkali pH 7.6 ↑ ↑ yellow grey-green blue change good ← ↑ of an indicator screened indicator alone since methyl should the orange also colour (a be distinct. change mixed is Methyl rather indicator) is orange indistinct. often used in preference. For an end indicator point happen, of the correspond The table some the to change working below correctly, titration rapid to work shows – the in the vertical pH in of the range some colour the part must of region the of change curve. the end sharply For this point at to must indicator . colour changes and the working range of indicators. Indicator Colour at pH in the lower range Colour pH at in the higher range pH orange red at pH 3.2 yellow at pH 4.4 3.7 methyl red red at pH 4.2 yellow at pH 6.3 5. 1 blue phenolphthalein yellow at pH colourless at 6.0 blue at pH deep 10. 1 reddish pH pink 7 .6 at at point methyl bromothymol 7 .0 pH 10 9.3 8.2 alizarin 114 the yellow yellow at pH at pH 13.0 12.5 end Chapter Choosing the pH correct 9 Acid–base equilibria pH indicator The selection depends on of a correct where the 14 14 12 12 indicator most rapid 10 change in pH occurs as the acid with a phenolph- thalein 8 titrated 10 phenolph- is bromothymol base. Most acid and indicators strong acid and strong can a be base used strong for a methyl base, 6 4 4 2 2 0 0 5 0 e.g. blue blue 6 Strong 10 15 20 25 30 35 40 45 50 –3 volume of because their the range vertical Phenolphthalein 0. 1 mol added (cm 10 15 20 25 30 35 40 45 50 –3 ) volume of a falls titration part is of often the of a strong acid with b a its 0. 10 titration base weak mol 3 base dm added (cm ) colour of a strong acid with a base curve. preferred pH because 5 3 base dm bromothymol strong within 0 orange, phenolphthalein, blue, thalein 8 change is pH more obvious. 14 Strong The acid sharpest and weak change in base the 14 12 12 10 10 thalein 8 curve base for is a strong between Methyl red, acid pH and 3.0 methyl a and orange phenolph- phenolph- pH bromothymol weak 7.5. blue 6 blue are 4 methyl orange because the their vertical part range of the falls 0 curve. 5 0 Phenolphthalein is 10 its colour to the 20 25 30 35 40 45 50 –3 change does vertical part 0. 10 mol dm 0 added (cm ) curve. It would slowly after only the 15 20 25 30 35 40 45 50 –3 0. 10 mol 3 base dm added (cm ) titration of a weak acid with d a titration of a weak acid with a of weak base change Figure 9.6.1 colour 10 volume of str the 5 3 base not c correspond 15 unsuitable volume of because methyl orange 2 0 within blue suitable 2 indicators bromothymol 6 or 4 bromothymol thalein 8 Selecting a suitable indicator for different acid–base titrations: a a strong equivalence acid and a strong base; b a strong acid and a weak base; c a weak acid and a strong base; point. d a weak acid and a weak base Weak The acid sharpest between its pH colour and not and fall colour change 7.0 and change methyl strong within the 11. pH the are not vertical before curve for a Phenolphthalein corresponds orange slowly in base the to the suitable part of weak is a vertical part indicators the equivalence curve. acid and suitable of the because They a strong indicator curve. their would Methyl range only base is because red does change Key points point. Weak No acid and indicator indicators changing is weak suitable would slowly. show For base for a a colour weak sudden example, Acid–base acid and change in a weak colour bromothymol blue base. None because starts the of pH changing indicators when there change in The range is change a sharp pH. the is the colour pH range of an (usually indicator about 2 is pH 3 when about of 24 cm alkali have been added and nishes changing units) over which the indicator 3 colour when 26 cm have been added. changes For in an an indicator acid–base colour to colour. the curve change place where in to work must the pH correctly titration, the correspond pH-titration changes suddenly. 115 9.7 Buffer solutions Learning outcomes What A On completion of this section, is buffer be able the buffer solution is solution? a solution which minimises pH changes when small you amounts should a of acids or alkalis are added. to: define term explain how control pH ‘buffer buffer solution’ Acidic buffers solutions An acidic base (the buffer salt solution of the is weak a mixture acid). An of a weak example is acid an and its aqueous conjugate solution of −3 describe the importance of buffer ethanoic acid ) (0.1 mol dm and its conjugate base, sodium ethanoate −3 solutions and in biological industrial systems ). (0.1 mol dm processes. + CH COOH(aq) Y CH 3 So a and If buffer the not ratio of change pH) On solution conjugate will base. conjugate contains We much, change addition acid say of acid the very to relatively that concentration very not COO (aq) + H (aq) 3 weak of high are conjugate hydrogen much this there (see buffer base ion also (salt) concentrations reserve base supplies (added salt) concentration Section of of both and (and acid base. acid does therefore the 9.3). solution: + ions H combine with CH COO ions. 3 So the position undissociated But because COO CH of , of equilibrium acid is the the shifts to the left and a little more formed. relatively ratio of high base to concentration acid changes of very added base little. 3 So On the pH addition hardly of changes. alkali to this buffer solution: + OH ions combine with H ions to form water . + The and removal a little of H more ions acid, shifts the position COOH CH of equilibrium dissociates to CH 3 But because COO CH , of the the relatively ratio of to the right COO 3 high base to concentration acid changes of very added base little. 3 A So the buffer pH may hardly be changes. made more effective by increasing the concentration of Exam tips both Remember the buffer and most base solutions completely salt that conjugate is in a of salt. ionised. You conjugate interchangeably. If acidic base t may used the does is see the acid ratio not of the amounts of position the conjugate concentration change therefore the and very pH) acid of concentration much, will or not alkali either the are the (salt). conjugate very added, has acid base hydrogen change equilibrium of of base been or so its ion much. the (added and concentration If, buffer far salt) however , will altered conjugate not as base to (and very work an large because lower by acid the unacceptable amount. Basic A buffers basic buffer solution is a mixture of a weak base and its conjugate acid −3 (salt). An example is an aqueous solution of ammonia (0.1 mol dm −3 its conjugate base, ammonium chloride (0.1 mol dm + NH (aq) 3 weak 116 + H (aq) ). + Y NH (aq) 4 base conjugate acid ) and Chapter This and On buffer solution conjugate addition acid. of contains There acid to relatively are this reserve buffer high concentrations supplies solution, of the added the both Acid–base equilibria base acid. aqueous + removes of 9 ammonia + ions. H A little more salt, NH , is formed. 4 On addition of alkali to this buffer solution, the aqueous ammonia + removes shifts the to the Buffers Many A left in of their of and will depend number H a on with hydrogencarbonate more work The pH at of specic the to the help buffer: in water . The ammonia systems buffers water form is position of equilibrium formed. systems only bodies. blood to little buffer Hydrogencarbonate combines ions natural enzymes animals parts added to pH keep blood is maintain Carbon blood form pH a near pH constant approximately this dioxide to values the (a 7. in So various pH 7.4. pH. product solution of respiration) of ions. + (g) CO + H 2 O(l) Y HCO 2 (aq) + H (aq) 3 + If the H ion concentration in the blood rises: + H The ions combine with HCO ions. 3 position of equilibrium moves slightly to the left. + The concentration There is buffer to still a of high ions H enough is reduced to concentration keep of the pH constant. ions HCO to allow the 3 function. + If the ions OH to ion form Phosphate blood concentration water . The buffers: plasma. The in the blood equilibrium Several type equilibrium of rises, moves slightly phosphate involved they ions 2 Y blood gets too acid, the right. present in the + H 4 acid If are the H + HPO 4 to with is: 2− PO H combine conjugate position of base equilibrium shifts slightly to the left 2− but there are still sufcient PO H 2 phosphates Protein buffers. the to act buffers: These amino as a acids or HPO 4 in solution in the for the 4 buffer . Some buffers and proteins can be and acidic amino or acids basic depending blood on the act as charges on Key points proteins: + PH Y + P + H or A buffer Y P protonated deprotonated protonated deprotonated form form form form in A industry solutions are used in electroplating, the manufacture of dyes a addition of buffer weak or Buffer on amount solutions minimises pH + H change Buffer solution + PH acid or solution acid weak and its base of a small base. is a mixture conjugate and its of a base conjugate and acid. treatment of leather . They are also present in some detergents, soaps and shampoos where it is important not to damage fabric, skin or Buffer solutions maintain pH by hair . keeping acid a fairly and Blood constant conjugate buffer ratio of base. solutions include − HCO ions, phosphate ions and 3 proteins. 117 9.8 Calculations Learning outcomes involving Equilibrium Consider On completion of this section, a buffer solutions aspects buffer solution consisting of a weak acid and its conjugate you base: should be able to: + CH know how to calculate the pH COOH(aq) buffer We data can write hydrogen know acid COO conjugate (aq) + H (aq) base solution from appropriate CH 3 weak a Y 3 of how to calculate the an equilibrium expression for this reaction in terms of ions: acid/ + base ratio required to make a COO [CH ][H ] _______________ 3 K = a buffer solution with a particular [CH COOH] 3 pH [CH [H COOH] [HA] ____________ 3 + So ] = K _____ + × or more generally [H ] = K a know how to conduct × a an [CH COO ] [A ] 3 experiment to determine the pH Since K is constant, the ratio of the concentration of acid to conjugate a of a buffer solution. + base (salt) controls the ion H concentration (and thus the pH). Dilution − of the buffer has no Deducing the We can the calculate value of effect on pH of its a since buffer the pH of of the weak K a pH buffer and A HA are diluted equally. solution solution if we know: acid a the equilibrium Worked Deduce concentrations example the pH of a of to weak acid and conjugate base. 1 buffer solution made 3 ethanoate the by adding 0.20 mol of sodium −3 of 500 cm 0.10 mol dm ethanoic acid. K of ethanoic acid a −5 = 1.7 Step × 1: −3 mol dm 10 Calculate the concentrations of the acid and its conjugate base: −3 [CH COOH] = 0.10 mol dm = 0.2 3 1000 _____ [CH COO ] −3 × = 0.40 mol dm 3 500 + Step 2: Rearrange the equilibrium expression terms of [H ] ions: COOH] [CH ____________ 3 + [H in ] = K × a [CH COO ] 3 Step 3: Substitute the values: 0.10 + [H ] _____ −5 = 1.7 × 10 × −6 = 4.25 × 10 −3 mol dm 0.40 + Step 4: Calculate pH: pH = −log [H ] −6 = −log 10 (to We can 2 (4.25 × 10 ) pH make buffer solution calculations easier to deal with if we expression: [salt] _____ pH = + pK log a 10 [acid] So if we are given the pK or calculate it from a calculation much more K , we can do the a simply. −5 Step 1: Convert K to a 118 = 5.4 10 s.f.) pK : a −log (1.7 10 × 10 ) pK = a 4.77 use the Chapter Step 2: Substitute into the 9 Acid–base equilibria equation: Exam tips log (0.40) __________ 10 pH = 4.77 + n buffer solution calculations, do (0.10) not fall pH = 4.77 + = 5.4 (to 2 concentration the of acid–conjugate base ratio same as conjugate hardly any an acidic buffer solution, the higher the acid–conjugate base lower with an a is the pH particular acid or in that [salt] Worked is pH, what concentrations. of we buffer . need volumes The the the example same example as to to But in know mix order below to exactly solutions shows [conjugate of thinking of hydrogen that ions make how of the how to a buffer much acid do this. Remember conjugate ionisation of that base the added as a acid. t salt. solution base and the concentration ratio, is the the base. of comes from For trap s.f.) is Calculating the the 0.6 the pH into to salt add of to known Remember base]. 2 3 How many solution of pH moles of containing sodium propanoate 0.10 mol of must propanoic be acid added to to make a 250 cm buffer of solution 5.30? −5 of K propanoic acid = 1.35 × 10 −3 mol dm a Step 1: Calculate the concentration of the propanoic acid: 1000 _____ [acid] = 0.10 −3 × = 0.40 mol dm 250 + Step 2: Calculate [H Step 3: Rearrange + ] the from pH: [H equilibrium −5.30 ] = 10 −5 = expression 5.01 with × [salt] 10 as −3 mol dm the subject: [acid] _____ [salt] = K × a + [H Step 4: Substitute the ] values: −5 (1.35 × ) 10 × 0.4 __________________ [salt] = −3 = − 5.01 × 1.08 mol dm 5 10 250 _____ Step 5: Calculate number of moles: 1.08 × = 0.27 mol 1000 Key points Determining the We can pH electrode into the from that nd pH of a connected buffer the the the pH of solution meter . The potential buffer to a and a solution pH the meter . value functioning difference buffer of (voltage) experimentally The of the solution glass the glass pH pH found electrode between the by using electrode by is direct based surface of a is glass the glass The pH can reading on the dipped be of a equilibrium fact weak and buffer calculated acid together solution using the concentrations and with its the of conjugate K the base value for the a the solution varies linearly with the pH. A hydrogen electrode (see section weak 10.1) can also be used to determine acid. pH. The acid/ solution the base can values and the K of ratio be of a buffer calculated pH value of of the the using solution weak acid. a The be pH using to of a buffer determined a a pH glass solution can experimentally electrode connected meter. 119 10 Electrode 10. 1 Introducng Learning outcomes potentals and standard Introducing equlbrum electrode electrode potental potentials 2+ When On completion of this section, a copper equilibrium should be able rod is placed in a solution is set the electrode (aq) ions, the following to: 2+ define Cu up: term − (aq) Cu of you + 2e Y Cu(s) ‘standard There potential’ are two opposing reactions: 2+ describe the standard hydrogen Cu ions in solution accept electrons and become Cu atoms. 2+ electrode The ions Cu get reduced. 2+ describe standard potentials for electrode metal–metal Cu atoms lose electrons and become Cu ions. The Cu atoms get oxidised. ion systems. For a relatively unreactive metal such as copper , the position of this 2+ equilibrium Cu atoms lies are to the right. relatively ions Cu difcult to are relatively easy to reduce and oxidise. Did you know? For The absolute metal and thought potential its to ions be in between solution caused by a a a relatively equilibrium reactive lies to the metal as magnesium, the position of this left. 2+ are build such − (aq) Mg + 2e Y Mg(s) up 2+ Mg of electronic charge on the ions easy of the of positive layer given metals. This attracts ions. An off charge by a more metal, difference a electrical is formed. The the to double greater, the the metal When a the metal ions in metal potential) solution. measure system. and relatively difcult to reduce and Mg atoms are relatively oxidise. layer electrons between are surface is It is the We is placed established not a this solution between possible difference call in to of the one its ions, metals measure between value the this a voltage metal–metal electrode voltage atoms and (absolute the directly. ion metal But system ions we and in can another potential solution. electrical metal surface – + – + – + – + – + – + – + – + – + The standard hydrogen electrode double layer If we we want need the a to standard The compare standard hydrogen standard hydrogen an a the at for of is electrode 101 kPa (1 different metals comparison. electrode hydrogen gas ability electrode by The denition consists to release electrode zero at all electrons, potential of temperatures. of: atmosphere) in equilibrium with −3 aqueous platinum the platinum gas which hydrogen temperature 298 of electrode hydrogen platinum the solution and hydrogen coated with hydrogen allows ions close at a concentration platinum ions. contact black (Platinum between in 1.00 mol dm contact black the of is with nely hydrogen both divided gas and ions.) K salt wire bridge hydrogen (101 gas kPa) finely divided Pt + H –3 (aq), 1 mol dm Figure 10.1.1 black on Pt The standard electrode hydrogen electrode The half equation for this electrode is: + H (aq) 1 __ − + e Y H 2 120 (g) 2 Chapter Measuring In order to standard measure the electrode electrode Zn place a rod of pure zinc potentials and equilibrium relating to the half equation: − (aq) + 2e Y Zn(s) −3 we Electrode potentials potential 2+ 10 into a 2+ 1.00 mol dm solution of Zn ions. 2+ This is the half Zn/Zn cell. We then connect this to a standard Ø hydrogen electrode . comparing the We voltage measure of the standard half cell with electrode the potential, standard , E by hydrogen electrode. V 0.76 V salt bridge H (101 kPa) 2 Exam tip Zn Pt It is the common bridge error to allows think that conduction 2+ + H a salt Zn (aq), (aq) of electrons. It connects two –3 1 mol dm ionic solutions, Ø Figure 10.1.2 Measuring E of ions the half between ionic other cells solution. affect is porous The of two half material when of electrode temperature 25 °C concentration pressure salt bridge. completing be salt in a high solution, a strip electrical be have potentials. conduction, ionic balance. movement circuit of potassium resistance we allows electrical is temperature So electrode This the bridge saturated potential. comparing a The soaked should pH by cells, maintained. ions, standard conditions connected voltmeter Concentration all the balance inert are must half cell maintaining two it 2+ for a Zn/Zn allowing The so lter so that paper or nitrate Remember a cell is that the sometimes ‘potential voltage called of its difference’. voltmeter . and to pressure use These of gases standard conditions are: (298 K) −3 of of any ions 1.00 mol dm gases involved 101 kPa. Ø The standard under other electrode standard half potential, conditions with E , the of a half standard cell is the hydrogen voltage measured electrode as the cell. 2+ For the the voltage half Zn/Zn cell connected to the standard hydrogen electrode, 2+ developed is – 0.76 V . If we replace the half Zn/Zn cell by a 2+ half Cu/Cu depends respect on to cell, the whether the voltage the standard half is + cell hydrogen 0.34 V . donates The or electrode sign of receives (see the cell voltage electrons Section with 10.3). Key points Electrode or ions The in potential solution standard measures to accept hydrogen the ability of an element to in H release electrons electrons. electrode is a half cell which gas at a pressure 2 −3 of 101 kPa is in equilibrium with a solution of 1.00 mol dm + H ions. Ø The standard half cell electrode under standard potential, E conditions , of a half compared cell with is a the voltage standard the hydrogen electrode. 121 10.2 Electrode Learning outcomes potentals A variety of and half cell potentals cells Ø There On completion of this section, by should be able are three main types of half cell whose E values can be obtained you connecting them to a standard hydrogen electrode: to: 2+ describe the standard measurement electrode potentials for non-metal–non-metal metal–metal non-metal–non-metal ion half systems the measurement electrode involving element oxidation define cells, the (aq) Zn(s)/Zn cells containing ion ions half half cell (see Section 10.1) of cells the same element in different oxidation of Half standard same half states. describe systems ion of in ions two of the different states ‘standard cells containing non-metals and non-metal ions potentials for cell If no metal using a is present, platinum reaction but solution of it its electrical electrode. must ions. be in connection The platinum contact Figure with 10.2.1 with is both shows the the inert. the solution It plays element chlorine/ is no made part and an chloride in the aqueous ion half potential’ Ø cell calculate standard cell connected to the standard hydrogen electrode. The value E for this potential, system is + 1.36 V . Ø E cell 1 Ø Cl (g) + e Y Cl in terms (aq) E = + 1.36 V 2 2 Ø Note: It does equation in not this matter way or as (g) Cl of + E values 2e Y 2Cl whether we write the half (aq) 2 V +1.36 V salt bridge Cl (g), 2 1 H atmosphere (g), 2 1 atmosphere Pt Pt 298 K Cl + H –3 1 mol –3 (aq), 1 mol dm dm Ø Figure 10.2.1 For a non-metallic bromide, in Measuring E the ions, the Half When a states, ions. sulphur/ cells in foil The cell the equilibrium should same be with half applies to in a its the ions, e.g. bromine solid in bromine/ liquid equilibrium and half with its sulphide. containing half both liquid platinum bromide e.g. for the chlorine/ chloride half cell ions contains ions ions must in different oxidation of have the the same same element standard in states different oxidation concentration of −3 . 1.00 mol dm A platinum electrode is used to make 3+ with the solution. Figure 10.2.2 shows the connection 2+ /Fe Fe electrical half cell connected to Ø the standard hydrogen electrode. 3+ Fe Some half cells MnO − (aq) may also − + e + 8H Fe + for acids or E = + 5e Y ions are included − The MnO , 4 −3 1.00 mol dm 122 is + 0.77 V . alkalis 0.77 V e.g. Mn Ø (aq) + 4H O(l) E = + 1.52 V 2 hydrogen reaction. system 2+ 4 The this Ø (aq) − (aq) value E 2+ Y contain + (aq) The . because 2+ Mn they are essential for the + and 8H are all present at a concentration of Chapter 10 Electrode potentials and equilibrium V +0.77 V salt bridge H (g), 2 1 3+ Fe –3 (aq), 1 mol 298 K dm 2+ Fe atmosphere –3 (aq), 1 mol dm + H Ø Figure 10.2.2 Standard We can 3+ Measuring E cell calculate for the Fe –3 (aq), 1 mol dm 2+ /Fe half cell potential the voltage of an electrochemical cell made up of two Ø half cells whose values E relative to the standard hydrogen electrode are known. Ø The voltage measured is the difference between the E values of the two Ø half cells. This is called the standard cell potential , . E The standard cell cell potential two is standard the half voltage cells are developed under standard conditions when joined. 2+ Figure 10.2.3 shows a 2+ ion Cu/Cu half cell connected to a Zn/Zn half cell. temperature 298 K + V –1. 10 V salt bridge zinc copper 2+ Zn 2+ 298 K (aq) Cu (aq) –3 1 Figure 10.2.3 mol –3 dm 1 mol dm Two half cells connected to give the standard cell potential Exam tips The two relevant half equations 2+ Cu − (aq) + Ø 2e 2+ Y Cu(s) E − (aq) Zn are: + = + 0.34 V Always = − 0.76 V the make sure that you include Ø 2e Y Zn(s) E you sign are of electrode doing potential calculations when involving Ø T o nd the value of E , we subtract the less positive (or more negative) cell electrode Ø E potentials. The negatives Ø value from the more positive (or less negative) E value. and positive can get confusing. Ø So in this case E = + 0.34 – (− 0.76) = + 1.10 V cell Key points The standard compared A electrode with platinum a potential standard electrode is of a hydrogen used with half cell is the voltage of the half cell electrode. half cells in which there is no metal electrode. The standard conditions cell when potential two is the standard voltage half cells developed are under standard joined. Ø Standard cell potential, E Ø , is the difference between the E values of the cell two half cells. 123 10.3 Redox reactons and cell dagrams Ø Learning outcomes E values On completion of this section, By convention, be able redox reactions electrode potentials refer to a reduction reaction, so you the should and electrons appear on the left of the half equations. to: Ø use standard electrode The more reduce to determine the direction in a (or less negative) the value of , E the ions or other species on the left of the the easier equation. it is They to are of better electron ow positive potentials at accepting electrons. So they are better oxidising agents. cell Ø construct The more negative (or less positive) the value of , E the easier it is to cell diagrams. oxidise the releasing weaker species on electrons. the They right are of the better half equation. reducing They oxidising stronger 2+ – (aq) Cr oxidising + Y (aq) + 2e Y Zn(s) + e –0.91 V = –0.76 V Y = 0.00 V power H (g) E 2 in species 3+ 2+ (aq) Fe + e Y Fe E = +0.77 V the + – (aq) Ag + e Ag(s) E = +0.80 V = +1.36 V (g) + e Y Cl (aq) E 2 2 stronger oxidising and release with the Zn weaker reducing agent list are above, good electrons ions Cl more readily) agent to e Ø Cl Cr – (lose 1 right Ø Y more readily) on Ø (aq) left Referring of Ø 2 e = reducing E 1 (aq) increasing E Ø of H Cr(s) – + (gain reducing Ø 2e 2+ Zn the at agent increasing species better agents. agent power are or we see reducing (the that: agents reaction is because further over they to are the more left) likely to compared Ag. 3+ Cl is a better oxidising agent than Fe because it accepts electrons 2 3+ more readily than ions. Fe 3+ Zn releases electrons more readily than H and Fe accepts electrons 2 + more readily than H The direction of When we connect electron flow two half in cells cells: Ø The half positive cell with the more with the less positive (less negative) value of E (more negative) value of E is the is the pole. Ø The half negative So when cell positive pole. the following two half cells + Reaction 1: Ag are – (aq) + e Ø Y 2+ Reaction 2: connected (aq) Cu Ag(s) E = – + 2e + 0.80 V Ø Y Cu(s) E = + 2+ the Cu/Cu 0.34 V + half cell is the negative pole and the Ag/Ag half cell is the Exam tips positive pole. Ø In order to deduce the direction of This is because according to values, E copper is better at releasing + electron ow, remember that: the electrons than silver and Ag ions are better at accepting electrons than 2+ more positive electrons. attracts the negative Cu goes ions. in the Therefore backward Reaction 1 direction. goes So in the the forward direction of direct electron 2+ external 124 wires is from the Cu/Cu and Reaction flow in + half cell to the Ag/Ag half cell. the 2 Chapter 10 Electrode potentials and equilibrium e salt Ag Cu bridge + Ag 2+ (aq) Cu (aq) + Ag Figure 10.3.1 2+ (aq) + e (aq) + 2e The direction of electron ow is from the negative to the positive pole Cell diagrams A convenient way of For a write cell diagram the outside of reduced the cell representing diagram. constructed as shown two below. metal/ The cells metal reduced is called ion a cell systems forms are on we the diagram. form oxidised form oxidised 2+ Zn(s) phase electrochemical from Zn form reduced 2+ (aq) Cu change Ø (aq) salt form Cu(s) E = +1.1 V bridge Ø The sign of E in these diagrams represents the polarity (+ or –) of the 2+ right hand electrode in the diagram. In this case the half Cu/Cu cell is 2+ + with respect to the half Zn/Zn cell. If we write the cell diagram the Ø other way round, the is E negative. 2+ Cu(s) The standard half Cu cell 2+ (aq) Zn diagram for Ø (aq) the Zn(s) E hydrogen = –1.1 V electrode is: + Pt [H (g)] 2H (aq) 2 The hydrogen gas and platinum electrode are regarded as being non-metals or ions a single phase. For more oxidation aqueous complex states, ions half the and a cells comma 2+ Pt If hydrogen bracketed or in Fe hydroxide the is involve shown separates 3+ (aq), Fe which platinum cell as the a separate different ions Cu appear diagram, the Cu(s) [Mn half E = + 4H O(l)], [MnO 2 reduced e.g. –0.43 V equation, they + (aq) the states are e.g. 2+ Pt different Ø (aq) in in from oxidation 2+ (aq) phase (aq) + + 8H (aq)] oxidised form Ag (aq) Ag(s) 4 form Key points The of electron ow (less negative electrode Cell ions direction negative In a diagrams which cell positive) species each take diagram oxidised the on a cell is from potential to the half cell with the most the half cell with the less potential. show may in electrode part of in the the reduced the oxidised and reduced species and other reaction. species are placed on the outside and the inside. 125 Ø 10.4 Usng E values to predct chemcal change Ø Learning outcomes Using Figure On completion of this section, E values to 10.4.1 compares predict some a reaction oxidising and reducing powers of some you Ø elements should be able and ions by comparing values: E to: Ø use E values to predict whether 2+ Mg a reaction is likely to occur or − (aq) + Zn species not. Ø 2e 2+ Y Mg(s) E Y Zn(s) E − (aq) + on (aq) + e −0.76 V = 0.00 V H (g) E = +0.34 V 2 gain right 2+ − (aq) Cu + lose Ø 2e Y Cu(s) E electrons electrons 3+ 2+ (aq) Fe more on Ø Y 2 left −2.38 V = species 1 + H = Ø 2e + e Y Fe Ø (aq) E = +0.77 V = +0.80 V = +1.36 V readily more + − (aq) Ag + e readily Ø Y Ag(s) E 1 Ø Cl (g) + e Y Cl (aq) E 2 2 Ø Figure 10.4.1 As the value of E gets more positive the species on the left are stronger oxidising agents and the species on the right are weaker reducing agents For each half equation, the more oxidised form is on the left. Ø The more reaction at positive to move accepting the in value the electrons of than the , E forward the greater direction. species is This above the is it. tendency because It is a it better of is the better oxidising agent. Ø The more reaction losing We can negative to move electrons use these the in the than ideas value reverse the to of the greater direction. species predict , E below whether This it. a It is is the is because a tendency better reaction will it is of reducing take the better at agent. place or Ø not. we If a say reaction that it is is likely feasible. to take For place using example: Will information zinc react from with values, E copper( II) ions? The relevant half reactions are: 2+ Zn − (aq) + 2e + 2e 2+ Zn(s) E Y Cu(s) E − (aq) Cu Ø Y = − 0.76 V = + 0.34 V Ø 2+ Cu Zn 2+ ions has a have a greater greater tendency tendency to lose to gain electrons electrons than than Cu. So Zn the ions and position of 2+ equilibrium of the Zn/ equilibrium of the Cu/Cu reaction Zn moves to the left and the position of 2+ reaction reaction moves 2+ In these predictions, as occur in 126 not the right. The overall is: Zn(s) do to a apply, we standard we have + Cu are 2+ (aq) → assuming electrochemical to consider Zn that cell other (aq) the + Cu(s) standard apply. If conditions standard information (see such conditions Section 10.5). Chapter The We anticlockwise can predict writing down negative the the the value direction pattern whether in reaction half from from potentials and equilibrium is feasible reactions by: with the one with the most top which starting pattern a the Electrode rule two at 10 the the the reaction bottom bottom left is feasible is given an anticlockwise left. is: reactant → product → reactant → product Ø E Ø E more negative (less positive ) reactant oxidised Example form reduced form 1 2+ Will product chlorine oxidise 3+ Fe ions to Fe ions? Ø E 3+ +0.77 Fe 2+ (aq) + e− Y Fe better (aq) reducing agent 1 +1.36 Cl can agent. see that Applying the the + e− Y Cl (aq) 2 2 better Y ou (g) oxidising better agent reducing anticlockwise 1 agent rule, reacts the with reaction 2+ Cl (g) + Fe the is better oxidising feasible: 3+ (aq) → Cl (aq) + Fe (aq) 2 2 2+ This because chlorine is better at accepting electrons than Fe ions and Exam tips 3+ Fe is better at releasing electrons than Cl ions. If Example you arrange the two half reactions 2 with the one with the most negative + Will aqueous iodine oxidise silver to silver ions Ag ? value be Ø at the the top, bottom the left reactants and top will right E 1 +0.54 species. I (aq) + e− Y I (aq) 2 2 better reducing agent + +0.80 Ag (aq) better Applying the + e− oxidising anticlockwise rule, Y Ag(aq) Key points agent the reaction is not feasible. The reaction Ø will be between silver ions and iodide ions rather than between silver A reaction value for iodide is feasible if the E and the reaction in the ions. forward direction is positive. Ø W ar ning! The feasibility of a reaction based on values E only tells us that a reaction is possible. It is no guarantee that a reaction will A happen. reaction Some reactions, although feasible, take place so slowly that they do value of to be happening. We say that these reactions are the half if the equation not involving seem is feasible Ø E the species being kinetically reduced is more positive than the controlled. Ø E the Ø Selecting E value of species the half being equation of oxidised. values The direction of a reaction can Ø Y ou must take care when selecting E values from a table of data. Some be half reactions may look very similar , especially if two ions are involved predicted from e.g. showing 2+ VO + + 2H + 2H 3+ + e Y V + e Y VO Ø + H + H O E = +0.34 V = +1.00 V 2 + + VO 2 2+ diagrams half equations and their Ø E values. Ø O E 2 127 10.5 Electrode potentals and electrochemcal cells Learning outcomes Concentration The On completion of this section, position of change equilibrium is and electrode affected by potential changes in concentration and you Ø temperature. should be able This also applies to redox equilibria. values E refer to a to: −3 concentration of ions of and 1.00 mol dm temperature of 298 K. What Ø describe how E values vary with concentration know how to happens when system? apply the T ake we for alter principles redox equilibria to concentration the half 2+ of + ions in a metal–metal-ion reaction: − (aq) Cr of the example Ø 2e Y Cr(s) E = −0.91 V energy 2+ storage devices If the concentration of Cr (aq) increases, the value of E becomes less including fuel negative. e.g. –0.83 V . cells. 2+ If the concentration more We can negative. apply reactions If the the If the are the value of E becomes principle to concentration changes in redox one: of shifts more the to ion the easily on the right as accepted, left of more the the equation electrons value of E are gets increases, accepted. more If positive negative). concentration equilibrium electrons less decreases, –0.99 V . concentration less the (or this (aq) Cr Chatelier ’s as equilibrium electrons (or Le such e.g. of are of shifts more the to ion the easily on left the as released, left more the of the equation electrons value of E are gets decreases, released. more If negative positive). 3+ For a system such as the concentrations of 2+ (aq) Fe both + ions e Y Fe equally (aq), has no increasing effect on (or the decreasing) value of E. 3+ This is because position of the increase equilibrium to in the the concentration right, but increase of in pushes Fe the the concentration of 2+ Fe pushes the Predicting If reactions difcult to a position of reaction are carried predict if a equilibrium under out the left to non-standard under reaction to non-standard is feasible. As a an equal extent. conditions conditions rough it guide is we more can say that: If the 0.3 V , electrode the potentials predicted of reaction two is half likely reactions to differ happen, e.g. by the more than reaction 3+ between copper difference is and ions Fe is likely to be 2+ + 2e Y Cu(s) 3+ the less difference than 0.3 V , E 2+ (aq) Fe If because in the + e Y electrode Fe potentials feasibility of the = +0.34 V = +0.77 V Ø (aq) E of two reaction half may reactions not be condence, cannot be e.g. predicted the reaction because the 3+ (aq) Fe between difference 2+ + e Y + e Y Fe (aq) 128 (aq) is 2+ ions and 0.03 V . Ø E + Ag Ag = +0.77 V = +0.80 V Ø Ag(s) E differ predicted + with the Ø (aq) Cu feasible 0.43 V . Fe ions by Chapter Batteries and Most batteries They often and cells have They are are They Some Button Many a solid cells The deliver can we use do not operate concentrations One equilibrium of under standard electrolyte. Modern conditions. batteries in size although high and lightweight. lightweight voltage for a ones Car are considerable batteries, being however developed. period of time. recharged. half cells type cell not such of so as cell there contain equations a those relies is used on no liquid the to power reaction spillage of electrolyte + The voltage the this of the type of negative pole and and electrolyte. electrolyte is a high iodine. Most a It is button paste. + e Y Li E + e Y I E cell cell should in use, are button the = −3.14 V = +0.54 V Ø conditions Another deliver 2 2 of the watch lithium Ø I voltage liquid or 1 The a of are: Li because and advantages: small a be potentials cells state do many heavy, button voltage. cells higher often quite Electrode cells and have 10 be however , not cell is silver +0.54 − is (−3.14) more = likely 3.58 V . to be about 3 V standard. the zinc−silver oxide is the oxide positive cell. The zinc is the pole. Did you know? Ø Zn(s) + Ag O(s) → ZnO(s) + 2Ag E = + 1.60 V cell 2 The Fuel use of fuel disadvantages In a cells have many cells fuel cell, a fuel (often hydrogen) releases electrons at one as well as advantages: electrode The storage of hydrogen in large Ø and for oxygen a fuel gains cell electrons with an at acidic the other electrolyte electrode. The relevant E values is a problem. Refuelling has are: + to be carried out Ø + 2H quantities 2e Y H E = 0.00 V 2 more + often than with a petrol Ø + 4H O + 4e Y 2H 2 O E = +1.23 V 2 engine. The overall reaction is: 2H + O 2 → 2H 2 O. 2 The fuel petrol or cells used diesel in some cars and buses have several advantages over W ater is hydrogen is produced from fossil fuels. engines: The the only product. No carbon dioxide or nitrogen oxides They do not work well at low temperatures. are released. They produce more They are efcient. more direct. very energy per The gram of fuel transmission than of petrol power to or the They are expensive. diesel. engine is Key points electron flow Ø V The ion value half cell negative negative positive electrode electrode the of ions Modern as in a metal/metal becomes the more concentration of decreases. batteries advantages H E in have terms of small size, O 2 low 2 electrolyte Fuel by Figure 10.5.1 cells using and high generate the voltage. a voltage energy from the Simplied diagram of a hydrogen−oxygen water mass reaction of oxygen with a fuel. fuel cell 129 Revson 1 For each reacton gven, wrte the questons specfied c A mxture of 1.40 mol and H 1.40 mol I 2 are 2 3 equlbrum expresson unts. Where K s and state requred, the assume assocated the placed partal and n the a 2 dm ask equlbrum pressures a i are 2NO measured Cl(g) Y n atmospheres. 2NO 2 Fe (g) + Cl 2 + I (g); Y Fe 0.36 mol K + I + equlbrum . What s (g) temperature, Y 2HI(g) s 2 mxture the value of K 2 contans at ths c temperature? __ (aq) H c 1 2+ (aq) establshed. The 2 – (aq) constant (g) 2 3+ ii at H p I (aq); K 2 2 c 3 d b i 3O (g) Y 2O 2 (g); (g) + ClF 3 Cl 2 (g) Y 3HF(g) + __ N 3 at 1 __ NH 6.00 mol SO are placed n a 3 dm ask 2 p 1 ii When K 3 (g) + Cl 2 2 a hgh temperature, 40% decomposes to SO 2 (g); 2 2 and Cl as shown by the equaton: 2 K p 2+ c i Ba(IO ) 3 (s) Y Ba + 2IO (aq); Al(OH) (s) Y Al 3 + 3OH (aq); i HF(aq) ii HCN(aq) + H O(l) Calculate K 3 K Y H 2 O Y SO (g) + Cl 2 at ths temperature. – (aq) + F (aq); K 3 a 4 a A mxture of 6.5 g H and 4.0 g N 2 + Y H – (aq) + CN (aq); a a HS + H O(l) Y H S(aq) + OH (aq); K 2 5.0 dm what vessel. What do you are the the total pressure Based on pressures of N s n 3.9 atm, understand by the term and H 2 For the equlbrum ? 2 system ‘dynamc 2NO(g) + Cl mxture at (g) Y 2NOCl(g), an equlbrum 2 equlbrum’? b If partal b b a placed – (aq) 2 2 are 2 3 K – e (g) 2 c sp + d (g) 2 sp – (aq) Cl 2 K 3 + ii SO – (aq) 2 Le Chateler’s prncple, explan a constant hgh temperature has how 3 partal pressures of NO = 9.6 × 10 Pa, Cl = 1.7 × 2 each of the factors lsted would affect the 4 10 4 Pa, NOCl = 2.8 × 10 Pa. Calculate K at ths p equlbrum system. temperature. PCl (g) Y PCl 5 (g) + Cl 3 (g); ∆H = +87 .9 kJ 2 c (Consder yeld, poston of equlbrum and value At a partcular contans only temperature, at N a a vessel partal ntally pressure of 2.50 atm 2 of equlbrum constant.) and O at a partal pressure of 1.50 atm. 2 i addng more PCl The 3 ii decreasng iii addng iv ncreasng v removng Cl a the equlbrum establshed pressure contans What the ? reacton 5 NH NO at t a s found partal the value the forward (aq) + H (aq) Y NH would be effect (aq), 4 reacton s Y of K that 2NO(g) the s pressure mxture of 0.76 atm. ? a What b The s meant by the term solublty? + 3 where the s and (g) p + For + O 2 temperature 2 c (g) 2 catalyst the N exothermc, what of: solubltes gven n What s i and the ii K n pure water, of the substances below were at temperature for ths determned at 25 °C. each sp substance? i ncreasng the pressure ii ncreasng the concentraton of i (aq) NH 0.710 g n of AgBrO –3 iii ncreasng the 400 ml of water. 3 3 ii temperature? 7 .52 × –3 mol dm 10 BaF . 2 c i The solublty product, for K MgCO sp 3 a For the system (g) 2NO Y 2NO(g) + O 2 –8 × 3 whch has reached equlbrum n a 2 dm 10 2 mol s 3.5 0.0222 mol NO , 6.34 g NO –6 dm at 25 °C. What would be the ask solublty, contanng (s) 3 (g), 2 at ths temperature n and 2 10.8 g O , what would be the value of K 2 , pure a water, along c –3 wth the approprate unts? soluton contanng 2+ Mg 0.015 mol dm ons? b For the system A(aq) + B(aq) Y R(aq), at a –2 partcular temperature, K = 4.5 × 10 3 dm –1 mol . ii Explan iii Would why the solubltes n i dffer. c In an equlbrum mxture at the same precptate MgCO f a soluton 3 –3 temperature, the known concentratons contanng n 0.055 mol dm magnesum ntrate –3 mol dm are [A] = 2. 10, [B] = 1.45. Calculate the s mxed wth a soluton –3 equlbrum concentraton of R. 1.22 × 10 Explan 130 contanng –3 mol dm your sodum answer. carbonate? Chapters 8–10 Revision 3 6 a The concentratons of HCl and C H 6 COOH are 8 A volume of questions –3 30 cm of 0. 15 mol dm KOH soluton 5 –3 both 0.025 mol dm . was placed n a concal ask 3 i Why s acds, the pH even dfferent for though they these have the 70 cm two and a total volume of –3 of 0. 10 mol dm HNO soluton added n 3 5.0 ml same alquots. The equvalence volume was found 3 to concentraton? ii iii What s What the s 25 °C, pH the of pH gven the of the HCl? the at K benzoc ths acd at temperature be 45 cm . a What s the ntal b What s the pH c What of s 20 ml of of × 10 What s HClO the The pH ? What of s a the of a s soluton? after the addton the pH of the soluton after addng 3 soluton 0.021 mol dm of 55 cm of HNO ? 3 pOH? 9 pH KOH 3 ? 4 c the soluton –3 mol dm –3 b of ? HNO a –5 6.3 pH the soluton was found to be a Draw a dagram of the apparatus whch would 11.4. What be used to determne the standard s electrode 4+ potental of the half cell Sn 2+ (aq)/Sn (aq). + i [H ], b i Draw a cell dagram for – ii [OH ] the two half cells Al 2+ a i Wrte base combnaton 3+ ? and 7 the a sutable equaton phenylamne, (C H 6 to show NH 5 ), how Zn of Ø (aq)/Al(s); E = –1.66 V Ø (aq)/Zn(s); E = –0.76 V the dssocates ii Calculate the standard iii Wrte balanced cell emf for ths cell. n 2 water. the equaton for the cell reacton. ii What s the conjugate acd of ths base? –10 iii The for C K b H 6 NH 5 s 4.3 × 10 –3 mol dm . 2 Calculate –3 the pH the pK of a 0.225 mol dm soluton, . b b i Explan why a mxture of H HC 3 acd) and NaC H 3 functon ntrc as acd a O 5 buffer and (sodum O 5 (lactc 3 lactate) would 3 whereas sodum a ntrate mxture would of not. –5 ii The for CH K a What by COOH s 1.8 × 10 –3 mol dm . 3 s the mxng pH of 15.5 g a buffer of CH soluton formed COOH (ethanoc acd) 3 wth 15.5 g of CH COONa (sodum ethanoate) 3 3 n a 250 cm volumetrc ask? 2– iii The K for CO b s the –4 s 2. 1 × 10 –3 mol dm . What 3 pH of a buffer formed by mxng –3 0. 105 mol dm –3 Na CO 2 NaHCO wth 0. 120 mol dm 3 ? 3 131 Exam-style Answers to all exam-style questions can questons be found on the Module 2 accompanying CD 6 Multiple-choice questions – From the data gven, whch of the followng are true? Ø E 20 1 A radoactve source has 1.6 × 10 atoms of a 4+ Sn radoactve sotope, wth a half-lfe of 3 days. 2+ (aq)/Sn (aq) +0. 15 V How 2+ V many atoms wll decay n 12 (aq)/V(s) – MnO 2– (aq)/MnO 4 19 A 1.0 × –1.2 V days? (aq) +0.56 V 4 19 10 C 4.0 × 10 4+ – i MnO s a stronger oxdsng agent than Sn (aq). 4 20 B 1.2 × 20 10 D 1.5 × 10 4+ 2 For the reacton X + Z → M, the concentraton ii V wll iii A reacton A i, B Only i and ii C Only i and iii D Only ii 2+ reduce Sn to Sn . 2+ of 2+ and V between Sn s feasble. –3 reactants tme s and products measured n s measured seconds. The n rate mol dm and equaton ii and iii are true. s are true. 2 Rate What concluson can be = k[Z] made from ths nformaton? 3 A The unt for the rate constant, k, s dm –1 mol B The unt for the C The reacton rate constant s . Consder the reacton, . (g) X + 2Y 2 rate doubles, f the concentraton The of reacton rate reactant X ncreases, as the For (g); ∆H = +58.0 kJ 2 of the followng the ncreases. reacton, (g) NO Step 1: NO (g) + NO Step 2: + CO(g) → mechansm (g) → NO 2 NO (g) combnatons NO(g) + CO the + CO(g) s correctly temperature? Position of equilibrium Yield Value of K shft decreases no to left change (g) B shft to rght ncreases ncreases C shft to rght decreases decreases D shft to left ncreases decreases proposed: + NO(g) → NO (g) + CO 2 rate the (g), slow (g) fast 2 8 s ncreasng 3 3 What of 2 two-step 2 effect c 2 the followng the concentraton A 3 Y2XY doubles. predcts D (g) 2 of Whch Z true. –1 s mol dm are true. –1 s 7 –3 s equaton for ths For propanoc acd, whch has reacton? –5 = K 1.35 × 10 –3 mol dm the dssocaton n water s a A Rate = k[NO ][CO] 3 CH 2 B Rate = k[NO CH 3 ] COOH(aq) + H 2 O(l) Y 2 2 – CH CH 3 C Rate = k[NO COO + (aq) + H 2 O (aq) 3 ][NO] 3 Whch D Rate = ][CO k[NO 2 of the followng statements are true? ] 2 – [CH CH COO + (aq)][H O (aq)] 3 2 3 ____________________________ i 4 Three monobasc acds X, Y and Z all have The K a = a [CH CH 3 COOH(aq)][H 2 O(l)] 2 –3 concentraton of 0.011 mol dm . The pK of X s 5.61, a –3 ii The pH of a iii Addng OH 0.010 mol dm soluton s 2. –7 the K of Y s 4. 13 × 10 and the pH of Z s 4.78. What a – s the order of the acds n terms of ncreasng would cause the equlbrum to shft acd to the rght. strength? iv A X < Y < Increasng the molarty of the acd would Z + result n a hgher concentraton of O H ons at 3 B Z < X < Y C Y < X < equlbrum. D 5 Z < Y Gven Z A i, ii and B ii, iii C iii and D i iii and iv are true. < X the followng standard electrode half cells and correspondng and iv iv are are true. true. potental values, and ii are true. Ø half cell E /V 9 3+ Mn – (aq) + Y 2+ Zn What s the solublty of 25 °C, f the solublty product Mn n Zn(OH) (aq) constant, + 2e Zn(s) –0.76 condtons s 3.0 Ø the correct cell dagram and E ? –6 cell 2+ A Zn(s) | Zn B Mn 3+ (aq) || Mn 2+ , Mn C Pt(s) | Mn D Zn(s) | Zn (aq), 2+ 132 (aq) 6.7 B 1.7 × –3 10 +0.73 V mol dm –8 × 10 –3 mol dm 2+ (aq) || Zn 2+ A 2+ (aq) | Mn 3+ (aq) (aq) | Zn(s) 3+ Mn –1.25 V –6 C 5.3 × 10 D 4.2 ×10 –3 mol dm 2+ (aq) || Zn 3+ (aq) || Mn at under sp Y –16 s K – (aq) water +1.49 these what pure 2 2+ e (aq), (s) | Zn (aq) +0.73 V 2+ Mn (aq) | Pt(s) +2.25 V –6 –3 mol dm × 10 3 mol –9 dm ? Module 10 The solublty of copper(i) bromde CuBr –5 water at 25 °C followng s 7 .28 × conclusons pure 12 Ths queston s based on Exam-style the two half questions cells gven –3 10 are n 2 mol dm correct, . Whch based on of the below, ths along wth ther standard electrode potental values. Ø data? E + –5 i The solublty wll be greater than 7 .28 ×10 , Ag f (aq) + e Y Ag(s) + CuBr s dssolved n a soluton of MnO KBr. The numercal value of the s K 5.3 +0.80 V 2+ (aq) + 5e Y Mn (aq) + 4H 4 –9 ii (aq) + 8H × a 10 O(l) +1.52 V 2 What do you understand by the term ‘standard sp 2 iii The unt of the would K be electrode –6 mol potental’? [2] dm sp b + iv The concentraton of on the Cu at n pure water, would be 7 .28 × 10 i Draw whch –3 i, ii B ii, and iii and iv are C ii D and and i, ii iii and iv are are iii be dagram used to of the set up apparatus, ths to determne the standard cell, cell n e.m.f. [3] true. ii iii labelled could mol dm order A a equlbrum, –5 On the Label dagram, Show true. the anode and the cathode. [1] electron ow. [1] true. are the drecton of true. c Calculate balanced the standard equaton for cell the e.m.f. cell and wrte a reacton. [2] Structured questions d Wrte e How the cell dagram. [2] –3 11 A student weak ppetted monobasc 25.0 ml acd of 0. 112 mol dm soluton nto a of a would ths was added an alkalne soluton, X, from each addton, the soluton was ensure complete mxng and the pH was shown on usng the a pH graph meter. The the student’s of n the respectve half cells affect the cell electromotve force? the why KCl would [2] Explan not be sutable for use n then the measured of swrled f to concentraton a value burette. After the concal ask. solutons To changng results salt brdge. [2] are 13 below: a i State ii Based the postulates on the of collson the collson theory, theory. [2] explan why 14 ncreasng the concentraton of a reactant 12 would generally cause an ncrease n the rate 10 Hp of a reacton. [2] 8 b The results shown n the table below were 6 obtaned for the reacton 4 A(aq) + B(aq) → C(aq) 2 where the ntal rate was determned by varyng 0 5 10 15 20 25 30 35 40 45 50 concentratons of A and B, at room temperature. 3 volume a Gve the name solution X or formula of an (cm ) alkal that could Experiment [A] [B] Initial –3 produce b the s results i What ii Determne the K shown of the on the graph. [1] acd? pK for the rate –3 ) (mol dm –3 ) (mol dm [3] a the (mol dm acd. –1 s ) –2 1 0. 125 0. 125 1.45 × 10 2 0.250 0. 125 2.90 × 10 3 0. 125 0.250 5.80 × 10 [1] a –2 c From the relevant equvalence volume data, determne the and any molarty other of the base. –2 [3] i d Explan greater why the than pH at the equvalence pont Sketch a graph rate a reacton 7 . of The soluton has ts maxmum bufferng reactant where V s be how the ntal determned from the the change n concentraton of a capacty V at llustrate could [2] montorng e to s over tme. [2] equvalence volume. 2 ii i ii What What s the do you ‘bufferng f The pH ndcator at ths pont? understand by the 6.8–8.4. State, red gvng sutable for use n a reasonng, determne the order wth respect to A the order wth respect to [2] B. [2] [2] has your a pH range reasonng, iii Wrte ths ttraton. the iv Calculate rate equaton for the reacton. [1] of whether a value for the rate constant, k, at t ths s your term capacty’? phenol Gvng [1] temperature, and state the unt. [2] [2] v What would be the rate of reacton, f the –3 startng concentraton of A was 0.30 mol dm –3 and B was 0. 14 mol dm ? [2] 133 11 The 11. 1 Physical elements Learning outcomes completon of ths Structure secton, be able explan 3 Period 3 and melting elements, their Period 3 elements points physical state at room temperature and you melting should Period properties of the The On of points in kelvin are shown below: to: the varaton n some Group physcal propertes elements and n terms bondng meltng wth ponts, conductvty, and of of Perod 3 I II III IV V VI VII 0 sodum magnesum alumnum slcon phosphorus sulphur chlorne argon 371 K 922 K 933 K 1683 K 317 K 392 K 172 K 84 K sold sold sold sold sold sold gas gas structure reference to electrcal electronegatvty densty. )K( Did you know? exsts n dfferent forms dfferent slghtly The meltng sulphur dfferent forms shaped dfferent (whch crystals meltng pont of has gniliob/gnitlem (allotropes). These have tniop Sulphur and ponts. monoclnc rod-lke 3000 crystals) 2500 boiling point melting point Si 2000 Al 1500 Mg 1000 Na P S 500 Ar s 392 K, but the meltng pont Figure 11.1.1 Melting and 0 of rhombc octahedral sulphur shaped Phosphorus also (whch crystals) has red, boiling points of the Period has s 11 It s the 13 whte 14 atomic 386 K. 15 16 17 18 3 elements number and We black forms. 12 can explain the differences in melting point in terms of structure and whte form bonding: whch melts at 317 K. Sodium, The magnesium ions sodium are to held and aluminium together aluminium by there a is sea an have of a giant delocalised increase in the metallic structure. electrons. number From of electrons + donated form. the to The Silicon, in lot of They b P increase forces with is to large metal ions number of between in melting lattice number ions order point similar of Mg of to strong forces. the to the So the Mg to Period diamond. covalent Al electrons, and Na 3+ and delocalised these the 2+ , Na the overcome increase highest It Al. 3 takes bonds. So high. chlorine points , has have because the a simple there of molecular are molecules. number contact S and covalent increasing sulphur , it between of the attraction points the and number So charge the very melting forces molecules. S low the is sulphur attractive with has giant break point W aals and c have to a when difcult boiling IV forms Phosphorus, ionic more and energy melting electrons the Group It of electrostatic the points the a the melting a is and elements. sea greater greater electrons the only The electrons points between higher melting structures. weak van in der the van der W aals molecule neighbouring point than 8 phosphorus, P . Chlorine is a gas since it is a diatomic molecule with 4 S a fairly small number of electrons. P Figure 11.1.2 The structures of: a sulphur; b phosphorus and; c chlorine 134 Argon these exists are only very as low, isolated so it has atoms. a very The low van der melting W aals point. forces between Chapter Density of The of table Cl Period below and Ar shows are temperature so those have Element 3 density their very The elements of Perod 3 elements the of 11 low of the liquids. Period They are 3 elements. gases at The density room densities. Na Mg Al S P S Cl Ar 0.97 1.74 2.70 2.32 1.82 2.07 1.56 1.40 −3 Density/g cm Density (iii) the depends way The and Si the density their has a on: (i) atoms the are increases mass lower mass of packed from the in Na atoms, their to Al (ii) the size of the atoms, lattice. as the size of the atoms decreases increases. density than Al because it has a more open lattice structure. The density Although together of the has Electrical P and atoms an S (and are Cl and heavier , Ar the as way liquids) the is relatively molecules are low. packed effect. conductivity of Period 3 elements −1 The electrical conductivity Period 3 elements hardly conduct as are (measured shown below. in siemens Chlorine per and metre, argon are Na Mg Al S Conductivity/ 0.218 0.224 0.382 2 P of the and S −10 × −17 10 1 × −23 10 1 × 10 −1 S m 10 Na, Mg metallic and Al are structure conductivity electrons Mg ) gases solids. Element 8 S m good free increases provided provides all are 2 and by Al conductors. to from each move Na to ‘atom’ provides 3). The when Al as a the increases The delocalised voltage is number (Na electrical electrons applied. of in the The delocalised provides 1 electron, conductivity of Si is poor Key points because (a it has substance no delocalised whose electrical electrons. It conductivity is classed as increases a metalloid with increase in temperature). Some of its electrons, however , can move out of Across of especially when ‘contaminating’ atoms are present in its Perod lattice. So it a semi-conductor . P and S hardly conduct electricity the elements because covalent molecules with no delocalised The 3 electronegativity of the atoms Na Mg increases Al a across Period S P 1.2 1.5 to atomc. meltng reects pont 1.8 2. 1 2.5 bondng the of the elements. 3. S and Across Perod 3 electrcal Cl alumnum 0.9 n perod conductvty Electronegativity to elements Element covalent molecular structure The changes from gant varaton across Period to electrons. Electronegativity of structure they smple are the is metallc called 3 position, ncreases then to decreases. 3.0 The densty Perod then 3 of the ncreases shows a to general Electronegatvty a elements n alumnum decrease. ncreases across perod. 135 11.2 Patterns Learning outcomes in Period Patterns The On completon of ths secton, covalent be able elements atomic radius is and often ionic used as radii a in measure Period of the 3 size of an atom. you This should in 3 is half the distance between the nuclei of two atoms of the same to: type. explan and the onc varaton radus of n atomc Perod 3 elements r descrbe the elements reactvty wth of oxygen, Perod 3 chlorne Figure 11.2.1 The covalent radius of an atom, r, is half the distance between the two nuclei and water predct the type of bondng n the The chlordes and oxdes of Perod atomic each elements by dfferences reference n onc to radus and there The the is type covalent or radus’ depends bond formed: van example, for decrease element into the additional in Period 3. As we move across the period, So the has one shielding charge size of more quantum the of proton shell outer from shell Na atoms in (the to its nucleus. third electrons Cl pulls decreases But electron by the across the the shell). inner electrons the added So shells. closer to period. Na Mg Al Si P S Cl 0. 156 0. 136 0. 125 0. 117 0. 110 0. 104 0.099 on metallc, der Waals. across same nuclear atom ‘atomc of no increase size the goes nucleus. Did you know? term successive electron electronegatvty. The radii 3 of atom For and nuclear charge sodum: Atomic covalent radus = 0. 134 nm metallc radus = 0. 191 nm (covalent) radius van der Waals radus = (nm) 0.230 nm Figure 11.2.2 Covalent when a rad are comparng generally atom szes The atomic radii of the elements from sodium to chlorine used across Metal shell. perod. ions So are their Non-metal shell. the ions There atom. which it formed ions is are by losing smaller formed more This is are by repulsion makes the electrons than gaining negative ionic each from their ion outer electron atoms. electrons between the in their electrons larger than outer than the electron there atom is radii successive decrease ion has from one 4+ to Na more Si . proton As in we its move across nucleus. But the period, electronic + structure pulls A the is the same. electrons similar The closer explanation increase to the (higher in nuclear nucleus. nuclear So charge the charge size and from of same the Na ions 4+ to accounts for the decrease in ionic radius from 3– P 2– S 0.20 Cl )mn( suidar + Na 0. 10 2+ cinoI Mg 3+ Al 4+ Si 0 11 12 13 14 atomic + Figure 11.2.3 136 The ionic radii from Na − to Cl 15 number 16 17 P to Si decreases. electronic 3− structure) in from derived. + The their Cl Chapter The reaction of Sodium reacts Period 3 elements with with steam reacts to + 2H very form O(l) → 2NaOH(aq) + slowly with Perod 3 H + does not O(g) H react (g) cold water but it reacts readily oxide: → MgO(s) + H 2 Aluminium of 2 magnesium Mg(s) elements water 2 Magnesium The vigorously: 2Na(s) 11 (g) 2 with hot or cold water but it reacts with steam: 2Al(s) + O(g) 3H → Al 2 Silicon, phosphorus and O 2 sulphur do (s) + 3H 3 not (g) 2 react. They are insoluble in water . Chlorine of dissolves hydrochloric slightly in chloric( i) and water acid and then reacts to form a mixture : + Cl (g) + H 2 The reaction of Sodium O(l) Y 2H (aq) + Cl (aq) + ClO (aq) 2 reacts Period vigorously 3 elements when heated to with oxygen form sodium oxide, Na O 2 (although the nal stable product is sodium peroxide, Na O 2 2Na(s) + O (g) → Na 2 Magnesium and aluminium O 2 react ): 2 (s) 2 vigorously with oxygen to form oxides: 2Mg(s) + O (g) → MgO(s) and 4Al(s) + 3O 2 Silicon reacts (g) → 2Al 2 slowly with oxygen: Si(s) + O (g) → SiO 2 Phosphorus reacts vigorously with oxygen O 2 (s) 3 (s) 2 to from phosphorus( v) Did you know? oxide: The 4P(s) + 5O (g) → 2P 2 O 2 bleachng chlorne Sulphur burns steadily in oxygen: S(s) + O (g) → SO 2 Chlorine The Apart and argon reaction of from argon do not combine Period (and 3 chlorine all of most with 3 chlorc(i) due to the formaton of acd. oxygen. with Period s (g) 2 directly elements itself) acton (s) 5 chlorine elements react with chlorine. Sodium, magnesium and MgCl AlCl 2 and and aluminium respectively, which react are vigorously ionic solids. to form Silicon, NaCl, phosphorus 3 sulphur react more slowly. with sodium: with silicon: with phosphorus: with sulphur: For example: 2Na(s) + Cl (g) → 2NaCl(s) 2 Si(s) + (g) 2Cl → SiCl 2 2P(s) + (l) 4 (g) 5Cl → 2PCl 2 2S(s) + Cl (s) 5 (g) → 2 S Cl 2 (l) 2 Key points The man atomic radius and the dstance of the outer Across Perod water nuences on charge tends Across 3, to Perod the reactvty of the n Perod shell 3 are the sze of the electrons from the elements wth oxygen, nuclear nucleus. chlorne and decrease. 3, the chlordes and oxdes change from onc to covalent compounds. 137 11.3 Properties Learning outcomes of Oxidation The On completon of ths Period secton, oxides be able descrbe and descrbe and n of the oxdes terms varaton states oxdes elements is in more the Perod 3 bondng of of of n Perod elements n 3 rises chlordes use oxidation across all leads the to the Period 3 all exist electronegative an in than positive any of oxidation these states elements. and in expanded of each element This corresponds their outermost octet for oxides in to Period the electron of P , S and 3 in their The oxides ability of the shells in bonding. elements to This Cl. chlordes elements electronegatvty radus state period. electrons Oxide Na O MgO Al 2 O 2 SO 3 P 2 O 4 SO 10 Cl 3 O 2 7 and Max. oxidation onc chlorides to: the oxdaton and states oxygen maximum oxides you because should of 3 +1 +2 +3 +4 +5 +6 +7 charge state explan the behavour trend of n oxdes acd–base and The hydroxdes across Perod non-metallic oxides can also form oxides in lower oxidation states 3. e.g. SiO, P O 2 O Cl 2 , SO 3 and several lower oxides of chlorine (Cl 2 O, ClO 2 , 2 ). 6 The chlorides because period. rising The to of Period chlorine is oxidation +5 in 3 more PCl . elements also exist electronegative states The of the chlorides maximum in that positive the other show oxidation a oxidation elements similar state for states in the pattern, sulphur in its 5 chlorides, however , Chloride is +2. NaCl MgCl AlCl 2 Max. oxidation +1 SCl 3 +2 PCl 4 +3 SCl 5 +4 2 +5 +2 state Some of the oxidation non-metallic states, e.g. elements , PCl S 3 Bonding Did you know? The Many onc compounds have in structure and of covalent character on small the polarsaton. Ths hghly-charged electron larger lkely anon. to cloud Ion occur s when caton charge hgh charge when s i the a ii has a hgh charge and in 2 and oxides of of these of the atoms Period compounds involved electronegativity, chloride will be and Mg, or Na, s are transferred can be 3 related the relative the in more bonding. likely it is The that greater the the oxide ionic. The electronegativity difference between or oxygen are therefore is ionic. so the great that metal one, atom two to the or three electrons non-metal atom. respectively These The other Period 3 oxides are covalently oxides bonded. has dioxide the s Al from most has a giant covalent structure (see Section 2.6). Although 4+ in anon lower much caton small, in dstorts of polarsaton and chlorides a Silicon a form because difference of bonding also a electronegativity degree 2 chlorides can Cl large. theory, that the an ion Si lattice can energy exist, cannot the fourth compensate ionisation for it to energy make is ionic so large stable. SiO 2 Oxides a of P , S and Electronegativity b Period + 3 Cl a differences chlorides. anhydrous have Sodium aluminium simple can also chloride chloride molecular is a be and used structure. to explain magnesium the structure chloride covalently-bonded are molecule, of ionic. Al Cl 2 But . 6 3+ Al Cl 3+ The separate ions electron ionic up is very small and the Al ion This high positive charge density tends to pull the chloride ion between towards the Al and it to Cl such atoms. an extent This is that called electron ion pairs are polarisation The polarisation of a large ion by a small, highly charged Al 138 highly electrons 3+ Cl is ‘atoms’ shared Figure 11.3.1 aluminium between larger the of density charged. builds radius ion Chlorides of Si, P and S have a simple molecular structure. in the Chapter Reactions of The oxides of Period sodium and 3 oxides magnesium and react 11 The elements of Perod 3 hydroxides with water to form hydroxides: Na O(s) + H 2 Magnesium sodium basic. hydroxide hydroxide. They react is less Oxides with + alkaline and acids MgO(s) O(l) → 2NaOH(aq) 2 to because hydroxides form 2HCl(aq) a of salt → it is less sodium and MgCl water . (aq) + oxide does not dissolve in water than magnesium are e.g. H 2 Aluminium soluble and O(l) 2 but reacts with acids and alkalis: Al O 2 Al O 2 (s) + 3H 3 (s) SO 2 + (aq) → Al 4 2NaOH(aq) (SO 2 + 3H 3 ) 4 O(l) → + 3H 3 2 substance which acts both as an acid (aq) 4 sodium A O 2 2NaAl(OH) and a base is aluminate said to be amphoteric Silicon dioxide is insoluble in water but like Al O 2 SiO (s) + 2NaOH(aq) → Na 2 SiO is an acidic SiO 2 oxide. It does not react reacts (aq) + H 3 with with hot alkali: 3 O(l) 2 acids. 2 Oxides acidic of P , S and solutions Cl and are all with SO acidic alkalis (g) + H + 6H 2 O P 4 Reactions of the Chlorides dissolve of in sodium water oxides. to form O(l) → H → 4H 2 (g) 6 they SO are water to form (aq) (aq) 3 Period and with 3 PO 3 magnesium because react e.g. 2 O(l) 2 chlorides of and They salts, 3 hydrated aluminium chloride Did you know? ionic. Anhydrous The chlorides of silicon and phosphorus are hydrolysed by water to has acidic solutions. Fumes of hydrogen chloride are also (l) + 2H + 3H 4 3 SiO + 4H 5 (s) + 4HCl(g) (aq) + 3HCl(g) → H PO 3 O(l) → 2 H PO (aq) + behaves chlorde 3 3 smple chlorde molecular dfferently alumnum 2 O(l) 2 (s) PCl → 2 (s) PCl O(l) a structure. released. It SiCl alumnum form 5HCl(g) water 4 to hydrated chlorde. Alumnum soluton to form an s hydrolysed acdc 3+ The hydrolysis of S Cl 2 thionic acid being is complex, several sulphur compounds including 2 [Al(H O) 2 ] 6 by soluton: 2+ → [Al(H O) 2 OH] + + H 5 formed. Key points The oxdaton rest, Chlordes chlordes Across Sodum s states apart from Ar, of Na are oxdes and Mg 3, the n to form oxdes water acdc of S, and P, S Na to Al are xed. The states. to form onc solutons. The other solutons. change from oxdes oxdes chlordes from oxdaton dssolve magnesum amphoterc. The and several hydrolysed Perod and of have basc to hydroxdes and Cl are amphoterc are (Al) to acdc. basc. Alumnum oxde acdc. 139 12 The 12. 1 Propertes chemstry Learning outcomes of Groups of Group Physical Elements On completon of ths secton, be able explan the of Group II structure descrbe araton n propertes elements n terms and the elements and VII elements properties of in the periodic orbitals, s, p, d table or f in can be their put outer into blocks shell. according Group II to elements the are in to: the IV you type should II II, s block outer reactons the Periodic T able. They have two s-type orbitals in their shell. of bondng wth in s-block of Group oxygen, water II p-block and d-block dlute state acds some uses of MgO, CaO, Figure 12.1.1 The s, p, d and f f-block Ca(OH) and CaCO 2 blocks n the Perodc Table 3 The table shows some physical properties of these elements. −1 Element Atomic Metallic Ionic Density/ Melting Ionisation energies/kJ mol −3 number radius/nm radius/nm g cm point/°C First beryllum, Be 0. 122 0.031 1.85 12 0. 160 0.065 1.74 694 763 calcum, Ca 20 0. 197 0.099 1.55 839 590 1150 strontum, 38 0.215 0. 113 2.60 769 548 1060 56 0.224 0. 135 3.51 725 502 966 magnesum, barum, 4 Mg Sr Ba Dd you know? Metallic The The metallc radus s half between two lled n a metallc s block radus s gant elements, greater than electron metallic 1450 radius ionic shells radii both increases. increase The down ionic the radius group is as much the number smaller radius because when forming ions the electrons in than the outer structure. shell For and 1760 neghbourng the nucle metallic ionic 900 the of dstance and 1278 Second the metallc the coalent have been lost. Density radus. The metallc dagram radus, shows the The r in density the Beryllium are r From and packed on the calcium magnesium more efciently mass to have in of the barium higher the atoms there is and an densities lattice with the way increase than less they in calcium wasted are packed density. because they space. r Melting The As points melting we go Ca, the So Sr and sea of there takes follow a the spoils Ba. are less high, group this Going delocalised is less points down Magnesium 140 depends lattice. similar to there trend down the is a are of general group between overcome trend. expected because electrons attraction energy as it has the these giant metallic decrease a away melting the and forces. structure. lattice which from electrons attractive in different electrons further these a contribute positive the The points. structure nuclei boiling to to nuclei. and it points Chapter First and go second As we further screened These is to Some When group, from more the the outweigh the these Group II they generally exhibit because chemical meals act and aluminium it as effect to other Reaction with oxygen a redox they reaction. greater the as II of Groups II, IV and VII are: the The inner nuclear down two From the electron charge shells. and so it group. electrons magnesium are outer II of metals properties Group go They the number increased we lose agents. lose physical the electrons of reactions. easier some than react, here. is the electrons considered This outer by reducing similar is chemistry nucleus effectively remove The energies reactions of Group So group the away factors easier shell. down ionisation 12 more which are reactivity with So down Beryllium more its oxygen their outer barium, reactive electrons. elements. from to like the has those reactions increases they are of not down the group. 2Mg(s) + O (g) → 2MgO(s) 2 The oxides are basic in character , CaO(s) + reacting H O(l) → with water Ca(OH) 2 Reaction with Magnesium steam to Calcium very slowly magnesium reacts vigorously Ca(s) white hydroxides: (s) 2 + with oxide with 2H cold and cold O(l) water water → precipitate is but hydrogen to more (see form Ca(OH) 2 A form water reacts form to (s) calcium + H 2 observed, but some vigorously Section with 11.2). hydroxide: (g) 2 Ca(OH) dissolves to form a 2 weakly alkaline similar manner alkaline down solution. with the The other increasing group as the elements vigour . The solubility of down the resulting the group react solutions hydroxides get in a more Key points produced increases. Reaction All the with members salt of of the acids the group metal. The react readily reactivity with acids increases to down form the hydrogen + 2HCl(aq) → MgCl (aq) + II Ba metals are elements from water to metal hydroxide. produce H SO H 2 (aq) → BaSO 4 The atomic (s) + H 4 compounds of Group Calcium oxide: making Calcium hydroxide: cement and mortar; making acidic soil; drying Group II limestone blocks for metals oxides making bleaching powder; The the metals group. burn which building; down reactivity elements dilute carbonate: of the in are air to more the group. agent limewater Calcium radii down II neutralising removing as SiO of Group with oxygen, hydrochloric down the Many compounds II water acid and increases group. 2 slag in the blast furnace for the extraction of iron; making calcium oxide for of Group II cement metals Magnesium oxide: for lining a (g) soluble and 2 form some hydrogen (g) Uses of to with 2 increases + Mg react group: 2 Ba(s) which and Mg(s) Group have important uses. furnaces 141 12.2 Group II Learning outcomes carbonates, Thermal decomposition of Group On completon of ths secton, be able II carbonates from form the oxide and explan the araton n barium, (s) and explan araton II The ntrates solublty of Group know the test for usng barum II of → CaO(s) + decompose CO 3 thermal of Group carbonates the to on heating dioxide: to: decomposton sulphates carbonates magnesium carbon CaCO and you to should ntrates temperatures shown at which (g) 2 these carbonates are 50% decomposed are below. the Carbonate sulphates sulphate MgCO CaCO 3 SrCO 3 BaCO 3 3 ons Temperature/K 813 1173 1553 1633 chlorde Going down Group decomposition. go down the The size The smaller charge of of Smaller So II a carbonates II cations cation cation, larger are the the C=O ionic become have the more same resistant charge to (+2) thermal but as we increases. better is at distorting the electron cloud ion. polarisers with it of large smaller ions ions (see have a Section greater 11.3). degree of bonding. degree bond the carbonate better carbonates in greater break metal the the Group The the ions covalence the Group group: II, All in of covalence, the the carbonate less is the energy required to ion. 2+ 2+ Mg Ba 2– 2– CO CO 3 3 2+ Figure 12.2.1 a The small Mg on s a good polarser of the carbonate on; b The larger 2+ Ba on s a poor polarser of the carbonate on Thermal decomposition of Group form II the nitrates oxide, from magnesium nitrogen 2Mg(NO ) 3 Going down Group decomposition. The better nitrate 142 it is (s) and barium, decompose the 2MgO(s) + 4NO (g) + O 2 nitrates the distorting on heating also become (g) 2 more resistant to thermal cation: the The greater The less the energy ion. degree of required electron covalence to break to oxygen: 2 smaller at → to cloud charge of the larger ion. nitrate II, The dioxide nitrates a in the ionic particular bonding. N−O bond in the Chapter Solubility of Group The solubility Magnesium of Group sulphate II II is The chemstry of Groups II, IV and VII sulphates sulphates very 12 in soluble. water Barium decreases sulphate down is the group. ‘insoluble’. Ø Solubility depends on the enthalpy change of solution, ∆H . As a rough sol Ø guide, the more endothermic the value of , ∆H the less soluble is the sol salt. The enthalpy change of solution depends on the relative values of Ø the lattice energy, and ∆H the enthalpy changes of hydration of the latt Ø aqueous ions, (see ∆H Section 6.3). hydr 2+ Ca 2– (g) + SO (g) 4 2+ H [Ca ] hyd 2– +H [SO hyd ] 4 H latt (aq) Ca SO 4 H sol Ca SO (s) 4 Ø Figure 12.2.2 An enthalpy cycle for calculatng ∆H sol Lattice energy decreases MgSO down > the CaSO 4 This the decrease large The But the is fairly sulphate enthalpy this size ion decrease of the small is because by of By Hess’s law: than So > order: BaSO it is 4 determined more by the size of cation. also larger the increases because it is in the same determined order . more by anion. Ø = ∆H Ø ∆H sol Ø SrSO the 4 hydration Ø the relatively cations > in 4 than change group − ∆H hyd latt Ø ∆H − ∆H hyd gets more endothermic down the group (the value of latt Ø gets ∆H more Testing for The relative following positive). sulphate insolubility test to detect of So the solubility decreases. ions barium sulphate sulphate is used as the basis for acid or hydrochloric the ions: Key ponts Acidify This acid the removes to form precipitate Add If solution to aqueous tested contaminating carbon of be dioxide. barium barium nitric carbonate If these carbonate chloride with would or ions, were not which interfere aqueous react removed with barium with acid. the insoluble the white Group are test. II more ions are present a white precipitate of Ths nitrate. barium sulphate resstant decomposton s because degree sulphate carbonates of and to down of the ntrates thermal the group. decreasng polarsaton of the is 2− large CO formed: or NO 3 2+ Ba (aq) SO (aq) 4 → BaSO by the 3 ncreasngly 2− + on large metal caton. (s) 4 The decreasng Group II group can of relate the enthalpy and to a be alues barum a of of n of the the terms ther hydraton energes. soluton sulphate, down explaned changes lattce When solublty sulphates chlorde s contanng whte added a precptate s formed. 143 12.3 Group IV Learning outcomes elements Some The On completon of ths secton, be able Group four explan the propertes elements structure descrbe Group descrbe araton of n n physcal the Group terms and the IV of in exsts the n bondng the of reacton wth to tn whch made by of the too tn cold cold may f their in the outer p block of principal the Periodic quantum T able. shell. They They are all all at the room temperature. Some of their physical properties are shown table. Structure Melting Electrical point/°C conductivity 3550 non-conductor the carbon (damond), C non-metal of Group slcon, S metallod 1410 sem-conductor germanum, Ge metallod 937 sem-conductor tn, metal 232 conductor metal 327 conductor IV has dependng as a lead, structure In countres become are Pb powdery a clmates, they Sn low exsts damond. hae in Element Trends whch are elements water. seeral forms, temperatures, smlar electrons bondng temperature. At grey form elements IV IV ther Dd you know? on properties of Group tetrahaldes tetrahaldes Tn IV tetrachlordes to: solids physical ther you have should and artcles and (see damaged left Si the melting Ge have Section fact outsde for in that a giant 2.6). lot points of covalent They all energy is structures have very needed similar high to to melting break the that of points many C(diamond) due to covalent the bonds in structure. long. The decrease decreasing in melting amount of point energy from C(diamond) needed to break to the Ge reects −1 E(C−C) = the bonds: −1 E(Si−Si) +350 kJ mol = +222 kJ mol −1 E(Ge−Ge) Sn and other the so Pb ions have There Carbon its does free relatively Si a in the move Ge melting of points metallic Section melting to are in 2.4). compared bonding Tin and with decreases lead have as most the fairly size large of ions, points. electrical diamond are some Section of Some different increase decreases does used its conductivity not in down conduct covalent electrons are the group. electricity bonding. because Graphite delocalised and so are 2.6). electricity electrons. conductivity metalloid of because (see with described trend electrons conduct ‘jumping’ increases 144 low form shell conduct delocalised by (see general outer and low strength +188 kJ mol conductivity is to relatively The increases Electrical all have metals. = to of places in a very their in small the lattice. temperature with increase semi-conductors . extent. electrons in Their (as can They move Their opposed is not of have position conductivity to temperature). structure do out metals, Si and described where Ge as are Chapter Sn and are is Pb are electrical delocalised. a greater electrons nuclear second trend force in All the all a conductors slightly attraction with from atomic Sn Sn. to This is This energies is its Sn. than in This a ions because results down their conductor greater than energies because better between Pb radius. ionisation The Group are of ionisation of is compared charge increase Sn is group and the the effect having opposite (see shell Pb. In of of chemstry of Groups II, IV and VII electrons lead there the to increased shielding higher Section The delocalised effect the lead outer than 12 the rst and and general 1.5). IV tetrachlorides elements in Group IV form tetrachlorides, XCl . The tetrachlorides 4 simple covalent molecules with tetrahedral structure. Cl Ge Cl Cl Cl Figure 12.3.1 All the This der The tetrahedral structure of germanum(iv) chlorde tetrachlorides is because W aals dipoles they forces cancel. are are volatile simple between their Although liquids covalent and CCl SiCl , SnCl 4 , and PbCl 4 , they They have are boiling only non-polar lower boiling points. weak van because points the than 4 there is no = 76 °C SiCl 4 simple pattern in = 57 °C GeCl 4 = SnCl the boiling points: PbCl = 87 °C 105 °C 4 Reactivity of the Group tetrahalides, = 4 114 °C 4 the low with 4 CCl All have molecules molecules. 4 GeCl − with IV tetrahalides the exception of CCl with , are water readily hydrolysed by 4 water to chloride the are oxide also in the +4 produced, (l) SiCl oxidation hydrolysis of the + 2H (l) + ease of hydrolysis → SiO fumes of hydrogen (s) + 4HCl(g) 2 tetrahalides 2H 4 The O(l) 2 heavier GeCl Acidic e.g. 4 The state. O(l) Y is GeO 2 reversible. (s) + For example: 4HCl(g) 2 increases down the group from SiCl to PbCl 4 the metallic chloride lead( ii) is nature of the hydrolysed, chloride also a Group little IV atom increases. decomposition of the When lead( iv) as 4 lead( iv) chloride to occurs. Key ponts C, S Down Group and and Ge a hae IV, general Group IV The Group gant the coalent elements ncrease n elements form excepton IV of show metallc are a Sn general character coalently tetrahaldes carbon structures; bonded hydrolysed and Pb are decrease and n metals. meltng electrcal pont conductty. tetrachlordes. rapdly by water wth the tetrachlorde. 145 12.4 Some propertes Learning outcomes The Group The On completon of ths secton, oxides be able descrbe and of IV oxides: the Group IV IV oxdes structure elements and exist in stability the +2 and +4 oxidation you states. should of Group The table shows the type of structure of these oxides. to: the thermal trends n stablty bondng of +2 oxidation state +4 oxidation state the CO smple molecular CO molecular 2 Group IV descrbe oxdes the trends character of Group oxdaton states n IV SO acd–base oxdes smple molecular SO gant coalent gant coalent gant coalent wth onc character gant coalent wth onc character 2 n GeO onc wth some GeO 2 +2 and +4. coalent SnO gant character onc SnO 2 PbO gant onc PbO 2 Down each character series of the of oxides, Group IV the structures elements get more increases. ionic and SnO as the PbO 2 degree from of covalent CO and character , CO are in solids their with ionic giant structure. structures metallic have some 2 All the and oxides, high apart melting 2 points. CO and are CO gases at room temperature, because they have a 2 simple molecular molecules in Carbon the structure only solid and liquid monoxide, CO has decompose on a weak forces between their states. strong triple bond and does not heating. Germanium( ii) with oxide disproportionates on 2GeO(s) heating: → GeO (s) + Ge(s) 2 Oxidation Tin( ii) oxide, heating oxides in in Carbon SnO, the the numbers and lead( ii) absence presence dioxide, of of air . +4 oxide, They PbO, are do 0 not readily decompose oxidised to on higher oxygen. has CO +2 strong double bonds and does not 2 decompose on The in oxides down the heating. the group. +4 Only oxidation lead( iv) state tend oxide, PbO to , decrease however , in stability undergoes 2 signicant thermal decomposition: PbO (s) → PbO(s) + 2 lead(iv) Acid−base The +2 + table and CO summarises oxidation 2 oxidation ery lead(ii) (g) oxide properties of the oxides below +4 O 2 oxide state weakly the acid–base properties of the oxides states. + acdc 4 oxidation CO state acdc 2 SO − SO ery weakly 2 GeO amphoterc GeO amphoterc 2 SnO amphoterc SnO amphoterc 2 PbO amphoterc PbO amphoterc 2 146 acdc in the Chapter The the oxides in the +2 corresponding CO only reacts oxidation oxides with hot in state the +4 are less acidic oxidation concentrated (more state. sodium For basic) 12 The chemstry of Groups II, IV and VII than example: hydroxide, but CO 2 forms an acidic solution in water: + (g) CO + H 2 CO also reacts O(l) Y HCO 2 with (aq) + H (aq) 3 dilute alkalis: 2 (g) CO + 2NaOH(aq) → Na 2 The oxides group as in the CO 2 both oxidation metallic states character of (aq) H O(l) 2 become the + 3 more Group IV basic atom down the increases. For example: Carbon silicon dioxide dioxide reacts only (s) SiO + with reacts aqueous with 2NaOH(aq) hot → alkalis (hydroxide concentrated Na 2 SiO 2 Reactions of the (aq) + H 3 sodium ions) but alkali: O(l) 2 silicate amphoteric oxides of Group IV Oxidation state +2 GeO, SnO character They For all and in PbO the react are order with all amphoteric, GeO acids to < SnO form < a but have increasingly salt with oxidation + 2HCl(aq) → (aq) SnCl + H 2 They For state +2. example: SnO(s) basic PbO. all react with alkalis to form oxo O(l) 2 ions with oxidation state +2. example: 2− GeO(s) + 2OH (aq) → GeO (aq) + H 2 O(l) 2 germanate(ii) ion 2− SnO(s) + 2OH (aq) → SnO (aq) + H 2 O(l) 2 stannate(ii) ion 2− PbO(s) + 2OH (aq) → PbO (aq) + H 2 O(l) 2 plumbate(ii) ion Oxidation state +4 GeO , SnO 2 They and PbO 2 all are all amphoteric. 2 react with acids to form a salt with oxidation state +4. For example: (s) SnO + 4HCl(aq) → SnCl 2 SnO reacts with dilute acid but PbO 2 They For + 2H O(l) 2 will only react with 2 concentrated (aq) 4 all acid, react oxidising with alkalis it to to chlorine form oxo (see ions Section with 12.5). oxidation state +4. Key ponts example: 2− (s) + GeO 2OH (aq) → GeO 2 (aq) + H 3 The relate stablty of Group O(l) 2 germanate(iv) IV ion oxdes n oxdaton state +4 2− SnO (s) + 2OH (aq) → SnO 2 (aq) + H 3 + 2OH (aq) → PbO the group. ion 2− (s) 2 down 2 stannate(iv) PbO decreases O(l) (aq) 3 + H The acdc character of the Group O(l) 2 IV plumbate(iv) oxdes n oxdaton state +2 ion and +4 and the Group state decreases basc IV +2 character oxdes are n more correspondng state down the group ncreases. oxdaton basc oxdes n than the oxdaton +4. 147 12.5 Relate stablty of Group IV catons and oxdes Learning outcomes Stability of Going On completon of ths secton, be able and descrbe Group IV , +4 oxidation the oxides the oxides of the the relate stabltes oxdes +4 the +2 oxidation oxidation state state become less become stable more with to the +2 state. of oxidation the in states to: respect and you stable should down +2 and aqueous state +2 oxidation state +4 catons CO decreasing CO decreasing 2 of the Group IV elements n stability and SiO stability SiO and 2 oxdaton states +2 and +4 wth Ge better GeO better 2 Ø reference to E alues reducing SnO SnO agent PbO PbO oxidising 2 descrbe based some on uses slcon(iv) of ceramcs oxde. agent 2 Carbon monoxide, readily, e.g. oxidised to at a CO, high carbon is a good dioxide which O Fe 2 Oxidation Carbon numbers: dioxide is a reducing temperature it is (s) more It iron( loses iii) electrons oxide to iron. It is stable. + 3CO(g) → 2Fe(l) + 3CO 3 (g) 2 +3 poor agent. reduces +2 reducing agent. 0 It loses +4 electrons with great difculty. Lead( iv) oxide, PbO , is a good oxidising agent. It accepts electrons 2 readily, e.g. it oxidises hydrogen chloride to chlorine: (s) + 4HCl(aq) → PbCl PbO 2 Oxidation Lead( ii) oxide is a –1 poorer (aq) + Cl 2 numbers: +4 oxidising agent in (g) + 2H 2 +2 O(l) 2 0 comparison with lead( iv) oxide. Reactions of some Group(IV) cations Ø In Section 10.4, feasibility of Group of a IV we showed reaction. metals in We how can E values apply oxidation can these states +2 be ideas and used to the +4. See to predict aqueous Figure the cations 12.5.1. E + increasing –0.30 Ge O + 4H 2+ + 2e Ge + 2H 2 O 2 oxidising increasing 4+ power of species +0. 15 Sn 2+ + 2e Sn reducing on the power left of (better ions on the at right 4+ accepting +1.69 Pb (better at 2+ + 2e releasing Pb electrons) electrons) Figure 12.5.1 Worked The reducng and oxdsng powers of some Group IV ons examples 1 to 3 together with Figure 12.5.2 show how we can Dd you know? 4+ compare the ability of Pb 4+ ions and Sn ions to oxidise other species. Ø Tables of E alues often hae tn E and lead ons wrtten n the form 2+ 4+ Pb –0.76 4+ and Sn . Many tn and Zn (aq) + 2e Zn(s) (aq) + 2e Sn(s) + 2e Sn lead 2+ –0. 14 compounds n the +4 Sn oxdaton 4+ +0. 15 state are nsoluble n water. So Sn 2+ (aq) 3+ hae to be dssoled n concentrated +0.77 Fe +1.69 Pb 2+ (aq) + e 4+ acds or alkals soluton. to get these ons nto Figure 12.5.2 (aq) Fe (aq) 2+ + 2e Pb (aq) An electrode potental dagram for determnng the feasblty of some reactons nolng Group IV catons 148 (aq) they Chapter Worked example 12 The chemstry of Groups II, IV and VII 1 Exam tps 4+ Will 2+ Pb ions oxidise an aqueous solution containing Fe ions to 3+ aqueous An ions? Fe easy way reacton Ø According to the to determne occurs or not s f a shown 4+ E values, Pb ions are better at accepting electrons 4+ 3+ than ions. Fe below. 4+ Pb ions are better oxidising For example: wll Pb (aq) agents. 2+ ons Ø According to the than The values, ions. Pb Fe ions Fe are reaction with the more 2+ direction, ions better are better reducing at releasing i.e. (aq) Fe negative → Fe Wrte down equatons E value goes in the ons? overall reaction is: the two noled half makng that you reerse the sgn Pb 2+ (aq) + 2Fe 2+ (aq) → Pb showng electron the loss 3+ (aq) + 2Fe (aq) ths case, equaton ii). 4+ i example of (aq) (n Worked sure reverse 3+ 4+ The (aq) electrons agents. equaton Fe 2+ Ø wth 2+ E 2+ react 2+ (aq) Pb + 2e Y Pb (aq) 2 Ø E 4+ Will Sn = +1.69V 2+ ions oxidise an aqueous solution containing Fe ions 2+ to ii Fe 3+ (aq) Y Fe (aq) + e 3+ aqueous ions? Fe Ø E Ø According to the values, Fe ions are better at accepting –0.77V electrons 4+ than = 3+ E Add the two Fe ions are better oxidising agents. +1.69 Ø According to the values, Sn ions are better at releasing Sn ions are better reducing agents. Ø The reaction with the more negative 2+ direction, So no i.e. = + 0.92 V E value goes in the If the alue poste, reverse of the the oltage reacton wll s occur. 4+ (aq) Sn reaction Worked 0.77 electrons 2+ ions. Fe – 2+ E 2+ than oltages 3+ ions. Sn → Sn 4+ (aq) So Pb (aq) 2+ reacts wth Fe (aq). occurs. example 3 4+ Will Zn reduce Sn ions? Ø According to the E values, Zn are better at releasing electrons than 4+ ions. Sn Zn is a better reducing agent. Ø The reaction with the more negative E value goes in the reverse 2+ direction, i.e. Zn(s) → (aq) Zn 4+ The initial reaction is: Zn(s) + Sn 2+ (aq) → Zn 2+ (aq) + Sn Ø But according to E (aq) 2+ values and the anticlockwise rule, Sn can be Ø reduced further by Zn to tin – (E 0.14 V). 4+ The overall Uses of reaction ceramics Fur nace linings: is: 2Zn(s) + based on SiO is a good Sn 2+ (aq) → 2Zn (aq) + Sn(s) silicon dioxide thermal insulator and has a very high 2 melting point Abrasives: due is SiO to the hard many and strong has a covalent high melting bonds. point (heat is given out 2 on grinding). Manufacture of glass and porcelain: is SiO relatively unreactive and 2 is easily moulded. The high melting point is useful for ovenware. Key ponts The relate stabltes of the oxdes and aqueous catons of the Group IV Ø elements Down Group agent can wth Ceramcs be IV explaned oxdes, respect based on to the the wth +4 +2 reference oxdaton oxdaton slcon(iv) oxde are to E state alues. becomes a better oxdsng state. used for furnace lnngs and abrases. 149 12.6 Group VII: Learning outcomes the Physical The On completon of ths halogens secton, elements be able explan the physcal and aratons propertes halogens n T able. terms of of n the They (molecules (the seven halogens) electrons in are in their the the relate halogens descrbe p outer block of principal the quantum are all non-metals having two which exist as diatomic the as reactty oxdsng use of molecules atoms). the structure −3 Halogen State at 20 °C Density/g cm Atomic Melting radius/nm point/°C 0.072 –220 of fluorne, VII have bondng explan the Group All to: shell. of you Periodic should properties F gas 1.51 (at gas 1.53 lqud 3. 12 sold 4.93 85 K) 2 agents bromne chlorne, Cl (at 238 K) 0.099 –101 2 water. bromne, Br (at 293 K) 0. 114 –7 2 odne, I (at 293 K) 0. 133 +114 2 Colour The and colour F : solubility of the gets yellow; darker Cl 2 Cl reacts and Br 2 is organic Br : reddish-brown; I water . called Chlorine ‘chlorine reacts water ’ : grey-black. 2 slightly and with ‘bromine water . water ’ Solutions respectively. iodide. in water . This such Iodine solution as is solution brown. cyclohexane, it is iodine When appears dissolved iodine is purple. in aqueous dissolved Iodine in vapour is purple. and atomic shells atomic radius increases electrons Melting the van with are solvents Density As yellowish-green; 2 insoluble potassium The group: 2 Iodine also the 2 Fluorine of : down halogens the there inner points W aals increases and halogens der points by get intermolecular is down the between down the forces. The boiling larger , group as the shielding density of number the increases outer down of electron shell the group. points increasing the group The the increased shells. and forces increase radius number molecules as boiling it takes points of electrons stronger . more show So energy a the to similar makes the melting break trend, these and F Cl 2 being solid. 150 gases at room temperature, bromine a volatile liquid and iodine 2 a Chapter Chemical The reactivity of the halogens halogen can solution. are less displace These Halogen are reactive a less redox Halide chemstry of Groups II, IV and VII down the halogen group. from an A more reactive aqueous halide reactions. ion (aq) chloride, Cl bromide, – Cl The halogens going reactive 12 turns Br (aq) iodide, orange turns I (aq) brown 2 no Br reacton – turns from orange 2 to no I reacton no reacton brown – 2 The reactions The cyclohexane can be can be conrmed layer by dissolves adding the cyclohexane halogen only, so to any solutions. changes veried. Aqueous chlorine potassium displaces bromine from an aqueous solution (aq) + 2KBr(aq) → 2KCl(aq) + Br 2 equation: Iodine will iodine is numbers: not 2Br (aq) → 2Cl (aq) + Br not as good –1 bromine an (aq) 2 0 displace (aq) 2 (aq) + Cl 2 Oxidation of bromide: Cl Ionic the colour –1 from oxidising 0 potassium agent as bromide because the bromine. Ø We can use oxidising values E agents, e.g. to explain does the chlorine relative oxidise reactivity iodide of ions halogens to as E iodine? /V I +0.54 (aq) + 2e 2I (aq) 2 Ø As E values get more positive (less negative), the halogens on the left +1.07 Br (aq) + 2e 2Br (aq) + 2e 2Cl (aq) 2 become better oxidising agents. They accept electrons more readily. +1.36 Ø As values E become better accepts Cl get less positive reducing electrons (more agents. more negative), They readily release than I halides electrons and 2 the I ions on more the (aq) 2 +2.87 readily. release Cl right (F + 2e 2F electrons Ø 2 Figure 12.6.1 more readily than Cl ) 2 Usng E alues to predct ions. the feasblty of reacton between Ø The reaction direction, with the more negative E value goes in the reverse halogens and haldes i.e. (aq) I → I (aq) 2 So according to the Cl anticlockwise (aq) + 2I (aq) rule, → the 2Cl reaction (aq) + I 2 is feasible: Key ponts (aq) 2 The use of Unsaturated Bromine When an bromine compounds water is used water to test for C=C contain to test for unsaturated compound bromine) the changes bromine) to solution one or double is more shaken colour C=C bonds from in with bonds double organic bromine orange (the bonds. Halogens Down the compounds. become water darker colour of exst as datomc molecules. (aqueous group, less n the olatle halogens and are colour. the The halogens are good oxdsng colourless: agents, CH =CH 2 + 2 Br → 2 CH Br – CH 2 Br decreasng 2 ethene bromine 1,2-dibromoethane (colourless) (orange) (colourless) the Bromne for oxdsng down water s unsaturated carbon the ablty group. used (C=C) to test bonds n compounds. 151 12.7 Halogens Learning outcomes and hydrogen Reaction of Sodium On completon of ths secton, thiosulphate, be able descrbe of iodine with O 2 by , sodium thiosulphate can be used to determine (see Section the redox reacton of of bleach. wth sodum Figure 12.7.1 shows that descrbe the halogens 3.6) this explan thosulphate reacton wth the stabltes to of reaction O S ions are better reducing agents than analyse of I ions feasible and I 3 2 are better oxidising agents than S O ions. 6 the hydrogen relate is 2− molecules E /V 2– and 2− because 4 the 3 titration to: 2 odne S 2 samples Na you concentration should halogens haldes S +0.09 thermal 2– O (aq) 6 4 + 2e 2S O 4 (aq) 3 the Group VII +1.54 I (aq) + 2e 2I 2 hydrdes. Figure 12.7.1 Dd you know? So Starch s a complex contanng chans the reaction 2Na carbohydrate of whch of may glucose be helcal form. Iodne forms O 2 (aq) + I 3 (aq) → Na 2 S 2 O 4 a and chlorine oxidation of are stronger sodium oxidising thiosulphate to complex s a these dark blue-black O 2 helces. Ths can complex stll sodum odne react s so weak wth speces thosulphate. has colourless been used starch that So up, agents than sulphate iodine and cause ions: 2− (aq) + 4Cl 3 (aq) + 5H 2 O(l) → 2SO 2 − (aq) + 8Cl + (aq) + 10H (aq) 4 colour. The The 2NaI(aq) weak 2− wth + n S complex (aq) 6 unts, arranged further a S 2 Bromine some is: estimation of chlorine in bleaches odne such when only as all the the Commercial commonly bleaches called determined by usually ‘sodium titration. contain sodium hypochlorite’. The procedure The chlorate( chlorine i), in NaClO. bleach This can is be is: remans. Dilute Add the bleach excess by acidied a known potassium − NaClO(aq) + with iodide distilled solution to water . liberate iodine: + (aq) 2I amount + 2H (aq) → I (aq) + NaCl(aq) + H 2 Titrate the liberated concentration The end point indicator halogens. is turns Hydrogen Hydrogen using with indicator . when the blue-black halides are formed reactions uoride, HF: when are colour hydrogen slower The chloride, down reaction (g) + F 2 Hydrogen thiosulphate of the of a known starch−iodine colourless. H sodium halides These Hydrogen iodine starch O(l) 2 (g) is → combines the halogen explosive even directly with the group. in cool conditions. 2HF(g) 2 HCl: The reaction is explosive in the presence of sunlight. (g) H + Cl 2 Hydrogen bromide, (g) → 2HCl(g) 2 HBr: H gas and Br 2 vapour react slowly on 2 heating. Hydrogen iodide, HI: gas H and 2 closed container to form H (g) 2 152 I vapour react slowly 2 an equilibrium + I (g) 2 Y mixture: 2HI(g) on heating in a Chapter Hydrogen uoride laboratory halides HF are All gases compared hydrogen at with at 19.5 °C may room the be and either so temperature. other its liquid The hydrogen state or gas. under The much halides is to chemstry of Groups II, IV and VII normal other higher due The hydrogen boiling its point of extensive bonding. hydrogen water boils conditions 12 halides are acidic – they form acids when they dissolve in e.g. + HCl(g)+ O(l) → H H 2 O (aq) + Cl (aq) Cl (aq) 3 + or more simply: The thermal The thermal atom HF and HBr aq stability of stability increases HCl(g)+ in HCl of the → H hydrogen hydrogen (aq) + halides halides decreases as the halogen size. are not decomposes decomposed slightly at at about temperatures 450 °C to form of 1500 °C hydrogen and bromine. HI decomposes rapidly at about 2HI(g) Y 450 °C (g) H to + 2 When a purple hot grey-black from a platinum vapour gas of solid or wire iodine on vice the is is side versa of in seen, the omitting hydrogen and iodine: (g) 2 placed rst I form a tube which tube. the of hydrogen then The liquid turns direct phase iodide directly formation is a to of a a solid called sublimation The the ease bond of thermal energies of decomposition the of the hydrogen hydrogen−halogen halides is related to bond: −1 Bond The bond The energy larger hydrogen So down nuclei So the the and going halogen H–F 562 431 H–Br 366 H–I 299 halogen halogen group the down energy/kJ mol H–Cl decreases and Bond this atom, way the because: greater is the distance between the nuclei. there bonding the in is a smaller attractive force between the electrons. group, it takes less energy to break the carbon– bond. Key ponts The reacton of odne wth sodum thosulphate can be explaned by Ø reference to E The halogens The relate group as alues. react wth thermal the hydrogen stabltes of hydrogen–halogen to form the bond hydrogen hydrogen strength haldes. haldes decrease down the decreases. 153 12.8 Reactons Learning outcomes of halde The reaction of When On completon of ths secton, silver nitrate be able descrbe ons the wth reactons aqueous of sler halde soluble descrbe by aqueous the ions in wth coloured cannot aqueous to an aqueous silver nitrate precipitates of solution silver of Cl halides , are Br or I ions, formed. be tested solution for and in no this way because precipitate is silver uoride is formed. procedure is: ammona reacton of halde Add dilute soluble ons added with ntrate The followed is ions to: Fluoride halide you characteristically should ons concentrated nitric acid to contaminating the solution carbonates under or test. This hydroxides removes which may any form sulphurc insoluble silver compounds. acd descrbe wth the cold solutons reacton and of hot of Add a few Record Add If drops of aqueous silver nitrate until a precipitate is seen. chlorne aqueous sodum hydroxde. the colour dilute the of aqueous precipitate the precipitate. ammonia does not to see if redissolve, the add precipitate redissolves. concentrated aqueous ammonia. Halide ion Colour of precptate Conrmatory test Cl Br I whte cream pale dssoles dssoles n concentrated concentrated NH NH dlute (aq) NH 3 If left exposed to The silver chloride The silver bromide The silver iodide each case, nsoluble (aq) 3 n (aq) 3 sunlight: In n yellow the precipitate precipitate precipitate equations AgNO for (aq) + turns turns does the grey grey not rapidly. very turn grey. precipitation KBr(aq) → slowly. reaction AgBr(s) + are KNO 3 similar e.g. (aq) 3 + Ionic The ion equation: precipitate is formed (aq) Ag of AgCl (See + Br dissolves Section (aq) in → AgBr(s) excess ammonia because a complex 14.3). Exam tps Reaction Remember between nole the haldes the aqueous that sold redox and sulphurc haldes solutons). with concentrated (not acd the The strength > Br ion, is Cl the less of with easier force of halide iodide it is to ions H lose H SO 2 . ions An are not acid−base reducing the an best and the solid agents reducing electron between SO 2 Chloride as being attraction Concentrated acid from follows agent. their nucleus outer and the The the pattern larger shell outer I the > halide because there electrons. NaCl 4 strong enough reaction reducing agents to reduce occurs: 4 NaCl(s) + H SO 2 154 sulphuric reactons (l) 4 → NaHSO (s) 4 + HCl(g) the S in Chapter Concentrated H SO 2 When bromide and solid ions react NaBr(s) with H + H SO 2 agents reaction to occurs reduce state S in chemstry of Groups II, IV and VII NaBr SO the is rst (l) → NaHSO 4 because from there the an acid–base sulphur reaction: 4 (s) + HBr(g) 4 bromide +6 ions are oxidation strong state in enough H SO 2 oxidation The 4 2 Further 12 dioxide, to reducing the +4 4 SO 2 2HBr(g) + H SO 2 OxNo: −1 Concentrated H SO 2 Hydrogen ones iodide above. They are strong oxidation is Iodide S. A state rst ions (g) + Br 2 solid formed in better (l) + 2H 2 O(l) 2 0 NaI to the an acid–base reducing reducing SO H of SO +4 and are enough in mixture → 4 2 H (l) 4 +6 agents −2 reaction agents to reduce oxidation than S state similar from in to bromide the the hydrogen the ions. +6 sulphide, 4 products may be formed: 2 2HI(g) + H SO 2 6HI(g) + (l) 2 OxNo: −1 + H The cold dilute + SO I (l) + 2H 2 + 3I (l) → H 4 + 4H O(l) 2 0 S(g) + 4I 2 (s) + 4H 2 −2 Sodium O(l) 2 (s) 2 chlorine alkali: (g) S(s) 0 +6 reaction of With → 4 2 −1 SO 2 (l) +6 8HI(g) OxNo: → 4 SO H with O(l) 2 0 alkalis chloride and sodium chlorate( i) are formed: Cl (aq) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H 2 OxNo: With are hot −1 concentrated alkali: Sodium +1 chloride and sodium chlorate(v) formed: (aq) 3Cl + 6NaOH(aq) → 5NaCl(aq) + NaClO 2 OxNo: Both a O(l) 2 0 −1 these single reactions substance reduced to a reduced to Cl Oxidation (aq) + 3H 3 0 are disproportionation becomes lower oxidised oxidation ions number and has state). oxidised been to In to a reactions higher these either abbreviated O(l) 2 +5 examples chlorate( i) to (reactions oxidation OxNo in state chlorine or in has chlorate( v) the above which and been ions. equations. Key ponts Haldes react precptates wth whch aqueous dssole sler to ntrate dfferent to form extents characterstcally n aqueous coloured ammona. − Halde as ons are exempled Chlorne and ncreasngly by reacts chlorate(v) Chlorde ons ther wth ons good ablty cold and to reducng reduce hot are formed n both these n the concentrated aqueous respectely. These agents solutons are order Cl sulphurc to form − < Br − < I acd. chlorate( i) dsproportonaton reactons. reactons. 155 Reson 1 Whch of of the followng decreasng best solublty of questons descrbes the Group the II order 5 metal Anhydrous bonded hydroxdes? Whch A Ba > Sr > Ca > Mg > Be B Ba > Sr > Ca > Be > Mg ths alumnum chlorde s a coalently compound. of the followng statements correctly explans statement? 3+ A The radus of B The charge on the Al s large. 3+ C Sr > Ca > Ba > Mg > Be densty of the Al on dstorts the – electron D charge of the Cl on. Be > Mg > Ca > Sr > Ba 3+ C 2 Whch of the followng statements relatng to The the of the +2 and +4 oxdaton states of dfference between Al ons s ery hgh. the – D Group IV elements A Tn(ii) B Lead(iv) C The s are readly oxdsed to are easly oxdsed to the +2 oxdaton state queston a Explan Germanum the As the exhbts elements on. after readng pages 162–7 terms to ‘monodentate’ lgands, and ge an and ‘bdentate’ example of each character. sem-conductor n Perod Classficaton s the followng statements 3 of the traersed from A Electronegatty B Meltng C Atomc D Screenng ponts s Perodc left to b rght, whch i Ge of use can of an arrow, the ste from whch of ii steadly. 4 occur. the metal true? decreases. rse by propertes. bondng 3 6 appled showng, D the Al ncreases as metallc polarses lead(ii). 6 wth effectely tn(iv). Answer compounds of 3+ on The Cl correct? compounds stablty and – Cl stablty electronegatty name atom and 6 of a complex exhbts a on n whch co-ordnaton the number respectely. Sketch the shape of the spatal arrangement of ons n i showng the bonds. – rad c decrease. of alence electrons The the ncreases. chlorde blue green 4 The relate HX, s acd strength represented by HF Whch of < of the hydrogen the followng HCl the followng < HBr < ths The thermal s hexaaquacopper(ii) i Wrte an dsplace on to water produce n the on. equaton for ii State the atom n ths dsplacement. co-ordnaton number of the copper HI statements prodes the each on. best Suggest a reason for the dfference n order? stablty of HX ncreases as the group number n the tetrachloro- complex. descended. Hydrogen odde has the largest standard a Suggest dssocaton bolng rse n the and ge maxmum examples oxdaton of, the states enthalpy. across The reasons for, bond steady C to haldes, 7 B able tetrachlorocuprate(ii) co-ordnaton A s order: iii explanaton for , on, Cl pont of HX ncreases as the group the Perod 3 elements. s b The pH of aqueous solutons of the chlordes ascended. of D Hydrogen bondng decreases as the group the first three elements of Perod 3 s gen s below: descended. Element Na Mg Al pH 7 7 5 Use a knowledge explan c the Descrbe the elements water, 156 of aboe ease sodum, gng structure and bondng to nformaton. of reacton alumnum releant between and equatons. the chlorne wth Chapters 8 a The table elements below shows n Group Element Na 3, the denstes sodum Mg Al to of the 10 a chlorne: S P The table below temperatures S Cl of Carbonate 11– 12 ndcates Reson the decomposton the four Group MgCO 1.0 1.7 2.7 2.3 1.8 2. 1 < 0. 1 Temperature/K (ii) CaCO 3 Density/ questons carbonates: SrCO 3 813 BaCO 3 1173 3 1553 1633 –3 g cm Carefully Draw a graph to ndcate the araton of b across Perod 3, elements sodum to explan the trend shown. densty i State the trend of solublty of the Group II chlorne. sulphates. b Refer to ther structure and bondng to explan ii Use nformaton proded by the releant the: enthalpy i steady rse of densty between sodum changes to explan ths trend. and c Lst one commercal use for each of the followng: alumnum ii large decrease n densty between sulphur i calcum oxde ii magnesum iii calcum and oxde chlorne. 9 State reasons wheneer a the and ge an possble, for tetrahaldes, example, the MCl , usng obseratons of the Group an n IV equaton d (a)–(c): are olatle, ii ncrease group b Carbon than s n non-polar ther elements: i wth of water as the descended. monoxde, CO, carbon s a better reducng carbonate knowledge the wth ii the solublty the ease and ts of II chemstry propertes water, resultng iii agent of Group the followng reacton compounds reacton your predct 4 i Use and of the to radum: subsequent pH soluton of the chlorde decomposton and of carbonate the ntrate(v) products. doxde, CO 2 11 c Slcon(iv) oxde has a ery hgh meltng pont a Explan wth reference the araton does not conduct electrcty. Use structure to structure and bondng but of meltng pont n the Group VII and elements. bondng to ge an explanaton for ths statement b about the i State be d Copy and what complete the followng table wth i the acd added ons formed and ii see f bromne were to to when the solutons oxdes chlorde of potassum odde and respectely. react ii Name iii How the type of reacton noled. alkal. would respect Oxide would by potassum nsertng you oxde. reactions i wth acd ii wth to you your explan answer your to obseratons part n ii? alkal c A black heated powder, wth P, conc. contanng HCl manganese eoled a greensh when gas ,Q, Germanum(ii) whch dssoled n water. A soluton of ths gas, S, oxde turned Tn(ii) oxde S when oer Lead(iv) ltmus paper allowed water n a to red before stand beaker n an beng bleached. nerted produced test bubbles of tube a gas, oxde T, on exposure i Suggest ii Name iii Wrte a to sunlght. name the and formula for P. gas Q. an equaton to explan an equaton for the producton of Q. iv Wrte the reacton producng soluton S. v Deduce the the oxdaton anons found vii Name viii Wrte the an type states of the halogen n n S. of reacton equaton for the whch produced S. producton of the gas T. 157 13 Transition 13. 1 An elements introduction Learning outcomes Electronic A On completon of ths secton, transition be able descrbe element is a d block element which ions with an incomplete d electron of the rst elements and of the rst row transition elements Element Electronc ttanum, T 1s the oxdaton states anadum, V 1s chromum, Cr 1s manganese, 1s 2 2s descrbe row the transton more are shown below: transton configuraton 6 2p 2 3s 6 3p 2 3d 2 4s of 2 or electron row 2 the rst one The ther ons descrbe forms sub-shell. electronc conguraton elements to: the transton elements configuration of transition congurations transition you stable should to 2 2s 6 2p 2 3s 6 3p 3 3d 2 4s elements characterstcs 2 of 2 2s 6 2p 2 3s 6 3p 5 3d 1 4s elements. 2 Mn 2 2s 2 ron, Fe 1s 2 2s 2 cobalt, Co 1s nckel, 1s 1s 6 6 2 2s 6 2p 6 7 2 4s 8 3d 6 3p 2 4s 3d 6 2 3s 6 6 3p 2 4s 3d 3p 2 3s 5 3d 3p 2 3s 2p 6 3p 2 3s 2p 2 2 copper, Cu 6 2 2s 2 3s 2p 2s 2 N 6 2p 2 4s 10 3d 1 4s Did you know? Scandum, Sc, and znc, Zn, are not transton elements although they are d 3+ block elements. Sc forms only one on, Sc , wth no electrons n ts d sub-shell. 2+ Zn forms The only one electronic pattern of on, Zn , and congurations lling the d has of sub-shell. a Cr complete and For Cu Cr , with one having electron one of the in d each orbital, orbitals sub-shell. not follow the arrangement of , gives lled. For a greater Cu, the d sub-shell greater stability electron When with in than which lost atom: 1s 2 2s 2 V atom: T ransition one 1s 2s stability in of each the d orbital, orbitals 1 4s [Ar]3d with than only a , gives a single 2+ : can 2 1s 2s 2 2p 6 3p 2 3s form 3s 2 3d 6 3p 6 electrons to form ions, it is the 4s e.g. 2 3s 6 2p lose rst, 6 2p 2 elements Fe 158 having elements are 2 Ti electrons sub-shell it. transition electrons paired d arrangement 10 a expected a 1 4s [Ar]3d completely do the 5 3d 2 4s 3 3d 4s more 2 3p 6 3d 2+ Ti 2 2 ion: 1s 3+ V than ion: one 6 : 1s type 3+ Fe 2 2s 2 2 1s 2s of 6 2p 2 2s 2p ion, 2 2p 6 3s 2 3s 6 6 3p 2 3s 3p e.g. 2 3p 6 3d 5 2 3d 6 3d 2 of Chapter The range of oxidation 13 Transton elements Ti states V T ransition elements may have several different oxidation states. Cr The most common The maximum oxidation state of transition elements is +2. Mn the oxidation involves all 4s From onwards, and state 3d of the transition elements up to Mn Fe electrons. Co Fe electrons become the +2 oxidation increasingly state harder to dominates remove as because the the nuclear 3d Ni charge Cu increases. Figure 13.1.1 Higher oxidation states of transition elements are found in complex The range of oxidation states in transition element compounds in 2− ions (see Section 13.3) or compound ions such as and MnO CrO 4 Period 4. All oxidation states are positive. 4 The commonest oxidation states are ringed. Characteristics of transition T ransition properties elements which have typical them apart They T ransition element ions form coloured T ransition element ions form complex T ransition elements often related with properties other is compounds metallic from (this form set elements and to different their the oxidation in are elements have very high density. T ransition elements have very high melting They Most when they compounds placed do not in a magnetic and ions of magnetic retain their the eld, transition they magnetism Co after and the align when S points good catalysts states). and boiling points. elements are themselves the paramagnetic : with magnetic eld the is eld. But removed. S N transition Fe, often properties: N Figure 13.1.2 states. oxidation T ransition typical some ions. have have compounds. compounds differences but metals: element A small rod of a transition element aligns itself in a magnetic eld Ni are magnetic fer romagnetic . eld has been They retain permanent magnetism withdrawn. Key points Transton of When the elements form one or more ons wth an ncomplete d sub-shell electrons. 4s a transton sub-shell Transton and elements compounds magnetc element forms then from exst (contanng propertes n the on, 3d arous complex and an electrons are rst lost from sub-shell. oxdaton ons), catalytc the and states, form often hae coloured characterstc actty. 159 13.2 Physical Learning outcomes On completon should be able of ths properties Ionisation secton, In Periods in the explan the changes radus of the n energy, and values 3 of atomic the Periodic and T able elements ionic there is radii of the rst ionisation energy and a considerable the atomic and variation ionic radii. to: relately atomc and rst small and values from elements, titanium to however , show only small changes in these copper . onc onsaton transton transition energy elements Element T V Cr Mn Fe Co N Cu 661 648 653 716 762 757 736 745 0. 132 0. 122 0. 117 0. 117 0. 116 0. 116 0. 115 0. 117 0.090 0.090 0.085 0.080 0.076 0.078 0.078 0.069 across Ø a transition you The 2 of –1 /kJ mol ∆H perod 1 descrbe qualtately physcal propertes the Atomc of transton radus/nm elements compared wth calcum. Ionc 2+ radus of ons/nm Ø Across this series of transition elements, the rst ionisation energy, ∆H , i1 generally increases, successive increase the case of d going the sub-shell. one ionic compared increased the element forces and to the of the caused the d is the block elements between of block Period them. transition s elements by the 4 The are elements, in metals, iron and s on the p II are some is an of which in block by the increased going sub-shell, ionisation the in atomic electrons The from electrons. is placed and marked properties of the just before transition differences calcium with nickel. ron nckel 839 1540 1450 Densty/g cm 1.55 7 .86 8.90 Atomc 0. 197 0. 116 0. 115 0.099 0.076 0.078 590 762 736 pont/°C –3 Ionc radus/nm radus/nm –1 Frst 160 onsaton energy/kJ mol one increased calcum Meltng the energy slightly elements. the 3d should added calcium calcium some d only the with a each electron Similarly, outer cancelled Group to but and repulsion there there of This since additional series, Although compares nucleus. difference elements 4. But smaller . the almost extra Period table the Atoms markedly electrons much in their the the nucleus is element in make along the by adding next next Comparing transition Calcium to in sub-shell. differences effect to is amount. electrons caused of increase large attractive transition repulsive with effect small proton outer same elements, tend very more the the a electrons The element radii by one of into outer transition between and on only have attraction are repulsion into element the electrons but two Chapter The transition calcium. This elements s compared electrons elements is a to with The form can greater charged) force atomic from the greater which of and the in ionic lling the d radii for of of attraction to pull the ionisation because and the this element The electronegativity with slightly less Electrical table can 4s of but 3d (and points elements than transition release and small sea in only its two outer transition sub-levels. often electrons of There highly in comparison are all compares the 8 the in outer especially better at to shielding So the is and outer is the the related to the elements. the there a electrons relatively outer electrons nucleus. elements greater than This transition nucleus transition have the than force are of calcium. smaller transition Ca = across higher attraction than that between 1.0, the radii This than elements Fe and series Ni from is = Ti is due to the fact calcium. signicantly 1.8. to higher The Cu as the elements character . electrical conductivity elements of calcium with some of good conductors electrical Ca T Mn Fe N Cu 0.30 0.024 0.007 0. 10 0. 15 0.63 –1 S m transition single the closer larger elements. Conductty/10 of the electrons. denser e.g. Element a in good p much elements. electrons. increases metallic are conductivity of transition transition Most of ions calcium, electronegativity less or relatively outer elements compared s between energies of transition good melting electrons’ the sub-shell electrons transition are and higher bonding the the transition are than that with of both calcium d sub-shell charge force rst of the The the ‘sea from ions have Calcium between typical tends nucleus The and metallic delocalised electrons element radii nuclear calcium get calcium. attraction transition Electrons the harder stronger Transton calcium. presence of the of are the with release corresponding The elements reects 13 s elements electron conductors. than are and a So it is too and conductors lled copper manganese conductor , good half or and to electricity. chromium nickel. reactive of completely Although be used in full are d Those shell relatively calcium is electrical a wiring. Key points Across the rst onc radus each electron row of and rst added transton onsaton goes nto elements, energy a d are the changes relately sub-shell and so n atomc small. Ths the s sheldng and because effect s mnmsed. The s physcal block propertes element of because transton of the elements presence of dffer from the d those of a typcal sub-shell. 161 13.3 Coloured Learning outcomes ions and The various oxidation T ransition On completon of ths secton, be able descrbe can states of vanadium is one such form ions element, in the arous oxdaton solid exhibit characteristic ammonium colours. vanadate, of than ions in one their V anadium VO NH 4 states more whose oxidation various state. oxidation to: states elements states you V anadium should oxidation . When ( v) is usually ammonium supplied vanadate as is 3 + anadum acidied the vanadium becomes part of a positive , ion,VO in which 2 explan the formaton ons transton by of coloured vanadium has an oxidation state of +5. elements. + +5 ↑ yellow VO 2 2+ +4 VO +3 V +2 V 0 V blue 3+ oxidation state green 2+ The redox potential chemistry of vanadium can purple/mauve be explained using an electrode chart. Ø E /v 2+ –0.76 – Zn + 2e Zn (aq) (s) 3+ 2 V –0.26 + e V (aq) (aq) 2+ –0. 14 – Sn + 2e Sn (aq) (s) 2+ + VO +0.34 + – 2H + (aq) 3 e V + e + (aq) (l) Fe (aq) + VO + 2 e (aq) + Figure 13.3.1 O 2 2 Fe +1.00 H (aq) 3+ +0.77 + (aq) – 2H (aq) + 2 e VO + (aq) H (aq) O 2 (l) An electrode potential chart showing the oxidation states of vanadium and other compounds Ø By reference to vanadium( v) E values, ions to a we lower can 2+ Addition of Fe select oxidation suitable compounds + ions to VO will reduce the VO 2+ ions to VO 2 2+ ions Fe reduce + ions 2 2+ ions. to state: are better reducing agents than VO + ions and VO 2 3+ ions are better oxidants than 2+ + ions. Fe So the reaction VO → VO 2 Ø goes in the forward direction and the reaction 2+ value will goes be in from the reverse yellow to direction with Fe ). Addition of Sn to The + ions VO will reduce the ions to V ions. 2 + the E change 3+ VO 2 Using higher colour blue. + the 3+ → (Fe anticlockwise rule, the reaction 2+ → VO VO 3+ → V goes 2 Ø in the forward direction and the reaction direction (Sn → with the higher value E 2+ goes in from the reverse yellow to blue to + + (aq) 2VO The colour change will be green. + Sn(s) ). Sn + 4H 2+ (aq) → 2VO 2+ (aq) + Sn (aq) + 2H 2 then: Sn(s) + 2VO O(l) 2 2+ + (aq) + 4H 3+ (aq) → 2V 2+ (aq) + Sn (aq) + 2H O(l) 2 + Addition of Zn to VO + ions will reduce the VO 2 2+ ions Using the anticlockwise to V ions. 2 2+ + rule, the reaction → VO VO 3+ → V 2+ → V 2 Ø goes in the forward direction and the reaction with the higher E 2+ value will 162 goes be in from the reverse yellow to direction blue to (Zn green to → Zn purple. ). The colour change Chapter The formation of Compounds appear coloured coloured when 13 Transton elements ions they absorb energy that corresponds 3+ to certain appear wavelengths purple spectrum. and red because Most of regions. of they the So light the absorb light the in light which solution visible mostly passes appears a spectrum. in the through is Aqueous green from region the ions Ti of violet, the blue purple. b noitprosba violet blue red green blue yellow violet appears purple red absorbs 400 500 600 700 green wavelength nm 3+ Figure 13.3.2 a The absorption spectrum of aqueous Ti ions; b Only violet, blue and red light are transmitted through the solution. What causes the colour? 3+ T ransition element number water (dative of covalent) molecules are ions, such molecules. bond called with as Ti Each the ligands , in water solution molecule transition and the are element resulting ion bonded forms ion. is a to a denite co - ordinate These called a water complex 3+ ion. The complex ion in the −water Ti complex has the formula 3+ [Ti(H O) 2 ] 6 The 3 d orbitals degenerate The presence orbitals pushed of to orbitals When of an of the an the isolated They into in a energy the complex ion. element same affects Orbitals levels than ion average the described electrons close those are as energy. to the further in the ligands d are away. The to orbital groups. moves light transition have element higher two electron energy, all ligands transition slightly split higher in orbitals . is from a d absorbed orbital in the of lower visible energy region of a d the Key points spectrum. The frequency of the light absorbed depends on the energy difference between the split d levels. Different ligands split the d energy levels An acded soluton ammonum different amounts. So different ligands may cause different colours anadate(v) can to react be of by wth reducng agents absorbed. to form anadum dfferent a ons oxdaton wth states b Ø dependng alues for A lgand on the s a the relate half E reactons. molecule or on Energy difference E = wth hn of a one or electrons 2+ a Degenerate orbitals in a Cu lone whch co-ordnate transton Figure 13.3.3 more bond metal par can form wth a on. 2+ ion; b The ligand (water) in a [Cu(H O) 2 ] 6 Transton element compounds ions causes d orbital splitting are often d-orbtal coloured splttng because caused of by lgands. 163 13.4 Complex Learning outcomes ions More A On completon of ths secton, and about ligand is be able explan the prncple of lgand know d that depends lgand on donate the n well exchange water , bonds from attracts stablty an are electrons and constants to covalent) exchange molecule or and ion complex with one or ions to a transition more element lone ion. pairs of Examples of electrons simple to: ligands ligands exchange you available should a ligand a in the lone ammonia are formed transition nuclear pairs complex of ion and the elements charge. So, electrons is chloride with formed. do the not Co - ordinate element shield relatively strongly Some ions. transition outer poorly enough examples the to are (dative ions because electrons shielded form nucleus co - ordinate shown in the very the bonds table equlbrum below. reacton where compete for the lgands bondng wth the Transton transton descrbe element some element on Lgand Formula of complex on examples of lgand 2+ 2+ water, Fe H O [Fe(H 2 O) 2 ] 6 exchange. 3+ 3+ ammona, Co NH [Co(NH 3 2+ 3+ NH ] 6 2– Cu 3 ) 3 chlorde on, Cl [CuCl ] 4 H 3+ N 3– Cr 3 hydroxde on, OH [Cr(OH) ] 6 + H + ammona, Ag NH Co N NH [Ag(NH 3 3 ) 3 ] 2 3 2+ 2+ damnoethane, N [N(NH CH 2 CH 2 NH 2 ) 2 ] 3 NH 3 NH CH 2 CH 2 NH 2 2 NH 3 3+ [Co(NH ) 3 Figure 13.4.1 Co-ordnaton ] number 6 A complex ion formed from The co - ordination number of a complex ion is the number of co - a cobalt 3+ ion and six ammonia molecules ordinate bonds a ligand Monodentate Bidentate forms ligands with form one the central bond per transition ligand, e.g. metal water , ion. ammonia. Exam tips These Remember that the charge ligands ligands form have two two bonds lone pairs per ligand, available to e.g. diaminoethane. form co - ordinate on bonds. a complex charges and all on the on s the the sum transton of the metal on Hexadentate ligands form 6 bonds per ligand, e.g. EDT A. lgands. a b H H C C N N H O H H O H H H Figure 13.4.2 164 Two bidentate ligands: a 1,2-diaminoethane; b benzene-1,2-diol Chapter Ligand If there Transton elements exchange is more transition ligands: 13 than metal O H one cation. and NH 2 transition . ligand For in a solution, example, The better they aqueous the ligand can compete ammonia is at for contains competing for a two the 3 element ion, the more stable is the complex formed. When we 2+ add a few drops of concentrated HCl to an aqueous solution of ions Cu 2+ (which contains the complex ions O) [Cu(H 2 equilibrium is set V ], following up: 2+ O) [Cu(H the 6 2 ] 2– (aq) + 4Cl (aq) Y [CuCl 6 ] (aq) + 6H 4 blue O(l) 2 yellow-green 2+ Addition of HCl to the [Cu(H O) 2 equilibrium to the right and ] complex shifts the position of 6 so the colour changes from blue to green as 2– the complex ] [CuCl (aq) is formed. Addition of water shifts the position 4 of equilibrium The to equilibrium constant, the left. constant for this reaction is called the stability K stab 2– [[CuCl ] (aq)] _________________________ 4 K = stab 2+ [[Cu(H O) 2 The larger complex Ammonia ammonia blue the and value the has a will complex of the more higher shift ion the is stability likely the ] 4– ] × [Cl (aq)] 6 constant, complex stability constant position of the will more stable than Cl equilibrium ions. to the + 4NH + 2H 3 O(l) Y [Cu(NH 2 ) 3 (H 4 occur as exchanges complexes are similar with to water , those of a of deep 2 ] ions and copper + 4Cl (aq) Y [CoCl + 4Cl (aq) ammonia. complexes: ] (aq) + 6H 4 O(l) 2 blue 2+ O) 2 (aq) 2– (aq) 6 pink [Co(H ] 2 blue Cl the 2+ O) [Co(H O) 2 deep ions ligand addition and 2+ (aq) 4 The the formed: yellow-green Cobalt( ii) So right 2– ] [CuCl is form. ] Key points 2+ (aq) + 6NH 6 (aq) Y [Co(NH 3 ) 3 pink ] (aq) + 6H 6 O(l) 2 yellow Transton elements form complexes Lgand exchange n one The red blood pigment haem is found as a group attached to the in red blood cells. The molecule has an Fe ion or more with number Lgands of 6. Four of the co - ordination positions are are is with a position atoms in a nitrogen is with complex atom an in oxygen ring the system. protein molecule. A fth molecule. The co - ordination The oxygen sixth molecule and carried to the cells for respiration. Carbon the element co-ordnate on by one or bonds. Lgand exchange depends on weakly the bound to position coordination is bonded to more nitrogen wth a transton co - ordination combnng lgands. protein 2+ haemoglobin by haem monoxide will stablty constants n an also equlbrum reacton where the 2+ bind to the ion. Fe It has a stability constant about 200 times higher lgands than O . In the presence of carbon monoxide, hardly any oxygen will the respiration is inhibited. The result is often death, even if of carbon monoxide is fairly transton element wth on. the concentration bondng bind 2 and compete for A colour change s often low. obsered when lgands are lgands n Examples a one or other complex. of lgand exchange 2+ nclude Cu more exchanged for 2+ (aq) or Co (aq) wth 2+ NH and CO/O 3 complex wth the Fe 2 n haem. 165 13.5 Complex Learning outcomes ions The The On completon should be able deduce ths secton, shapes planar, of complexes tetrahedral redox shapes of shape of a reactions complex complex ion ions depends on: you to: the (square of and its co - ordination the type of number ligand which bonds to the transition element ion. or The shape of the complex cannot be predicted using the VSEPR theory octahedral) because their 3+ descrbe the use of Fe MnO /Mn and Cr as O 2 redox inuence on in d orbitals differ from those in s and p orbitals in structure. , 3+ 2– 4 electrons 2+ /Fe 2+ the /Cr The most common co - ordination numbers are 2, 4 and 6. 7 systems. Co-ordnaton Complexes number involving gold, 2 silver or copper in oxidation state +1 usually + have a linear structure, e.g. ) [Ag(NH 3 ] , CuCl 2 2 + [H N → Ag ← NH 3 Co-ordnaton These shape may is be more number either 4 square common. It ] 3 planar is or found tetrahedral. in many The square ii), nickel( planar copper( ii) and 2– platinum( ii) complexes. The tetrahedral form is found in the [CoCl ] 4 2– ion and the molecule. Ni(CO) The reason why the [CoCl 4 coordination chloride ion transition is of 4 relatively element rather large. than So 6 fewer (see Figure ligands 13.4.1) ion can t is has round that a the the central ion. 2 a ] 4 number 2 b N Cl N C C Cl Ni Co Cl C C N CO O 2 Cl N C 2 Cu 2– Figure 13.5.1 a [Ni(CN) ] 2– has a square planar shape; b [CoCl 4 CO O 2 A square planar complex of copper with 2-hydroxybenzoate ions Square Each planar ligand complexes forms Co-ordnaton This Did you know? is the Examples sometmes use two can also coordinate number commonest be formed bonds to excreted n EDTA transton posonng. The the complex urne. ligands. element ion. type of coordination. It is always octahedral. O) ] , 3– [Fe(CN) 6 ] 2+ and [Ni(NH 6 ) 3 ] 6 as metal ons are An octahedral bond with the Octahedral Each 166 bidentate transition 6 3+ antdote the the 2 lgand using with are: [Co(H hexadentate an has a tetrahedral shape 2 Figure 13.5.2 Doctors ] 4 C ligand shape is central complexes forms usually formed transition two can also if element be coordinate the ion formed bonds ligand is using with atom relatively three the forming the small. bidentate transition ligands. element ion. Chapter 13 Transton elements CH 3 a H 2 b CH O NH 2 2 2 H H O N NH 2 2 2 CH 2 Co H OH Co O CH 2 2 2 H N NH 2 OH 2 CH 2 NH 2 2 CH OH 2 2 Figure 13.5.3 Octahedral complexes of cobalt: a with a monodentate ligand, water; b with a bidentate ligand, diaminoethane Transition When compounds reagents, the Different useful ions in a of state states demonstrating as way. a A solution few manganate( vii) of of the iron( ii) of of are iron( ii) agent ions to + MnO (aq) + 8H may sulphate such as iron( iii) → 5Fe change. are often containing can be acidied iron( used ii) in potassium ions. 3+ (aq) suitable complexes solution + (aq) with element element A redox treated ammonium oxidising oxidise and transition reactions. 2+ 5Fe elements transition redox drops will compounds transition oxidation oxidation such such element 2+ (aq) + Mn (aq) + 4H 4 O(l) 2 2+ The colour of the solution changes from the light green of the Fe ions to 3+ the reddish-brown products conrms precipitate agents colour is such the formed. as zinc of ions. Fe presence The react of iron( opposite iii) to of when change ions 3+ 2Fe addition ions colour iron( iii) with The + Zn(s) → 2Fe a to the red-brown occurs form when iron( ii) 2+ (aq) alkali reducing ions e.g. Did you know? 2+ (aq) + Zn (aq) Potassum The to iron( ii) the ions solution formed when a in this reaction grey-green can be precipitate tested is for by adding alkali formed. usually manganate(vii) because Potassum managnate(vii) hgh MnO ions present in a solution of acidied t can degree prmary The dchromate preferred to n be of redox ttratons obtaned purty. standard s potassum It s to a used because t as a s also to ge potassium 4 stable manganate( vii) dioxide and can be hydrogen used as an sulphide oxidising as well as agent being to test useful for in redox titrations 2+ (see to page 45). indicate It the permanent is reduced end pink point to of a colouration Mn ions. redox titration. is seen due The to colour At the the change end presence can point of n ar, readly soluble sulphur be used a stable mass soluton, and ttrates has a hgh molar reproducbly. a excess MnO 4 Key points ions. acid), An example C H 2 is in estimating the concentration of ethanedioic (oxalic O 2 4 + 2– 5C O 2 (aq) + 2MnO 4 (aq) + 16H The shape → 10CO 4 (g) + 2Mn (aq) + 8H 2 O(l) complexes 2 number Potassum dchromate(vi) Potasium dichromate( vi) is a bonds good oxidizing agent. 2– the orange O Cr 2 of transton element 2+ (aq) In acidic conditions, The depends and to type the shapes of metal of on the lgand whch caton. complexes are 3+ ions are reduced to green Cr ions. It is used as in lnear, 7 tetrahedral, square planar 2+ volumetric analysis 2– Cr O 2 to determine 2+ (aq) + 6Fe the concentration + (aq) + 14H of 3+ (aq) → 2Cr ions. Fe and octahedral. 3+ (aq) + 6Fe – (aq) + 7H 7 O(l) 2 MnO used A redox indicator such as barium diphenylamine sulphonate is rst drop of excess dichromate converts this to a deep as O 2 can be 7 oxdsng agents and added. to The 2– and Cr 4 test for reducng agents n bluish the laboratory. Durng these solution. reactons changes characterstc colour occur. 167 14 Qualitative tests for 14. 1 dentifying cations Learning outcomes Qualitative There On completon should be able of ths secton, are descrbe to: used to how a ame dentfy test partcular can descrbe catons how usng soluton of to dentfy an analysis stages in analysing the elements present in a compound: Preliminary the tests: action of these heat include on the the appearance of the solid, flame test compound. be metal catons (1) you and four ions specc Making a solution of the T esting for the cations T esting for the anions substance under test. present. present. (See Section 14.3) aqueous carbonate ons, know The flame test the prncples based these and on wrte whch ths s equatons for descrbe a test for A s reactons ammonum flame block test of Clean ons. can the a used platinum hydrochloric a be Periodic Place Hold the Note any Observe identify or The some cations, procedure Nichrome wire by especially those in the is: dipping it in concentrated acid. sample wire of on colour the to T able. the compound the in edge the coloured of a on the end of non-luminous the wire. Bunsen flame. flame. flame through a diffraction grating or spectroscope. non-luminous flame Pt wire sample Figure 14.1.1 The flame colours I sodum for selected Group – orange- – llac visible to the yellow 168 cations – you look see the region most lines at the (see brck the lines of 1.3). in sodium – red copper greensh- red apple-green the The the – element blue crmson through lines Section obvious in flame coloured – are: Transition strontum barum will metal (orange-red) potassum you loop II calcum yellow When on Carrying out a flame test typical Group with a diffraction line colours line spectrum seen emission are grating emission in or spectroscope, spectrum the spectrum. particularly in flame For distinct. the test are due example, the Chapter Making a Attempt If it does If it is Many So it to dissolve not for the dissolve, insoluble tests is solution of the in water , cations important to solid warm and know Qualtatve tests for ons solid in a few cubic centimetres of water . gently. try to anions the 14 dissolve depend solubility it in on nitric acid. solubility rules for and precipitation. common inorganic compounds: + Common salts of Group I cations and are NH soluble. 4 All Sulphates nitrates Chlorides, are soluble. 2+ are soluble except those of 2+ , Ca Ba 2+ , Sr 2+ and Pb + bromides and iodides are soluble except those of Ag and 2+ Pb Carbonates and cations NH sulphites are insoluble except those of Group I + and 4 + All hydroxides are insoluble except those of Group I cations, NH , 4 2+ 2+ and Sr Using Ba sodium Aqueous sodium carbonate to test for carbonate is sometimes used cations to conrm the presence or + absence of a Group I cation or . NH Most carbonates are insoluble in 4 water . + An aqueous solution of Group I cations or will NH not form a 4 precipitate on are soluble. An aqueous precipitate addition solution because of sodium containing their carbonate other carbonates because metal are cations insoluble. these will carbonates give Group II Did you know? a carbonates When give a white precipitate and transition element carbonates may n coloured precipitates, e.g. is MnCO ammonum salts are heated give a test tube they almost always pink. 3 decompose. But when the vapours 2+ Mg ions can be distinguished from other Group II ions by adding are ammonium carbonate in the presence of ammonium chloride. cooled further sold these conditions, no precipitate is up the salt s reformed, + HCl → NH 3 ammonium These ions off, ammonium salts decompose on warming. Ammonia gas is by Cl(s) 4 compounds pured Many the e.g. formed. NH A test for tube, Under can therefore be sublmaton. given e.g. Key points Cl(s) NH → NH 4 (NH ) 4 (g) + HCl(g) 3 SO 2 (s) → 2NH 4 (g) + SO 3 (g) + H 3 O(l) 2 Ammonium oxide All rather nitrate than ammonium is an exception. It decomposes to form nitrogen( i) a ammonia. salts give off ammonia when warmed with dilute When colours dependng the put n are man seen, colours n aqueous Cl(aq) + NaOH(aq) → NH 4 lne emsson (g) + NaCl(aq) + H 3 spectrum. Specc metal ammonia can be detected by the fact catons can be O(l) 2 dented that it turns damp red usng an aqueous litmus soluton paper on are Bunsen ame, hydroxide: NH The elements characterstc the sodium some non-lumnous of carbonate ons. blue. When a compound ammonum aqueous ons sodum ammona s s contanng warmed wth hydroxde, released. 169 14.2 dentifying Learning outcomes cations Tests Many On completon of ths secton, using metal be able descrbe catons how to usng dentfy sodum metal aqueous explan (if aqueous may any) be formed solution redissolve identied by the by observing addition of the dilute colour sodium of the hydroxide in of the excess substance sodium under test. Some of the to precipitates hydroxide. If no precipitate is formed: A cation from Group I or ammonium ions ammona may can hydroxide hydroxde or cations sodium to: an aqueous you precipitate should (2) the prncples on be present. whch (The Group I cations can be distinguished by flame + tests and ions NH by warming the alkaline solution.) 4 the above test s based deduce onc symbols for equatons the wth reacton of state If a white precipitate metal 2+ wth () sodum aqueous is insoluble in excess sodium Ca 2+ , 2+ Sr or Ba may be present hydroxde (these () which 2+ , Mg catons forms hydroxide: can be distinguished by a flame test). ammona. If a white precipitate forms which is soluble in excess sodium hydroxide: 3+ 2+ , Al 2+ Pb or Zn ions may be present. Exam tips If a coloured the A lot of memory work s precipitate is formed the colour may be used to identify cation: requred 3+ to learn all these qualtatve Cr tests. → grey-green precipitate → pale → green → rusty-brown → cream → green 2+ One way of dong ths s to wrte Cu the blue precipitate 2+ on and test on a sheet of paper Fe n precipitate (turning brown) 3+ separate columns, then cover Fe one (reddish-brown) precipitate 2+ of the columns up and see how Mn well precipitate (turning dark brown) 2+ Ni you can remember the precipitate tests. The precipitates insoluble in formed water are apart of from metal those hydroxides. of Group I or Metal hydroxides ammonium are ions. Did you know? T ypical equations are: 2+ The reacton of some transton (aq) Fe + 2OH (aq) → Fe(OH) + 2OH (aq) → Cu(OH) (s) 2 2+ metal catons wth sodum (aq) Cu hydroxde (s) 2 3+ s qute complcated because complex formaton. The (aq) Cr of + 3OH (aq) → Cr(OH) (s) 3 actual The precipitates of aluminium, zinc and lead hydroxides dissolve in 2+ reacton of Cu ons wth OH ons excess may be more accurately sodium hydroxide because soluble aluminates, zincates represented plumbates are formed: as: (s) Al(OH) + NaOH(aq) → NaAl(OH) 3 (aq) 4 2+ [Cu(H O) 2 ] +2OH → sodium 6 aluminate 2+ [Cu(H O) 2 (OH) 4 ] + 2H 2 O(l) 2 Zn(OH) (s) + 2NaOH(aq) → 2 Na Zn(OH) 2 Pb(OH) (s) 2 Exam tips + 2NaOH(aq) → Na and of plumbates alumnates, are – wrtten AlO , 2– ZnO 2 respectvely. You to use 170 zncates sometmes 2– and PbO 2 2 should these formulae as be prepared well. zincate Pb(OH) 2 sodium The formulae (aq) 4 sodium (aq) 4 plumbate(ii) and Chapter Tests The using effect of aqueous conrmatory formed by tests the substance the precipitate the under test allows reactions ammonia some are of contains and the metal The aqueous as ions of ammonia effect for due cations colour of distinction same OH on ions. some the Qualtatve tests for ons ammonia ammonia for addition the of aqueous 14 to to an excess between sodium the in the solution provides precipitate aqueous aqueous cations hydroxide (if any) solution ammonia to be made. because of on Most aqueous reaction: + NH (aq) + H 3 The concentration O(l) Y NH 2 of (aq) + OH 4 hydroxide ions in dilute aqueous ammonia is + low in comparison with NaOH, so test solutions containing Ca 2 + ions, + and Sr Ba 2 even no ions may form only slight white precipitate or precipitate. 3+ a 2 2+ can Al be distinguished from Zn by the use of aqueous ammonia. 3+ ions Al form a white precipitate which is insoluble in excess 2+ aqueous ammonia. redissolves which is in excess soluble ions Zn in aqueous water is form a white ammonia. A precipitate colourless which complex ion formed: 2+ (aq) Zn + 2OH + 4NH (aq) → (aq) → Zn(OH) (s) 2 2+ in excess: (s) Zn(OH) 2 Copper which hydroxide are soluble [Zn(NH 3 dissolves in water . in excess The ) 3 ammonia complex ion is ] – (aq) + 2OH (aq) 4 to a form deep complex blue ions colour . 2+ (aq) Cu + 2OH (aq) → Cu(OH) (s) 2 in excess: 2+ Cu(OH) (s) + 4NH 2 often (aq) + 2H 3 simplied O(l) → [Cu(NH 2 ) 3 (H 4 O) 2 ] (aq) + 2OH (aq) 2 to: 2+ (s) Cu(OH) + 4NH 2 Other (aq) → [Cu(NH 3 transition element ) 3 hydroxides ] (aq) + 2OH (aq) 4 such as and Co(OH) Ni(OH) 2 will but also redissolve in does Mn(OH) excess not aqueous ammonia to 2 form complexes, redissolve. 2 Key points Some metal precptates catons on n aqueous addton of soluton form NaOH(aq) or NH characterstcally coloured (aq). 3 Dependng n excess on the metal NaOH(aq) or caton, NH these precptates may or may not dssolve (aq). 3 Alumnum and complex and copper(ii) Ionc ons znc hydroxde hydroxde redssolve redssolves n n excess excess sodum ammona hydroxde because soluble are formed. equatons can be wrtten to show precptaton and complex on formaton. 171 14.3 dentifying Learning outcomes anions Testing for When On completon should be able of ths secton, an acid carbonates is added to a carbonate, carbon dioxide is released: you to: CaCO (s) + 2HCl(aq) → CaCl 3 (aq) + CO 2 (g) + H 2 O(l) 2 2– descrbe tests to dentfy CO , 3 2– NO , SO 3 Carbon dioxide turns limewater (a dilute solution of calcium hydroxide) 2– , SO 3 , Cl , Br , I and 4 milky. The cloudiness in the solution is a ne precipitate of calcium 2– CrO ons carbonate: 4 explan the prncples on whch Ca(OH) (aq) + CO (g) 2 these tests deduce are onc equatons wth CaCO (s) + H 3 O(l) 2 state Testing for symbols for → 2 based these nitrates reactons. There are several tests for nitrates: 3 W arm a little of the solid with a 2 cm of concentrated sulphuric acid. Exam tip Then , NO You wll not be expected add are a small piece formed, a of copper . nitrate is If brown probably as the you for to complex ones for should smple work () the equatons ntrate know Cu(s) such () test. be But + 2NO (s) + 4H A conrmatory suspected (or able → present: Cu sutable Secton + 2NO (g) + 2H 2 test nitrate Devarda’s (aq) NO is to and alloy). add then On aqueous either sodium zinc warming, hydroxide powder ammonia or gas to the aluminium is O(l) 2 powder released: + 4Zn(s) + 7OH (aq) + 6H O(l) → NH 2 (g) + 4Zn(OH) 3 (aq) 4 nformaton The (see (aq) 2– equatons from rst gven dioxide, 2+ (aq) 3 prncples nitrogen 3 the formulae compounds out of 2 to + remember fumes ammonia can be identied by using damp red litmus paper . If 4.2). ammonia is Testing for The solution present, the to be tested is carbonates barium are of barium turns blue. sulphates contaminating nitrate litmus then sulphate is acidied present. added. formed If a (see nitric Section acid barium sulphate 2+ is present, remove a any or white aqueous precipitate 12.2). + SO (aq) → BaSO 4 Testing for to chloride 2– (aq) Ba with Aqueous (s) 4 sulphites 2– The sulphite ion has the formula SO . Sulphites are more unstable to 3 heat than sulphite. If the sulphates Sulphur sulphite solution gas are and is indicates dioxide in the so simple gas solution, solution the a is presence is hydrochloric a to heated. acid The dioxide conrmed by Bubbling solution Soaking The If through turns a + 2H choking a (aq) acidic solution from piece of ‘dichromate sulphur is rst release → SO (g) + 2 a sample of the solid added of to the sulphur dioxide smell. H O(l) 2 Its identication can be either: purple lter paper ’ dioxide green-blue. 172 has a + (aq) 3 Sulphur heat sulphite: 2– SO is released. then of test is of to paper is in then present, potassium manganate( Vii). The colourless. potassium placed the paper over dichromate the turns mouth from solution. of the orange to test tube. Chapter Testing for The test The with to see either silver procedure suspected is tests for ons aqueous in full Aqueous the give in nitrate white Section silver precipitate a silver or aqueous lead nitrate. nitrate given halide. whether Chlorides Qualtatve halides involves Testing 14 12.8. nitrate dissolves precipitate is in of Nitric then added aqueous silver acid is added followed by to the testing ammonia. chloride: + (aq) Ag This precipitate formation of a dissolves complex + Cl (aq) readily in → AgCl(s) aqueous ammonia due to the ion: + AgCl(s) + (aq) 2NH → [Ag(NH 3 Bromides give a cream ) 3 precipitate of ] (aq) + Cl (aq) 2 silver bromide: + Ag The silver bromide (aq) + precipitate Br (aq) → dissolves AgBr(s) only in excess aqueous ammonia. Iodides in give aqueous a pale yellow precipitate of silver iodide which is insoluble ammonia: + Ag Testing Nitric then with acid is lead added (aq) + I (aq) → AgI(s) nitrate to the suspected nitrate. A solution of lead nitrate is added. Chlorides give a white precipitate of lead( ii) chloride: 2+ Pb (aq) + 2Cl (aq) → PbCl (s) 2 Bromides Iodides The give give addition a of a pale deep iodide yellow yellow ions precipitate precipitate to a of of solution lead( lead( of lead ii) ii) bromide. iodide. ions can also act as a 2+ conrmatory test Testing for for Pb ions. chromates 2– The chromate ion has the formula, CrO . We add a few cubic 4 centimetres chromate. chromate If of dilute the ions sulphuric chromate turns ion orange is to the present, due 2– to the solution the + 2H of yellow formation + (aq) 2CrO acid of the suspected colour of the dichromate ions: 2– (aq) → Cr 4 O 2 chromate (aq) + H 7 O(l) Key points 2 dichromate Ions When a drop of dilute a colouration hydrogen peroxide is added to the can be followng blue is seen which quickly turns dented usng the dichromate, aqueous solutons: green. 2– CO usng hydrochlorc acd and 3 then lmewater – NO usng Cu and H 3 SO 2 or Al 4 – and OH 2– SO usng Ba(NO 4 ) 3 2 2– SO usng H 3 SO 2 Haldes 4 usng AgNO or Pb(NO 3 ) 3 2– CrO usng 4 acd and H O 2 2 173 2 Exam-style Answers to all exam-style questions can questions be found on the Multiple-choice questions 1 Which of the following are chemically ii very high iii high tensile Structured questions properties of ceramic 6 a i Define ii State four inert the term characteristics melting point iii Write the strength electronic i, ii B i, ii, D i, transition configuration for Mn of ions, showing electrons in the only iii, Use the information provided in part 3+ ions Fe are easily oxidised to Fe represents the correct order of – 3– , P 3+ , Al resist oxidation to Mn [2] ions. [2] increasing When iron(III) chloride B P C Al D Na – , Cl is dissolved in water a red- Na 3+ Na + – , Cl 3+ , Al chloride gives a in above P ii PbCl B SnCl above Write an [2] equation to represent the reaction. – acid white [1] solution: or grey 7 precipitate with a aqueous chloride aqueous the following potassium chlorides manganate(VII). will undergo b the Define the terms i co-ordination ii ligand complex concentrated added to is [2] [2] When sulphate reactions? A the , Cl decolourises of explanation for 3– , P mercury(II) Which an observation. 3– , Give 3+ , Al + , is formed. Na + , precipitate + , i 3– ii ions radius? Cl i ions. 3+ Mn brown A explain iv c 3 to iv ii A iii observations: 2+ ionic the [2] 2+ ii, Which of iii i 2 the orbitals shell. the following ii, a compression b C of [2] 3+ and Fe valence A metal’. [4] the occupation resists ‘transition metal. 3+ iv 3 accompanying CD material? i Module a pale until blue hydrochloric solution present in of acid is slowly copper(II) excess, a yellow solution produced. 4 i Write the formulae of the species responsible 2 for C PbCl D SnCl the pale blue and yellow solutions 2 respectively. [2] 4 4 Which of the A Carbon B Lead(IV) statements monoxide below turns is blue ii Write the equation for iii Using the concept litmus is more stable to heat The chemistry affected D The is 5 An by not very i The ii t of the screening element germanium inert-pair effect of is i significantly t chloride does Which A Be B Mg C Sr D Ba 174 the above colour change. what would the inner is lone easily not electrons in properties: hydrolysed. pairs forming react element at all best fits to lead ii Write co-ordination readily the with above an is be observed slowly aqueous the formula for formed. the following ammonia [3] when added solution of until copper(II) sulphate. compounds. iii explain Describe excess effect. effective. exhibits accepts [2] constant, chloride. aqueous C reaction. stability than c silicon(IV) of pink carefully chloride the correct? water. properties? [3] the final copper species [1] Module 8 a Explain the difference in the reaction (if any) of and SiCl 4 with water and write Exam-style questons 10 Element/symbol CCl 3 b.pt. density equations for 4 any reactions involved. The standard electrode [5] F b 4+ Ge potential for the 2+ (aq)/Ge (aq) values for Sn and system Pb to is –1.6 . Use deduce the Cl similar relative Br stability Group of the +2 and +4 oxidation states of the elements. [3] I c Chlorine into a disproportionates hot concentrated when solution bubbled of potassium a i hydroxide. i Write ii an the ionic and chlorine Write two equation state species ionic process the and to show what oxidation physical number of involved. half equations thereby a Give i explanations for is BaSO less group [2] to explain the soluble than term Explain and b A properties indicate trends of the of by the the use above elements as the descended. your [2] answers and to bonding part with i in terms regards to of b.pt. [4] crystalline solid A sulphuric white fuming gas when acid treated readily with evolves a MgSO . B as well as a reddish vapour. [2] 4 Magnesium carbonate strontium temperature Explain and the density. white decomposes at when the is trends carbonate’s dissolved in water forms a solution C 350 °C which b table arrows concentrated B while above statements: 4 ii is structure [5] the following ii describe ‘disproportionation’. 9 the of vertical happens the Copy produces a colourless gas, D, with sodium decomposition 1340 °C. carbonate, which when solution calcium bubbled through a [4] of hydroxide produces a white in precipitate. i atomic ii ionic radius [2] When radius cream of aqueous A the Group compound heating oxide, of is a brown containing element gas and 10.46% a (M) on added to C dissolves a in white to produce oxygen a by ammonia. i State the ii State the formulae names of the two gases B and D. [2] metallic mass. colourless aqueous strong of the species responsible N for dissolves which elements. a Group evolves N, readily nitrate precipitate forms concentrated c silver [1] the production of the gases. aqueous dentifying the observations underlying the solution, S. above S when reacted with an appropriate dilute M on crystallising from the [2] acid iii produces answers. Write two ionic equations to explain the resulting observations associated with the reaction solution. of C i State the formula ii Deduce of the brown gas. c the atomic mass of the with element M. State iv Write the name and formula of Use relevant M. equation formation of which M from S. represents [2] to information explain the involving relative electrode oxidising [1] power the nitrate. [2] potentials iii silver [1] of the halogens. [3] the [2] 175 Data Selected sheets bond energies Selected electrode potentials Ø Diatomic molecules Polyatomic molecules –1 Electrode –1 Bond 436 C–C 350 Mg N≡N 994 C=C 610 Al O=O 496 C–H 410 V F–F 158 C–Cl 340 Zn Cl–Cl 244 C–Br 280 Fe Br–Br 193 C–I 240 V I–I 151 C–N 305 N H–F 562 C–O 360 Sn H–Cl 431 C=O 740 Pb H–Br 366 N–H 390 2H Bond H–H energy/kJ mol energy/kJ mol E /V + Bond Bond reaction + K e Y K –2.92 2+ + 2e Y Mg –2.38 3+ + 3e Y Al –1.66 2+ + 2e Y V –1.2 2+ + 2e Y Zn –0.76 2+ + 2e Y 3+ Fe –0.44 2+ + e Y V –0.26 2+ + 2e Y N –0.25 + 2e Y Sn –0. 14 + 2e Y Pb –0. 13 2+ 2+ + + 2e Y H 0.00 2 2– H–I 299 N–N 160 S O 4 2– + 2e Y 2S 6 O 2 +0.09 3 2+ O–H 460 Cu + O–O 150 VO 2e Y Cu 2+ +0.34 + + 2H 3+ + e Y V + H O +0.34 2 + I 2e Y 2I +0.54 2 3+ 2+ + Fe e Y Fe +0.77 + + Ag e Y Ag + +0.80 + + VO 2H 2+ + e Y VO + H 2 + Br O +1.00 2 2e Y 2Br +1.07 2 2– O Cr 2 + Cl + + 14H 3+ + 6e Y 2Cr + 7H 7 O +1.33 2 2e Y 2Cl +1.36 2 – 4 176 + + MnO 8H 2+ + 5e Y Mn + 4H O 2 +1.52 sheets Data 0.571 301 rL muicnerwal ]262[ 0.371 201 oN muilebon ]952[ 9.861 101 dM muivelednem ]852[ 3.761 001 mF muimref ]752[ 9.461 99 sE muinietsnie ]252[ 5.261 89 fC muinrofilac ]152[ 9.851 79 kB muilekreb ]742[ 3.751 69 mC muiruc ]742[ 0.251 59 mA muicnema ]342[ 4.051 49 uP muinotulp ]442[ ]541[ 39 pN muinutpen ]732[ 2.441 29 U muinaru 0.832 9.041 19 aP muinitcatorp ]132[ eC munec 1.041 09 hT munoht 0.232 901 tM muirentiem ]86 2 [ muimso 2.091 801 sH muissah ]962[ muinehr 2.681 701 hB muirhob ]462[ netsgnut 8.381 601 gS muigrobaes ]662[ mulatnat 9.081 501 bD muinbud ]262[ fH muinfah 5.871 401 fR muidrofrehtur ]162[ 75 aL munahtnal 9.831 98 cA muinitca ]722[ 65 aB muirab 3.731 88 aR muidar ]622[ 74.58 55 sC muiseac 9.231 78 rF muicnarf ]322[ 85 rP muimydoesarp muidiri 2.291 aT 27 95 dN muimydoen 1.591 W 37 06 mP muihtemorp munitalp eR 47 16 mS muiramas dlog 0.791 sO 57 26 uE muiporue yrucrem 6.002 rI 67 36 dG muinilodag muillaht 4.402 tP 77 46 bT muibret dael 2.702 uA 87 56 yD muisorpsyd htumsib 0.902 gH 97 66 oH muimloh ]902[ lT 08 76 rE muinolop bP 18 86 muibre ]012[ iB 28 96 mT muiluht enitatsa oP 38 07 bY muibretty nodar ]222[ tA 48 17 uL muitetul nR 58 73 bR 68 rS 01.93 muidibur 83 26.78 Y 80.04 muitnorts 93 muirtty rZ 69.44 19.88 04 22.19 bN 78.74 muinocriz 14 muiboin oM 49.05 19.29 24 49.59 cT 00.25 munedbylom 34 ]89[ uR 49.45 muitenhcet 44 1.101 hR 58.55 muinehtur 54 9.201 dP 39.85 K muissatop muidohr 64 aC muiclac 4.601 gA 96.85 cS muidnacs muidallap 74 iT muinatit revlis dC 55.36 V muidanav 9.701 84 rC muimorhc 4.211 nI 93.56 nM esenagnam muimdac 94 nori muidni nS 27.96 eF 8.411 05 oC tlaboc nit bS 16.27 iN lekcin 7.811 15 uC reppoc 8.121 eT 29.47 cniz ynomitna 25 nZ 6.721 I 69.87 aG muillag muirullet 35 eG muinamreg enidoi eX 09.97 sA cinesra 9.621 45 eS muineles nonex 0 8 . 38 rB enimorb 3.131 rK notpyrk 13.42 02 210.9 muihtil 149.6 3 iL eB 4 muillyreb 11 91 99.22 12 92 22 03 cimota 21 32 13 42 23 52 33 62 43 89.62 72 53 90.82 82 63 79.03 evitaler 70.23 aN muidos yeK )notorp( cimota eman cimota rebmun lobmys ssam 54.53 gM muisengam 800.1 59.93 lA muinimula 31 iS nocilis 41 P surohpsohp norob 18.01 51 S nobrac 10.21 61 ruflus 10.41 71 lC enirolhc negortin 81 rA nogra negyxo B 00.61 C eniroufl N 5 00.91 O 6 noen F 7 81.02 eN 8 9 negordyh 01 muileh 1 H 300.4 2 eH 177 Answers to revision questions z b Module 1 Chapter 1 1 All z y (page atoms of the x 12) same element are exactly cannot be atoms of atoms combne broken dfferent down elements hae more orbital x dfferent complex orbital p any further, masses, 2 13 to form x alke, s atoms y a 1s b 1s structures 2 2s 2 6 2p 2 2s 2 6 3s 6 2p 2 3p 2 3s 2 4s 6 3p 10 3d c 1s d 1s 2 2 2s 2 4s 6 2p 2 2s 2 3s 6 2p 6 3p 2 3s 6 3p 5 3d (compounds). 14 a The rst onsaton energy, ΔH , s the energy needed 1 Atoms are not ndestructble. Atoms are splt n nuclear to reactons, and they are made up of een smaller remoe all are of the same dentcal), element that s, can they hae dfferent possess masses Electrons b are deflected to the poste plate because they because they not deflected are postely charged and of at of all because they deflecton of an because an electron hae no electron s gaseous lghter nuclear The absolute Ths s due to the gaseous state mole to form of one masses and charges of small, charges so t when s much makng easer more than n ons. + e use comparsons charge, dstance of the outer electrons from the ncrease n nuclear charge as one goes that a The electron n Al s remoed from the hgher energy of orbtal compared to Mg where the electron to be mass. sub-atomc to (g) charge. s comes from relate and the lower energy 3s orbtal. It partcles requres less energy to remoe the electron masses n Al, and ts perod. therefore are ery n sheldng remoed 3 element are 3p proton one negate neutrons 17 a an → Ca nucleus, across degree of of Ca(g) Sze the 16 The n sotopes. are negately charged, protons are deflected to the plate atom + (not 15 2 each partcles). mole Atoms electron from partcles atoms (subatomc one therefore t has a lower rst onsaton energy. dong b The electron to be remoed from S comes from the 3p x calculatons. orbtal 4 a 13 protons, 13 electrons, 14 b 19 protons, 19 electrons, 20 c 53 protons, 53 d 94 protons, 94 (10 × electrons, neutrons neutrons 78 electrons, neutrons 145 + (11 × t The repulson ths electron whch s also s spn-pared between s n easer the these to 3p wth two remoe orbtal another electrons than but s the not electron. means electron that n P, spn-pared. neutrons 18 18.7) where Group II 81.3) ______________________ 5 = 10.8 Chapter 2 (page 32) 100 234 238 6 a i U → b i 4 Th + Th → Ra Pa → + e ii + He C → N + a i Sgma Iodne-131 s used to study s used for datng objects whch were once s used n Uranum-235 used to bonds) are formed by the end-on orbtals. P bonds (π smoke of bonds) p are formed atomc by the sdeways orbtals. lng b Amercum-241 (σ atomc thyrod functon oerlap Carbon-14 of –1 ii 7 bonds oerlap e 7 6 1 2 0 14 14 –1 4 Rn 86 88 0 91 90 ii 2 234 218 222 He 90 92 234 Sgma bonds are stronger because the electron densty s detectors symmetrcal about a lne jonng the two nucle, whereas s generate energy n nuclear reactors n a p bond the electron densty s not symmetrcal. 8 Electrons are arranged n energy leels whch are 2 a lone pair quantsed. of When an electron absorbs partcular frequency moes The to a electron moes As up t seen back does as a or hgher so, lne the of the emsson a Preously to a t of energy becomes of a excted and leel. loses lower energy that quantum energy eentually down a waelength electrons that ths quantum energy t of energy and bond leel. loses s gen partcular frequency or out and s waelength of pair electrons n spectrum. P 9 energy excted electrons fall back to the n = 1 electrons fall back to the n = 2 O leel. Cl b Preously energy 10 ΔE = ΔE s excted leel. hν 3 the energy dfference h s Planck’s ν s the frequency 11 2, 12 a 6 and A between the two energy constant. of the a leels. A co-ordnate both the + b radaton. 10 O regon of space where there s a hgh probablty of H ndng 178 an electron bond s formed electrons for the when coalent one atom bond. prodes Answers to 3+ + 10 2– 2– Al K Iodne a s a smple Weak an molecular revision questions sold: der Waals forces exst between the O molecules, S + allowng the sold to break easly. 3+ b There c Because are no free ons or delocalsed electrons. Al K 2– of attracton between the 3+ 2– + 2 [2,8] [2,8,8] 2 [2,8,8] the forces molecules O to turn are the so sold weak nto that a not lqud much and the energy lqud s needed nto a apour. 2– 11 Water has a much hgher bolng pont due to the fact that O t s extensely whch only hydrogen possess an bonded unlke der Waals forces the other between hydrdes ther 2– 3 [2,8] 5 a Because the element has one electron n ts outer molecules. shell 12 PCl would hae a hgher meltng pont than BCl 3 t s a metal, therefore the bondng s of In ths type of bondng a lattce of metal ons the greater by a ‘sea’ of delocalsed The meltng pont would be of electrons t possesses, produce more stronger an energy to der Waals forces break. Also t can of PCl s greater than that of BCl 3 strong electrostatc attracton between the and the delocalsed electrons acts n all results n the bond beng strong and hard to As a result a the meltng magnesum, pont wll be n on a Electronegatty noled n bondng b s the coalent par of As you go ablty the bondng par from the sze of of a partcular towards to ts an an der Waals forces. – the there are oxygen three bond pars and one lone atom. on – the there are sulphur two bond pars and two lone atom. atom attract c Trgonal d Tetrahedral planar e Pyramdal – there are three bond pars. the – there are four bond pars. tself. of the electrons attracton of electronegatty atoms ncrease beng the on – the there are three phosphorus bond pars and one lone atom. decreases resultng n the Chapter 3 (page 46) ncreasngly further 1 nucleus. a FeCl (aq) + 3NaOH(aq) → Fe(OH) 3 a non-polar c non-polar b polar d polar (s) + 3NaCl(aq) 3 3+ Fe 7 of decreases. down Group VII, because strength Non-lnear par c the causng hgh. bond formaton electrons Electronegatty agan alumnum pars 6 the Pyramdal b sodum, , 3 break. par c whch that drectons 13 and sad metal ncrease ons be hgh. mass Ths whch electrons. requre b number s therefore surrounded because 3 metallc. – (aq) + 3OH (aq) → Fe(OH) (s) 3 b Mg(s) + 2HCl(aq) → MgCl (aq) + H 2 + Mg(s) + 2H (g) 2 2+ (aq) → Mg (aq) + H (g) 2 8 a Permanent dpole–dpole forces, an der Waals forces, c Na S 2 hydrogen O 2 (aq) + 2HCl(aq) → 2NaCl(aq) + H 3 O(l) + 2 bondng S(s) + SO (g) 2 b i an der Waals forces 2– S O 2 ii hydrogen bondng iii hydrogen bondng 2 a i + (aq) A mole permanent Magnesum a gant → H O(l) + S(s) + SO s the amount of (g) 2 of substance speced partcles whch as has there the are dpole–dpole forces chlorde has a gant onc structure, coalent structure and alumnum n exactly 12 g of the carbon-12 sotope. damond ii has (aq) number atoms 9 2H 2 same iv + 3 s Molar mass s the mass of 1 mole of a substance n metallc. grams. a All three solds hae hgh meltng ponts. In magnesum –1 b chlorde, forces of strong onc attracton bonds due between to the the must bonds be broken. between the In oppostely damond, carbon strong atoms 331 g mol 250 g mol must –1 ii 310 g mol ii 20/100 –1 charged 3 ons i iii electrostatc a i 7 . 1/142 b i 0.025 ii 2.0 = 0.05 mol = 0.2 mol coalent be broken × 84 = 2. 1 g and, –3 n of alumnum, attracton delocalsed b make or up the fact free to between chlorde dssoled the Damond metallc electrons Magnesum molten strong t metal must wll n be are conduct possesses ons to and because 4 the to electrcty ons or Balanced C H 3 the ons any 10 × 36.5 (g) equaton for + 5O 8 = 0.073 g (g) → reacton: 3CO 2 Number when of From whch (g) + of C balanced of O = 5.5/44 equaton, 1 mole C electrcty state electrons n both 0. 125 mol H reacts wth 5 moles 8 . due that So to number of moles of O = 0. 125 the × 5 = 0.625 mol 2 Mass are of O = 0.625 × 32 = 20 g 2 a Molecular formula s N O 2 conduct = 8 2 5 wll O(g) 2 H 3 moe. Alumnum 4H 2 moles 3 moe. n × the forces electrcty now free no due broken. conduct water structure cannot that the bonds sold Emprcal formula s . 4 NO . 2 and molten free to state because the delocalsed electrons are b Molecular formula s S Cl 2 c moe. Magnesum Emprcal formula chlorde s soluble n water due to the fact 6 that the ons present can form on–dpole . 2 s SCl. bonds C H wth 0.48 0.08 _____ Number water of moles _____ = 0.04 12 Damond s not soluble n water because the atoms 0.08 = 2 1 are 0.08 0.04 _____ held = molecules. together to form bonds wth Mole water. rato _____ = 1 0.04 Alumnum s not soluble n water because the force 0.04 of Therefore emprcal formula s CH . 2 attracton between the metal ons and the delocalsed Emprcal formula electrons wth s too strong to allow the ons to form mass s 12 + 2 = 14 bonds water. 179 Answers to revision questions Therefore 56 ___ Dde molar mass by emprcal formula mass = = 0.0055 mol potassum manganate(vii) reacts 4 2+ wth 14 Multply each Therefore atom n emprcal formula molecular formula s C 7 a Number of moles = 6/24 = by × 0.0055 = 0.0275 mol Fe 3 4. Ths s the number of moles 3 H 4 5 Conertng 8 Therefore 0.25 mol molar n 25.0 cm of ron(ii). 3 25.0 cm to dm 3 : 25.0/1000 concentraton: = 0.025 dm 0.0275/0.025 = –3 Mass = 0.25 × 32 = 1. 10 mol dm 8 g 3 b Conertng 3 120 cm to dm 3 120/1000 Number Mass = = of Module 0. 12 dm moles 0.005 × at 64 r.t.p. = = 0. 12/24 Volume of Chapter 3 56 cm of moles 7 (page 88) 3 of butane n dm = of butane at s.t.p. 56/1000 = 0.056 dm 1 Number 2 0.005 mol 0.32 g 3 8 = = 0.056/22.4 a Set up apparatus smlar to Fgure 7 . 1. 1, substtute the = dfferent reactants and measure the olume of N at 2 0.0025 mol specc Balanced equaton for tme nterals. reacton: b The a I I (aq) s b Determne brown n colour. 2 H 2C 4 (g) + 13O 10 (g) → 8CO 2 (g) + 10H 2 O(g) 2 2 From balanced equaton, 2 H moles C 4 moles of H react to produce and III 10 10 at least the rst two half-les. For a rst constant. For a O. 2 10 ___ Therefore 0.0025 moles C H 4 produce × 0.0025 mole of reacton, the half-les are order reacton, the half-les become second = 10 2 0.0125 order longer. O H 2 3 Volume of steam at s.t.p. = 0.0125 × 22.4 = 3 9 C H x (g) + O y (g) → xCO 2 (g) yH The a Second graph b Frst c Thrd d Rate would be an upward cure. order O(g) 2 3 3 160 cm 1 ol + 2 3 20 cm c 0.28 dm order 100 cm 8 ol order 5 ol 2 C H 5 (g) + 8O y (g) → 5CO 2 (g) + yH 2 = k[X] [Y] O(g) 2 Rate 10 moles of O reacts wth C so the other 6 moles _______ must e k 6 = = –2 mol 9200 dm –1 s 2 react so 6 wth [X] H, moles H O are formed whch contan 12 atoms of H. 2 Therefore pentane must be C 4 a H 5 Draw may 12 a [Y] tangent to be aratons tangent s the n cure the at the answer tme a Number of moles of KNO –2 = = 0.033 mol [There on how drawn.] 3.33 _____ 10 gen. dependng i For tme = 0 s, Intal rate = 1. 1 × 10 ii For tme = 20 s, Rate = 7 .6 For tme = 60 s, Rate = 3.4 –3 mol dm –1 s 3 –3 101 × 10 250 3 Conertng dm 3 = = 1000 b i 0.033 concentraton = ii of moles of Na CO 2 = = 5 _____ 3 to 75 cm dm 3 : = a Change Frst a Knetc b Refer c = = Number Mass: of to dm moles: 0.00025 × Change 6 750 cm 25/1000 0.01 3 b × = 0.025 = = 0.025 dm 0.00025 mol 0.015 g 3 to t = 35 s, Second order, Mass: of 0.20 moles: × 80 = s energy a Burette dm 0.27 i Rate ii Refer a D b C c C d = the alue. 35 s constant. partcles (y-axs) 7 .3. would decrease. to Secton 7 .3. A 3 : 750/1000 = 0.75 dm 7 a × 0.75 = 0.20 mol 16.0 g readings/ t (x-axs), fracton of to Secton i NO(g) ii Step 1 Rate = + SO (g) + O iii 1 2 3 4 k[SO Final volume 27 . 15 25.80 37 .90 26.50 Initial volume 0.50 0. 15 12.50 1.00 Actual volume 26.65 25.40 25.50 (g) → SO 2 ][NO 2 12 ntal ½ half-lfe 2 Number the half 3 : 58.5 to 0.33 mol dm 3 25 cm taken for decrease –3 0.075 3 11 Frst c 0.025 ______ concentraton tme to 0.075 dm 1000 Molar the 3 106 Conertng –1 s 0.025 mol 75 3 s ½ 2.65 _____ Number –3 mol dm 0. 13 mol dm 0.25 b Half-lfe concentraton –3 = 10 0.25 dm ______ Molar × –1 s –3 _____ 3 to 250 cm –3 mol dm ], (g) + NO 3 the NO 2 (g) 2 functons as a 2 catalyst. 3 b cm i Mechansm ii The Chapters 25.65 (25.40 + 8–10 Aerage olume of acd I would (page 25.50) double. 130) 2 _______________ b rate [NO (g)] [Cl (g)] 2 2 ________________ = 1 a i K –3 = Unts: mol dm Unts: mol Unts: atm Unts: atm Unts: mol Unts: mol c 2 2 [NO 2 3 = Cl(g)] 25.45 cm 2+ [Fe ½ (aq)][I (aq)] 2 ________________ 13 Number of moles of potassum manganate(vii) ii used: K 3+ [Fe 1½ dm – (aq)][I 27 .50 (aq)] 2 ______ 0.2 ½ = c × = (p 0.0055 mol ) O 3 ______ b 1000 i K –1 = p 3 (p ) O From balanced 2 equaton: 3 2+ (aq) MnO + 5Fe (p + (aq) + 8H ½ ) ( p HF (aq) ) ½ (p N ) Cl 2 2 __________________ → 4 ii K 2 = p 2+ Mn 3+ (aq) + 5Fe (aq) + 4H (p O(l) ) (p NH 2 – 1 mol of potassum manganate(vii) reacts wth 5 mol ron(ii) c i K = [IO sp K sp 180 3 2 (aq)] 2+ [Ba 3 (aq)] –9 dm 3 3+ ii ) ClF 3 = [Al – (aq)][OH 3 (aq)] 4 –12 dm the Answers to + O [H 2– (aq)][F (aq)] iii 3 _________________ d i K K Unts: –4 of CO b –3 = revision questions = 2. 1 × –3 10 mol dm 3 mol dm a pH [HF(aq)] + (aq)][CN [H 8 (aq)] ________________ ii K = 10.3 a pH = 13.2 b pH = 12.7 c pH = 1.9 a Dagram –3 = Unts: mol dm a [HCN(aq)] S(aq)][OH [H (aq)] 2 __________________ e K –3 = Unts: mol dm 4+ 9 b [HS smlar to Fgure 10.2.2, wth the Sn (aq)/ (aq)] 2+ Sn 2 a Dynamc equlbrum s the state that exsts for system, where the rate of the forward connected to the a 3+ b closed (aq) i Al(s) reacton Al standard hydrogen electrode. 2+ (aq) Zn (aq) Zn(s) Ø ii E = +0.90 V iii 2Al(s) cell or process s equal process. The at equlbrum represented K s only an s N by (g) Yield i decrease of of + a and the system 3H (g) reerse and the system can Y temperature n or 2+ 2NH + 3Zn 3+ (aq) → 2Al (aq) + 3Zn(s) be of Module of 3 the dynamc 2 Chapters 11–12 (page 156) (g). 3 Position of Value of equilibrium equilibrium shft no to reacton reactants constant K. The alue changng of the products constant 2 b rate equlbrum example equlbrum the reman by affected system. An to concentraton left constant 1 A 6 a effect 2 C 3 Monodentate central C lgand atom, e.g., 4 can C only form ammona 5 one B bond molecule, H to the N:→ 3 ii ncrease shft iii no no effect to rght effect no effect no effect bdentate atom; lgand e.g., can form two bonds ethane-1,2-damne, H to iv c ncrease v ncrease i No of ii effect shft to rght ncrease shft to rght no on poston equlbrum Yeld of shft n effect equlbrum, b yeld or alue i tetrachlorocuprate(ii) ii to NH 2 2 rght, no and ↓ hexaaquaron(iii) 2 Cl poston central CH 2 ↓ constant ncreases, on alue of the NCH 2 3 OH 2 Cl H effect OH O 2 2 constant Cu(II) iii Yeld decreases, shft to the left, alue of Fe(III) equlbrum constant decreases –3 3 a K b [R] c K = Cl Cl 15.3 mol dm c –3 = 0. 14 mol dm H O OH 2 = 2 33.4 c OH –3 d K a p = 0.53 mol dm 2 c 2+ 4 = 3.7 atm , p H = c K [Cu(H O) 2 ] – (aq) + 4Cl 2– (aq) → CuCl 6 (aq) + 6H 4 O(l) 2 2 –4 K i N 2 b c 0.2 atm = 5.0 × = 0.24 –1 10 Pa ii Co-ordnaton iii The no. 6; co-ordnaton no. 4 p chlorne on s much larger than the water p molecule 5 a The maxmum amount of a solute that dssoles n and the copper gen olume of i ×10 a solent –5 K = 5.67 = 1.70 at 2 a partcular therefore of these only ons. temperature –6 mol can a accommodate four b atom 7 dm a In Perod 3, oxdaton states rse steadly as the sp –6 ii K × i For pure water, For a 3 10 –9 mol dm elements ether lose or gan In electroposte all ther alence electrons. sp –4 c solublty s 1.9 × 10 –3 soluton of ii The iii MgCO s 2.4 common 2+ 0.015 mol dm –6 solublty Mg mol dm effect 2+ would precptate, [Mg 2– ][CO 3 ] > case of oxdaton ons, –3 ×10 on –3 mol dm electrons from to +4 a i HCl s a strong acd and dssocates are +1 to shared Na–Al, +3; from n poste slcon coalent bonds to rsng +7 . Na O Al 2 sp oxdaton 6 rse from chlorne e.g. K 3 states elements, number O 2 +1 P 3 O 2 +3 Cl 5 O 2 +5 7 +7 completely, b For NaCl and MgCl : structure onc crystal lattce. 2 whereas C H 6 COOH s a weak acd and dssocates 5 Bondng – strong onc (electroalent) bonds, ons held partally. together ii pH = 1.6 iii pH = 2.9 On pH c i = [H 1.7 , ii [OH pOH = –12 = 3.98 × = 2.51 wth –3 10 of a water neutral the crystal soluton lattce s contanng destroyed hydrated a pH of ons 7 . mol dm –3 ] n 12.3 + ] electrostatc forces. addton resultng b by For –3 ×10 alumnum chlorde: sold lattce structure. mol dm Bondng – onc wth apprecable coalent character. + 7 a i C H 6 NH 5 (aq) + H 2 O(l) Y C 2 H 6 NH 5 (aq) + OH (aq) 3 On addton of water the small on (large charge and hghly charged + ii C H 6 iii NH 5 pH = 3 alumnum 9.0, pK = i The weak hexaaquaalumnum(iii) acd HC H 3 added and ts O 5 would neutralse would neutralse any bond base. C H any acd O 5 from strong acd and the NO added. , from the be Ntrc salt 3 functon as a whch weakens the the O–H n the water molecules thus enablng protons to the 3 acd extracted does . Ths results n acd behaour as s exhbted a on, base 3 conjugate 3 salt whch forms 9.4 b b densty) by a pH of 5. not One of the possble equlbra s represented by the base. equaton: –5 ii K of CH a COOH = 1.8 × 10 –3 mol dm 3 3+ [Al(H –3 [CH COOH] = 1.03 mol dm O) 2 (aq) + H O(l) –3 COO ] = → 2 2+ [Al(H [CH ] 6 , 3 0.756 mol dm O) 2 OH] 5 (aq) + + H O (aq) 3 , 3 ∴pH = 4.61 181 Answers to c Sodum revision questions reacts explosely wth water producng 10 a hydrogen: As the group s temperatures 2Na(s) + 2H O(l) → 2NaOH (aq) + H 2 Carbonates, (g) descended, of the MCO , the decomposton carbonates ncrease. decompose as ths: 3 2 MCO (s) → MO(s) + CO 3 Alumnum s protected by an oxde lm and does The followng react wth water, howeer at eleated (g) 2 not ponts are useful n explanng the trend: temperatures 2– The steam wll conert the metal to ts smaller 2Al(s) + 3H O(g) → Al 2 has a yellow-green contans a low soluton, mxture Cl (g) + H 2 8 a The O 2 solublty of n called (s) → + the oxde on, O , allows t to (aq) + Hence water, and a attached to the metal be caton. makng the crystal structure more thermally stable. whch chlorc(i) HClO strongly (g) 2 and forms chlorne HCl 3H 3 water hydrochlorc O(l) of oxde: more Chlorne sze The acds: resultant further (aq) larger strengthens The tendency for wll decrease charge the these densty bondng of to the the carbonates oxde metal to form on on. the oxde 2 correct graph should be drawn from the down the group and therefore these gen carbonates become more resstant to decomposton coordnates. as b i Rse n densty from sodum to the group s descended. alumnum: Or Structure: gant metallc structures Large Bonding: metallc bondng – atomc whle charge ncreases charge resultng the large metallc bonds leadng to close packng II metals seen as a steady ncrease n Decrease n densty between sulphur and sulphur possesses structure. Chlorne exsts as a sold polarsaton the decrease as the group s catons of descended, to ther ablty produce to carbon polarse small, doxde the and larger the carbonate oxde results n the hgher on thermal stablty of coalent smple the carbonates i Solublty lower down the group. coalent b gaseous of chlorne: decreases. Ths Structure: allow for densty densty. on ii catons charge of therefore atoms of anons. The n Group stronger denstes radus of decreases of the Group II sulphates decrease as the molecules. group Bonding: sulphur’s sold structure coalent bondng whle descended. ndcates ii stronger s only The solublty depends on the enthalpy change of weak soluton, ΔH . soln ntermolecular an der Waals forces hold the By applyng Hess’s law ; ΔH . = ΔH soln gaseous chlorne molecules together. Ths results Both ΔH and ΔH hyd the sgncant reducton n – ΔH hyd latt n decrease as the group s latt densty. descended, howeer the decrease by ΔH s more hyd 9 a i All tetrahaldes, MCl , exst as coalently bonded, 4 sgncant than that of ΔH (ΔH latt smple tetrahedral molecules wth weak an ntermolecular bonds, hence on the der sze Waals depends hyd of the caton). Therefore as the group s ther olatlty. descended ΔH becomes more poste resultng soln The symmetry of the tetrahedral structure leads to n a net dpole moment of zero, thus ther decrease solublty of the sulphates. non-polar c i calcum oxde: nclude reducng acdty of sols, nature. paper ii The reactty of the tetrahaldes ncrease as s descended – the elements to metallc, and hence become more magnesum oxde: unstable, ndgeston), thus of hydrogen chlorde calcum gas along wth the oxde of oxdaton state 4 (CCl carbonate: + 2H 4 + d In → SO + O → i the + + 4HCl +2, 4 thus CO readly state s oxdsed more to CO stable . the better reducng glass, producton reducng acdty n of sols should react ery gorously wth water, 13. Radum chlorde – soluble, radum carbonate nsoluble Radum ntrate(v) would be most dfcult to s decompose. agent. 11 Slcon(iv) the than Hence CO 2 c (blast furnace), Radum – oxdaton s n (g) 2 iii the (concrete), housng), (g) ii PbO 2 carbon 4HCl 2 2H 4 b O 2 PbCl cement, (road, unreacte). pH SCl (relee cables s 4 beng medcaton are formed steel exceptonal, laxate, nsulaton for wth constructon water, fumes nsectcdes these iii compounds hdes, change from acd non-metallc de-harng the ii group manufacture, a Meltng ponts of Group VII ncrease as group s oxde: descended. Structure: macromolecule wth each slcon atom Structure: Changes from surrounded by four others coalent Bonding: a Very strong tetrahedral aboe molecules coalent bonds wth slcon atom Intermolecular forces become stronger as enronment. group The coalent solds. Bonding: n smple (damond-lke). propertes lead to ery hgh meltng s descended, – sze of molecules ncrease. pont; b i Br (l) added to KI (aq) – brown soluton formed. 2 the absence of any unpared electrons account for ts Br (l) added to KCl (aq) – reddsh brown colour. 2 nsulatng d propertes. Oxides Reactions With acid With ii Redox reacton iii Iodne dsplaced odde ons. by bromne, bromne – 2I – (aq) + Br (l) → I 2 2+ Germanum(ii) oxde (aq) Ge oxdses alkali (aq) + 2Br (aq) 2 2– GeO (aq) c 2 i P = ii Q MnO 2 2+ Tn(ii) oxde Sn 2– (aq) SnO = Cl 2 (aq) 2 iii MnO iv Cl + 4HCl → MnCl v Ox. vi Dspropotonaton vii 2HClO(aq) 2 2+ Lead(iv) oxde Pb (aq) + Cl 2 + 3 (aq) (g) + H 2 O(l) → HClO(aq) O 2 + HCl(aq) 2 state +1 → 2HCl(aq) –1 + O (g) 2 182 2H 2 2– PbO to Glossary Born–Haber cycle An enthalpy Dipole cycle A separaton of charge n a A used Acid A proton Acid–base (hydrogen indicator changes colour acd–base A when ttraton Acid dissociation on) donor. substance the pH changes constant, Boyle’s whch n an The equlbrum partcles to allow Alkali A constant for energy must a The hae reacton soluble a weak mnmum when to they law nersely lattce The olume of proportonal proton donor. A base a to gas ts An s molecule. One energy. a has s pressure. acd s acd. energy collde Buffer solution alkals A soluton changes are n added to when the acds example, of the are allotropes of The wth an acd as wthout change well whch A postely charged as The olume negately charged The same of atoms chemcal A redox type oxdsed The nto of and reacton atom n n a reduced breakng ons, up of especally dssocatng a gas a referrng nto H and A wth splitting proportonal to energy system A system n splttng hgher and of lower leel, lgands caused energy s not around a by the central presence transton whch element or a kelns. on. number nto the an n The s d-orbtals matter (Z) a on. of of number s acds d-orbital law can Closed A n ons. alkal. Atomic reacton group + temperature Anion A the to atoms decay. oxde and charge. the to drectly react ncreases nucle Charles’ An reacton charge tself. He radoacte the molecule Cation emtted when Amphoteric oxide that carbon. 2+ Alpha radiation (α) substance and of or another molecule negate reaction atom Dissociation A poste the smultaneously. same damond of partal Disproportionation speces rate graphte one whch soluton. occur. Catalyst For a end reacton. that pH base. Dfferent forms element. end where C Allotrope partal replaces mnmses or a other Displacement a proton acceptor. a Activation calculate Brønsted–Lowry theory rapdly. K to ganed on. or of Dot and cross diagram A dagram lost. protons n the nucleus of the atom of space showng Common Atomic orbital A regon the nucleus where there s probablty of ndng one or solublty s, p, types d of electron. Orbtals salt Half orgn on n the the same nucle type n a of of the compound outer showng of the electrons by dots or common. Dynamic ion An on contanng a (equilibrium) element bonded to an products to reactants as are well as bonds. two reactants A reacton, lgands conerted co-ordnate In central the by between of a has equlbrum radius n soluble whch can Compound atoms sparngly or f. radius/covalent dstance a another electrons crosses. transton Atomic of addng arrangement n the by Complex be reducton two an specc The a salt good effect shell the outsde ion the substance to products. contanng coalent atoms from two or more dfferent bond. E elements Avogadro constant The number bonded of Condensation atoms, ons or molecules atoms, ons or molecules. n a mole Conjugate law Equal olumes of the same number change n state when changes pair An to a acd lqud. Electrical conductivity and substance base on metals dffer alence to The and conduct ablty graphte electrcty. of a In delocalzed gases each contan The of a apour Avogadro’s together. sde of an equaton whch electrons are responsble for of + from each other by Co-ordinate bond A a H on. electrcal conducton. In molten and molecules. where both by same the coalent electrons bond are aqueous proded B Base A proton (hydrogen on) the radiation (β) Electrons gen the nucleus when a Bidentate the number The number bonds formed central Lgands whch can form a of transton by lgands element on complex. electrode co-ordnate bonds wth the Electron afnity afnty element bond the Boltzmann distribution the the fracton of one (single) sharng curve A mole A bond formed partcular partcles energes of electrons mole of gaseous two of a par of electrons atoms. aganst radius See atomc radus. mole of gaseous partcular off the humped of energy a Electronic Dative covalent bond mole of needed partcular gaseous n molecules (under Degenerate orbitals the same energy Delocalised condtons). polarisation the A bondng coalent Atomc orbtals equally due bond electrons to of a bond to tself. of The electrons e.g. 1s are not dfferences 2 2s n shells and 5 2p are free to seen n a more adjacent n Diatomic A seres of spectroscope brght due to leel. electrons moe spectrum at Electrons between electrons fallng from hgher to whch three energy leels. or The formula showng atoms. the shared to conguration Empirical formula where ablty coalent bond. to bond lower Bond a 2 excted standard ons. See Co-ordnate lnes one one one energy. gradually. The mole to to form cure Emission a n electrons sub-shells energy 1+ The atom arrangement break when added wth D characterstc ‘tals’ s atoms Electronegativity attract Bond electron change graph Covalent whch The rst enthalpy on. between a s central by transton has See Standard potental. two Covalent It conducton. potential decays. n of are radoacte to sotope responsble for Electrode ons off co-ordnate from moble acceptor. Co-ordination Beta atom. salts, Elements or smplest rato of atoms n a compounds molecule. electronegatty. Shown by an arrow whch hae molecules contanng only Endothermic + and the sgns δ and δ two atoms. For example, energy from carbon A reacton whch absorbs chlorne, the surroundngs. monoxde. 183 Glossary Energy level The regons Hybridisation at arous The combnaton of M dstances from electrons the hae nucleus dened where amounts the mxed of prole diagram Shows enthalpy change from products along Enthalpy n a chemcal a the change reacton The pathway. energy transferred Shows chemcal a to H A atom , The compound calculated from A K c present gas A K the gas whch obeys the gas law under equation and n xed of pressure and number products and back Mole acd–base ndcator reacton whch to the Half of dstance two atoms whose ncreases n n a wth electrcal an temperature. The n mass of a mole of a grams. The amount of number substance of there are atoms carbon-12 speced hang bond The electrostatc exactly partcles 12 g of oppostely charged each type of atom Molecular orbital An equaton those n ons and molecules are The the by atoms. In a mxture of molecules equlbrum (or constant for orbtal formed orbtal’s from shown. surroundngs. water An atomc Mole fraction product of number molecule. takng dfferent reacton the a whch out the n ons. reactons equation Shows attracton of n n sotope. Molecular formula Ionic the nucle molecules. are part the electrons. substance mass same the ges metal Weak forces n only energy A combnng A a to ndcator. equal. Exothermic n ons and the Ionic metal nRT. See as reacton of of of concentratons. The the forward the number nucleus. condtons. relatonshp between are radius a bond formed delocalsed substance Ionic rate all concentratons A A total n Relates olume, Molar PV = The attracton ncrease Ideal redox reaction bond the of (A) protons law equlbrum. reactants present Boyle’s reacton. to by Metallic water. equlbrum A Metallic sea a or between whch wth wth atom. + metal. Intermolecular forces Equilibrium of N Metalloid p at reacton or number neutrons conductty constant, expression or molecules F, O proceed. moles. K a I Indicator lnkng to breakdown by temperature, a ntermolecuar between routes p expresson for weak catalyst. constant Equilibrium wth ΔH c K orbtal Mass A bonded and Charles’ Equilibrium an between alternate reacton to proten bond Hydrolysis Ideal Enzyme to form character. force formed the reactants reacton, Enthalpy cycle for orbtals Hydrogen energy. Energy atomc onsaton of atoms), the number of moles of a water, partcular F molecule dded by the total K w number of Feasibility The lkelhood or not of Ionisation a energy The rst reacton occurrng as predcted by E moles of Monodentate energy s the energy needed to 1 alues. mole of electrons from 1 mole placed n a magnetc eld, lnes tself up gaseous atoms (under molecules. whch form bond wth the one central of transton A substance whch, when Lgands remoe co-ordnate Ferromagnetic all the onsaton o element on. standard condtons). N wth the eld and retans ts magnetsm Ion polarisation The dstorton of the Non-polar electron when the magnetc eld s remoed. cloud of an anon by a A Isotopes hang radiation charged) (γ) Radaton Atoms of dfferent the of wth no charge. caton. Nucleon G Gamma hghly molecule (small separaton and/or same numbers number See Mass number. element of protons. gen O off an sotope decays and a proton s K conerted an Gas to a neutron by capturng Kinetic theory electron. constant gas Giant The equaton, ionic Order constant n the deal R structure A gant gases are those n n Partcles constant n lquds and moement and solds brate. structure (with respect to substance) the power In to a rate whch rased. The oerall s the these of particular equaton the s sum a ths s concentraton order of reacton powers. L made Giant up of poste structure dmensonal metallc or and These hae network onc negate of a Oxidation ons’. Lattice three- of coalent, A regular atoms, ons H The half energy when 1 The mole of an an element or on oxdaton state another oxdaton and an on An the an ts onc gaseous ons s (under a oxdaton redox principle affectng When the equaton or are reducton equlbrum balanced changed, shfts to the concentraton the decrease to taken for the An on of a lmtng lone or molecule half ts wth The co-ordnate one s enthalpy change for or whch ndependent the reacton can bonds element wth a central on. of the takes exchange lgand The replacement of by another n a transton complex. A par of electrons n the outer place. shell of an atom whch n s not bonded. a A wth ts magnetsm eld of s. ncreases the another s substance the eld whch, but when does the lnes not when tself retan magnetc remoed. pressure, nddual gas p n The a pressure mxture of of an gases. Permanent dipole–dipole Intermolecular force molecules whch dpoles. + pH –log [H 10 184 t that magnetc eld, up forces route pair Paramagnetic Partial a Lone by whch elements. element reacton electrons ntal alue. The Group VII law of reactant one Hess’s par amount Ligand Halogens (or whch reduced P placed transton to number) poston mnmse part form or or reactant electrons number of wth more tme oxdsed A A atom showng Ligand The how agent remoes ncrease reactant. the poston electrons. Half-life Oxidising partcular oxdaton change. of a descrbes change compound state/number to or state. equaton of ether Oxidation electrons n equlbrum equation of enthalpy condtons) condtons Half three of state. n Le Chatelier’s one n Loss oxdaton electrochemcal standard contanng molecules n gen formed from cell cell, or arrangement dmensons. ‘bonds’. Lattice Half repeatng ] hae between permanent Glossary pH range about (of 2 indicator) pH ndcators unts The oer change range whch of partcular of the carbon-12 exactly Relative colour. 12 sotope has a mass of unts. molecular mass (M ) The mass Standard enthalpy change of hydration The 1 speced mole of a enthalpy change gaseous when on s r pi bond A coalent sdeways pK –log a oerlap –log w s A a of molecule whch an sotope 12 measured atom has a of the mass on a scale on ons carbon-12 of conerted (under Standard exactly to an aqueous standard enthalpy soluton of condtons). unts. change of neutralisation The when water enthalpy change w or molecule separaton of Principal quantum outsde whch orbtals. a a bond space atomc of K 10 Polar of nolng K 10 pK bond may the up whch there A regon nucleus to a a Resonance of charge. level contan electrons n of an certan a hybrid compound between of atom number maxmum two Reversible The s or products the reactants. structure somewhere the n can A be 1 reacton n changed whch back mole addton (under more forms. reaction the actual Standard the of an standard The speced standard S Q Standard Salt Quantum Energy of xed alues bridge An nert support soaked acd to an alkal change of enthalpy molar stochometrc number. s formed from condtons). enthalpy reaction to of change amounts equaton when n react the (under condtons). enthalpy change of n solution The 1 a enthalpy change when only, KNO used to make electrcal contact 3 whch can be absorbed or emtted mole between two half substance s dssoled n cells. excess an of by water (under standard atom. Semi-conductor Substances whch hae condtons). low electrcal conductty at room Standard R temperature Radioactive decay an atomc The nucleus, breakng emttng up of partcles conductty Shielding but as ncrease temperature (screening) The hydrogen cell under has a standard process. nner shells of electrons half to whch Pt electrode dppng nto of –3 the A condtons ncreases. ablty + 1 mol dm n electrode n reduce H ons n equlbrum wth the H gas. 2 Radioactive isotopes Isotope of effecte nuclear charge on other Stoichiometry elements whch hae nucle that break electrons, especally those n the reactant down spontaneously and constant The proportonalty Sigma bonds Bonds formed by n the rate equaton. end-on oerlap of atomc of n a step The slowest step Solubility product (K ) equaton. acid/base Acds or bases whch orbtals. are Rate determining rato shown the Strong constant mole products shell. balanced Rate The outer An completely onsed n soluton. equlbrum sp Sublimation n a reacton mechansm. expresson showng the The sold from Rate equation An equaton showng the of each on n a sparngly soluble between the soluton rased to the power of state of reactants whch relate the rate of reacton and the rate Spectator ions Ions whch play no constant. n The amount or a Regons or change of a specc constant The shell or reactant per expresson second. of less showng reagents the noled mechanism A seres of In showng the stages n prncpal hae a chemcal n the reacton contaner, the concentraton and the ar surroundng the lgand mxture as well as exchange. nstruments steps the electrons energy. nclude reacton Reaction wthn where equlbrum solent product a reacton. Stability these concentraton a beng formed. Surroundings reaction of wthout part more Rate of or ce ersa concentratons. quantum affect gas ther Sub-shell concentratons a salt lqud relatonshp drect formaton concentraton a Standard cell potential The dppng nto the reacton dfference mxture. reacton. n the standard electrode potental of System Real gas A gas whch does not obey the two half The reactants and products n a cells. chemcal reacton. 5 gas laws and low especally at hgh pressure temperature. Standard and conditions temperature Pressure of 10 Pa 298 K. T Redox reaction A reacton where Standard electrode potential The Transition reducton and oxdaton occur at the electrode potental of a half elements whch forms same tme. compared wth a standard agent donates A reactant electrons oxdaton (or number) of that stable d block ons wth element an hydrogen ncomplete Reducing A cell d sub-shell. electrode. decreases the another Standard change enthalpy whch change takes An place at enthalpy V pressure 5 reactant. of 10 Pa and temperature 298 K. Valence Reduction Gan of electrons or decrease Standard enthalpy shell electron (VSEPR) theory n oxdaton state. atomisation The enthalpy atomic mass (A ) The weghted when 1 mole of gaseous atoms mass of the par atoms of an formed from ts repulsion element order repulson s of lone par-lone s r aerage The change decreasng Relative pair change of > lone-par bond-par > bond-par (under bond-par. element measured on a scale on whch standard condtons). van der Waals forces an atom of the carbon-12 sotope has Standard enthalpy attracton a mass of exactly 12 unts. combustion The enthalpy isotopic proporton or abundance percentage of The when a 1 mole completely of a substance combusted between molecules of caused change by Relative Weak forces change of temporary dpole-nduced dpoles. s (under W partcular sotope n a mxture of standard sotopes. Relative Standard isotopic mass partcular sotope measured on a The mass of of an element scale on whch an a The condtons). enthalpy enthalpy substance atom (under change of formation change when s formed from standard 1 ts mole of a Weak are acid/base partally Acds onsed or n bases whch soluton. elements condtons). 185 Index Key n terms the are n bold and are also lsted glossary. beta radiation bidentate blood A absolute acid 104, bond equlbra Brønsted–Lowry dssocaton pH pH–ttraton acid–base acd–base theory indicators ttratons acd–conjugate bond bond energies 2, coalent perod 114–15 bond 112–13 rato 119 ) energy curve 62–3, covalent 3 elements 108–11 Boyle’s law cycle and coalent 7 bonding 18–19 number bonding covalent 18–19 cross conductty 99 lmt co-ordination 68 electrcal 105 Process co-ordinate 62 16–17 polarisation see pairs conergence 80 energes bondng 14, Born–Haber (k Contact 7 dot co-ordnate constant conjugate 117 bondng 108–11 112–13 43, base acid dissociation 104–5 106–7 cures model dssocaton see also 104–19 constants calculatons solutons atomc 59 conductty Boltzmann distribution 53 141 acd–base buffer Bohr’s temperature condense 5 164 164, dagrams radius 166 14 see structures, 16–19 atomic gant radius 25 137 D 22 70–3 Dalton’s 52 dative atomc theory covalent bond 2 18–19 a acdc buffers activation acte ste alzarn alkali buffer 99, 165 ons carbon 104 rule 127 aqueous sodum 171 hydroxde emsson spectra number (Z ) atomic orbitals radius 3 17 4 160 2–3 142, monoxde cell potentals dot 125 onc 129 Charles’ law 53 chemcal equatons chemcal equlbra 23, energes 10–11 calculatons 94–5, 4–5 equilibrium constant (K equilibrium constant (K solubility atomc 8–9 theory atomsaton change chlordes product 2 see of standard cobalt law 40 ) 92–3 ) 96–7 (K ) 100–1, base seres change common 104 basc buffers basc dssocaton complex 116–17 constant (K ) 110 redox structures double bond ductle 25 18, 20 19 of 80 102–3 electrical group conductivity IV perod 3 elements elements redox electron standard enthalpy effect ions 163, reactons 103 164 166–7 2, electron cells benzene 186 129 31 compounds 2 concentraton 80, 98, 128 120–9 chemcal 3, 126–7 6 decent electron ow 128–9 change 124–5 70 molecules electronegativity electron 128–9 cells 22, 135 124–5 conguration b batteres 161 3 afnity electronic 135 elements reactons electron 77 144–5 potentials predctng combuston ion 90–91 E electrode 162–3 see 16 16–19 (equilibrium) electrochemcal 76 ons 155 163 bondng electrc elds 91 theory 51 110–11 electrochemcal combuston 7 155 165 coloured Balmer 137 , 45 19 splitting transton 173 colormetry B 139 system collson 40–1 134, chromates closed 14 constant Avogadro’s enthalpy atomsaton attracte forces Avogadro chlorne 138, reaction 108–9 sp orbtals ttratons 90–103 p sotopes (VI) cross diagrams dynamic 35 c onsaton and coalent 122–3 135 22 d-orbital 168–9 dagrams 149 elements 150 dissociation cell 31 140 disproportionation 148 80 ceramcs 3 dsplayed formula 148 146, 21, 150 displacement 172 146, electrons elements dchromate dipole equlbrum 6–7 perod 64 carbon cells 136 elements leel 6–7 2 structure energy II diatomic doxde cations 8–9 elements transton atomc group carbon catalysts 134 perod 129 144 carbonates 172–3 bonding 163 densty 160–1 calormetry 169 ammona atomic delocalised C calcum aqueous atomic 104–5 116–19 5 134 antclockwse atomc cells co-ordinate degenerate orbitals halogens amphoteric oxide argon solutions button ammonum anions see also 114 134 radiation ammona 151 blue Brønsted–Lowry theory 114 104 alumnum water bromothymol 80 81 yellow allotropes alpha bromne 116 energy transfer electroplatng 48 117 9 17 Index emission spectra endothermic energetcs bond half 6–7 empirical formula reactions 60 halde energies 62–3, changes 63, 64–5, Hess’s Hess’s law 63, law lattice hydraton 66–7 energy 68, 6–7 energy prole diagrams enthalpy cycle enzymes 81 equilibrium equlbrum 66 94–5, equilibrium constant (K equilibrium constant (K equilibrium expression equilibrium reactions 108–9 ) 92–3 ) 96–7 152–3 30 standard enthalpy change bonding 23 buffers hydrogen electrode 120 hydrogen haldes 84, 117 152–3 86, on 113, 145 104 ethene 30, excted exothermic expanded 92 90–1 113 gas 54–57 ideal gas equation nsoluble 60 octet 55–7 processes ame test meltng pont group II group IV odne ion 127 159 168 (A) 14, effect (common) ionic bonding ionic equations ionic product of 45 58 144 150 bonding radius 25, 140 elements 21 140 135 136 30 methyl orange methyl red 114 114 mass 35 35 concept 34–45 law calculatons 40–1 36–7 empirical formula 103 38–9 molecular formulae 20–1 23, 4 elements Avogadro’s ttratons 159 metallic mole 22–3 26 72 3, metallic mole intermolecular forces 17 F ferromagnetic 59 molar 99 100 odne–thosulphate feasibility number metals 114 ndustral 6 theory 25 meltng methane ideal indicator 63 state malleable metalloids I p pont oxde halogens c equalence 134 magnesum mass 30 hydrogencarbonate hydroxonum magnesum magnetc elds hydraton hydrolysis 90 calculatons Brønsted–Lowry M 150–1, see hydrogen 61 60 see 7 173 66–7 hydrocarbons levels enthalpy of 70–3 energy seres 154 hybridisation 68–9 theory Lyman 152–3, halogens 68 60–1, Lowry 121 85 ons haldes 60–73 enthalpy equation half-life 38 34, water soluton 37 (K ) 107 39 concentratons molecular formula 38, 42–3 41 w forces of attracton formaton see 14–15 standard onc enthalpy of formaton fuel cells change 3 129 ionic G gas radiation constant gases 55 knetc laws isotopes theory ionic giant structures ground structure 14, 21, 24 24–5, 134 25 analyss state 103 II IV group IV elements group IV oxdes group IV tetrachlordes elements catons of of non-polar matter standard nuclear 55–7 172 molecules 28 136 22, charge nucleon 58–9 enthalpy neutralsaton 3 142, non-metals equaton see non-lnear, V-shaped 52–9 52–3 gas states 23, 27 10 number see L lattice 144–5 146–7 150–1 Process 99 lattices lead 144–5 energy 21, ntrate mass number lnear 122–3 orbtals atomic degenerate principle exchange 163, 98–9, 165 14, pairs 128 164–5 see also 58 order 16, 29 91 163 hybrdsaton order equlbrum 8–9 molecular 28 lquds lone O 70–3 24 Le Chatelier’s ligand 68, 173 lqud-apour 165 cells ntrates 54–5 laws ligands half neutralsaton change 140–1 148–9 H haem N 35 6 group group VII gases deal 18 168–73 23, 4–5 kinetic theory gas 164 bonds neutron group Haber tests ions K giant grametrc multple 28–31 96 monodentate 160 15 14 shapes mole fraction 54–5 25 solds molecules, 24 138 52–3 structures 26 162–3 spectator 52–3 53 coalent graphte 73, qualtate heatng gant ons 21, 10 elements polarisation molecular 160 giant energy lattces molecular orbital 136 elements structure, transton ion 5 14 compressng gas elements transton onsaton gamma molecular rad perod 30 15 d-orbital splitting 82 of reacton 82–3, oxidation 48 oxidation numbers 84–5 148 187 Index oxidation states coloured perod ons 3 oxdaton 48, 122, 146-148 elements states redox ttratons reducing 162–3 reduction 137 (oxdaton numbers) state 45 agents 51, symbols Stock 154 35 nomenclature stoichiometry 48 mass (A strong relative atomic relative isotopic abundance relative isotopic mass relative molecular ) 4 41, 49 44 acids/bases 105, 113, 115 r 48–9 oxdes 138, 139, 146–7 (M ) structure 4 and covalent 4 bondng bonds 14–31 14, 16–17 r oxidising oxygen agents 137 , 51 141 resonance reversible P paramagnetic partial perod ( p) elements phenolphthalen salt pH range pH scale polar 117 cures 112–13 22 equlbrum 98–9 potassum dchromate (vi) potassum manganate (vii) dfference precptaton pressure 52, 102, 80, 167 45, 167 121 134, doxde matter of sulphtes 154, sulphur 173 compounds sodum carbonate sodum hydroxde sodum thosulphate 3 153 8 172 acd area system 99, 154 80 surroundings 169 17 28 partcles 134 surface 134 21 14–15 172 sulphurc 26–7 134 22–3 81 sulphates 149 sodum 60 60 152 T 14 group 98 states substrate molecular 24–5, 20–1 bonding sub-shells 10 144 ntrate solds bonding sub-atomc 144 15 slcon solublty 170 (screening) bonds smple 6 135, 14, metallic VSEPR theory 121 slcon sler 15 potental ionic 2 sublimation shielding 114 of theory 8 sigma constant poston bridge semi-conductor shells 106–7 bonds Planck’s 114 134 pH-ttraton pi 23 structures 22 intermolecular forces 90 S 134–9 buffers phosphorus atomc 96 permanent dipole–dipole forces phosphate giant 31 reactions Rutherford’s electronegativity 56–7 159 pressure 3 hybrid mass 27 , II solubility 100 temperature sulphates product 143 (K ) 80, tetrachlordes 100–1, 102–3 99 145 tetrahaldes 145 tetrahedron 28 sp principal quantum proten proton buffers level 6 soluton see 117 soluton shape, molecules 28 concentratons solutons Q quantum analyss 103, stability 168 radioactive decay 82, equations rates of order reacton of of theory see enthalpy redox reactions 188 ons standard 50–1 48–51, 166–7 124–5 starch states redox 60 potential change 120–1 60–1, 43, 63, 112–13 ons 158–61, 167 163 164, exchange reactons anadum trgonal 44–5, elements ons 142 153 166 164–5 167 162 planar 28 V change of valence change of anadum change of change of 162 23, 26 58 apour pressure VSEPR see valence 91 shell electron pair V-shaped molecules water 27 , 28 change of W 64 enthalpy 64, repulsion 28 repulsion 64 enthalpy pair (VSEPR theory) aporsaton 68 enthalpy electron van der Waals forces 66–7 enthalpy shell theory change of change of 69 hydrogen weak electrode 152 of lgand 64 enthalpy solution 54 stablty complex 123 70 enthalpy reaction standard 51 equatons complex enthalpy neutralisation standard 151 seres gases redox 86–7 63 halogens reactty standard standard 76 mechanisms reactty enthalpy hydration knetcs reaction change thermal 68–9 formation 84–5 reacton reacton real standard standard combustion 82–3, decomposton coloured 165 conditions standard 80 thermal transition potential electrode standard 76–87 reacton cell atomisation 86 44 163 standard standard step 82–3 reaction collson 5 42–3, 35 standard 64–5, 84 rate determining rate 5 isotopes constant 23, constant standard R rate of ttratons ions splitting, d-orbital 6 radioactive change 169 spectator qualtate enthalpy neutralsaton 3 pyramdal standard matter 14–15, 58–9 120–121 23, 107 , 137 , acids/bases 141, 105, 145 108–9, 113, 115 Chemistry for Unit 1 CAPE® Achieve your potential Developed guide in will CAPE® Written exclusively provide by an designed ● ● in learning to Engaging the with additional experienced syllabus information ● the Caribbean Examinations support to Council®, maximise your this study performance Chemistry. Chemistry key you with an and team enhance activities teachers examination, easy-to-use outcomes of from your that this study double-page the study help syllabus of you the and guide format . and the in such the covers Each contains subject , develop experts a CAPE® all topic range the essential begins of with features as: analytical skills required for examination Examination tips with essential advice on succeeding in your assessments Did You Know? boxes to expand your knowledge and encourage further study This study choice also questions examiner sample to build a fully interactive examination skills Examinations Thornes subjects includes and answers confidence at to CSEC® produce and a Council series (CXC®) of of incorporating with in O x f o rd has Study multiple- accompanying preparation for the CAPE® worked Guides exclusively across a wide with range of CAPE®. How Pa r t CD, examination. Caribbean Nelson and feedback, Chemistry The guide University P re s s to get in touch: web www.oup.com/caribbean email schools.enquiries.uk@oup.com tel +44 (0)1536 452620 fax +44 (0)1865 313472