7Strurtu,r,o, bo@ al4d, t l r wgroryMtry Atomic structure f& Inffoduction Much of chemistry depends upon Coulomb's Law which states that the electrostatic force of attraction, F, is given by: F*(q..Q)*r, where q- and q- are the charges on the objects (e.g. the nucleus, an electron, ions etc.) and r" is the square of the distance between their centres. This means that the bigger the charge, the bigger the force, and the further the centres are apart, the weaker the force. <ta W Things to learn iJ Atomic numbet (4 of an element is the number of protons in the nucleusof its atom. [l wtass number of an isotope is the number of protons plus the number of neutrons in the nucleus. E Isotopes are atoms of the sameelement which have the samenumber of protons but different numbers of neutrons. They have the same atomic number but different mass numbers. E nehtive atomic mass (A) of an element is the average mass (taking into account the abundance of each isotope)of the atoms of that element relative to llIzth the mass of a carbon-LZ atom. f nehtive isotopic mass is the massof one atom of an isotope relative to lll2th the mass of a carbon-L2 atom. f] Rehtive molecular mass (MJ of a substance is the sum of all the relative atomic masses of its constituent atoms. E Molar mass is the mass of one mole of the substance. Its units are gramsper mole (g mol-'), and it is numerically equal to the relative molecular mass. E fst lonisation energy is the amount of energyrequired per mole to remove one electron from each gaseous atom to form a singly The relative molecular mass isalso called the relative formula mass especially for ionic substances. lonisation energies are always endothermic and relate tothe formation ofa positive ion. positive ion E(g)*E.(g)+e- f Zna ionisation energy is the energychangeper mole for the removal of an electron from a singly positive gaseous ion to form a doubly positive ion E.(g)-E'.(g)+e-
Transcript
7 Strurtu,r,o, bo@ al4d,t l rwgroryMtry
Atomic structuref& Inffoduction
Much of chemistry depends upon Coulomb's Law which states that theelectrostatic force of attraction, F, is given by:
F * ( q . . Q ) * r ,where q- and q- are the charges on the objects (e.g. the nucleus, anelectron, ions etc.) and r" is the square of the distance between theircentres. This means that the bigger the charge, the bigger the force, andthe further the centres are apart, the weaker the force.
<taW Things to learn
iJ Atomic numbet (4 of an element is the number of protons in thenucleus of its atom.
[l wtass number of an isotope is the number of protons plus thenumber of neutrons in the nucleus.
E Isotopes are atoms of the same element which have the same numberof protons but different numbers of neutrons. They have the sameatomic number but different mass numbers.
E nehtive atomic mass (A) of an element is the average mass(taking into account the abundance of each isotope) of the atoms ofthat element relative to llIzth the mass of a carbon-LZ atom.
f nehtive isotopic mass is the mass of one atom of an isotope relativeto lll2th the mass of a carbon-L2 atom.
f] Rehtive molecular mass (MJ of a substance is the sum of all therelative atomic masses of its constituent atoms.
E Molar mass is the mass of one mole of the substance. Its units aregrams per mole (g mol-'), and it is numerically equal to the relativemolecular mass.
E fst lonisation energy is the amount of energy required per moleto remove one electron from each gaseous atom to form a singly
The relative molecular mass is alsocalled the relative formula massespecially for ionic substances.
lonisation energies are alwaysendothermic and relate to theformation of a positive ion.
positive ionE ( g ) * E . ( g ) + e -
f Zna ionisation energy is the energy change per mole for theremoval of an electron from a singly positive gaseous ion to form adoubly positive ion
E. (g ) -E ' . (g )+e-
A common error is to miss out the+ charge on the formula of a speciesresponsible for a l ine in a massspectrum.
f l = t * u c r u R E , B o N D t N G A N D M A t N G R o u p c H E M t s r R y
*i lst electron affinity is the energy change per mole for the additionof one electron to a gaseous atom to form a singly negative ion
E ( g ) + e - - - E ( g )
J Znd, electron affinity is the energy change per mole for the additionof an electron to a singly negative gaseous ion to form a doublynegative ion
rG) + e- -- E'-(g)iJ s block elements are those in which the highest occupied energy level
is an s orbital. They are in Groups I and 2.Similar definitions apply to p block (Groups 3 to 7 and 0) and dblock (Sc to Zn) elements.
^_______-t::It/a
ixj Things to understandU !
Mass spectra
t An element is first vapourised and then bombarded by high-energyelectrons that remove an electron from the element and form a positive
ion. This ion is then accelerated through an electric potential, deflected
according to its mass and finally detected.
O Metals and the noble gases form singly positively charged ions in the
ratio of the abundance of their isotopes.
t Non-metals also give molecular ions. For example Br, which has two
isotopes 'nBr
15oo/o) and "Br 150o/o), will give three lines at m/e values of
158, 160 and L62 in the ratio I:Z:L These are caused by ('nBr-'nBr;+,
ltnBr-ttBr;+ and (ttBr-ttB1;+'
I The relative atomic mass of an element can be calculated from mass
spectra data as follows:
A, = the sum of (mass of each isotope x percentage of that isotope)/l0O
Work"ei, oxaloLpb
Electron strrrcture
O The first shell only has an s orbital.
o The second shell has one s and three p orbitals.
o The third and subsequent shells have one s, three p and five d orbitals.
o Each orbital can hold a maximum of two electrons.
o The order of filling orbitals is shown in Figure 1.1 below.
Boron was analysed in a mass spectrometer.Calculate the relative atomic mass of boron using the results below.
Peaks at mle ot Abundance (%)
10.0 18.7
1 1 . 0 81 .3
Answer ' . A,= ( 10.0 x 18.7 + 11.0 x 81.3 ) / 100 = 10.8
@otoMrc s rRucruRE
Fig l . lorbitals
The order of filling of atomic
ln an atom, the outer electrons areshielded from the pull of the nucleusby the electrons in shells nearer tothe nucleus (the inner electrons).
Fig 1.2 The first ionisation energies(kl mol-') of the elements up to krypton
Electron structures can be shown in two ways:
o The S, p, d notation. For vanadium (atomic number 23) this is:
rs", 2s' zpu, 3s' 3pu 3d" 4st.
o The electrons in a box notation. For phosphorus (Z = 15) this would be:
ruuuuuutr t r t r1s 2s 2p, 2p, 2p, 3s 3p, 3p, 3p,
Sizes of atoms and ions
O The atoms become smaller going across a period from left to right,because the nuclear charge increases, pulling the electrons in closer,though the number of shells is the same.
o The atoms get btgger going down a group, because there are moreshells of electrons.
O A positive ion is smaller than the neutral atom from which it was made,because the ion has one shell fewer than the atom.
O A negative ion is bigger than the neutral atom, because the extrarepulsion between the electrons causes them to spread out.
lst ionisation enerry
O There is a general increase going from left to right across a period (see
Figure 1.2). This is caused mainly by the increased nuclear charge(atomic number) without an increase in the number of inner shieldingelectrons.
240022002000180016001,4001,2001000800600400
15 20Atomic number
tsI
&'xbok(u(u
(€cn
You should be able to sketch thevariation of 1st ionisation energieswith atomic number for the first 20elements.
The firs! six ionisation energies/kJ mol of an element are:1st 7862nd 15803rd 32304th 4360sth 160006th 20000There is a big jump after the fourth,so the element is in Group 4.
I s t * u c r u R E , B ' N D T N G A N D M A r N c R o u p . H E M r s T R y
O There are slight decreases after Group 2 (this is because, for Group 3, itis easier to remove an electron from the higher energy p orbital), andafter Group 5 (this is because, for Group 6, the repulsion of the twoelectrons in the p, orbital makes it easier to remove one of them).
O There is a decrease going down a Group. This is caused by the outerelectron being further from the nucleus. (The extra nuclear charge isbalanced by the same extra number of inner shielding electrons.)
Successive ionisation energies
o The 2nd ionisation energy of an element is always bigger than the first,because the second electron is removed from a positive ion.
O When there is a very big jump in the value of successive ionisationenergies, an electron is being removed from a lower shell, e.g. if thisiump happens from the 4th to the 5th ionisation energy, four electronshave been removed from the outer shell during the first fourionisations, and so the element is in Group 4.
Electron affinity
o The 1st electron affinity values are always negative (exothermic), as anegative electron is being brought towards the positive nucleus in aneutral atom. They are the most exothermic for the halogens.
o The 2nd electron affinity values are always positive (endothermic),because a negative electron is being added to a negative ion.
ChecklistBefore attempting the questions on this topic, check that you can:
iJ pefin e A, , M, , and relative isotopic mass.
,J Calculate the number of neutrons in an isotope, given the atomic andmass numbers.
iJ Calculate the relative atomic mass of an element from mass spectradata.
[l Oefine first and subsequent ionisation energies.
[l mplain the changes in first ionisation energies for elements across aPeriod and down a Group.
f Deduce an element's Group from successive ionisation energies.
E Work out the electronic structure of the first 36 elements.
E Oefine the Lst and 2nd electron affinities.
@ o t o M r c s r R U c r u R E
The answers to the numberedquestions are on page 121.
5
Testing your knowledge and understanding
For the following set of questions, cover the margin, write down youranswer, then check to see if you are correct. (You may refer to the PeriodicTable on the inside back cover)
I State the masses and charges (relative to a proton) of:a proton (p)a neutron (n)an electron (e).
o How many neutrons are there in an atom of ?iXa ?
o What is the atomic number of an element which has an atom of massnumber 99 and contains 56 neutrons ?
I Write equations with state symbols for:a the 1st ionisation energy of sodiumb the 1st ionisation energy of chlorinec the 2nd ionisation energy of magnesium.
O Using the ls, 2s 2p. . notation, give the electronic structures of theelements:a u Cb ,rM8c ,uF€.
Explain the difference between relative isotopic mass and relativeatomic mass. Illustrate your answer with reference to a specific element.
Lithium has naturally occurring isotopes of mass numbers 6 and 7.
Explain why its relative atomic mass is 6.9 not 6.5.
Magnesium was analysed in a mass spectrometer.Peaks were found at three different mle ratios.
mle Abundance (%)
24.0 78.6
25.0 1 0 . 1
26.0 1 1 . 3
Calculate the relative atomic mass of magnesium.
Gallium has a relative atomic mass of 69.8. Its mass spectrum shows twopeaks atmle of 69.0 and 71.0.
Calculate the percentage of each isotope in gallium.
The successive ionisation energies of an element X are given below:
State, giving your reasons, which group the element X is in.
6 a Sketch the variation of 1st ionisation energy against atomic numberfor the first L L elements.
b Explain why the ionisation energy is less for:
i the element of atomic number (4 of 5 than that for element of
Z = 4
ii the element of Z -- 8 than that for element of Z = 7
I To be able to calculate the number of moles from data.
t To set out calculations clearlv.
to learn
,-ldt r iH(H
1 mol of NaOH has a mass of 40 g1 mol of 0, has a mass of 32 g.
I I t t * u c r u R E , B . N D T N G A N D M A I N G R o u p . H E M r s r R y
iii the element of Z = 11 than that for element of Z = 3.
7 Using the electron in a box notation, give the electronic structure ofchlorine (Z = I7).
8 Explain, in terms of electronic structure, why the chemical properties oflithium, sodium and potassium are similar but not identical.
9 a Give the equation, with state symbols, that represents the 1stelectron affinity of i lithium, ii chlorine, iii oxygen.
b Give the equation that represents the 2nd electron affinity ofoxygen.
c Why is the 2nd electron affinity of oxygen endothermic whereasthe 1st electron affinities of lithium, chlorine and oxygen are allexothermic?
iJ The Avogadro Constant is the number of carbon atoms in exactly 12 gof the carbon-I} isotope. Its value is 6.O2 x 10" mol-'.
iJ O.re mole of a substance is the amount of that substance that contains6.O2 x 1023 particles of that substance. This means that one mole of asubstance is its relative atomic or molecular mass expressed in grams.
t] fne molar mass of a substance is the mass (in grams) of one mole.
[J Amount of substance is the number of moles of that substance.
[] fne empirical formula is the simplest whole number ratio of theelements in the compound.
Things to understandCalculation of empirical formulae from percentage data
It is best calculated using a table.
Things
Element Percentage of element Percentage divided by A,of element Divide each by lowest
@ r o * M U L A E , E e u A T r o N S A N D M o L E s
lonic equations must also balancefor charge.
Avoid wr i t ing mol = 0.11. Insteadstate the name or formula of thesubstancee.g. amount of Na = 0.11 mol .
I
As yolume in dm- = volume incm-/1000, the following formula canbe used:moles = MxW1000or concentration = moles x1000lVwhere M is the concentration andt/is the volume in cm-.
The last column gives the empirical formula, but if any value in this
column comes to a number ending in .5 or .25, you must multiply all thevalues by 2 or 4 to get integers. So here the empirical formula is C3H6O,.
Equations
These must balance. The number of atoms of an element on one side ofthe equation must be the same as the number of atoms of that element onthe other side.
Ionic equations
There are three rules:o Write the ions separately for solutions of ionic compounds (salts, strong
acids and bases).
o Write full 'molecular' formulae for solids and all covalent substances.
o Spectator ions must be cancelled and so do not appear in the finalequation.
Moles
There are three ways of calculating the amount of substance (in moles):
I For a pure substance X
The amount of X (in moles) = mass of X (in grans)/its molarrnASS
Workpi, oxaloLpb
t For solutions:
The amount of solute = concentration (in mol d--') x volume(in dm')
Therefore concentration = moles / volume in dm'
Workd,oxiloLpb
Calculate the amount of NaOH in 22.2cm3 of 0.100 mol dm-' solution
Answer:0.100 x 0.0222 = 2.22x 10-' mol of Na0H
O For gases:
Amount of gas (in moles) =
volume.
The molar volurne of a gas is 24temperature and pressure (RTP).
volume (in dm')/molar
dm3 mol-r, measured at room
Calculate the amount of H,0 in 1.1 g of water
Answer:1.1 gl18 g mol-' = 0.061 mol of H,0
I I t t * u c r u R E , B . N D I N G A N D M A I N G R o u p c H E M I s r R y
Work^ei, oxiloL\-lx'
Calculation of number of particles
The number of particles can be calculated from the number of moles.
r The number of molecules = moles x Avogadro's constant.
o The number of ions = moles x Avogadro's constant x the number ofthose ions in the formula.
Workpi, oxilo7-h,
Calculations based on reactions
These can only be done if a correctly balanced equation is used.
Reocting moss questions
First write a balanced equation for the reaction.Then follow the route:
Step 1 Step 2 Step 3
Mass A - Moles A-Moles B +Mass B
For Steps 1 and 3 use the relationship:
amount of A or B (in moles) = mass/molar mass
For Step 2 use the stoichiometric ratio from the equation:
moles of B = moles of A x ratio B/A
Work^ei,, oxilo7-h'
In Step 2, the stoichiometric ratio is211 as there are 2Na0H moleculesfor each 1Si0, molecule in theequation.
Calculate the amount of H,(g) in 3.2 dm" at RTP.
Answer'. 3.2124 = 0.13 mol of H,(g)
Calculate the number of carbon dioxide molecules in 3.3 g of C0,.
Answer: Amount of C0, = 3.3144 = 0.075 molNumber of molecules = 0.075 x 6.02 x 10" = 4.5 x 10"
Calculate the number of sodium ions in 5.5 g of Na,CO,
Answer: Amount of NarCO, =5.51106 = 0.0519 molNumber of Na- ions = 0.0519 x 6.02 x 10" x2 =6.2x 10"
Calculate the mass of sodium hydroxide required to react with 1.23 g of silicondioxide.Answer. Equation: SiO, + 2Na0H ---l Na,Si0, + H,0
Step 1 amount of SiO, = 1.23160 mol = 0.0205 molStep 2 amount of NaOH = 0.0205 mol x 211 = 0.0410 molStep 3 mass of NaOH = 0.0410 x 40 = 1.64 g
@ r o * M u L A E , E e u A T r o N s A N D M . L E S
In Step 2 the stoichiometric ratio is112as there is 1H2S0o moleculefor every 2Na0H molecules inthe equation.
Titrotions (this will only be exomined ot AS in Unit test 38)
The route is much the same:
For Steps 1 and 3 use the relationship:
amount (in moles) = M x V/1000
For Step 2 use the stoichiometric ratio from the equation:
moles of B = moles of A x ratio B/A
Work^ei, oxalo7.b
Concentrotion of solutions
This is either: amount of solute (in molqs) units: mol dm-"volume of solution in dm''
or: mass of solute (in grams) units: g dm-'volume of solution in dm''
Gos volume colculotions
1 For reactions where a gas is produced from solids or solutions,follow:
Step 1 Step 2 Step 3Mass of A*Moles of A+Moles of gas B + Volume of gas B
Step 1 use the relationship:
moles = mass/molar mass
Step 2 use the stoichiometric ratio from the equation:
moles of A = moles of B x ratio of B/A
Step 3 use the relationship:
volume of gas B = moles of B x molar volume
concentration and SteP 1
volumeof A +
Step 2 Step 3^ answer
M o l e s A + M o l e s B + a b o u t B
25.0 cm'of a solution of sodium hydroxide of concentration 0.212 mol dm-'was neutralised by 23.4 cm'of a solution of sulphuric acid. Calculate theconcentrat ion of the sulphuric acid solut ion.Answer. Equation: 2Na0H + H,SO. ---) Na,SOo + 2Hr0
Step 1 amount of Na0H = 0.212x25.011000 = 5.3 x 10-'molStep 2 amount 0f H2S0o = 5.3 x 10-3 x1l2 =2.65 x 10- 'molStep 3 concentration of H2S0o = 2.65 x 10-' mol/0.0234 dm3
= 0.113 mol dm- '
ln Step 2 the stoichiometric ratio isl l2as there is 1C0, molecule forevery 2NaHCO, molecules in theequation.
o You must show all the steps inyour calculat ions
o Don't cut down to two or threesignif icant f igures in the middleof a calculat ion
o Check every calculation to ensurethat you have entered the datacorrectly
l I t t * u c r u R E , B . N D T N G A N D M A t N G R o u p c H E M l s r R y
Workpj, oxu,u?h,
For calculations involving gases only, a short cut can be used. Thevolumes of the two gases are in the same ratio as theirstoichiometry in the equation.
Workti, oxilov-l"o
Significant figures
You should always express your answer to the same number of significantfigures as stated in the question or as there are in the data.
If you cannot work this out in an exam, give your answer to 3significant figures (or 2 decimal places for pH calculations), and you areunlikely to be penalised. Do not round up numbers in the middle of acalculation. Any intermediate answers should be given to at least 1 moresignificant figure than your final answer.
ChecklistBefore attempting questions on this topic, check that you can:
iJ Calculate the empirical formula of a substance from theo/o composition.
t-J Write balanced ionic equations.
[i Culc.rlate the number of moles of a pure substance from its mass, of asolute from the volume and concentration of its solution, and of a gasfrom its volume.
i] Catculate reacting masses and reacting gas volumes.
L] Use titration data to calculate the volume or the concentration of onesolution.
i-,',-.,;r:,,'' L
Calculate the volume of carbon dioxide gas evolved, measured at roomtemperature and pressure, when 7.8 g of sodium hydrogen carbonate is heated.The molar volume of a gas is 24 dms mol-' at the temperature and pressure ofthe experiment.Answen Equation: 2NaHC0, - NarCO, + HrO + COr(g)
Step 1 amount of NaHCO' =7.8184 mol = 0.0929 molStep 2 amount of C0, = 0.09286 mol x 112 = 0.0464 molStep 3 volume of C0, = 0.0464 mol x 24 dms mol-' = 1.1 dm'
i , r yL$.:i*'q'ffr''
r:
What volume of oxygen is needed to burn completely 15.6 cm'of ethane?Answen Equation: 2C,Hu(g) + 70,(g) - 4C0,(g) + 6Hr0(l)
Calculation:volume of oxvqen oas = 7 = 3.5volume of ethane gas 2volume of oxygen g?s = 3.5 x 15.6 = 54.6 cm'
@ a " R M U L A E , E e u A T r o N s A N D M o L E s 11
Testing your knowledge and understanding
For the first set of questions, cover the margin, write down your ans\\-er,then check to see if you are correct.
r The table below contains data which will help you.
Substance Solubility
Nitrates All soluble
Chlorides All soluble except for AgCl and PbCI,
Sodium compounds All soluble
Hydroxides All insoluble except for Group 1 andbarium hydroxides
Write ionic equations for the reactions of solutions of:
a lead nitrate and potassium chlorideb magnesium chloride and sodium hydroxidec sodium chloride and silver nitrated sodium hydroxide and hydrochloric acid.
t Calculate the amount (in moles) of:
a Na in I.23 g of sodium metalb NaCl in 4.56 g of solid sodium chloridec Cl, in 789 cm'' of chlorine gas at room temperature and pressure
d NaCl in 32.1 cm'' of a 0.111 mol dm-' solution of sodium chloride.
I Calculate the volume of a 0.222 mol dm-'' solution of sodiumhydroxide, which contains 0.0456 mol of NaOH.
t Calculate the number of water molecules in 1.00 g of HrO.
o Calculate the number of sodium ions in 1.00 g of NarCO,,.
) 4.44 g of solid sodium hydroxide was dissolved in water and thesolution made up to 250 cmS. Calculate the concentration in
a g dm-'
b mol dm-'.
The answers to the numberedquestions are on pages 121-122.
An organic compound contains 82.760/o carbon and 17.24o/ohydrogen by mass. Calculate its empirical formula.
It was found to have a relative molecular mass of 58. Calculate itsmolecular formula.
Balance the equations:
a N H . + O , - N O + H r O
b Fe'.(aq) + Sn'.(aq) -- Fe'.(aq) + Snn.(aq)
What mass of sodium hydroxide is needed to react with 2.34 g ofphosphoric(V) acid, H3PO*, to form the salt Na,POn and water?
What volume of 0.107 mol dm' potassium hydroxide, KOH, solution isneeded to neutralise 12.5 cmt of aO.O747 mol dm-' solution ofsulphuric acid?
What volume, measured at room temperature and pressure, of hydrogensulphide gas, HrS, is required to react with 25 cm" of a 0.55 mol dm-'
B o N D t N G A N D M A t N G R o u p c H E M t s r R y
solution of bismuth nitrate, Bi(NO3)r? The molar volume of a gas is 24
dmt mol-' under these conditions. They react according to the equation:
3HrS(g) + 2Bi(NO3).(aq) 'BirSr(s) + 6HNO,(aq).
6 What volume of hydrogen gas is produced by the reaction of 33 dm'of
methane gas when it is reacted with steam according to the equation:
CHn(g) + H,O(g)--' CO(g) + 3H,(g) ?
andIntroduction
ould be able to distinguish between:
o Chemical bonds
Ionic
Covalent
Metallic
ffiltryYou sh
O Intermolecular forces
Hydrogen bonds
Dispersion forces
between separate ions
which are divided into two tyPes:
polar covalent where the bonding pair of electronsis nearer to one atom
pure covalent where the bonding pair of electronsis shared equally
bonding caused by electrons delocalisedthroughout the solid.
(between covalent molecules)
between 6- H in one molecule and 6 F, O orN in another molecule
between all molecules. Their strengthdepends upon the number of electrons in the
molecule.
between 6- atoms in one molecule and 6
atoms in another molecule.
n-bond
l . 3 o a n d n b o n d s .
Dipole/dipole forces -
Things to learnE en ionic bond is the electrostatic attraction that occurs between an
atom that has lost one or several electrons - the cation - and one that
_ has gained one or several electrons - the anion.
c-I e covalent bond occurs when two atoms share a pair of electrons. It
results from the overlap of an orbital containing one electron belonging
to one atom with an orbital that contains one electron that belongs to
the other atom. The overlap can be head on, which results in a o bond
or, side by side, which results in a n bond (see Figure 1.3). A double
bond is a o and a n bond, with two pairs of electrons being shared.
tr n dative covalent bond is a covalent bond formed when one of the
_ overlapping orbitals contained two electrons and the other none.
U n metallic bond is the force of attraction between the sea of
delocalised electrons and the positive ions which are arranged in a
_ regular lattice.iJ the electronegativity of an element is a measure of the attraction its
atom has for a pair of electrons in a covalent bond.
o-bond
Fig
@ s r R u c r u R E A N D B . N D T N G
In hydrated cations, such as[Mg(HrO)r]' ', the water moleculesare bonded to the magnesium ion bydative covalent bonds. Each oxygenatom bonds with its lone pair intoempty s and p orbitals in the Mg'ion.
lonic substances will be watersoluble i f the energy required toseparate the ions in the lattice(lattice energy) is compensated forby the exothermic nature ofhydration. This is the energyreleased when the very polar watermolecules are attracted to the + and- ions.
In HCI there are no hydrogen bonds,the dispersion forces count forabout 80% of the totalintermolecular forces, withdipole/dipole accounting for theremaining 20%.
J n hydrogen bond is an intermolecular force that exists between a 6-hydrogen atom in one molecule and a 6- fluorine, oxygen ornitrogen atom in another molecule.
"J van der Waals forces are between covalent molecules and are causedby dipole/dipole and induced dipole/induced dipole (dispersion) forces.
Things to understandIonic bonding
o An ionic bond is likely if there is a large difference (greater than about1.5) in the electronegativities of the two atoms.
o Cations with a small radius and/or high charge have a large chargedensity, and so are very polarising. Anions with a large radius and/orhigh charge are very polarisable. If either the cation is very polarising orthe anion is very polarisable, the outer electrons in the anion will bepulled towards the cation and the bond will have some covalentcharacter.
I Ionic bonding gives rise to an ionic lattice, which is a regular three-dimensional arrangement of ions.
* An ionic bond is, on average, the same strength as a covalent bond.
Covalent bonding
Covalent substances are either:
i giant atomic, such as diamond, graphite and quartz (SiOr)
ii simple molecular, such as I, and many organic substances
iii hydrogen-bonded molecular, such as ice, and ethanol
iv non-crystalline, such as polymers like poly(ethene).
Polar covalent bonds may result in polar molecules, but for linearmolecules of formula AB, planar molecules of formula AB, tetrahedralmolecules of formula AB, and octahedral molecules of formula ABu, thepolarities of the bonds cancel out, and the molecule is not polar.
Intermolecular forces
o The strongest intermolecular force is hydrogen bonding. In moleculeswith many electrons, such as l, the next strongest are induceddipole/induced dipole forces (sometimes called dispersion forces) and, inmost cases, the weakest are permanent dipole/permanent dipole (ordipole/dipole) forces.
o The strength of dispersion forces depends mainly upon the number ofelectrons in the molecule. This is why I, with 106 electrons is a solid,whereas Cl, with 34 electrons is a gas.
o This explains the trend in boiling temperatures of the noble gases, asthe dispersion forces are less in helium than in neon than in argon etc.Likewise the boiling temperatures of the Group 4 hydrides increase inthe Group from CHn to PbH4.
o The boiling temperatures of the hydrides of Groups 5, 6 and 7 (seeFigure 1. 4) can also be explained. Because there is hydrogen bondingbetween HF molecules but not between the other hydrides in the group,HF has the highest boiling temperature. After the drop to HCl, there is asteady upward trend in boiling temperatures from HCI to HI (in spite ofa decrease in the dipole/dipole forces) because the dispersion forcesincrease.
400350300250
K ZOO150100500NHs PH: AsH3
Group 5
Fig 1.4 Boiling temperatures ofGroups 5, 6 and 7 hydrides
Do not say that the particles start tovibrate, as they are vibrating even atvery low temperatures.
The stronger the force betweenparticles, the more energy needed toseparate them and hence the higherthe melting temperature.
o lt is the bonds not the atoms thatrepel,
o Don't forget to count the numberof lone pairs.
4003503002so
K 200150100500
sbH3 Hzo HzS HrSeGroup 6
I I = t * u c r u R E , B . N D T N G A N D M A ' N G R o u p c H E M r s r R y
4003s0300250
K 200150100500
H2Te HF HCI HBT HIGroup 7
f If the number of electrons in two different substances is about equaland neither has hydrogen bonds, then dipole/dipole forces cause adifference in boiling temperature. This is the case between butane(34 electrons, non-polar, boiling temperature - 0.5 'C) and propanone(32 electrons, but polar, so boiling temperature + 56.2'C).
Effect of heat (melting)
When a solid is heated from room temperature until it melts:
O The particles (ions, molecules or atoms) vibrate more.o As the temperature rises the vibrations increase until they become so
great that the forces between the particles are overcome, and the regulararrangement in the lattice breaks up. The substance is then a liquid.
O In an ionic solid, such as NaCl, the vibrating particles are ions whichare held by strong forces of attraction, and so the ionic solid has a highmelting temperature.
o In a simple molecular solid, such as iodine (Ir) or ice, the vibratingparticles are molecules which are held by weak van der Waals orhydrogen bonding forces of attraction, and so the solid has a lowmelting temperature.
o In a giant atomic solid, such as diamond, the vibrating particles areatoms which are held by strong covalent bonds, and so the solid has avery high melting temperature.
Shapes of molecules
O These are explained by the electron pair repulsion theory which states:i The electron pairs arrange themselves as far apart from each other as
possible in order to minimise repulsion.
ii The repulsion between lone pairs is greater than that between a lonepair and a bond pair, which is greater than that between two bondpairs.
o The number of o bond pairs of electrons and lone pairs in the moleculeshould be counted.
o Any r bond pairs should be ignored when working out the shape of amolecule.
@ s r R u c r u R E A N D B o N D T N G
2 bond pairslinear
bond pairstriangularplanar
bond pairstetrahedral
h:rLg0'
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o The total number of pairs of electrons indicates the arrangement ofthe electrons.
2 = linear
3 = trianSular planar
4 = tetrahedral
5 = trigonal bipyramid
6 = octahedral
But the shape will differ if any of the pairs are lone pairs (seeFigure 1.5).
4 p a i r s : 3 b o n d + 1 l o n e :
molecular shape is pyramidal, e.g. NHr, or PCl,
2bond + 2lone:
molecular shape is bent or V-shaped, e.g. HrO
6 p a i r s : 5 b o n d + 1 l o n e :
molecular shape is a square-based pyramid.
4 b o n d + 2 l o n e :
molecular shape is square planar.
Shapes of ions
Negative jons have gained 1 electron for each negative charge.Thus SO;- has 4 o bonds (and 2 n bonds) and no lone pairs around thesulphur, and so the ion is tetrahedral.CO.'- has 3 o bonds (and 1 z bond) and no lone pairs around the carbon,and so it is triangular planar.NO; has one single covalent, one dative covalent and one double covalentbond (i.e. 3 o and 1 z) and no lone pairs around the nitrogen and so isalso triangular planar.Positive ions have lost 1 electron for each positive charge.Thus NHn. has 4 o bonds and no lone pairs around the nitrogen, and so istetrahedral
-Hi ) J / / l
r Checklist- ? ' - ^ , . ' -f'-
Before attempting the questions on this topic, check that you understand:
lJ the nature of ionic, covalent and dative covalent bonds
LJ tfre effect of difference in electronegativity on type of bonding
iJ that polar bonds may not give rise to polar molecules
J the effect of size and charge of ions on the type of bonding
J intermolecular forces such as hydrogen bonding and van der Waals
,J the different structures of solids and their properties
J the trends in boiling temperatures caused by intermolecular forces
,J the nature of hydration of ions
iJ the changes in motion and arrangement of particles on change of state
J a metallic bond
J the shapes of molecules and ions.
bond pairs +1 lone pairpyramidal
bond pairs +2lone pairsV-shaped
o ?s bond pairs Il
noo ^
Trigonal 120"\ lrebipyramid
o/l
bond pairsoctahedral
Fig 1.5 Shapes of molecules ond ions
90"and 180'
The answers to the numberedquestions are on pages 122-123.
l l " t * u c r u R E ,
B . N D T N G A N D M A ' N G R o u p c H E M r s r R y
Testing your knowledge and understandingFor the first set of questions, cover the margin, write your answers, thencheck to see if you are correct.
O State which types of bond are present in a double bond as in O=C=O.
I State the value of the charge on the ions of the following elements:
Al, Mg, Na, R N, O.
o State which of the cations above:
a has the smallest radiusb has the largest charge densityc is the most polarising.
O State which of the anions above is the most polarisable.
O Which ion has the largest radius: Li'or Na- or K-?
o Which of these three ions, Li*, Na* or K*, has:
a the largest surface charge densityb the largest polarising power?
Which is the most electronegative element in each list:a C , O , N , Fb Cl, Br, I?
Which is the least electronegative element of:a Be, Mg, Cab Na, Mg, A1?
a Which is the molecule in each group with the most polar bond:
Explain the processes of melting and boiling in terms of thearrangement and motion of the particles.
State, for each of the following pairs, which has the stronger forcesbetween particles and hence has the higher boiling temperature, andexplain why in terms of the types of force present:
NH, or PH,HCI or HBrCH3COCH.. (propanone) or CnH,o @utane)Pn or SuNaCl or CCln
In order to work out the shape of a molecule or ion, you should firstevaluate the number of o bond pairs of electrons and the number oflone (unbonded) pairs of electrons around the central atom.
Construct a table similar to that shown below, and use it to deduce the
shapes of the following species.
Explain why solid sodium metal conducts electricity whereas solidsodium chloride does not.
abcde
SiH4, BFr, BeCl2, PCI3, SFu, XeFn, NHr*, PCl6-
Molecule/ion
Number ofo bond pairs
Number oflone pairs
Total numberof electron pairs
Shape
NH, 3 1 4 pyramidal
I t t * u c r u R E , B . N D T N G A N D M A r N G R o u p q H E M I s r R y
Theffitry
Table IIntroduction
r The melting and boiling temperatures of the elements depend on thetype and strength of the bond or intermolecular force between particles.
a There are trends in physical and chemical properties across a Period anddown a Group. For instance
1 Elements show decreasing metallic character across a Period.2 Elements show increasing metallic character down a Group.
learn and understandElectronic structure
O The elements in the Periodic Table are arranged in order of atomicnumber, so each element has one more proton and hence one moreelectron than the previous element.
o Elements in the same group have the same number of electrons in theirouter shells. These are called the valence electrons.
o Elements in the same Period have the same number of shells containingelectrons so their outer or valence electrons are in the same shell.
o The order in which electrons fill the orbital types is:
1s; 2s,2p; 3s, 3p; 4s,3d, 4p; 5s, 4d, 5p; 6s (see Figure 1.1 on page 3)
Melting (and boiling) temperatures of the Period 3elements (Na to Ar)
O The melting temperature depends upon the strength of the forcesbetween particles that separate during melting (or boiling).
These particles may be:i metal ions in a sea of electrons in metalsii covalently bonded atoms in a giant atomic structureiii molecules with intermolecular forces between them in simple
molecular solids.
o To understand the trends in melting temperatures, you should firstdecide what type of bonding or force is between the particles.
o If the solid is metallic: the greater the charge density of the ion in thelattice, the stronger the force holding the lattice together and so thehigher the melting temperature.
o If the solid is a giant atomic lattice: the covalent bonds throughoutthe lattice are strong and so the solid has a very high meltingtemperature.
o If the solid is a simple molecular substance: the melting (or boiling)temperature depends upon the strength of dispersion (induceddipole/induced dipole) force between the molecules.
o In Period 3, sodium, magnesium and aluminium are metallic, siliconforms a giant atomic lattice, and phosphorus, Pn, sulphur, S' chlorine,Cl, and argon, Ar, all form simple molecular solids.
Periodic
Things to
There are two exceptions with thefirst 40 elements. Because of theextra stability of a half fi l led or fullset of d orbitals, chromium is [Ar]4s', 3du, and copper is [Ar] 4s', 3d'o
The more electrons in the molecule,the stronger the dispersion forceand the higher the melting or boilingtemperature.
@ t " E P E R r o D r c T a e L E I
The answers to the numberedquestions are on page 123.
Electrical conductivity
Solids conduct electricity by the flow of delocalised electrons.
Thus metals conduct electricity. Graphite also conducts, but only in theplane of the layers. This is due to the n-electrons that are delocalised aboveand below the lavers.
^.HflK ' Checklistl / H
H
Before attempting the questions on this topic, check that you can:
i-i Write the electronic structure for elements numbers 1 to 36.
lJ Explain the variation of melting and boiling temperatures for theelements in Period 3.
iJ Explain the change in electrical conductivity of the elements inPeriod 3.
lJ Explain the variation in ionisation energies of the elements in Period 3.
)Hj - l J ) / l -
T)() Testing your knowledge and understandingH
L What, in terms of electronic structure, are the features that thefollowing have in common:
a members of the same group , €.8. Group 2,
b members of the same period, e.g. Period potassium to krypton?
2 State the type of solid structure of the elements listed. Give youranswer as one of:
3 Explain the difference in melting temperatures of the followingelements:
Explain the meaning of: i nuclear charge, ii screening (or shielding)by inner electrons.
Use the concepts explained in (a) to explain why: i The 2ndionisation energy of sodium is very much more than its 1stionisation energy. ii The lst ionisation energy of sodium is less thanthe 1st ionisation energy of magnesium.
Element Na Mg Si Pn s8 CI,
Melting temperature/O6 98 650 1410 44 1 1 3 -101
4 The ionisation energies of sodium and magnesium are listed below:
Element lst ionisation energy/kJ mol-' 2nd ionisation energy/kJ mol-'
Sodium 494 4560
Magnesium 736 1450
I I t t * u c r u R E , B . N D T N G A N D M A r N G R o u p c H E M r s r R y
Introduction to oxidation and
Introductiono Redox reactions are those which involve a transfer of electrons
o Remember OIL RIG (Oxidation Is Loss, Reduction fs Gain of electrons)
4A'W Things to learnE Oxiaation occurs when a substance loses one or more electrons.
There is an increase in the oxidation number of the element involved.
[l nn oxidising agent is a substance that oxidises another substanceand so is itself reduced. The half equation involving an oxidising agenthas electrons on the left-hand side, i.e. it takes electrons from thesubstance being oxidised.
l] neduction occurs when a substance gains one or more electrons.There is a decrease in the oxidation number of the element involved.
|J A reducing agent is a substance that reduces another substance andso is itself oxidised. The half equation involving a reducing agent haselectrons on the right-hand side, i.e. it gives electrons to the substancebeing reduced.
^ -- ,J,,+ f"/ |
W Things to understand!---.-.-l'
Oxidation number
I The oxidation number is the charge on an atom of the element in acompound calculated assuming that all the atoms in the compound aresimple monatomic ions. The more electronegative element is given anoxidation number of -1 per bond.
O There are some rules used for calculating oxidation numbers. Theyshould be applied in the following order:
1 The oxidation number of an uncombined element is zero.
2 A simple monatomic ion has an oxidation number equal to itscharge.
3 The oxidation number of Group 1 metals is always +1, and ofGroup 2 metals is +2.
4 Fluorine always has an oxidation number of -1, hydrogen (exceptin metallic hydrides) of +1, and oxygen (except in FrO andperoxides) of -2.
5 The sum of the oxidation numbers in a molecule adds up to 0, andthose in a polyatomic ion (such as SOr*) add up to the charge onthe ion.
O When an element is oxidised, its oxidation number increases.
red uction
orl Rtc
In an overall equation, the totalincrease in oxidation number of oneelement (increase x number ofatoms of that element) must equalthe total decrease in oxidationnumber of another element.
@ l N T R o D U c r r o N T o o x r D A T r o N A N D R E D U c r r o N
Remember that an oxidising agentbecomes reduced ( loses electrons),and a reducing agent becomesoxidised.Half equations must balance foratoms and for charge. .For example, both sides of the Cr"'equation add up to -5.
21
Work^ei, oxal44f-{x,
Half equations
These are written:either as reduction with electrons on the left side of the half equation,
e.g. Cl, (g) + 2e ^- zCl- (aq)
here chlorine is being reduced and so is acting as an oxidising agent.
or as oxidation with electrons on the right side of the equation,
e.g. Fe'- (aq) -.- Fe'- (aq) + e-
here iron(ll) ions are being oxidised, and thus are acting as a reducingagent.
.- Many oxidising agents only work in acid solution. Their half equationshave H. ions on the left hand side and HrO on the right. This is likelywith oxdising agents containing oxygen (such as MnOr-),
e.g. MnO.,-(aq) + 8H.(aq) + 5e- + Mn'.(aq) + 4HrO(l)
r If a redox system is in alkaline solution, OH-side and HrO on the other,
e.g. Cr'.(aq) + BOH-(aq) * CrOn2- +
Overall redox equations
t Overall redox equations are obtained by adding half equations together.
I One half equation must be written as a reduction (electrons on the left)
and the other as an oxidation (electrons on the right).
o When they are added the electrons must cancel. To achieve this it maybe necessary to multiply one or both half equations by integers,
e.g. for the overall equation for the oxidation of Fe'. ions by MnOn-
may need to be on one
4H,O(l) + 3e-
lf the question asks for the overallequation for the reaction of A with B,make sure that both A and B appearon the left-hand side of the finaloveral l equation. lons
Calculate the oxidation number of chlorine in Clr, Cl-, MgCl, and Cl0r-.
Answers:.in Cl, = 0 (uncombined element: rule 1)in Cl- = -1 (monatomic ion: rule 2)in MgCl, = -1 (Mg is 2+; (+2) +ZCl = 0; therefore each Cl is -1:
rules 3 and 5)i n C l 0 ; = 1 5 ( C l + 3 x ( - 2 ) = - 1 : r u l e s 4 a n d 5 )
I I = t *ucruRE,
Checklist
B O N D I N G A N D M A I N G R O U P C H E M I S T R Y
The answers to the numberedquestions are on pages 123-124.
Before attempting the questions on this topic, check that you can:
[] Define oxidation and reduction.
lJ oefine oxidising agent and reducing agent.
E Calc.tlate oxidation numbers in neutral molecules and in ions.
fl Write ionic half equations.
I-l Combine half equations and so deduce an overall redox equation.
Testing your knowledge and understandingFor the first set of questions, cover the margin, write your answer, thencheck to see if you are correct.
O In the following equations, state which substance, if any, has beenoxidised:a 2Cen. (aq) + 2l- (aq) + zce'* (aq) + I, (aq)b H. (aq) + OH- (aq) + H,O (l)c Zn (s) + 2H. (aq) + Zn'- (aq) + H, (g)d zFe'* (aq) + 2Hg" (aq) + 2Fe'. (aq) + Hgr2. (aq)
o Calculate the oxidation number of the elements in bold in thefollowing:so2, H2s,CtOn'-, CtrOr'-,HrO", H2SO4, FerOn
1 Construct ionic half equations for:
a Sn'* ions being oxidised to Snn* ions in aqueous solution
b Fe'* ions being reduced to Fe2* ions in aqueous solution.
c Now write the balanced equation for the reaction between Fe3* andSn'* ions in aqueous solution.
2 Construct ionic half equations for:
a hydrogen peroxide being reduced to water in acid solutionb sulphur being reduced to hydrogen sulphide in acid solution
c Now write the balanced equation for the reaction betweenhydrogen peroxide and hydrogen sulphide in acid solution.
3 Construct ionic half equations for:
a PbOr(s) being reduced to PbSOr(s) in the presence of HrSOn(aq)b PbSO.(s) being reduced to Pb(s) in the presence of water
c Now write the balanced equation for the reaction between PbO,and lead in the presence of dilute sulphuric acid.
a iodide ionsb nonec zincd i ron( l l ) ions
+4 -2+6 +6-1 +6 Z!1t at+2 and 2 ar +3;
average 2i)
@ e R o u P I A N D G R o u p 2
@ Group 1 and Group 2fntroduction
J fne properties of elements and their compounds change steadily downa Group.
J fnis means that the answer to a question about which is the most/leastreactive, easiest/hardest to decompose, most/least soluble, etc. will bean element, or a compound of that element, either at the top or at thebottom of the Group.
J Down a Group the elements become increasingly metallic in character.Thus:o their oxides become stronger basesO they form positive ions more readilyO they form covalent bonds less readily.
-HGt'\ t lz ' l
it/''Ea'
Physical properties of the elements
I Group 1: All are solid metals; their melting temperatures and hardnessdecrease down the group; all conduct electricity.
o Group 2: AII are solid metals; their melting temperatures and hardnessdecrease down the group (except magnesium which has a lowermelting temperature than calcium); all conduct electricity; their meltingtemperatures are higher than the Group 1 element in the same period.
Flame colours of their compounds
O Group 1: lithiumsodium
carmine redyellow
,A l*light out
potassium lilacO Group 2: calcium brick red
strontium crimson redbarium green.
o These colours are caused because:I Heat causes the compound to vaporise and produce some atoms of
the metal with electrons in a higher orbital than the ground state(e.g. in the 4th shell rather than the normal 3rd shell for sodium).
2 The electron falls back to its normal shell and as it does so, energy inthe form of visible light is emitted. The light that is emitted is of acharacteristic frequency, and hence colour, dependent on the energylevel difference between the two shells (see Figure 1.6).
Ionisation energies
o The value of the lst ionisation energy for Group 1 and of the lst and2nd ionisation energies for Group 2 decreases down the group. This is
3rd
Fig 1,6 Emission of a spectrol line
The vigour of the reaction increasesdown the group.
The rate of reaction increases downthe group.
S T R U C T U R E , B O N D I N G A N D M A I N G R O U P C H E M I S T R Y
because as the atoms get larger, the outer electrons are further from thenucleus and so are held on less firmly. The increase in nuclear charge iscompensated for by an increase in the shielding by the inner electrons.
Reactions of the elements with oxygen
O Group 1: All burn.Lithium forms an oxide: 4Li * O, - zLizOSodium forms a peroxide: 2Na + Or - NarO,Potassium and the others form a superoxide: K + O, - KO,
o Group 2: AII burn to form ionic oxides of formula MO, except that inexcess oxygen barium forms a peroxide (BaOr).
ZCa + Or'ZCaO
Reaction of the elements with chlorine
o Group 1: All react vigorously to form ionic chlorides of formula MCl.These dissolve in water to produce hydrated ions, e.g.
NaCl(s) - Na* (aq) + Cl (aq)
I Group 2: All react vigorously to produce ionic chlorides of formulaMCl, except that BeCl, is covalent when anhydrous. All Group 2chlorides are soluble in water producing hydrated ions of formula[M(H,O).]2-. Beryllium chloride gives an acidic solution because ofdeprotonation:
[Be(H,O)u]'* +H,O + [Be(H,O)'(OH)]. + H,O*
Reaction of the elements with water
o Group 1: All react vigorously with cold water to give an alkalinesolution of metal hydroxide and hydrogen gas, e.g.
2Na(s) + 2H,O(l) -- 2NaOH(aq) + H,(g)
t Group 2: Beryllium does not react but magnesium burns in steam toproduce an oxide and hydrogen:
M g + H r O - M g O + H ,
the others react rapidly with cold water to form an alkaline suspensionof metal hydroxide and hydrogen gas:
Ca + zH,O - Ca(OH)r+H,
Reactions of Group 2 oxides with water
O BeO is amphoteric and does not react with water.O MgO is basic and reacts slowly with water to form a hydroxide.a All the others react rapidly and exothermically to form alkaline
suspensions of the hydroxide, which have a pH of about 13.CaO + H,O-- Ca(OH),
Solubilities of Gro$p 2 sulphates and hydroxides
O Sulphates: their solubilities decrease down the group. BeSOo and MgSOoare soluble; CaSOo is slightly soluble, SrSOo and BaSOo are insoluble.
o Hydroxides: their solubilities increase down the group. Be(OH), andMg(OH), are insoluble, Ca(OH), and Sr(OH), are slightly soluble, andBa(OH), is fairly soluble.
O Addition of aqueous sodium hydroxide to solutions of Group 2 saltsproduces a white precipitate of metal hydroxide. (Barium produces afaint precipitate):
M'.(aq) + 2OH-(a9) - M(OH)'(s)
@ e R o u P I A N o G n o u p 2
The compound is more likely to bedecomposed on heating if the cationpolarises the anion. ThusGroup 2 compounds (cation 2+)decompose more readily than Group1 compounds (cation only 1+).Compounds of metals higher in agroup (smaller ionic radius)decompose more easily thancompounds of metals lower in theGroup.
Lithium
K is +1, Ca is +2
It is extremely insoluble.
O Addition of aqueous sulphate ions to a solution of Sr'* or Ba'* ionsproduces a white precipitate of metal sulphate:
Ba'.(aq) + SOn'-(aq) - BaSOn(s)
Thermal stability of nitrates and carbonates
o Thermal stability increases down both groups:o Group 2 nitrates all decompose to give a metal oxide, brown fumes of
nitrogen dioxide, and oxygen:2Ca(NOr)r- 2CaO + 4NO, + O,
O Group 1 nitrates, except lithium nitrate, decompose to give a metalnitrite and oxygen:
2NaNO, -- 2NaNO, r O,
but 4LiNO,, -- ZLirO + 4NO, + O,o Group 2 carbonates all decompose (except barium carbonate which is
stable to heat) to give a metal oxide and carbon dioxide:C a C O ' , - C a O + C O ,
t Group 1 carbonates are stable to heat except for lithium carbonate:L i ? C O , - L i ' O + C O .
H' 9 ^ r
f__- Checklist^ a / - ^ - ' -H
Before attempting the questions on this topic, check that you know:
tJ ttre physical properties of the Group 1 and Group 2 elements
r-l ttre flame colours caused by their compounds
iJ the trends in ionisation energies within a group
iJ ttre reactions of the elements with oxygen, chlorine and water
J the reactions of their oxides with water
,J the oxidation states of the elements in Group 1 and in Group 2
,J the trends in solubilities of Group 2 sulphates and hydroxides
.J tfre reason for the trend in thermat stability of Group 1 and Group 2nitrates and carbonates.
Testing your knowledge and understandingFor the first set of questions, cover the margin, write your answer, thencheck to see if you are correct.
o Which Group 1 metal has the highest melting temperature?
o What is the oxidation number of:i potassium in KrCrrO, andii calcium in CaCOr?
o Barium compounds normally are poisonous, but barium sulphate isgiven to people in order to outline their gut in radiography (X-rayimaging). Why is barium sulphate not poisonous?
The answers to the numberedquestions are on page 124.
I I " t * u c r u R E ,
B . N D I N G A N D M A I N G R o u p c H E M t s r R y
You are given sotid samples of three chlorides. One is lithium chloride,one is potassium chloride and the other is barium chloride. Describe thetests that you would do to find out which was which.
2 Explain why sodium compounds give a yellow colour in a flame.
3 Exptain why the Lst ionisation energy of sodium is larger than the Lstionisation energy of potassium.
Write balanced equations for the reactions of:
a calcium with oxygen
b calcium with water
c potassium with water
d magnesium with steam.
Explain why the addition of dilute sodium hydroxide to a solution ofmagnesium chloride produces a white precipitate, but little or noprecipitate when dilute sodium hydroxide is added to a solution ofbarium chloride.
State and explain which Group 2 element forms the least thermallystable carbonate.
Write balanced equations for the thermal decomposition of thefollowing, but if there is no reaction at laboratory temperatures, say so:
a lithium nitrate, sodium nitrate and magnesium nitrate
b sodium carbonate, magnesium carbonate and barium carbonate.
4
6
@ c R o u P 7 ( c a L o R r N E T o r o o i N r e )
Group 7 (chlor ine to iodine)<taiAl Introductiony-v
Group 7 chancteristics are that:
r The halogens are oxidising agents with their strength decreasing downthe group.1 Chlorine is the strongest oxidising agent and iodine is the weakest.2 Chloride ions are the weakest reducing agents, and so are the hardest
to oxidise, whereas iodide ions are the strongest reducing agents, andso are the easiest to oxidise.
I They form halide ions, such as Cl-, and oxoanions, such as ClO,-.o Solutions of the hydrogen halides are strong acids.
ffi Things to learn and understandLry
Disproportionation
This occurs when an element is simultaneously oxidised and reduced. Itfollows that there must be at least two atoms of that element, with thesarne oxidation number, on the left of the equation, and that the elementmust be able to exist in at least three different oxidation states.Chlorine disproportionates in alkali:
Cl, + zOH- - Cl- + OCI- + H,O(0) (-1) (+1)
Physical properties (at room temperature)
o Chlorine is a greenish gas.O Bromine a brown liquid.o Iodine is a dark grey lustrous solid which gives a violet vapour on
heating.o A solution of iodine in aqueous potassium iodide is red/brown, and in
non-oxygen-containing organic solvents it is violet.
Tests for the elements (only examined at AS in UnitTest 3B)
o First observe the colour of the halogen or its solution.O Chlorine rapidly bleaches damp litmus paper. It will displace bromine
from a solution of potassium bromide (then test for bromine by addinga suitable organic solvent such as hexane, which will show the brownbromine colour).
o Bromine is brown and will (slowly) bleach litmus. It will displace iodinefrom a solution of potassium iodide (then test for iodine with starch orhexane as below).
o Iodine turns a solution of starch blue-black, and it forms a violetsolution when dissolved in suitable organic solvents such as hexane.
Chlorine is a stronger oxidisingagent than bromine, which isstronger than iodine.
I I " t * u c r u R E ,
B . N D T N G ^ N D M A I N G R o u p c H E M I s r R y
Halides
o All hydrogen halides are covalent gases, but are soluble in water becausethey react with water to form ions. Their solutions are strongly acidic:
HCI + HrO + H,O* (aq) + Cl-(aq)
Hydrogen fluoride is a weak acid and is only partially ionised.o All metal halides are soluble in water, except for silver and lead halides.
Test for halides (only examined at AS in Unit Test 38)
To a solution of the halide add dilute nitric acid to prevent carbonatesfrom interfering with the test. Then add silver nitrate solution followed byammonia solution.
Addition of concentrated sulphuric acid to the solidhalide
o Chlorides produce steamy acid fumes of HCl.O Bromides produce steamy acid fumes of HBr with some brown bromine
and some SO, gas.a Iodides give clouds of violet iodine vapour.
This is due to the fact that HBr is just powerful enough as a reducing agentto reduce some of the concentrated sulphuric acid to sulphur dioxide anditself be oxidised to bromine.The HI initially produced is a very powerful reducing agent. It reduces theconcentrated sulphuric acid and is itself oxidised to iodine.
Oxidation numbers
o Chlorine is 0 in Clr.o Chlorine is -1 in chlorides.o Chlorine is +1 in CIO- ions.O Chlorine is +5 in CIO; ions.
Redox
o Chlorine is a powerful oxidising agent and is reduced to the -1 state.o The half equation is:
Clr(aq) + 2e- .. 2Cl-(aq)
and is similar for the other halogens.o Chlorine disproportionates in alkali at room temperature:
Cl,(aq) + 2OH-(a9) - Cl-(aq) + OCI-(aq) + H,O(l)
(0) (-1) (+1)o Chlorate(I) compounds disproportionate when heated:
3OCf+2CI -+C lOr -
3x(+1) Zx(-L) (+s)o Bromine is extracted from seawater by bubbling in chlorine gas which
oxidises the Bf ions.
Chloride Bromide IodideAddition of Ag-(aq) white precipitate cream precipitate yellow precipitate
Addition ofdilute NH, precipitate dissolves no change no change
The answers to the numberedquestions are on pages 124-125.
ffil4),
Befo
Checklistre attempting questions on this topic, check that you can recall:
iJ the physical properties of the elements
lJ ttre tests for the elements and for the halides
iJ ttre reactions of concentrated sulphuric acid with the halides
iJ examples of -1, 0, +1 and +5 oxidation states of chlorine and thedisproportionation of chlorine and chlorate(I) ions
I ttre relative strengths of the halogens as oxidising agents.
<ffi,rftn Testing your knowledge and understandingtH'
For the first set of questions, cover the margin, write your answer, thencheck to see if you are correct.
Name the halogen that is a liquid at room temperature.
,r Describe the test for iodine.
* Which of Cl, Br, and I, is the strongest oxidising agent?
O Which of HCl, HBr and HI is the strongest reducing agent?
Why is hydrogen chloride gas very soluble in water?
It is thought that a sample of solid sodium carbonate has beencontaminated by some sodium chloride. How would you test for thepresence of chloride ions in this sample?
Explain why hydrogen chloride gas is produced but chlorine is notwhen concentrated sulphuric acid is added to solid sodium chloride,whereas iodine and almost no hydrogen iodide is produced whenconcentrated sulphuric acid is added to solid sodium iodide.
Define disproportionation and give an example of a disproportionationreaction of chlorine or a chlorine compound.
t2
I t t * , r c r u R E , B o N D T N G A N D M A r N G R o u p c H E M r s r R y
Practlce Test: Unit LTime allowed Lhr
All questions are taken from parts of previous Edexcel Advanced GCE questions.
The answers are on pages IZ5-t26.
b
L a
2 a
Complete and balance the following equations.i C a + O ,ii NarO + HrO +iii NarO + HCI +State and explain the trend in thermal stability of of the carbonates of the Group 2 elements as thegroup is descended.
t1ItlIl2l
t3l
bc
State the structure of, and the type of bonding in, the following substances. Draw labelleddiagrams to illustrate your answers.i Graphiteii Sodium chlorideExplain why both graphite and sodium chloride have high melting temperatures.i Explain why graphite is able to conduct electricity in the solid state.ii Explain why sodium chloride conducts electricity in the liquid state.
(Totat 7 marks)[May 2002 Unit Test 1 question 2]
t4It3It3lr2ltlI
(Totat 13 marks)lMay 2002 Unit Test L question 6l
t2ltlI
I2lI2l
3 a i Define the terms atomic number and mass numberii Identify the particle that contains 35 protons, 44 neutrons and 34 electrons.
i the PH. moleculeii the AIH; ion.
Bromine consists of two isotopes of mass numbers 79 and 81. A sample of bromine gas, Br, wasexamined in a mass spectrometer. The mass spectrum showing the molecular ions is given below.
Relativeintensity
b i Explain how the molecules in the sample are ionised. t2lii State how the resulting ions are accelerated. tlliii Identify the species responsible for the peak at mle = 160. tl]
(Total 7 marks)llanuary 2001 CH1 question L & lune 2001 Unit Test L question 3l
Hydrogen forms compounds with most non-metallic elements and with some metals.Calculate the empirical formula of the compound used in the manufacture of artificial rubber, whichhas the following composition by mass:hydrogen tI.Lo/o carbon 88.9o/o
b The boiling temperatures of hydrogen chloride and hydrogen iodide are:hydrogen chloride -85 'C Hydrogen iodide -35 "CExplain why hydrogen iodide has a higher boiling temperature than hydrogen chloride l2l
c Draw and explain the shapes of:
100
t3l
158 160 162
P n l c t r c e
5 a
6 a
d Calculate the number of molecules in 8.0 cm' of gaseous phosphine, PH. at room temperatureand pressure.
(The molar volume of a gas at room temperature and pressure should be taken as 2.4 x lOn cm' mol-t.The Avogadro constant is 6.0 x 10" mol-'.) t2l
(Total L1 marks)
ffanuary 2001 CH1 question 2]
Hydrogen sulphide is produced when concentrated sulphuric acid is added to solid sodium iodide, butsulphur dioxide is produced when concentrated sulphuric acid is added to solid sodium bromide.i Complete the following table:
Compound Formula Oxidation state of sulphur in the compound
Sulphuric acid HzSO'
Hydrogen sulphide H,S
Sulphur dioxide SO,
tslii Use your answers to a part i to suggest which of the ions, iodide or bromide, has the greater reducing
power. t2li Write an ionic half-equation to show the oxidation of chloride ions, Cl-, to chlorine, Cl,. tUii Write an ionic half-equation to show the reduction of chlorate(l) ions, OCf, to chloride ions, in
acidic solution. I2liii Bleach is a solution of chlorate(l) ions and chloride ions. Combine the two ionic half-equations
above to produce an equation which shows the effect of adding acid to bleach. tll(Total 9 marks)
[May 2002 Unit Test 1 question 7l
b Write one equation in each case to represent the change occurring when the following quantities aremeasured.i The first electron affinity of sulphurii The first ionisation of sulphur.
Complete the following for the element sulphurNumber of electrons in the 1st shell =Number of electrons in the 2nd shell =Number of electrons in the 3rd shell =
o 2 4 6 8 1 0 1 2 1 4 1 6 1 8
Number of electron removed
One of the chlorides of sulphur is sulphur dichloride, SClr. It is a liquid at room temperature;electronegativity of S is 2.5 and that of chlorine is 3.0i Draw a dot and cross diagram to show the bonding in SClr.ii The shape of the molecule is bent (V-shaped). Explain why the molecule has this shape.iii State, with reasons, whether SCl, is a polar molecule overall.
(Total
fJanuary 2002 CHl question 2 & May 2002 Unit Test 1
t1l
r2ltll
c The graph below shows the logarithm of the successive ionisation energies of sulphur plotted against thenumber of the electron removed. Account for the form of the graph in terms of the electronic structureof sulphur.
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$ s.soH s.oo! +.so€ 4.oo
A 3.soI E.oo3o z.so
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[nerget ics IIntroduction
o State symbols should always be used in equations in this topic.
O In definitions in this topic, there are three points to note:
1 the chemical change that is taking place2 the conditions3 the amount of substance (reactant or product).
o It always helps to add an equation with state symbols as an example,because this may gain marks lost by omission in the word definition.
O It is assumed for simplicity that the value of LH is an indicator of thedirection of a chemical reaction. The more exothermic the reaction, themore likely it is to take place, but other factors, such as activationenergy (see page 44), also determine whether the reaction happens.
Things to learnJ Standard conditions are:
O gases at a pressure of L atmosphere
o a stated temperature (usually 298 K)
o solutions, if any, at a concentration of 1.00 mol dm-'
o substances in their most stable states, e.g. carbon as graphite notdiamond.
[-l Standard enthalpy of formation, AI{'is the enthalpy change, understandard conditions, when one mole of a compound is formed fromits elements in their standard states, e.g. AFII'for ethanol is theenthalpy change for the reaction:
2C(graphite) + 3H,(g) +'/,Or(g) - C,H.OH(l)
[f Standard enthalpy of combustion, AF/,'is the enthalpy change, understandard conditions, when one mole of a substance is completelyburned in oxyger, €.g. AI1.'for ethanol is the enthalpy change for thereaction:
C,H'OH(I) + 3O,(g) - ZCO,(g) + 3H,O0)
E Standard enthalpy of neutralisation AI/:.,, of an acid is the enthalpychange, under standard conditions, when the acid is neutralised bybase and one mole of water is produced, e.g. A.Fif;",, of sulphuric acid,by sodium hydroxide solution, is the enthalpy change for the reaction:
'/,HrSOn(aq) + NaOH(aq) -'l,Na,SO,(aq) + H'O(l)
It follows from this definition thatthe enthalpy of formation of anelement, in its standard state, iszero.
You should always give an equationas an example for this and otherenthalpy definitions.
O E n e R G E T r c s r
Route 1
Route 2
Fig 2.3 A reaction cycle
for hydrochloric acid the enthalpy change is for the reaction:
HCI(aq) + NaOH(a9) - NaCl(aq) + H,O(l)
for any strong acid being neutralised by any strong base, theenthalpy change is for the reaction:
H-(aq) + OH-(aq) - H,O(l) Aff" = -57 kJ mol-'
E Aoerage bond enthalpy. This is the average quantity of energyrequired to break one mole of covalent bonds in a gaseous species(at one atmosphere pressure). It is always endothermic, e.g. theC-H bond enthalpy in a gaseous compound is the enthalpy change forthe reaction:
C-H(in compoundxs) - C(8) + H(g)
E nn exothermic reaction gets hot, so that heat is then given out to thesurroundings. For all exothermic reactions A^tI is negative (seeFigure 2.1). This means that chemical energy is being converted into
_ thermal (heat) energy.[J An endothermic reaction gets cold, so that heat is then taken in from
the surroundings. For all endothernic reactions AII is positive(see Figure 2.2).
1 *..ctantst -
tIII
EnthalpyEnthalpy
Products
Fig 2.1 Enthalpy diagrom for anexothermic reoction
Products
Fig 2.2 Enthalpy diogram for onendothermic reoction
E Hess's Law. The enthalpy change for a given reaction is independentof the route by which the reaction is achieved (see Figure 2.3).Thus the enthalpy change proceeding directly from reactants toproducts is the same as the sum of the enthalpy changes of all thereactions when the change is carried out in two or more steps.
NI = A.FI, + NIr+ AJI.,
ffi Things to understandLry
Calculation of A.ff*.ooo from A^tI, dataYou can use the expression:
Afl"uoio,, = the sum of Aflo,rnation of products - the sum of AI{o,-u,ion of reactants
O Remember that if you have two moles of a substance, you must doublethe value of its AId.
O The enthalpy of formation of an element (in its standard state) is zero.
LH2= - LH" because aH, representsthe reverse of combustion.l f an equation is reversed, the signof i ts lH must be changed.
The mass in this expression is thetotal mass of the system (usuallywater) not the mass of the reactants.
H " R G A N r c
c H E M r s r R y , E N E R G E T T . ' , K r N E T r c s A N D E e u r L I B R r u M
Workri, oxiloq-lr'Calculate the standard enthalpy change for the reaction:
4NH,(g) + 50,(g) -- 4N0(9) + 6H,0(g)given the following AHelkJ mol-':
NH,(g) = -46.2, N0(g) = +90.4, H,0(g) = -242.
AnSwer. AHf,o", = SUITI Aff of produCtS - Sum AH,F of reactants= 4 x(+90.a) + 6 x (-242)-4 x (-46.2) -5x (0) = -906 kJ mol-'
Calculation of LII, from A^f[" data
To do this you use the alternative route:
ElemenLH,
1 mol of substance
where AH, = the sum of the enthalpies of combustion of the elementstaking into account the number of moles of each element,and A^H, = - the enthalpy of combustion of I mol of the substance.
Thereforer Aflo,-u,,on = LH, + A,H,
Workz,oL oxiloLf./*,Calculate the standard enthalpy of formation of propan-1-ol, C2H'CH,0H, giventhe following aH+,..ou*on /kJ mol-1 :propan-1-ol(l) = -2010, C(graphite) = -394, H,(g) = -286
Answer. 3C(s) + 4H,(g) + 10,(g) - C,H'CH,OH(|)
3 x AH. (sraphite) I \+ " m"fi1c,HscH,0H)
3C0,(g) + 4H,0(l) '
AH*, =4H", of elements - aH", of compound= 3 x (-394) + 4 x(-286) - (-2010) = -316 kJ mol- '
Calculation of AII values frorn laboratory data (onlyexamined at AS in unit test 38)
I Calculation of heat change from temperature change:
heat produced = rl?SS x specific heat capacity x rise in temperature
I Then heat produced per mole = heat produced in experiment
the number of moles reactedo Then enthalpy change, AFI, is minus the heat produced per mole (for an
exothermic reaction A/{is a negative number).
O For an endothermic reaction, calculate the heat lost per mole. Thisequals A,H, a positive number.
\ /o , .L H ' \
/combustion products of the elements
:
- t r l
@ e N E R G E r r c s I
Work^ei, exiloLf.b
Calculation of A^tI values fromCalculation of A^tI values from average bond enthalpies
O Draw the structural formulae for all the reactants and all the products.
il Decide what bonds are broken in the reaction and calculate the energyrequired to break the bonds (endothermic).
Decide what bonds are made in the reactionreleased by making the bonds (exothermic).
Add the positive bond breaking enthalpy toenthalpy.
H H
t lC - C - H
r lH H
H
+
H
H_ H --__________+ H_
make (exo)c-c -3482 x C-H -824
Given the following average bond enthalpies/kJ mol-':C-C is +348, C=C is +612, H-H is +436, and C-H is +412.
Answer.
A 0.120 g sample of ethanol was burnt and the heat produced warmed 250 gof water from 17.30 'C to 20.72 'C.The specific heat capacity of water is 4.18J gto0-t. Calculate A{omustion of ethanol.
Answer. Heat produced = 250 g x 4.18 J gj o6-t x3.42 'C = 3574 J.Amount of ethanol = 0.120 g146 g mol-' = 2.61 x 10-'molHeat produced per ffi010 = 3574 J12.61 x 10-3 mol = 1370 x 10'J mol-'.A4omuustion = -1370 kJ mol-t
Workpi, oxiloLplr,Calculate the enthalpy of the reaction:
H
H
break (endo)C=C +612H-H +436
+1048 -1172
A4,,,rion = +1048 - 1172 = -124 kJ mol-'
ChecklistBefore attempting questions on this topic, check that you:
i Kno* the standard conditions for enthalpy changes.
J Kno* the signs of A,Hfor exothermic and endothermic reactions.
J Can define the standard enthalpies of formation, combustion andneutralisation.
l-l Understand an energy level diagram.
J Can define and use Hess's Law.li Know how to calculate values of A/{from laboratory data.iJ Can use bond enthalpies to calculate enthalpies of reaction.
and calculate the energy
the negative bond making
The answers to the numberedquestions are on pages 126-127.
A,H = +90.3 kJ mol-'
A,H = -57.L kJ mol-'
ffitry
E l o * o A N r c c H E M I s r R y , E N E R G E T T c s , K r N E T r c s A N D E e u r L r B R r u M
Testing your knowledge and understandingState the conditions used when measuring standard enthalpy changes.Give equations, with state symbols, which represent the followingenthalpy changes :
i the enthalpy of formation of ethanoic acid, CH,COOHQ)
ii the enthalpy of combustion of ethanoic acid
iii the enthalpy of neutralisation of an aqueous solution of ethanoicacid with aqueous sodium hydroxide.
Draw an enthalpy level diagram for the following sequence of reactions:
1
2
I'N,(g) + O,(g) -.' NO(g) +'t,O,
NO(g) +'1,O.(g)' NO,(g)
and hence calculate the enthalpy change for the reaction:'/,Nr(g) + or(g) * Nor(g)
4 Given that the standard enthalpies of formation of NOr(g) and NrOn(g)are +33.9 kJ mol-' and +9.7 kJ mol-' respectively, calculate the enthalpychange for the reaction:
2NO,(g) - N,O,(g)
5 The standard enthalpy of combustion of lauric acid,CH3(CHr),'COOH(s), which is found in some animal fats, is -7377 kImol-'. The standard enthalpies of combustion of C(s) and Hr(g) are -394kJ mol-' and-286 kJ mol-1.
Calculate the standard enthalpy of formation of lauric acid.6 100 cm' of L.00 mol dm-'HCI was added to 100 cm'of 1.00 mol dm-'
NaOH in a polystyrene cup, both solutions being initially at 2L.IO'C.On mixing the temperature rose to 27.9O oC. Determine the enthalpy ofneutralisation and state whether the reaction is exothermic orendothermic. You may assume that the polystyrene cup has a negligibleheat capacity, the solution has a density of L.00 g cm-t and that the finalsolution has a specific heat capacity of 4.18 J g-' oC-'.
7 Calculate the enthalpy change for the reaction:
CrH,(g) + HrO(g) - CHTCHTOH(g)given the following average bond enthalpies in kJ mol-':
When you write structural formulae,check that:oevery carbon atom has four bondsoevery oxygen has two bonds andoevery hydrogen and every halogen
have only one bond.
daLN
Introduction
0rganic chemistry I
i.J Organic compounds consist of a chain of one or more carbon atomsand contain functional groups (see Table 2.1). The functional groupgives the compound certain chemical properties. For instance the C=Cgroup (except in benzene rings) reacts in a similar way in allcompounds. Thus knowledge of the chemistry of ethene, H,C=CH,enables you to predict the reactions of all compounds containing theC=C group.
Toble 2.1 Functionol groups. *These will only be met at A2. R is o carbon-contoining group.
E you must learn the equations and conditions for the reactions in thespecification.
Substance Alcohol Aldehvde Ketone Acid Ester
*Acid
chloride
*
Amide Nitrile Amine
GroupI- C - O HI
H.tC:o
R/
R .tc:o
R ' /
. . O_ C
\oHarO
R_Cto-R'
, a O_ C
tcl
,ro_ C
\NHZ_ C = N
I-C-NHzI
€tfaLN Things to learn
o Homologous series: a series of compounds with the same functionalgroup, the same general formula, and where one member differs fromthe next by CHr.
r Emplrical formula: shows the simplest whole number ratio of theatoms present in one molecule.
I Molecular formula: shows the actual number of atoms present inone molecule, e.g. propane is CrHr.
o Structural formula: shows each atom in the molecule separately andhow it is bonded. A more condensed way is to show the groupingsaround each carbon atom, e.g. propane can be written as CHTCH,CH3and propan-2-ol as CH'CH(OH)CH3.
fl Isomers: two or more compounds with the same molecular formula.
f Homolytic fission: when a bond breaks with one electron going toeach atom (forming radicals).
ff Heterolytic fission: when a bond breaks with the two electrons goingto one atom.
E Substitution: a reaction in which an atom or group of atoms in onemolecule is replaced by another atom or group of atoms.
[l laaition: a reaction in which two molecules react together to form asingle product.
There are three structural isomers offormula CrHu0:o propan-1-ol CH3CH,CH,OHr propan-2-ol CH'CH(0H)CH3I methoxyethane CH3OCH,CH3
E " R G A N r c
c H E M r s r R y , E N E R G E T T . ' , K r N E T r c s A N D E e u r L r B R r u M
U nnmination: a reaction in which the elements of a simple moleculesuch as HBr, HrO etc. are removed from the organic molecule and notreplaced by any other atom or group of atoms.
[l nyArolysis: a reaction in which water (often catalysed by aqueous acidor aqueous alkali) splits an organic molecule into two compounds.
f Free radical: a species which has a single unpaired electron, e.g. Cl'
f Nucleophile: a species which seeks out positive centres, and musthave a lone pair of electrons which it donates to form a new covalentbond.
[l nlectrophile: a species which seeks out negative centres, and accepts alone pair of electrons to form a new covalent bond.
Things to understand
Isomerism
O Structural isomers may have:
i different carbon chains (straight or branched)
tt the functional group in different places in the carbon chainiii different functional groups.
O Geometric lsomerism
I Geometric isomerism occurs when there is a C=C in the molecule,and both carbon atoms have two different atoms or groups attached.
ii Geometric isomers are not easily interconverted because of thedifficulty of rotating about the n bond.
Alkanes (e.9. methane, CHJ
O Alkanes burn in excess oxygen to form carbon dioxide and water:
CHn + 2Or-' CO, + zHzO
O Alkanes react with chlorine (or bromine) in the presence of UV light ina stepwise substitution reaction:
CHn + Clr- CH3CI + HCI
CH3CI+Cl r -CHrC l r+HCl
Nkenes (e.g. propene, CIITCH=CH,)
O Alkenes add hydrogen:
CH3CH=CH, + H, - CH:CHTCH'
conditions: pass gases over a heated nickel catalyst
O Alkenes add bromine (or chlorine):
CH3CH=CH, + Br, - CHTCHBTCHTBT
conditions: at room temperature with the halogen dissolved in hexane.
O Alkenes add hydrogen halides, e.g. HI
CH3CH=CH, + HI' CH3CHICH3conditions: mix gases at room temperature.
cHa cH,\ /
c i s C-C/ \
H H
cHs H\ /
trans C - C/ \
H CHs
Ftg 2.4 Geometric isomers ofbut-2-ene
The decolourisation of a brownsolution of bromine is a test for aC=C group in a molecule.
With unsymmetrical alkenes, thehydrogen goes to the carbon thatalready has more hydrogen atomsdirectly bonded to it. This isMarkovnikov's rule.
@ o R G A N r c c H E M r s r R Y I
lLIIH H
t lC - C
l lH H
O Alkenes are oxidised by potassium manganate(Vll) solution:
CH,CH=CH, + [O] + HrO - CH'CH(OH)CHTOH
conditions: at room temperature, when mixed with sodium hydroxidesolution it produces a brown precipitate.
o Alkenes can be polymerised. Ethene forms poly(ethene) (Figure 2.5)and propene forms poly(propene) (Figure 2.6):
n-CHr-Qlf, + -(CH r-CHr)-,conditions either: a very high pressure (about 2000 atm)and a temperature of about 250 "C
or: a catalyst of titanium(Iv) chloride and triethyl aluminiumat 50 oC and pressure of about L0 atm.
Halogenoalkanes (e.9. 2-iodopropane, CHTCHICH,)
o They substitute with aqueous sodium (or potassium) hydroxideto give an alcohol:
CH3CHICH, + NaOH - CH.CH(OH)CH. + NaI
conditions: boil under reflux with aqueous sodium hydroxide.
o They eliminate with ethanolic potassium hydroxide to give analkene:
CH3CHICH, + KOH - CH,CH-CH'+ KI + H,O
conditions: boil under reflux with a concentrated solution of potassiumhydroxide in ethanol, and collect the gaseous propene over water.
o They substitute with potassium cyanide:
CH3CHICH. + KCN' CH,CH(CN)CH, + KI
conditions: boil under reflux with a solution of potassium cyanide in amixture of water and ethanol.
O With ammonia they form amines:
CH3CHICH, + 2NH, - CH,CH(NHr)CH, + NH'I
conditions: heat excess of a concentrated solution of ammoniain ethanol with the halogenoalkane in a sealed tube.
o The test for halogeno compounds:
1 Warm the substance with aqueous sodium hydroxide.2 Add dilute nitric acid until the solution is just acidic.3 Add silver nitrate solution.
Result:t white precipitate, soluble in dilute ammonia indicates chloro-
compound
O cream precipitate, soluble in concentrated ammonia indicates bromo-compound
o yellow precipitate, insoluble in concentrated ammonia indicates iodo-compound.
AIcohols
Primary (1') alcohols contain the CHrOH group.
Secondary (2') alcohols have two C atoms attached to the CHOH group.
Tertiary (3') alcohols have three C atoms attached to the COH group (seeFigare 2.7).
Fig 2.5Poly(ethene)
Ftg 2.6Poly(propene)
The rate of all these reactionsincreases C-Cl to C-Br to C-|,because the bond enthalpy andhence the bond strength decreasesC-Cl to C-Br to C-|. The weaker thebond the lower the activation energyand hence the faster the reaction.
H
I- C - O H
H - c -1 0 l l- c - c - o H
t l- C -
H
l i 2 0- c - c - o H
t l- c -
I3o
Ftg 2.7 1", 2" and 3" alcohols
ln equations in organic chemistry,an oxidising agent can be written as[0] and a reducing agent as IH], butthe equation must stil l balance.
Note that this reaction works for allalcohols, and isomers of the alkenemay be produced.
This is a test for an 0H group.Steamy acidic fumes are given offwhen phosphorus pentachloride isadded to the dry organic substance.
E } R G A N r c c H E M r s r R y , E N E R G E T I . ' , K t N E T t c s A N D E e u t L t B R t u M
o Oxidation reaction with aqueous orange potassium
dichromate(Vl), KrCrrO, acidified with dilute sulphuric acid.
a 1' alcohols are oxidised via an aldehyde to a carboxylic acid. The
solution turns green ([Cr(HrO)u]'. ions formed).
CH,CH,OH + [O] - CH,CHO + H,O
then CH3CHO + [O] - CH,COOH
conditions: to stop at the aldehyde: add potassium dichromate(Vl)in dilute sulphuric acid to the hot alcohol and allow the aldehyde to
distil off.
to prepare the acid: boil the alcohol with excess acidified oxidising
agent under reflux.
b 2" alcohols are oxidised to a ketone. The solution turns green.
CH3CH(OH)CH3 + [O] 'CH.COCH, + H,O
conditions: boil the alcohol and the acidified oxidising agent underreflux.
c 3'alcohols are not oxidised and so do not turn the solution green.
o Dehydration of an alcohol to an alkene:
CTH'OH-HrC=CHr+HrO
conditions: add excess concentrated sulphuric acid (or concentratedphosphoric acid) to the alcohol and heat to 170'C.
o Ilalogenation Io,2" and 3'alcohols react with:
1. Phosphorus pentachloride, PCI' to form a chloroalkane:CTH.OH + PCl, - CrHrCl + HCI + POCI,
conditions: at room temperature and the alcohol must be dry.
Solid sodium bromide and 50o/o sulphuric acid to give abromoalkane:
NaBr + HrSOn -- HBr + NaHSO,
then HBr + CTH'OH - CrHrBr + HrO
conditions: add the sulphuric acid mixed with the alcohol to solid
sodium bromide at room temperature.
Phosphorus and iodine to give the iodoalkane:2P + 3lr- 2Pl,
then 3CrHrOH + PI, -- 3CrHrI + H.PO,
conditions: add the alcohol to a mixture of moist red phosphorus
and iodine at room temperature.A summary of reactions is shown in Figure 2.8.
@ o R c n r u r c c H E M r s r R Y I
zCOz+ 3HrO
o2Spark
cH3cH2cN
Fig 2.8 Summory oforgonic reoctions
The percent yield is less than 100%because of competing reactions andhandling losses.
CH2BTCHTBT cH2oHCH2OH cH3cooH
KMnOn Boil under refluxKtCtrOrlHrSOn(aq)NaOH(aq)
conc H2SO4
cH3cH2NHz
Bonding and reactivity
Bond strength is the dominant factor that determines reactivity.
a A n-bond between two carbon atoms is weaker than a o bondbetween two carbon atoms. Thus alkenes are more reactive thanalkanes and react by addition, whereas alkanes react by substitution.
b A C-I bond is weaker than a C-Cl bond, and so iodoalkanes arerapidly hydrolysed by aqueous sodium hydroxide, unlikechloroalkanes which react very slowly.
Quantitative organic chemistry
un This involves calculation of empirical formulae from percent data (seepage 6).
w This involves calculation of percentage yield.
First calculate the theoretical yield from the equation (see page B),
then the o/o yield = actual yield in grams x 10oo/o
theoretical yield in grams
Applied organic chemistry
* Liquid versus gaseous fuels. You should consider the followingpoints:
L the ease with which gaseous fuels can be piped into the home2 the easier handling of liquid fuels at a filling station for cars3 the extent and type of pollution produced:
a the quantity of carbon dioxide and any other greenhouse gasesproduced per kilojoule of energy
b the emission of oxides of nitrogen and sulphur, and the way inwhich these can be limited
c the emission of particulates
4 the energy produced per unit mass of fuel for aeroplanes5 the extent to which the world's supply of fossil fuels is limited.
41,
Brz
CrzW
at 100'C at l7O"C
NaOH
NH, in ethanol
cH3cH2cl
DDT is a chlorine-containingpesticide and it has eradicatedmalaria from some countries, but itsoveruse has led to the destruction ofwildl i fe. l t is a very inert chemical(owing to the strength of the C-Clbond) and so persists in anorganism, being stored in fattyt issue.
ethene and but-1 -ene
2-chlo ro-3-methylpentan-1 -ol
E " R G A N r c
c H E M r s r R y , E N E R G E T T . ' , K r N E T r c s A N D E e u r L r B R r u M
a Polymers. You should realise that simple polyalkenes are resistant tobreaking down under environmental conditions. This is because of thestrength of the C-C and the C-H bonds.
* Organic halogen compounds. There are three main uses of these:
1 As polymers. PVC is used as a waterproofing material, as anelectrical insulator and as a stable and maintenance free material forwindow frames. When burnt it produces harmful fumes ofhydrogen chloride. PTFE is used as a non-stick coating for saucepansetc. These polymers are very stable owing to the strength ofcarbon-halogen and C-C bonds.
2 As herbicides. Complicated chlorine compounds, such as 2,4-D and2,4,5-T are used as selective weedkillers. They are stable and sopersist in the environment.
3 Chlorofluorocarbons (CFCs) make excellent refrigerants, but theyare so stable in air and water that they diffuse to the stratospherewhere they are broken down by light energy and form chlorineradicals. These catalyse the destruction of ozone. New substanceshave been developed which contain hydrogen as well as chlorine,fluorine and carbon, and these are less stable and are broken downby atmospheric oxidation.
Checklist
Before attempting the questions on this topic, check that you:
tJ Can name simple organic molecules.
tf Can write structural formulae of structural and geometric isomers.
lJ Know the reactions of alkanes with air and halogens.
Ll Know the reactions of alkenes with hydrogen, halogens, hydrogenhalides, potassium manganate(Vll) and in polymerisation.
[J Kno* the reactions of halogenoalkanes with potassium hydroxide (bothaqueous and ethanolic), potassium cyanide and ammonia.
[f Kno* the oxidation, dehydration and halogenation reactions ofalcohols.
'J Can relate reactivity to bond strength and polarity and to the stabilityof intermediates.
if Can calculate the empirical formula of a compound.
iJ Understand the advantages and disadvantages of the use of organiccompounds.
Testing your knowledge and understanding
For the first set of questions, cover the margin, write your answer, thencheck to see if you are correct.
o Which of the following compounds are members of the samehomologous series:
The answers to the numberedquestions are on pages 127-128.
O What product is obtained if a large excess of chlorine is mixed withmethane and exposed to diffused light?
O A compound X contained 54.5olo carbon, 36.40/o oxygen and9.Io/ohydrogen by mass. Calculate its empirical formula.
1 Write the structural formulae of:
a 1, l-dibromo-I,2-dichloro-2,3-difluoropropane
b 1-chlorobutan-2-ol.
2 Write out the structural formulae of the isomers of:
a C3H8O (alcohols only)
b CrH,,
c CoH, (no cyclic compounds).
3 Define:
a free radical
b homolytic fission.
4 Define:
a an electrophile
b heterolytic fission.
5 Write equations and give conditions for the reaction of:
a ethane + chlorine
b ethane + oxygen.
6 Write equations and give conditions for the reaction of propene with:
a hydrogen
b bromine *
c hydrogen iodide
d potassium manganate(Vll) solution *
For reactions marked *, state what you would see.
7 Write equations and name the products obtained when 2-iodopropaneis:
a shaken with aqueous dilute sodium hydroxide
b heated under reflux with a concentrated solution of potassiumhydroxide in ethanol
c heated in a sealed tube with a concentrated solution of ammonia inethanol.
8 Write the equation and give the conditions for the reaction ofpropan-Z-ol with sulphuric acid.
9 Write the structural formula of the organic product formed, if any, giveits name and say what you would see when:
a propan-l-ol is heated under reflux with dilute sulphuric acid andexcess aqueous potassium dichromate(Vl)
b 2-methylpropan-2-ol is heated under reflux with dilute sulphuricacid and aqueous potassium dichromate(Vl)
c phosphorus pentachloride is added to propan-2-ol.
1O Why does bromine react rapidly with ethene at room temperature butonly slowly with ethane?
Kinetics I
Things
O R C A I \ | I C C H E M t S T R y , E N E R G E T T C S , K I N E T I C S A N D E Q U I L I B R I U M
L1 Why does L-bromopropane react more slowly than l-iodopropane withwater?
12 When 5.67 g of cyclohexene, CuH,o, reacted with excess bromine, 15.4 gof CuH,oBrz was obtained. Calculate the theoretical yield of CuH,oBr, andhence the percentage yield of the reaction.
13 Explain why compounds such as CClrF, are harmful to theenvironment.
o The rate of a reaction is determined by:
1 the rate at which the molecules collide
2 the fraction of the colliding molecules that possess enough kineticenergy 'to get over the activation energy barrier'
3 the orientation of the molecules on collision.
to learn
[] ^Lctivation energy is the minimum energy that the reactantmolecules must have when thev collide in order for them to formproduct molecules.
[l Factors which control the rate of a reaction are:1. the concentration of a reactant in a solution
2 the pressure, if the reactants are gases
3 the temperature
4 the presence of a catalyst
5 the surface area of any solids
6 light for photochemical reactions.
to understand
Collision theory
o The effect of an increase in concentration of a solution, or theincrease in pressure of a gas, is to increase the frequency of collisionof the molecules, and hence the rate of reaction.
O The effect of heating a gas or a solution is to make the molecules orions:
1 move faster and so have a greater average kinetic energy
2 which increases the fraction of colliding molecules with a combinedenergy greater than or equal to the activation energy
3 which results in a greater proportion of successful collisions.
o The effect of heating can be shown by the Maxwell-Boltzmanndistribution (see Figure 2.9) of molecular energies at two temperatures T,and T, where T, is > 7,.
Things
An increase in temperature alsocauses an increase in the rate ofcollision, but this is unimportantcompared with the increase in thefraction of molecules with thenecessary activation energy.
@ K r N E r r c s I
Note that the peak has moved to theright for f,, and that the total areasunder the curves are the same, butthe area under the curve to the rightof E equals the fraction of themolecules with energy 2 E^, which isgreater for Trthan for I,.
Fig 2.9 Moxwell-Boltzmann distribution
Energy E^^. EL d t d
Fig 2.lO Effect of a catalyst
Fig 2.1| Energy profile diagrams
tttq,)
I(u
(J
f r .
(t)(u
Iq)
(J
Catalysts work by providing an alternative route with a loweractivation energy. Thus, at a given temperature, a greater proportion ofthe colliding molecules will possess this lower activation energy byfollowing the route with the catalyst, and so the reaction will be faster.This is shown on a Maxwell-Boltzmann diagram (see Figure 2.lO).
The enthalpy profile diagrams for an uncatalysed and a catalysedreaction are shown in Figure 2.I1.
Do not say that a catalyst lowers the activation energy.
Note that the A.Elvalues are the same for both paths.
Thermodynamic stability
o This is when the enthalpy level of the products is much higher than theenthalpy level of the reactants.
o Thus the substances on the left-hand side of a very endothermicreaction are said to be thermodynamically stable relative to those onright.
o The substances on the left-hand side of a very exotherrnic reactionsaid to be thermodynamically unstable relative to those on the right.Whether a reaction will then take place depends upon its kineticstability.
Kinetic stability
If a reaction has such a high value of activation energy that no moleculespossess sufficient energy on collision to react, the system is said to bekinetically stable. An example is a mixture of petrol and air, which is
r r l
the
are
:------- Transi t ion
Uncatalysed
Intermediate
The answers to the numberedquestions are on page 128.
E " R G A N r c
c H E M r s r R y , E N E R G E T T . ' , K r N E T r c s A N D E e u I L r B R r u M
thermodynamically unstable but kinetically stable. No reaction occursunless the mixture is ignited.
ChecklistBefore attempting the questions on this topic, check that you:iJ Can recall the factors which effect the rate of reaction.
J Can explain these rate changes in terms of collision theory.
[i Can draw the Maxwell-Boltzmann distribution of molecular energies attwo different temperatures.
,J Can use this to explain the effect of a change of temperature and theaddition of catalyst.
J Understand the concepts of thermodynamic and kinetic stability.
Testing your knowledge and understandingFor the following questions, cover the margin, write your answer, thencheck to see if you are correct.O For a reaction involving gases, state three factors that control the rate of
the reaction.
O For a reaction carried out in solution, state three factors that control therate of the reaction.
e For the reaction of a solid with a gas or with a solution, state one otherfactor that controls the rate of reaction.
o Give an example of a solid catalyst used in a gas phase reaction.
L Define
i activation energy
ii catalyst.
2 Explain, in terms of collision theory, why changes in temperature andin pressure and the addition of a catalyst alter the rate of a gas phasereaction.
3 Draw the Maxwell-Boltzmann distribution of energy curve for a gas:
i at room temperature (mark this T,) and
ii at 50 "C (mark this [).
Now mark in a typical value for the activation energy of a reaction thatproceeds steadily at room temperature.
Explain, in terms of activation energy, why animal products such asmeat and milk stay fresher when refrigerated.
Draw energy profile diagrams of:
i an exothermic reaction occurring in a single step
ii the same reaction in the presence of a suitable catalyst
iii a reaction where the reactants are thermodynamically stable
iv a reaction where the reactants are both thermodynamically andkinetically unstable.
Pressure, temperature, and catalyst
The surlace area of the solid
Fe (Haber), or V,Ou (Contact), or Nior Pt (addition of hydrogen to C=C)
@ G H E M T c A L E e u r L r B R r A I
Le Chatelier's principle states thatwhen one of the factors governingthe position of equilibrium ischanged, the position will alter insuch a way as if to restore theoriginal conditions.
Do not say'the side with a smallervolume', as a gas will always fill itscontainer.
Chemical equilibria I
<ffiW Introduction
Many reactions do not go to completion because the reaction is reversible.As the rate of a reaction is dependent on the concentration of thereactants, the reaction will proceed up to the point at which the rate of theforward reaction equals the rate of the reverse reaction, when there is nofurther change in concentrations. The system is then said to be atequilibrium.
ffi Things to learn and understandLry
Dynalnic equilibrium
O At equilibrium, the rate of the forward reaction equals the rate of thereverse reaction.
O Both products and reactants are constantly being made and used up,but their concentrations do not change.
O This can be demonstrated by using isotopes. For the reaction:
CH3COOH(I) + C,HsOH(l)+CH,COOC,H,(I) + H,O(l)
Mix the four substances in their equllibrium concentrations, buthave the water made from the isotope tto. After some time the "Oisotope will be found (by means of a mass spectrometer) in both theethanoic acid and the water, but the concentrations of the foursubstances will not have changed.
Effect of changes in conditions on position ofequilibrium
O Le Chatelier's principle may help you to predict the direction of thechange in the position of equilibrium, but it does not explain it.
O Temperature. An increase in temperature will move the position ofequilibrium in the endothermic direction. Likewise a decrease intemperature will move the equilibrium in the exothermic direction:
N,(g) + 3H,(g) =. 2NH,(g) LH - -92.4 kJ mol-'
As this reaction is exothermic left to right, an increase in temperaturewill cause less ammonia to be made, thus lowering the yield.
O Pressure. This applies only to reactions involving gases. An increase inpressure will drive the equilibrium to the side with fewer gas molecules.Thus, for the reaction above, an increase in pressure will result in moreammonia in the equilibrium mixture, i.e. an increased yield. This isbecause there are only two gas molecules on the right of the equationand four on the left.
O Concentration. This applies to equilibrium reactions in solution. If asubstance is physically or chemically removed from an equilibriumsystem, the equilibrium will shift to make more of that substance:
All industrial processes that involvepassing gases through a bed ofcatalyst must work above1 atmosphere pressure in order toforce the gases through the system.
The answers to the numberedquestions are on page 129.
E ' R G A N r c c H E M r s r R y , E N E R G E T T . ' , K r N E T r c s A N D E e u r L r B R r u M
2CrOn'z-(aq) + 2H.(aq) + Cr,O,'-(aq) + HrO(l)
Addition of alkali will remove the H'ions, causing the equilibrium tomove to the left.
O Catalyst. This has no effect on the position of equilibrium. What itdoes is to increase the rate of reaching equilibrium, thus a catalystallows a reaction to be carried out at a reasonable rate at a lowertemperature.
Optimum industrial conditions
Temperoture
Many industrial reactions, such as the Haber process for manufacturingammonia, are reversible and exothermic. For such reactions:o If a high temperature is used, the yield at equilibrium is small, but the
rate of reaction is fast.
r If a low temperature is used, the theoretical yield is higher, but the rateof reaction is slow.
O In systems such as this, a catalyst is used to allow the reaction toproceed rapidly at a temperature at which the yield is reasonably good.This is often called a compromise temperature, balancing yield withrate. Any unreacted gases are then separated from the products andrecycled back through the catalyst chamber.
Pressure
O High pressures are extremely expensive, and are only used if the yield atlower pressures is too small to be economic.
o Two examples of manufacturing processes that use very high pressuresare the Haber process and the polymerisation of ethene.
ChecklistBefore attempting the questions on this topic, check that you:
J Understand that equilibria are dynamic.
J Can deduce the effect of changes in temperature, pressure andconcentration on the position of equilibrium.
lJ Can predict the economic conditions for an industrial process.
ffi Testing your knowledge and understandingConsider the reversible reaction at equilibrium:
A(g) + B(g)
Which statements are/is true about this system?
i There is no further change in the amounts of A or B.
ii No reactions occur, now that it has reached equilibrium.
iii The rate of formation of B is equal to the rate of formation of A.
Consider the equilibrium reaction:
NrOn(B) + 2NO,(g) NI = +58.1 kJ mol-'
State and explain the effect on the position of equilibrium of:
@ I N D U ' T R T A L I N . R G A N T c c H E M I ' T R Y
ii What will happen to the position of equilibrium if thetemperature is increased from 25 "C to 35 'C?
White insoluble lead(Il) chloride reacts reversibly with aqueous chlorideions to form a colourless solution:
PbCl,(s) + 2Cl-(aq) + PbCl,'(aq)
State and explain what you would see when concentrated hydrochloricacid is added to the equilibrium mixture.
In the manufacture of sulphuric acid, the critical reaction is:
2SO,(g) + O,(g) -- 2SO,(g) Nf = -196 kJ mol-'
The reaction is very slow at room temperature. Why are conditions of725 K and a catalyst of vanadium(V) oxide used?
@ Industr ial inorganic chemistry
ffi InffoductionIndustrial chemists always need to keep manufacturing costs down. Theydo this by:
L making the reaction as fast as possible
2 ensuring a high yield
3 keeping the temperature and pressure as low as possible.
ffi Things to learn and understandLry
The Haber process for the manufacture of ammonia
N,(g) + 3H,(g) + 2NH,(g) NI=-92.4 kJ mol-'
O Because this reaction, left to right, is exothermic and the number of gasmolecules decreases, the yield and kinetics are affected according to thetable below:
O Thus the following conditions are found to be the most economic:
Temperafire: 400 "C. A higher temperature would reduce the yield,
and a lower one would make the reactionuneconomically slow.
Reaction conditions Effect on yield Effect on rate
Increase in temperature Decrease lncrease
Increase in pressure Increase Very slight increase
Addition of catalyst None Very large increase
Uses of ammonia:I manufacture of nitric acidt manufacture of fertilisers such as
ammonium nitrate and ureao manufacture of polyamides such
as nylon.
Uses of sulphuric acid:I the manufacture of fertilisers
such as ammonium sulphater the manufacture of phosphate
fertiliserr the manufacture of paints and
detergents.
Pressure:
Catalyst:
Yield per cycle:
Pressure:
Catalyst:YieId:
Manufacture of aluminium
O Aluminium is too reactive for its oxide to be reduced by carbonmonoxide or other cheap reducing agents, so the expensive method ofelectrolysis of a molten ionic compound has to be used.
o The ore contains aluminium oxide (amphoteric) with large impurities ofiron oxide (basic) and silicon dioxide (weakly acidic). The ore is treatedwith a hot 10olo solution of sodium hydroxide, which reacts with theamphoteric aluminium oxide to form a solution of sodium aluminate.Iron oxide does not react as it is a base, and silicon dioxide does not
E " R G A N r c
c H E M r s r R y , E N E R G E T T . ' , K r N E T r c s A N D E e u r L r B R r u M
The oxidation
200 atm. The yield at 1 atm is too low, and so a highpressure is necessary even though it is very expensive.Iron.This allows the reaction to proceed at a fast, andhence economic, rate at a moderate temperature.15o/o. The gases from the catalyst chamber are cooled inorder to liquefy the ammonia, and then the unreactednitrogen and hydrogen are recycled, giving a finalconversion of nearly lOoo/o.
of ammonia to nitric acid
This happens in three stages:1 The ammonia and air are passed over a platinum/rhodium catalyst at
900 "c:
4NH.(g) + 5O,(g) + 4NO(g) + 6H,O(g)2 On cooling, the nitrogen(Il) oxide reacts with more air:
aNO(g) + 2O,(g) -* 4NO,(g)
3 The nitrogen(IV) oxide and air are then absorbed into water:
4NO,(g) + O,(g) + 2H,O(l) + 4HNO,(aq)
The contact process for the manufacture of sulphuricacid
O This takes place in three stages:
1 The combustion of sulphur:
S( l )+O, (g ) -SO, (8 )
2 The reversible oxidation of the sulphur dioxide:
SOr(g) +'lrOr(g) + So,(g)3 Absorption by the water in 98o/o sulphuric acid.
I The conditions for stage two, which is exothermic and has fewer gasmolecules on the right, are:
Temperature: 425 "C. A higher temperature would reduce the yield,and a lower one would make the reactionuneconomically slow.2 atm. This is enough to force the gases through theplant. A higher pressure is not necessary because theyield is high under these conditions.Vanadium(v) oxide.960/o per cycle. The gases from the catalyst chamber arepassed into the absorber containing the 980/o sulphuricacid, and all the SO, is removed. The gases then goback through another bed of catalyst, giving a finalconversion of 99.8o/o.
I N D U S T R I A L I N O R G A N I C C H E M I S T R Y
Aluminium is protected fromcorroding by its layer of oxide whichreforms when it is scratched. lt isalso less dense than iron. Theseproperties make it ideal as a materialfor aeroplane wings and fuselageand for drink cans (even though it ismore expensive than iron). lt is alsoan excellent conductor of electricity,and so is used in overhead powercables because of these threeproperties.As the electrolytic method ofmanufacture is very expensive, it iseconomically sensible to recyclealuminium objects.
Chlorine is used for:i water sterilisationii the manufacture of organic
chlorine compounds, such as theplastic PVC and the herbicide2,4-D
iii the manufacture of HCl.
Sodium chlorate(l) is used:i as a domestic bleachii as a disinfectant.
react because of its giant atomic structure. These solids are filtered off,and carbon dioxide is blown through the solution precipitatingaluminium hydroxide. This is obtained by filtration and heated toproduce pure aluminium oxide.
o The purified aluminium oxide is dissolved in molten cryollte, NarAlFu,at 900'C.
O The solution is electrolysed using carbon anodes dipping into a steelcell lined with carbon, which is the cathode.
a At the cathode, aluminium ions are reduced:
A l ' *+3e- -A l ( l )
O The molten aluminium sinks to the bottom of the cell and is siphonedoff.
o At the carbon anode, oxygen ions are oxidised and react with theanode:
2O'- + C(s) -' CO'(g) + 4e-
o Because of the expense of this method of manufacture, it is economic torecycle aluminium drink cans.
The production of chlorine
o This is done by the electrolysis of an aqueous solution of sodiumchloride (brine).
At the titanium anode (+): NaCl contains Cl- ions and these areoxidised to Clr:
zCl-(aq) -Cl,(g) +2{
chlorine gas is produced.
o At the steel cathode (-): water is ionised in an equilibrium reaction:
H,O[) + H-(aq) + OH- (aq)
Na* ions are very hard to reduce, and so H* ions are preferentiallyreduced:
2Hr(aq)+2e--Hr(g).
The removal of the H- ions drives the water equilibrium to the right,producing OH ions. The overall equation for the reaction at thecathode is:
zHrO + 2e- - ZOtf(aq) + Hr(g)
Thus sodium hydroxide and hydrogen are produced.
The anode and cathode compartments are separated by an ion-exchangemembrane which allows Na. ions to pass through but keeps thechlorine separated from the hydroxide ions that are produced atthe cathode.
Manufacture of sodium chlorate(I)
If the electrolysis of sodium chloride solution is carried out with thesolution being stirred, sodium chlorate(I) and hydrogen are obtained. Thechlorine produced at the anode disproportionates when in contact withthe alkali from the cathode:
Cl,(aq) + 2OH-(aQ) - OCI-(aq) + Cl-(aq) + H,O(l)
The answers to the numberedquestions are on page 129.
O n c e N I c c H E M I S T R Y , E N E R G E T I C S , K I N E T I C S A N D E Q U I L I B R I U M
ChecklistBefore attempting questions on this topic, check that you:
."j Know the conditions for the manufacture of ammonia.
*i Know the conditions for the manufacture of nitric acid.
.-- Know the conditions for the manufacture of sulphuric acid.
.-; Can iustify the conditions in terms of the economics and the chemistryof these processes.
; Know some of the uses of ammonia, nitric acid and sulphuric acid.
-.,1 Know the conditions and electrode reactions used in the manufacture
of aluminium.
J Can recall the details of the production of chlorine and sodiumchlorate(l) and their uses.
lHo - q r r l-
,i.: . Testing your knowledge and understandingt-----!
For the following questions, cover the margin, write your answer, thencheck to see if you are correct.
o State the conditions used in the manufacture of ammonia.
O State two uses of ammonia.
o State the conditions used for the oxidation of ammonia in themanufacture of nitric acid.
O State the conditions used for the oxidation of sulphur dioxide in theContact process.
O State two uses of sulphuric acid.
o What property of aluminium oxide is the basis of its purification frombauxite?
f What is the essential condition used in the manufacture of aluminium?
O Write the equations for the reactions at the cathode and at the anode inthe manufacture of aluminium.
Explain the economic reasons for the choice of the conditions used inthe Haber process.
In the Haber process why is the gas mixture cooled after it has left thecatalyst chamber?
Write the equations for the reactions at the anode and at the cathode inthe manufacture of chlorine.
A temperature of 350 to 450 'C, apressure of 200 to 250 atm and aniron catalyst
To make fertllisers, and to makenylon (or nitric acid)
A temperature of 900'C and aplati n u m/rhod i u m catalyst
A temperature of 400 to 450'C, avanadium(V) oxide catalyst and apressure of 2 atm
To make fertilisers, and paints (ordetergents)
It is amphoteric.
Cathode Al'- + 3e- - Al
Anode: 20 ' -+C-COz+4e-
P R A G T I c E :
Practice Test: Unit 2Time allowed Lhr
All questions are taken from parts of previous Edexcel Advanced GCE questions.
The answers are on pages I29-I3O.
La Propene, C.Hu, and but-2-ene, CH.CH=CHCH3, are in the same homologous series. Explain the termhomologous series.
b Draw a representative length of the polymer chain of poly(propene)c i Draw the geometric isomers of but-2-ene.
ii Explain how geometric isomerism arises.(Total 8 marks)
llune 2001 Unit Test 2 question 2 & May 2002 Utrit Test 2 qtestion 5l
The rate of any chemical reaction is increased if the temperature is increased.Draw a diagram to represent the Maxwell-Boltzmann distribution of molecular energies at aa
btemperature T, and at a higher temperature Tr.Use your diagram and the idea of activation energy to explain why the rate of a chemical reactionincreases with increasing temperature.
Consider the following reaction scheme:
step 1
KMnOnin alkali
CrHrBr (maior product)SI
I aqueous NaOH
It
c3H8oP
| *.0 3 (oxidation)
VcH3cocH3
t3lr2lr2lt1I
t3l
t4l(Total 7 marks)
llune 2001 Unit Test 2 questiott 3)
r2ltutlI
r2lr2lr2lt3ltlI
(Total 14 marks)IMay 2002 Unit Test 2 question 4l
I2lt3I
CrHu
propene
III*a
a i Give the reagent and the conditions needed for step 1.ii Give the structural formula of S.
b i Give the structural formula of P.ii State the type of reaction in:
step 1the conversion of S to P.
c i Give the reagent and the conditions needed for step 2.ii Give the reagent and the conditions needed for step 3.
d Give the structural formula of compound Q.
4a State Hess's Law.b Define the term standard enthalpy change of combustion.c The equation for the combustion of ethanol in air is
CrHsOH[) + 3O,(g) - 2CO,(E) + 3H,O0)
E l o * . A N I c c H E M r s r R y , E N E R G E T T . ' , K r N E T r c s A N D E e u r L r B R r u M
And the structural representation of this is:
+ 3O:O # 2O-C:O + 3H-O -H
Calculate the enthalpy change for this reaction using the average bond enthalpy values given below.t3I
tt Draw and label an enthalpy level diagram to represent this reaction. I2lGotaf 1O marks)
lJune 200L Unit Test 2 question 5f
The reaction in the Haber Process that is used to produce ammonia is:N,(g) + 3Hr(g) : 2NH,(B) N:I=-92k1mol-'
i What temperature is used in the Haber Process? tl]ii Justify the use of this temperature. t3Ilii Name the catalyst used in the Haber process. tlliv How does a catalyst enable a reaction to occur more quickly? t2lAnother industrial process is the one which recovers chlorine from HCl, which often is a by-product inorganic preparations
H Ht l
H - C - C - O - Ht l
H H
5a
b
rhisisahomog.-X.:l13)oi'*!?.il,u1*11i1f,1;"3,?:1t'State the meaning of the termsi homogeneous
NI = -115 kJ mol-'
ii dynanic equillbriumState and explain the effect on the position of equilibrium of the reaction in b of:I decreasing the temperature.ti decreasing the volume of the reaction vessel
a Give the ionic equation for the reaction taking place at the anode.b Give the ionic equation for the reaction taking place at the cathode.c State which of these reactions is an oxidation process.d Explain why the anodes need to be replaced frequently.e Explain why an electrolyte of pure molten bauxite is not used.
(Total 14 marks)lMay 2002 question 3 Unit Test 2 & lanuary 2001 question 4 Module Test 2l
Aluminium metal is manufactured by a process in which purified bauxite, dissolved in molten cryolite,is electrolysed at 800'C. Graphite electrodes and a current of about 120 000 amperes are used.
tlIl2l
l2lr2l
tlItlItlltzlr2l
Gotaf 7 marks)lJune 2001 Unit Test 2 question 7f
[Paper total 6O marks]
Bond Average bond energy/kJ mol-' Bond Average bond energy/kJ mol-'C-H + 4L2 C-C + 348c-o + 360 o-H + 463Ot:O + 496 c:o + 743
s AS laborator/ MtryLaboratory chemistry Io The specification for these tests includes all of Units L and 2.
O Unit Test 38 will contain many of the calculations for AS level, and soTopic t.2 (pages 6 to 12) must be thoroughly revised.
ffi Unit Test 3A: Assessment of practical'Lb skilts I
iJ tfris is either internally assessed or a practical exam.
iJ Notes and textbooks are allowed in the tests.
J the practical exam will contain some quantitative work, probably atitration or an enthalpy change experiment.
J You should know the following gas tests:
o H, burns with a squeaky pop
I O, relights a glowing spill
o CO, turns lime water milky
o NH3 turns red litmus blue
O Cl, bleaches damp litmus; turns KBr solution brown
a NO2 brown gas which turns starch/iodide paper blue-black
I SO, turns acidified potassium dichromate(Vl) solution green.
E you should know the tests for the following ions:
o COr'- Add acid. Test for COr.
o HCO3- Add to almost boiling water. Test for COr.
o SOn'- Gives white precipitate when dil HCI and BaClr(aq) areadded to the solution.
o HSO4- Solution is acidic to litmus, then test as for SOn2- above.
o SO32- Add dil acid to solid and warm. Test for SOr.
o Halide ion: To solution add dil HNO3 then AgNOr(aq). Whiteprecipitate soluble in dil ammonia indicates chloride;cream precipitate insoluble in dil but soluble in concammonia indicates bromide; pale yellow precipitateinsoluble in conc ammonia indicates iodide.
a NO3- Heat solid. All nitrates give off Or. All but sodium andpotassium nitrates also give off NOr. Alternatively addaluminium powder and sodium hydroxide solution.Nitrates give off ammonia gas.
a NHn. Add dil NaOH and warm. Test for ammonia.
o Mg'. Gives white precipitate when dil ammonia is addedto a solution.
O You must know how to carry out a flame test and the colours obtained.
Ion Flame colour
Li' camine red
Na' yellow
K' lilac
Ca'* brick red
Sr'* crimson red
Ba"* pale green
C=C group Decolourisesbrown brominein hexane
C-OH group Steamy fumeswith PCl.
Organic halides Warm withsodiumhydroxidesolution. Acidifywith dilute nitricacid and thentest as for ionichalides as onpage 55
L A B o R A T O R Y C H E M I S T R Y I
ffi []nit Test 38: Laboratory chemistry I
J this is a written paper, taken by alt candidates.
J tt is designed to assess a candidate's ability (related to the topics inUnits 1 and 2) to:
evaluate information generated from experimentsdescribe and plan techniques used in the laboratory.
J Tests
o You should know the tests listed in Unit 3A above.
o You should also know the tests for alkenes, the OH group and thehalogen in halogenoalkanes.
o You should be able to deduce the identity of a compound from theresults of a series of tests.
J Techniques
You should be able to describe techniques used in:o titrations and enthalpy change measurements
O simple organic procedures such as distillation and heating underreflux.
[l Phnning
You should be able to:O Plan a series of tests to determine the identity of an inorganic or
organic compound.
O Describe how to make up a solution of known concentration fortitrations.
O Plan an experiment to determine the enthalpy of a reaction such asthe combustion of a liquid, or the neutralisation of an acid.
o Plan an experiment to follow the progress of a reaction in whichthere is a change in physical state, such as the production of a gas.
[l calculations
You should be able to calculate:o empirical formulae
O reacting masses
o results from titration data
o enthalpy changes from experimental data.
iJ Evaluation
You should be able to criticise:o an experimental plan or apparatus
o the results of an experiment in terms of significant figures, accuracyor experimental error etc.
[J sarety
You should be able to suggest specific safety precautions when a substanceis flammable, toxic or irritating.
I
ll
Practice Test: Unit 38
Time allowed t hr
All questions, except 5, are taken from parts of previous Edexcel Advanced GCE questions.
The answers are on pages l3O-I32.
1 Complete the table below.
Gas Reagents or test Observation expected for a positive result
Hydrogen Burning splint
Oxygen Glowing splint
Carbon dioxide
Sulphur dioxide Potassium dichromate(VI )solution acidif ied with dilutesulphuric acid.
Solution turns from orange to .
Moist blue litmus paper Turns red and is then bleached white
t6I(Total 6 marks)
llune 2002 Unit Test 38 question 2l
In a series of experiments to investigate the factors that control the rate of a reaction, aqueoushydrochloric acid was added to calcium carbonate in a conical flask placed on an electronic balance.The loss in mass of the flask and its contents was recorded for 15 minutes.
CaCO,(s) + 2HCl(aq) CaCl,(aq) + H,O(l) + CO,(g)Four experiments were carried out.. Experiments L, 3 and 4 were carried out at room temperature (20"C).o The same mass of calcium carbonate (a large excess) was used in each experiment.o The pieces of calcium carbonate were the same size in experiments 1, 2 and 4.
a The results of experiment I give the curve shown on the graph below.
Mass loss/g
1.00
0.50
15 Time/min
1.50
Experiment Calcium carbonate Hvdrochloric acid
1 Small pieces 50.0 cm' 1.00 mol dm-'z Small pieces 50.0 cm' 1.00 mol dm-' heated to 80 "C3 One large piece 50.0 cm' 1.00 mol dm-'4 Small pieces 50.0 cm'2.00 mol dm
El to "oRAroRY cH EM rs rRY I
i Explain why there is a loss in mass as the reaction proceeds. t2lii Explain the shape of the curve drawn for experiment 1. t2lDraw curves on the graph above to represent the results you would expect for experiments 2, 3and 4. tsli Calculate the mass of calcium carbonate that exactly reacts with 50.0 cm3 of 1.00 mol dm-'
aqueous hydrochloric acid. (Molar mass of CaCOr = 100 g mol-') t3lii Based on your answer to c part (i), suggest a suitable mass of calcium carbonate to use in the
experiments. Explain your answer. t2l(Total 1.2 marks)
[anuary 2002 Unit Test 38 question 2]
A student was required to determine the enthalpy change for the reaction between iron and coppersulphate solution. The student produced the following account of their experiment.
Suggest two improvements you would make to this experiment. Give a reason for each of theimprovements suggested. t4l
b In an improved version of the same experiment a maximum temperature rise of I5.2 oC occurredwhen reacting excess iron with 50.0 cmt of 0.500 mol dm-' aqueous copper sulphate solution.i Using this data and taking the specific heat capacity of all aqueous solutions as 4.181 g-' deg',
calculate the heat change.ii Calculate the number of moles of copper sulphate used.iii Calculate the enthalpy change for this reaction in kJ mol-'
A student carried out an experimentof limestone following his own plan.of the mean titre are given below.
tur2l
(Total 8 marks)[une 2002 Unit Test 38 question 4]
to find the percentage of calcium carbonate, CaCO, in a sampleThe student's account of the experiment, results and calculations
A piece of iron, mass about 3 g, was placed in a glass beaker. Then 50 cm'of 0.5 mol dm-'aqueouscopper sulphate solution was measured using a measuring cylinder and added to the beaker. Thetemperature of the mixture was measured immediately before the addition and every minute afterwardsuntil no further change took place
Fe+CuSOo - FeSOo +Cu
Timing Beforeaddition
1 min 2 mins 3 mins 4 mins 5 mins
Temperature/'C 1 6 27 29 26 24 22
AccountI Mass of piece of limestone = 5.24 gll A pipette was used to transfer 50 cm'of 2.00 mol dm-'aqeous hydrochloric acid (an excess) to a
100 cm' beaker. The piece of limestone was placed in the beaker and left until there was no moreeffervescence.
Ill The acidic solution in the beaker was filtered into a250 cm3 volumetric flask. A small amount of thesolid impurity remained in the filter paper. The solution in the volumetric flask was carefully made upto 250 cm'with distil led water.
lV A pipette was used to tranfer 25.0 cm3 portions of the acidic solution to conical flasks. The solutionwas then titrated with 0.100 mol dm-'aqueous sodium hydroxide.
The accuracy of the student's method was judged to be poor by his teacher. The teacher suggested thatthe procedure in II could be improved, and that the titres used to calculate the mean were incorrectlychosen.i Suggest, with a reason, one improvement to the student's procedure in II.ii Recalculate a value of the mean making clear which titres you choose.
b i Using your answer to a part (ii), calculate the amount in moles of sodium hydroxide in the meantitre. tlI
ii Hence state the amount in moles of hydrochloric acid in a 25.0 cm' portion of the acidic solutiontransferred in IV. tllHence calculate the amount in moles of hydrochloric acid remaining after the reaction in II. tllCalculate the amount in moles of hydrochloric acid transferred to the beaker in II. tUHence calculate the amount in moles of hydrochloric acid used in the reaction. tl]Hence calculate the amount in moles of calcium carbonate and the mass of calcium carbonate inthe sample of limestone lM, (CaCOr) = 100) I2l
vii Hence calculate the percentage of calcium carbonate by mass in the sample of limestone. tUThe burette used in the titrations had an uncertainty for each reading of +0.05 cm'.i Which of the following should be regarded as the actual value of the titre in titration 3?
A between 14.80 and 14.90 cm':B between L4.825 and 14.875 cm3:C between 14.75 and 14.95 cm'.
ii Suggest one reason why a student may obtain volumes outside the uncertainty of the burette whencarrying out a titration. tll
(Total L4 marks)fJune 2001 Unit Test 38 question 4]
Propan-Z-ol (molar mass 60 g mol-1 and boiling temperature 82 "C) can be prepared by the reactionbetween 6.15 g of 2-bromopropane (molar mass I23 g mol-' and boiling temperature 59 "C) and excessaqueous 2.0 mol dm-' sodium hydroxide. This reaction is slow at room temperature.Describe the procedure, identifying the apparatus that you would use, to prepare a pure sample ofpropan-2-ol from 6.15 g of 2-bromopropane. tsl
b Calculate the minimum volume of sodium hydroxide solution that must be taken to ensure complete
lll
lv
v
vt
I2lr2l
tlI
reaction of the 2-brompropane.Your teacher suggested that an 80o/o yield would be an excellent result. Calculate the mass of purepropan-Z-ol that you would need to prepare to obtain this yield.Suggest one reason why your yield will be below 1,00o/o.
(Total 1O marks)
r2l
r2ltu
4 ?erinry, qMttatun'e4 rulib r rn, arud, Win nAl
lroryMnyEnerget ics l l
H \C ' I- .l Introduction
.a- t(H
J This topic extends the use of Hess's Law to reactions involving ions.
iJ yon must be able to construct a Born Haber cycle.
f] Lattice energy is crucial to the understanding of ionic bonding, becauseit is this release of energy that makes the formation of an ionicsubstance thermodynamically favourable.
fl Oefinitions of enthalpy changes must include:i the chemical change taking placeii the conditionsiii the amount of substance (reactant or product)iv an example of an equation with state symbols, as this may gain
marks lost through omission in your word definition.
Things to learn
,J fne sign of the value of AIltells you the direction of the movement ofheat energy:o positive (endothermic) for heat flowing into the system
o negative (exothermic) for heat flowing out of the system.
[J nntnalpy of atomisation, A.f,Iu, is the enthalpy change for theproduction of one mole of atoms in the gas phase from theelement in its standard state. It is always endothermic.
;J nntnApy of hydration, A.f,foru, is the enthalpy change when onemole of gaseous ions is added to excess water. It is always exothermic.
[i fattice energy, LII""', is the energy change when one mole ofionic solid is made from its separate gaseous ions. It is alwaysexothermic.
<ffiLH Things to understand
Born-Haber cycle
o This relates the enthalpy of formation of an ionic solid to theenthalpies of atomisation of the elements concerned, the ionisation
a,H^tor chlorine is the enthalpychange for:%Cl,(g) - Cl(g)AHn,o for the Na- ion is the enthalpychange for:Na-(g) + itQ--+ Na-(aq)Af{.n for sodium chloride is theenthalpy change for:Na-(g) + Cl-(g) -- NaCl(s)
O E N e R G E r r c s t l
lf a bond is partially covalent, theexperimental lattice energy (fromBorn Haber) will be bigger (moreexothermic) than the valuecalculated from a purely ionicmodel.The greater the difference betweenthe experimental and theoreticalvalues, the greater the extent ofcovalency. This will happen if thecation is very polarising (small orhighly charged) or the anion is verypolarisable (large). See page 13.
energies for the formation of the cation, the electron affinity for theformation of the anion and the lattice energy of the ionic substance.
O This can either be drawn as a Hess's Law cycle, or as an enthalpy leveldiagram (see Figures 4.1 and 4.2).
Mg'.(g) + 2cl(g)
4,tizcr(g) |
l * ll * . 1I N I f I
t osnI vg2.(g) + ctr(gt
NrzMg(S) + Clz(g)
+ Clz(g)-MgClr(s)
Fig 4.2 Enthalpy level diagram
AH, of MgCl, = NI^ of Mg (AH,) + lst ionisation energy of Mg (NIr) + 2nd ionisation energv of Mg (aH,) + 2 xA,F / "o f C l (AHr) +2xe lec t ronaf f in i t yo f C l (AHr)+A/ { , , , ,o f MgCl r (AF l r )=AH, +A,H.+AH,+ lH,+ lH.+- \H"
Any one value can be calculated when all the others are known.
Lattice enerry
O Lattice energy is always exothermic.o Its value depends upon the strength of the force of attraction
(sometimes called the ionic bond) between the ions.O The strength of this force depends upon the value of the charges and
the sum of the radii of the ions:i the larger the charge on either or both of the ions, the larger (more
exothermic) the lattice energy
ii the larger the sum of the ionic radii, the smaller (less exothermic)the lattice energy.
o Theoretical values of lattice energy can be calculated assuming that thesubstance is 100o/o ionic.
Voriotion of lottice energies in o Group
O Group 2 sulphates. As the radius of the cation increases down theGroup, the lattice energy decreases. But because the radius of thesulphate ion is much larger than the radii of any of the Group 2cations, the sun of the radii of the cation and the anion alters onlyslightly, and so the lattice energy only decreases slightly.
O Group 2 hydroxides. As the radius of the cation increases down theGroup, the lattice energy decreases. But because the anion is small, andmatches the size of the cation (the ionic radii of Ba'* and OH- are aboutthe same), the surn of the radii alters significantly, and so the latticeenergy decreases by a large amount.
Mg-(g) + Clr(g)+ zCf (g)
MgClr{s)
NIl
Mg(s) + Clr(g)
You must not use aHhyd whenreferring to a compound. lt refers toseparate ions. The enthalpy changefor NaCl(s) + ?Q + Na+(aq) + Cl-(aq)is called the enthalpy of solution.
Note that both lattice and hydrationenthalpies are negative numbers, so-Aff.n is positive
E l r r R r o D r c r r y , q u A N T r r A T r v E E e u r L r B R r A A N D F U N c T t o N A L G R o u p c H E M I s T R y
Enthalpy of hydration
o Enthalpy of hydration is always exothermic.O For a cation it is the result of the force of attraction between the ion
and the 6 oxygen in the water.O For an anion it is the result of the force attraction between the ion and
the 6- hydrogen in the water.I The value of enthalpy of hydration will depend upon the value of the
charge on the ion and its radius:i the larger the charge, the larger (more exothermic) the enthalpy of
hydration,
ii the larger the ionic radius, the smaller (less exothermic) theenthalpy of hydration.
Vorlotion of entholpies of hydrotion in o Group
The value will become less exothermic as the radius of the cation increasesdown the Group.
Solubility
The direction of a chemical reaction is partially determined by the value ofthe enthalpy change. When discussing solubilities, it is assumed that themore exothermic the enthalpy of solution, the more soluble the substanceis likely to be. The enthalpy of solution can be estimated from aconsideration of lattice energies and enthalpies of hydration of the twoions as can be seen from a Hess's law diagram.
M-(g) + rG)
l tl*'a l*ul t
MX(s) + aq (aq) x-(aq)
A H,"rn = - (lattice energy) + (hydration enthalpy of each cation + hydration enthalpy of each anion)
o Solubility of Group 2 sulphatesi The enthalpy of hydration becomes much less as the cations
become bigger.
ii Because the radius of the sulphate ion is much bigger than theradius of any of the cations, the lattice energy only decreasesslightly - see above.
iii (i) and (ii) together mean that the enthalpy of solution becomesconsiderably less exothermic and causes the solubility of thesulphates to decrease down lhe Group.
r Solubtlity of Group 2 hydroxidesi The enthalpy of hydration becomes less as the cations become
bigger.
ii Because the radius of the hydroxide ion is a similar size to that ofthe cations, the lattice energy decreases considerably.
iii (i) and (ii) together mean that the enthalpy of solution becomesconsiderably more exothermic and causes the solubility of thehydroxides to increase down the Group.
O E N e R G E T T c s r l
The answers to the numberedquestions are on page 132.
Checklist
Before attempting questions on this topic, check that you:
:J Can define enthalpy of atomisation.
iJ Can define enthalpy of hydration.
;J Can define lattice energy.
iJ Can construct a Born-Haber cycle.
IJ Cun use a Born-Haber cycle to calculate lattice energy or electronaffinity.
[] U.tdetstand why the value of AF/** from Born-Haber might be moreexothermic than the theoretical value.
[,] Can state and explain the change in solubilities of the Group 2sulphates and hydroxides.
#--xG T
L v v e
For the following questions, cover the margin, write your answer, then
check to see if vou are correct.
I In each of the following state which compound has the larger (more
exothermic) lattice energy:a MgCO. or BaCO,b NarCO. or MgCO,c NaF or NaCl.
I Arrange the following ions in order of decreasing (less exothermic)
enthalpy of hydration: Na*, Mg'*, Bat* and Alt*
o State the formula of the most soluble Group 2 hydroxide.
The following data, in kJ mol-', should be used in this question.
Enthalpies of atomisation: calcium +193; chlorine +LZL.
Ionisation energies for calcium: 1st +590; 2nd +1150
Electron affinity for Cl(g): -364
Lattice energies: for CaClr(s) -2237; for CaCl(s) -650 (estimated).
a Construct a Born-Haber cycle for:
i the formation of CaCl(s)ii the formation of CaClr(s).
b Calculate the standard enthalpy of formation of:
i CaClr(s)ii CaCl(s)iii hence calculate the enthalpy of the reaction:
2CaCl(s) * Ca(s) + CaCl'(s)
and comment on the fact that CaCl(s) does not exist.
Comment on the values of the lattice energies, in kJ mol-', given
below:
Experimental Theoretical
NaF -918 -912
MgI, -2327 -t944
a MgCO,b MgCO,C NAF
A l * > M g ' . r B a ' * > N a *
Ba(0H),
E I " r * r o D r c r r y ,
e u A N T r r A T r v E E e u r L r B R ' A A N D F U N . T T ' N A L G R o u p c H E M I s r R y
The Periodic Table l l'&
Inffoduction
;J you will need to look again at Topic 1.4.
iJ fne elements become more metallic down a Group. This means thattheir compounds, such as their chlorides, become more ionic and theiroxides become more basic.
f fn. elements become less metallic from left to right across a Period.This means that their chlorides become more covalent and their oxidesbecome more acidic.
[-] There are many formulae and equations that must be learnt inthis Topic.
[l fne properties of elements and their compounds change steadily fromthe top to the bottom of a Group and from left to right across a Period.
Things to learn and understand
Variation of properties across Period 3
A table is given below showing the formulae of the products of thereactions of the elements with oxygen, chlorine and water:
Notes: (') with excess oxygen sodium peroxide is formed, ''' aluminium chloride is covalent, (" with excesschlorine, (n) with steam.
The hydroxides become less basic andmore acidic as the surface chargedensity of the cation increases.
Silicon dioxide is such a weak acid thatmolten sodium hydroxide is required.The oxides of the non-metals formincreasingly strong acids as thenumber of oxygen atoms in the acidincreases.
@ T H E P E R r o D r c r A B L E r l
lonic chlorides dissolve in water toform hydrated ions.These hydrated ions maydeprotonate, if the surface chargedensity of the metal ion is high.Covalent chlorides (of Period 3)react with water to form HCI andeither the oxide or the oxoacid.
I SO, weakly acidic: SO, + HrO --* H2SO3 + H.(aq) + HSO.-t SO,, strongly acidic: SO. + HrO -- HzSOn -- H'(a9) + HSO,-
iJ cntorideso NaCl ionic solid, dissolves in water:
NaCl(s) f ?q--+ Na-(aq) + Cl-(aq)o MgCl, ionic solid, dissolves in water:
PCl,( l ) + 3H.O(l)-- H,PO,(aq) + 3HCl(aq), PCl. covalent solid, reacts with water:
PCl.(s) + 4HrO(l) - H,PO.,(aq) + SHCl(aq)
Variation of properties down Group 4
-l The metallic character increases down the Group:carbon and silicon are non-metallic (not malleable; only graphiteconducts electricity)germanium is semi-metallic (semi-conductor).tin and lead are metallic (malleable and electrical conductors).
iJ fne +2 oxidation state becomes more stable relative to the +4state. Lead(IV) is oxidising and is itself reduced to lead(Il), whereastin(Il) is reducing and is itself oxidised to tin(IV). This is illustrated by:o lead(IV) oxide oxidises concentrated hydrochloric acid to chlorine:
PbO, + 4HCl - PbClz + Cl, + HrOo tin(Il) ions are oxidised to tin(IV) ions by chlorine:
Sn'* + Cl, -- Sn'. + zcrlJ Tetrachlorides. Both CCl, and SiCln are covalent and are tetrahedral
molecules owing to the repulsion of the four bond pairs of electrons.SiCl4 is rapidly hydrolysed by water. A lone pair of electrons from theoxygen in the water forms a dative bond into an empty 3d orbital inthe silicon atom. The energy released is enough to overcome theactivation energy barrier involved in the breaking of the Si-Cl bond.Carbon's bonding electrons are in the 2nd shell and there are no 2dorbitals. The empty orbitals in the 3rd shell are of too high an energy tobe used in bonding, and equally important is that carbon is such asmall atom that water molecules are prevented from reaching it by thefour much larger chlorine atoms arranged tetrahedrally around it.
if .q.ciaity of the oxides. This decreases down the Group:
o Carbon dioxide is weakly acidic, and reacts with metal oxides andalkalis:
CO,(g) + 2OH-(a9) - CO,'-(aq) + H,O(l)O Silicon dioxide is very weakly acidic, and reacts with molten sodium
The trend in metallic character iscaused by the decrease in ionisationenergy as the radius of the atomincreases, making it energeticallyfavourable for the formation of ionicbonds.
PbO and Pb0,
Pb0
The answers to the numberedquestions are on pages1 32-1 33.
E I t r R r o D r c r r y , e u A N T r r A T r v E E e u T L T B R I A A N D F U N . T I . N A L G R o u p c H E M I s T R y
Before attempting the questions on this topic, check that, for the Period 3
elements, you:
i Can write equations for the reactions of the elements with oxygen and
with chlorine.
J Can write equations for the reaction of the elements with water.
J Can state and explain the acid/base nature of the metal hydroxides.
J Can state and explain the acid/base nature of the non-metal oxides.
J Know the formula of their chlorides and their reactions with water, and
can relate the reactions to their bonding.
Before attempting the questions on this topic, check that, for the Group 4
elements, you:
iJ Kno* why the elements become more metallic down the Group.
[J Kno* that the +2 oxidation state becomes more stable down the
Group.
l-f Know why SiCln is rapidly hydrolysed by water, whereas CCln is
unreactive with water.
E Can write equations to show the acid/base character of the oxides of
carbon, silicon and lead.
-.________.-O , t 1 7 i --]fl,
Testing your knowledge and understandinga . t ' -
For the following questions, cover the margin, write your answer, thencheck to see if you are correct.
o State the formulae of lead(Il) oxide and lead(IV) oxide.
o State the formula of the most basic Group 4 oxide.
1 a Write equations for the changes caused by the addition of water to:
bC
i sodium chlorideii silicon tetrachlorideiii phosphorus pentachloride.Relate these reactions to the bonding in the chlorides.Explain why silicon tetrachloride, SiCln, reacts rapidly with water,
but tetrachloromethane, CCln, does not react even when heated with
water to 100 oC.
Write two ionic equations to show that aluminium hydroxide is
amphoteric.Write ionic equations for the reactions, if any, of:aqueous acid with:i magnesium hydroxideii lead monoxideiii sulphur dioxideaqueous alkali with:i magnesium hydroxideii lead monoxideiii sulphur dioxide.Write an equation to show lead(IV) oxide acting as oxidising agent.Write an equation to show aqueous tin(Il) ions acting as a reducingagent.
2
3
4 ab
@ c H E M T c A L E e u r L r B R t u M l l
The value of ( is specific to areaction - and to the equation usedto represent that reaction - and canonly be altered by changing thetemperature.
Chemical equilibria llffi IntroductionLry
[l Concentration, in the context of equilibrium, is always measured inmol dm-', and the concentration of a substance A is written as [A].
E fne common errors in this topic are to use moles not concentrationsand initial values not equilibrium values when substituted inexpressions for K..
iJ Calculations of K must be set out clearly, showing each step.
J A homogeneous equilibrium is one in which all the substances are inthe same phase, whereas in a heterogeneous equilibrium there are twoor more phases.
Things to learn and understand
The equilibrium constant K"
The equilibrium constant, measured in terms of concentrations, is foundfrom the chemical equation.
o For a reactiont *A + nB+ rC + yD
where ffi, fl, x and y arc the stoichiometric numbers in the equation:K. = [C]'"q [D]/"q
[A]'"0 [B]'",
All the concentrations are equilibrium values.
O In a heterogeneous equilibrium a solid substance does not appear in theexpression for K..
o K. is only equal to the quotient [C]' [D]'
tAr tBrwhen the system is at equilibrium.
O If the quotient does not equal K., the system is not in equilibrium andwill react until it reaches equilibrium when there will be no furtherchange in any of the concentrations.
o K. has units. For the reaction:N r + 3 H r + 2 N H .
K. = [NHJr"o
Nrl"o [HJ'"'
the units of K. are (mot d--')' 1= mol-2 dm6
(mol dm*) (mot d--')'
E I " t * t o D l c t r y '
e u A N T I T A T t v E E e u r L r B R r A A N D F U N . T T . N A L G R o u p c H E M r s r R y
Colculotion of K,
This must be done using equilibrium concentration values, not initialconcentration values. The calculation is done in 5 steps:
i Draw up a table and fill in the initial number of moles, the changefor each substance and the equilibrium number of moles of eachsubstance.
ii convert equilibrium moles to concentration in mol dm-r.iii State the expression for K..iv Substitute equilibrium concentrations into the expression for K..v work out the units for K. and add them to your answer.
WorkpoL oxiloLf.bAn important reaction in the blast furnace is the formation of carbon monoxidefrom carbon and carbon dioxide. This reaction is an example of aheterogeneous equilibriu m.when 1.0 mol of carbon dioxide was heated with excess carbon to atemperature of 700 'c in a vessel of volume 20 dm', g5% of the carbondioxide reacted fo form carbon monoxide. calculate K,tor
C(s) +CO,(g) +2C0(g)o C(s) does not appear in the K.
expression because it is a solid.o The amount of C0 made is twice
K = t!91,* = (0.095), (mol dm-s), = 3.6 mot dm-3reo;i; ffi
Partial pressure
o The partial pressure of a gas A, po, in a mixture is the pressure that thegas would exert if it alone filled the container. It is calculated from theexpression:
partial pressure of a gas = mole fraction of that gas x total pressurewhere the mole fraction = the number of moles of that gas
the total number of moles of gas
o The total pressure, P, is equal to the sum of the partial pressures of eachgas in the mixture.
The equilibrium constant K"
The equilibrium constant, measured in terms of partial pressure, onlyapplies to reactions involving gases. solids or liquids do not appear inthe expression for K".For a reaction:
mA($ =* xB(g) + ZC(g)
where m, x and y are the stoichiometric numbers in the equation:K, = (p).-t_(pJ *_
(Po)^ "n
O . H E M T . A L E e u r L r B R r u M r l
As carbon is a solid it is ignored inthe expression.
Konly equals the quotient when thesystem is at equi l ibr ium. l f thequotient is greater than K, thereaction will move to the left until_the two are equal. lf the quotient isless than K the reaction will moveto the right until the two are equal.Temperature changes alter the valueof K(unless AH= 0). Concentrat ionand pressure changes alter the valueof the quotient.
All partial pressures are equilibrium values.
So for the reaction:CO,(g) + C(s) +2CO(g)
160 = p(CO)'"' and its units are atm2= atm
P(co,)*
Colculotions involving K,
o The calculation of Ko fromthose involving K.:i Draw up a table and fill in the initial number of moles, the change
for each substance, the equilibrium number of moles of eachsubstance and the total number of moles at equilibrium.Convert equilibrium moles to mole fraction.Multiply the mole fraction of each by the total pressure.State the expression for K'.Substitute equilibrium concentrations into the expression for Ko.Work out the units for K' and add them to your answer.
oh oxal,u?12
Variation of K with conditions
o Temperature. This is the only factor that alters the value of K.
i If a reaction is exothermic left to right, an increase in temperaturewill lower the value of K. This means that the position of
equilibrium will shift to the left (the endothermic direction).ii If a reaction is endothermic left to right, an increase in temperature
will increase the value of K. This means that the position ofequilibrium will shift to the right.
O Catalyst. This neither alters the value of K nor the position ofequitibrium. It speeds up the forward and the reverse reactions equally.
Thus it causes equilibrium to be reached more quickly.
atm
equilibrium data is done in a similar way to
ll
iiilv
Y
vl
0.080 mol of PCluwas placed in avessel and heated to 175 "C. Whenequilibrium had been reached, it was found that the total pressure was 2.0 atmand that 40o/o of the PClu had dissociated. Calculate K,tor the reaction:
PCl,(g) + PCl,(g) + Cl,(g)
Answen As 40% had dissociated, 60% was left. Thus at equilibrium:0.60 x 0.080 = 0.048 mol of PClu was present and therefore0.080 - 0.048 = 0.032 mol of PClu had reacted.
e u A N T r r A T r v E E e u r L r B R r A A N D F U N . T T . N A L G R o u p c H E M r s r R y
O Concentration. A change in the concentration of one of thesubstances in the equilibrium mixture will not alter the value of K, butit will alter the value of the quotient. Therefore the reaction will nolonger be at equilibrium. It will react until the value of the quotientonce again equals K. If the concentration of a reactant on the left handside of the equation is increased, the position of equilibrium will moveto the right.
o Pressure. A change in pressure does not alter K. If there are more gasmolecules on one side than the other, the value of the quotient will bealtered by a change in pressure. Therefore the reaction will no longer beat equilibrium, and will react until the value of the quotient once againequals K. If there are more gas molecules on the right of the equationand the pressure is increased, the position of equilibrium will move tothe left. For example, if the pressure is increased on the equilibriumNrOn(B) + 2NOr(g), the position of equilibrium will shift to the left asthere are fewer gas molecules on the left side of the equation.
ChecklistBefore attempting questions on this topic, check that you:
[i Can define the partial pressure of a gas.
[l Can deduce the expression for K. and its units given the equation.
E Can deduce the expression for Ko and its units given the equation.
E Can calculate the value of K. given suitable data.
|J Ca.r calculate the value of K, given suitable data.
fl Kt o* not to include values for solids and liquids in the expressionfor Ko.
I] Knot" that only temperature can alter the value of K, and how it willaffect the position of equilibrium.
Testing your knowledge and understandingFor the following questions, cover the margin, write your answer, thencheck to see if you are correct.
o Dry air at a pressure of 104 kN m-'contains 78.Io/o nitrogen 2I.Oo/ooxygen and 0.9o/o argon by moles. Calculate the partial pressures ofeach gas.
O Write the expression for K., stating its units, for the following reactions:^ 2SO,(g) + O,(B) + 2SO,(g)
b zHcl(g) +'1,o,(g) + H,o(g) + cl,(g)
O Write the expression for Ko, stating its units, for the following reactions:a CO($ + 2H,(g) + CH,OH(g)
b Fe,O,(s) + CO(g) + 2FeO(s) + CO,(g)
L When a 0.0200 mol sample of sulphur trioxide was introduced into avessel of volume I.52 dm3 at 1000'C, 0.01,42 mol of sulphur trioxidewas found to be present after equilibrium had been reached.
P(N,) = 0.781 x 104 = 81.2 kN m-'P(0,) = 0.210 x 104 =21.8 kN m-'P(Ar) = 0.009 x 104 = 0.9 kN m-'
8 K= [S0J' = mofl dmg
tsortoJb K= [H,0] [Cl,] = mofz dmil2
lHctit ro,r
d 6=p(CH,0H) atm-'p(CO) x p(H,))'
b K,= 499) no.unitsp(c0)
A C I D - B A S E E Q U I L I B R I A
The answers to the numberedquestions are on page 133.
Calculate the value of K for the reaction:
2SO,(g) + 2SO,(g) + O,(g)
2 I.O mol of nitrogen(Il) oxide, NO, and 1.0 mol of oxygen were mixed ina container and heated to 450 "C. At equilibrium the number of molesof oxygen was found to be 0.70 mol. The total pressure in the vessel was4.0 atm. Calculate the value of K, for the reaction:
2NO(S) + O,(g) -- 2NO,(g)
3 This question concerns the equilibrium reaction:
2SO,(g) + O,(g) -- 2SO.(g) AH = -796 kJ mol-l
K. = 3 x 10* mol-t dm3 at 450'c
a 2 mol of sulphur dioxide, I mol of oxygen and 2 mol of sulphurtrioxide were mixed in a vessel of volume l0 dmt at 450 "C in thepresence of a catalyst. State whether these substances are initially inequilibrium. If not, explain which way the system would react.
b State and explain the effect on the value of K. and on the position ofequilibrium of:
i decreasing the temperature
ii decreasing the pressure
iii adding more catalyst.
(g Acid-base equilibria<ffiW Introduction
o It is essential that you can write the expression for K" of a weak acid.
O Make sure that you know how to use your calculator to evaluatelogarithms, and how to turn pH and pK" values into [H-] and K" values.
I This means using the lg key for log,o and 10'key for inverse log,o.
o Give pH values to 2 decimal places.
O Buffer solutions do not have a constant pH.They resist changes in pH.
O You may use H-, H-(aq) or HrO- as the formula of the hydrogen ion.
d-aLN Things to learn
[| A Brsnsted-Lowry acid is a substance that donates an H- ion (aproton) to another species.
tr n monobasic (monoprotic) acid contains, per molecule, one hydrogenatom which can be donated as an H" ion. A dibasic acid contains two.
tr e Bronsted-Lowry base is a substance that accepts an H* ion fromanother species.
E K. = [H-(aq)] [OH-(ag)] = 1.0 x 1O" mol' dm* at 25 "C.
f A neutral solution is one where [H-(aq)] = [OH-(aq)].
f pH = -log,o[H.(aq)],'or more accurately = -log,o([HrO.(aq)]/mot dmt).
lf the acidity of a solution increases,[H-] increases but the pH decreases.
The 2nd ionisation of a dibasic acidsuch as H2S04 is weak. So [H-] for0.321 mol dm-'H,Soo is not 0.642mol dm-', but is only 0.356 mol dm-'
For an alkali with 2 0H per formula,the [0H-] is 2 x the [alkali].The pH of a 0.109 mol dm-' solutionof Ba(OH), ispOH = - log,o(2 x 0.109) = 0.66Therefore pH = 14 - 0.66 = 13.34Remember that acids have a pH < 7and alkalis a pH > 7.
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E pOH = -log,olOH-(aq)1.
E pH + pOH - !4 at Z|'C.
tr e strong acid is totally ionised in aqueous solution, and a weak acidis only partially ionised in aqueous solution.
f K, for a weak acid, HA, = tHl tAl / tHAl.
f PK. = -log'o K".
tr e buffer solution is a solution of known pH which has the ability toresist changing pH when small amounts of acid or base are added.
Base + H* - its coniugate acid:NH, (base) + H* + NHn* (coniugate acid)
E When sulphuric acid is added to water, it acts as an acid and the wateracts as a base:
HrSOn+HrO- HrO* + HSO*-acid base coniugate acid of HrO conjugate base of HrSOn
J Uydtogen chloride gas acts as an acid when added to ammonia gas:
H C I + N H r - N H n ' + C facid base conjugate acid of NH. coniugate base of HCI
pH scale
f Water partially ionises
HrO+ H'(aq) + OH-(aq)
As the [HrO] is very large and so is effectively constant, its value can beincorporated into the K for the reaction:
K* = tH-] tOH-] = L.0 x LO'n mol' dm* at 25'C
or pH + pOH = \4at25"C
f So at 25 "C:
o A neutral solution is one where [H-] = [OH-], and the pH = 7.o A n a c i d i c s o l u t i o n i s o n e w h e r e [ H ] > [ o H l , a n d t h e p H <o A n a l k a l i n e s o l u t i o n i s w h e r e [ H ] < [ o H l , a n d t h e p H >
E pH of strong acids and bases
I For a strong monobasic acid:pH - -log,o[acid]
Therefore the pH of a O.321 mol dm-' solution ofHCI = -log,o (0.321) - O.49
O For a strong alkali with one OH- ion in the formula:pOH = -log,o[alkali] and' p H = L 4 - p O H
@ A c r D - B A s E E e u r L r B R r A
At half way to the end point (12.5cm' if the end point is at 25 cm3),the curve for a weak acid and astrong base will be almost horizontalat a pH of about 5. This is when thesolution is buffered and [HA] = [A ].Thus at the half way PH, FK. = PHfor the weak acid.
Therefore the pH of a 0.109 mol dm-' solution of NaOH is calculated:pOH = -log,o(0.109) = 0.96
fherefore pH = 14 - 0.96 = 13.04
pH of weok ocids
o Weak acids are partially ionised. A general equation for this is:H A ( a q ) + H - ( a q ) + A - ( a q )
o The acid produces eqrral amounts of H- and A-, so that:
lH'(aq)l = [A-(aq)]
O This enables either the pH of a weak acid solution or the value of K. of aweak acid to be calculated.
Worktl' oxarorylt,
Titration curveso These are drawn either for the addition of base to 25 cm' of a 0.1 mol
dm-' solution of acid until the base is in excess, or for addition of acidto a base (see Figure 4.3).
a If the acid and base are of the same concentration, the end point is at25 cm'.
t There are several pH values to remember.
1 3
1 L
9
7
5
3
1 3
1 1
9
7
5
3
0 2 5Volume of alkali/ cm3
0 2 sVolume of alkali/ cm3
0 2 5Volume of alkali/ cm3
Calculate the pH of a 0.123 mol dm'solution of an acid HA which hasK. = 4.56 x 10'mol dm{:
lH'(aq)l = 7'j( JHA(aq)l)=/+.SO x 10* x 0.123) = 2.37 x 10-'mol dm-3
pH = - log,o(2.37x10- ' ) = 2.63
StartingpH
End pointpH
Vertical pHrange
FinalpH
Strong acid/strong baseWeak acid/strong baseStrong acid/weak base
131
795
3.5 to 10.57 to 10.53.5 to 7
13131 1
Strong acid/ Weak acid/
Fig 4.3 Titration curves
-|l
5 i l P e n r o D r c r T y , e u A N T t T A T t v E E e u r L r B R | A A N D F U N G T T o N A L G R o u p c H E M T S T R Y
Indicators
O For use in a titration, anthe vertical part of the
indicator must change colour within the pH oftitration curve.
Buffer solutions
o These consist of an acid/base conjugate pair, e.g. a weak acid and its saltsuch as CH.COOH/CHTCOONa, or a weak base and its salt, e.g.NH3/NH4CI.
o To be able to resist pH changes, the concentration of both the acid andits conjugate base must be similar.
I Consider a buffer of ethanoic acid and sodium ethanoate. The salt isftrlly ionised:
CH3COONa(aq) - CH'COO-(aq) + Na-(aq)
The weak acid is partially ionised:CH3COOH(aq) + H'(aq) + CH,COO-(aq)
The CHTCOO- ions from the salt suppress most of the ionisation of theacid, and so both [CH.COOH(aq)] and [CH,COO-(aq)] are large.
If H' ions are added to the solution, almost all of them are removed byreaction with the large reservoir of CH.COO- ions from the salt:
H-(aq) + CH,COO-(aq) - CHsCOOH(aq)
If OH- ions are added, almost all of them are removed by reaction withthe large reservoir of CHTCOOH molecules of the weak acid:
OH-(aq) + CH,COOH --' CH,COO-(aq) + H,O(l)
Calculotion of the pH of o buffer solution
Consider a buffer made from a weak acid, HA, and its salt NaA.The acid is partially ionised:
H A ( a q ) + H . ( a q ) + A ( a q )
Thus K^ = [H-(aq)] [A (aq)]
IHA(aq)l
The salt is totally ionised, and suppresses the ionisation of the acid,Therefore [A-(aq)] = [salt]
and [HA(ag)] = [weak acid]
Thus 16. = [H-(aq)] [salt]tweak acldf
or [H-(aq)] = K. trygg&gql[salt]
pH = -loglH.(aq)l
Vertical range of pH Suitable indicator
Strong acid and baseWeak acid/strong baseStrong acid/weak base
3.5 to 10.57 to 10.53.5 to 7
Methyl orange or phenolphthaleinPhenolphthaleinMethyl orange
lf the concentrations of the weakacid and its salt are the same, then:
[H- (aq)] = K^ , andpH = pK,.
@ AcrD-BA 'E Eeur . - rBR 'A
Workpi, oxiloLf.h'
Enthalpy of neutralisation
When a strong acid is neutralised by a solution of a strong base, A.FI,,",,, =-57 kJ mol-', because the reaction for strong acids n'ith strong bases is:
H-(aq) + OH-(aq) = H,O(l lThe value for the neutralisation of a weak acid is less because energv has tobe used to ionise the molecule to form H- ions.
a -
E IJ)'I
;i //,-r.--t---)a
H
Before attempting the questions on this topic, check that you:
lJ Can identify acid/base coniugate pairs.
*l Can define pH and K-.
u.J Can define K" and pK" for weak acids.
U Understand what is meant by the terms strong and weak as applied toacids and bases.
iJ Can calculate the pH of solutions of strong acids, strong bases and weakacids.
iJ Can recall the titration curves for the neutralisation of strong and weakacids.
tl Can use the curve to calculate the value of K" for a weak acid.
if Understand the reasons for the choice of indicator in an acid/basetitrations
[J Can define a buffer solution, explain its mode of action and calculate its pH.
j.ldt, XE), Testing your knowledge and understandingL / . - | , -
H
For the following questions, cover the margin, write your answer, then checkto see if you are correct.
o Identify the acid/base conjugate pairs in the reaction:HrSOn + CH,,COOH + CH.COOH2- + HSO,-
O Calculate the pH of:a 0.11 mol dm* HCIb 0.1 1 mol dm-' LiOHc 0.11 mol dm-' Ba(OH),
Calculate the pH 0f 500 cm3 of a solution containing 0.121m01 of ethanoic acidand 0.100 mol of sodium ethanoate.pK^tor ethanoic acid = 4.76Answer. gK, = 4.76,
[weak acid] = 0.121+ 0.500 = 0.242 mol dm-3[salt] = 0.100 + 0.500 = 0.200 mol dm-3.[H-(aq)] = (. [weakacid] = 1.74 x 10 u x0.242=2j0 x 10-u mol dm-'
Isalt]pH = -log,o[H-(aq)] = 4.68
0.200
Acid H,S0o: its conjugate base HSO.-Base CH,COOH: its conjugate acidcH,c00H,-
abc
0.9613.0413.34
The answers to the numberedquestions are on pages 133-134.
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e u A N T r r A T r v E E e u r L r B R r A A N D F U N c T t o N A L G R o u p c H E M t s r R y
Calculate the pH of O.22 mol dm-' C2H'COOH which
25 cm' of a weak acid HX of concentration 0.10 molwith 0.10 mol dm-' sodium hydroxide solution, andintervals. The results are set out below:
L
2
has pK. = 4.87.
dm-t was titratedthe pH measured at
a Draw the titration curve and use it to calculate pK, for the acid HX.
b Suggest a suitable indicator for the titration.
3 a Define a buffer solution and give the name of two substances that
act as a buffer when in solution.
b Explain how this buffer would resist changes in pH, if smallamounts of H- or OH- ions were added.
4 Calculate the pH of a solution made by adding 4.4 g of sodium
ethanoate, CHTCOONa, to 100 cmt of a0.44 mol dm-'solution of
ethanoic acid. K" for ethanoic acid = I.74 x 10-s mol dm-'.
Volume NaOH/cm3 5 10 T2 20 23 24 25 26 30
pH 4.5 4.8 4.9 5 .5 6.5 7.0 9.0 12.o 12.5
@ 0rganic chemistry l l..H
" ',"'"\ fhings to learn and understandH(H
Isomerism
iJ Structural. There are three types:
a Carbon chain. Here the isomers have different arrangements of
carbon atoms in a molecule,
e.g. butane, CH.CH TCHTCH, and methyl propane, CH.CH(CHr)CH..
b Positional. Here a functional group is on one of two or more
different places in a given carbon chain,
e.g. propan-l-ol, CHTCHTCHTOH, and propan-Z-oI, CH.CH(OH)CH..
c Functional group. Here the isomers have different functional groups,
e.g. ethanoic acid, CH3COOH and methyl methanoate, HCOOCH..:J Stereoisomerism. There are two types:
a Geometric. This is caused by having two different Sroups on eachcarbon atom of a >C=C< group (see Figure 4.4).
H,C CH,\ /
C - C
cis-but-2-ene trans-but-Z-ene
The two isomers are notinterconvertible at ordinarytemperatures because there is norotation about a double bond. H:C H
\ /C - C
/ \H CH.
/ \H H
Fig . cis ond trans isomers of but-2-ene
Iis the
@ o R c A N t c c H E M t s r R Y t l
Having four different grouPsattached to one carbon atom willcause this. This carbon atom iscalled a chital centre.
A 50/50 mixture of the two isomerswill have no effect on Planepolarised light. This type of mixtureis called a racemic mixture and isoften the result when chiralsubstances are produced by achemical reaction.
Methanal produces a PrimarYalcohol.
Optical. Optical isomers are defined as isomers of which one
non-superimposable mirror image of the other.
Thus lactic acid (2-hydroxypropanoic acid), cH3cH(oH)cooH'exists as two optical isomers. These must be drawn three
dimensionally so as to show that one is the mirror image of the
other (see Figure 4.5).
COOH
I, t \
HO \\ CH,
H
COOH
I. F-.-
Hsc /l oHH
Fig 4.5 Opticol isomers of loctic ocid
One isomer will rotate the plane of polarisation of plane polarised
monochromatic light clockwise and the other will rotate it anti-
clockwise.
Further reactions
Grignord reagenfs
These have the formula R-Mg-halogen such as C,HuMgl. They are used to
increase the carbon chain length, because they are nucleophiles and
contain a 6-carbon atom which will attack a 6'carbon atom in other
compounds.
PreparationHalogenoalkane + magnesiumConditions: dry ether solvent (flammable, therefore no flames), under
reflux with a trace of iodine as catalyst: e.g.C2H' I+Mg-CrHrMgI
ReactionsO + aldehydes to produce (after hydrolysis with dilute hydrochloric acid) a
O + ketones to produce (after hydrolysis) a tertiary alcohol:C,H,MgI + CH.COCH, - CH3C(CH3)(OH)CrH'
2-methyl butan-2-ol
O + solid carbon dioxide (dry ice) to form (after hydrolysis) a carboxylic
acid:CrHrMgI + COr(s) -'CTHTCOOH
propanoic acid
O Grignard reagents react with water to form alkanes:CrHrMgI+HrO-CrHu
which is why they must be prepared and used in dry conditions.
Corboxylic ocids
These have the functional grouP:
o/- C
o - H
Test for a carboxylic acid. Whenadded to aqueous sodiumhydrogencarbonate (or sodiumcarbonate), it gives a gas, C0,,which turns lime water milky.
The acid is a catalyst. This reactionhas a low yield because it is areversible reaction.
This reaction has a high yieldbecause it is not reversible.
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PreparationThey can be made by the oxidation of a primary alcohol. The alcohol isheated under reflux with dilute sulphuric acid and excess potassiumdichromate(Vl).
conditions: heat under reflux with aqueous sodium hydroxide.
Corbonyl compo,unds (oldehydes ond ketones)
These contain the C=O functional group. Aldehydes have the generalformula RCHO, and ketones the general formula RCOR where R and R arealkyl or aryl groups, and may or may not be different.
o/
R - C\
H
Aldehyde
o/
R - C\
R'
Ketone
@ o R G A N r c c H E M r s r R Y r l
The formation of an orange or redprecipitate with 2,4-DNP is a test forcarbonyl compounds, i.e. for bothaldehydes and ketones.
The formation of the silver mirrorwith ammoniacal silver nitrate is thetest which distinguishes aldehydes,which give a positive result, fromketones, which have no reaction.Aldehydes are also oxidised bypotassium manganate(Vl l) oracidified potassium dichromate(Vl).
The only aldehyde to do this isethanal, CH3CH0.
This reaction is often used inorganic'problem' questions. Thepale yellow precipitate produced onaddition of sodium hydroxide andiodine is the clue. The organiccompound that produces thisprecipitate will either be a carbonylwith a CH,C=Q group or an alcoholwith a CH.CH(0H) group. lt is not ageneral test for ketones.
Reactions in commona Addition of hydrocyanic acid, HCN, to produce a hydroxynitrile:
CH,CHO + HCN - CH,CH(OH)CN2 -hy dr oxypr o p a n e n i t ri I e
conditions: there must be both HCN and CIS present,
e.g. HCN + a trace of KOH, or KCN(s) + some dilute sulphuric acid.
o Reduction with lithium tetrahydridoaluminate(Ill) to produce an
alcohol:CH3CHO +2[H] -CHTCHTOH
CH'COCH, + 2[H] - CH.CH(OH)CH3
conditions: dissolve in dry ether, followed by hydrolysis with H-(aq).
Or reduction with sodium tetrahydridoborate(Ill), also to produce an
alcohol:CH3CHO + 2[H] 'CHTCH,OH
CH3COCH. + 2[H] - CH,CH(OH)CH,
conditions: aqueous solution, followed by hydrolysis with H-(aq).
a Reaction with Z,4-dinitrophenylhydrazine (2,4-DNP) to give anorange/red precipitate:
NOt NO,
2,4.DNP
Specific reaction of aldehydesAldehydes are oxidised to carboxylic acids by a solution of silver nitrate indilute ammonia. On warming the silver ions are reduced to a mirror ofmetallic silver.Aldehydes are also oxidised by warming with Fehling's solution, which isreduced from a blue solution to a red precipitate of copper(I) oxide:
CH.CHO+[O] -CH'COOH
Iodoform reoction
o This is a reaction which produces a pale yellow precipitate of iodoform,CHI3, when an organic compound is added to a mixture of iodine anddilute sodium hydroxide (or a mixture of KI and NaOCI).
t The group responsible for this reaction is the CH'C=O group, and thereaction involves breaking the C-C bond between the CH, group andthe C=O group.
CHTCOCTH, produces CTHTCOONa + CHI.(s)
a All ketones containing the CH.C=O group will give this precipitate.O This reaction is also undergone by alcohols containing the CH.CH(OH)
group. They are oxidised under the reaction conditions to the CH.C=Ogroup, which then reacts to give iodoform.
Acid chlortdes
These have the general formula RCOCI, and the functional group is:
e.g. ethanoyl chloride, CH.COCI, and propanoylchloride, CrHrCOCl.
o/
- C
CI
The amine can be regenerated fromthe salt by adding a strong alkali.
Nitriles can be prepared by thereaction of a halogenoalkane withKCN in aqueous ethanol.Hydroxynitriles can be prepared bythe addition of HCN to carbonylcompounds in the presence of base.
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Reactions of ethanoyl chlorideO + water to produce the carboxylic acid, ethanoic acid:
CHTCOCI + HrO - CHTCOOH + HCI(g)
observation: steamy fumes of HCl.
o + alcohols to produce an ester:CH3COCI + CTHTOH - CHTCOOCTH' + HCI
observation: characteristic smell of the ester, ethyl ethanoate.
o + ammonia to produce the amide, ethanamide:CHTCOCI + 2NH. - CHTCONH, + NH4CI
I + primary amines to produce a substituted amide:CHTCOCI + CrHrNH, - CH3CONH(CrH') + HCI
Nitrogen compounds
Primory omines
o They contain the -NH, group, e.g. ethylamine C2HrNHz.O They are soluble in water (if the carbon chain is fairly short) because
they form hydrogen bonds with water molecules.o They are weak bases.
Reactions of ethylamineO + acids, to form an ionic salt that is soluble in water:
C2H'NH2 + H.(aq) + CrHrNHr.(aq)
| + acid chlorides, to form a substituted amide:C2H'NH2 + CHTCOCI -'CH'CONHC2H, + HCI
Nitriles
They contain the C=N group, e.g. propanenitrile C,H'CN.
Re ations of prop anenitrileO Hydrolysis either with acid:
CTH'CN + H* + zHrO - CTHTCOOH + NHn*
conditions: heat under reflux with dilute sulphuric acidor with alkali'
crHscN + oH- + Hro'crHrcoo- + NH,
conditions: heat under reflux with dilute sodium hydroxideO Reduction by lithium tetrahydridoaluminate(Ill):
C'H'CN + a[H] - C2H'CHNH,
conditions: dry ether, followed by addition of dilute acid
Amldes
They contain the CONH, group, €.8. ethanamide CHTCONHT.
o/
c H g - Q\
NH,
Reactions of ethanamideO Dehydration with phosphorus(V) oxide:
CH3CONH,-HzO-CH'CN
conditions: warm and distil off the ethanenitrile.
@ o R G A N r c c H E M r s r R Y r l
Amides can be prepared from acidchlorides + ammonia.Amides can be converted into anamine either with the same numberof carbon atoms (reduction) or withone less (Hofmann degradation).
Amino acids can be prepared fromchloroacids by heating them in asealed tube with an ethanolicsolut ion of ammonia.
conditions: add liquid bromine to the amide at room temperature, andthen add concentrated aqueous sodium hydroxide solution and warm.The amine distils off.
Amino ocids
These contain an NH, and a COOH group (usually bonded to the samecarbon atom). They react both as acids and bases.They are water-soluble ionic solids because they form a zwitterion.
NHTCH.COOH + NH,.CHTCOO-
Reactionso With acids:
NHTCHsCOOH + H'(aq) - NH,'CH.COOH(aq)
t With bases:NH'CH'COOH + OH-(aq) - NH.CH.COO-(aq) + H.O
Figure 4.6 summarises the reactions of organic substances.
GrignardCrHrMgBr
AcidscH3cooH
CO2.-----------C2H'COOH
HCHO ---------------- C2H5CH2OH
CH3CHO -------+ C2H'CH(OH)CH3
CH3COCH3 + C2H'C(CH,)zOH
H z O - C r H u
C2H5OH - CH3COOCTHs
LiAlHn- CH3CH2OH
PCl, -----_+ CH3COCI
AcidchloridescH3cocl
Aminesc2HsNH2
H2O-CH3COOH
C2H'OH ---) CH3COOCzHs
NH3-CH3CONH2
CH3NH2+ CH3CONHCH3
H*-C2HsNH;
CH3COCI- CH:CONHC,H'
H*---+CH3COOH
OH--CH3COO-
LiAlHa--+ CH3CH2NH,
Aldehydes andketones>C=O
Esters l- H- CH3COOH + CTHTOH
CH"COOC"H. IJ ' OH- ___---_-+ CH3COO- + CTHTOH
HCN->C(OH)CN
2,4-DNP .-+ red precipitate
lHl - >CH(OH)
Aldehydes only l- Ag'lNH3---+ CH3COO- + silver mirror
CH.CHO IFehling's --------------- CH3COO- + red precipitate
t- P+Oto- CH3CNAmides ICH.CONH, IL Brr(r)lNaoH-+ CH3NH2
NitrilesCH3CN
Fig 4.6 Summary of organic reactions
E I t r * r o D r c r r y , e u A N T r r A T r v E E e u r L r B R ' A A N D F U N . T T o N A L G R o u p c H E M t s r R y
ChecklistBefore attempting questions on this topic, check that you:
i Can recognise stereoisomerism (geometric and optical) in organiccompounds.
iJ Know the effect of an optical isomer on plane polarised light.
iJ Understand the nature of a racemic mixture.
li Can recall the preparation and reactions of Grignard reagents,
.J Can recall the reactions of carboxylic acids.
lJ Can recall the reactions of esters.
'J Can recall the reactions of carbonyl compounds (aldehydes andketones).
iJ Can recall the reactions of ethanoyl chloride.
iJ Can recall the reactions of amines with acids and with acid chlorides.
J Can recall the hydrolysis and the reduction of nitriles.
J Can recall the reactions of amides with PnO,o and with BrrlNaOH.
iJ Know that amino acids are both acids and bases.
ffi Testing your knowledge and understandingFor the following questions, cover the margin, write your answer, thencheck to see if you are correct.
* Write the formulae of the products of the reaction of CH,MgI with:a propanal
b butan-2-one
c methanal.
o State the names of the reagents and give the conditions for thefollowing conversions:a C,H'COOCH, to C,H'COOH + CH,OH in a high yield
b C2H'COOH to CHTCH,CH,OH.
O State the names of the reagents needed to convert ethanoyl chloride to:a ethanoic acid
b methyl ethanoate
c ethanamide.
o State the structural formulae of the product obtained by reacting:a ethylamine with dilute hydrochloric acid
b ethvlamine with ethanovl chloride.
a CH,CH,CH(0H)CH3b cH,cH,c(0H)(cH3)cH3c CHTCHTOH
a Heat under reflux with dilutesodium hydroxide, then adddilute sulphuric acid.
b Add lithiumtetrahydridoaluminate(l I l) in dryether, followed by dilutesulphuric acid.
a waterb methanolc ammonia
a CH3CH2NH'- Cl-b cH,c0NHcH2cH3
O o R G A N r c c H E M r s r R Y r l
a Heat under reflux with NaOH(aq)then acidify.
b Add LiAlH4 in dry ether, followedby dilute alkali.
c Warm with liquid bromine andconcentrated aqueous sodiumhydroxide.
Because it forms the,ryitterion-NHrCHS00-
The answers to the numberedquestions are on page 134.
O State the conditions for the conversion of:a ethanenitrile, CHTCN, to ethanoic acid, CH3COOH
b ethanenitrile, CH3CN, to ethylamine, CzHsNHz
c propanamide, CTHTCONH2, to ethylamine, C2HsNHz.
O Explain why aminoethanoic acid is a solid that is soluble in water.
L Draw the stereoisomers of:
a CH,CH=C(CI)CH,
b cH.cH(oH)F.
2 Outline how you would prepare 2-methylpropanoic acid from2-iodopropane.
3 Describe the tests that you could do to distinguish between:
a ethanoic acid and ethanoyl chloride
b propanal and propanone.
E I " r * r o D r c r r y ,
e u A N T r r A T r v E E e u r L r B R r A A N D F U N c T t o N A L G R o u p c H E M I ' T R Y
Practice Test: Unit 4Time allowed 1. hr 3O min
All questions are taken from parts of previous Edexcel Advanced GCE papers
The answers are on pages 134-136.
1 a Draw a Born-Haber cycle for the formation of magnesium chlodde, Mgcl, and use the values below to
calculate the lattice energy of magnesium chlodde. [5]
AIflkJ mol-'
lst electron affiniW of chlorine -364
1st ionisation energy of magnesium +736
2nd ionisation energy of magnesium +1450
Enthalpy of atomisation of chlorine +LZL
Enthalpy of atomisation of magnesium +150
Enthalpy of formation of MgClr(s) -642
b Use the following to answer the questions in this section.
AfflkJ mol-'
Enthalpy of hydration of Sr'. -1480
Enthalpy of hydration of Ba'. -1360
Enthalpy of hydration of OH- -460
Lattice energy of Sr(OH)'(s) -r894
Lattice energy of Ba(OH)r(s) -L768
Explain why the lattice energy of strontium hydroxide is different from that of barium hydroxide.[2]
fxplain why the hydration *tfrutpy of a cation is exothermic. l2l
Use the lattice .n.igy and hydration enthalpy values to explain why barium hydroxide is more
soluble in water than strontium hydroxide. t4l(Total 13 marks)
lJanuary 2001 CH 3 question 4l
I
ll
lll
2 a i Give the structural formula of a nitrile, CnHrN, that has an unbranched chain
ii primary amines can be made by reducing nitriles. Suggest a reagent that could be used for this
purpose.iii Draw the structural formula of the amine produced by reducing the nitrile given in a part i.
i What feature of an amine molecule makes it both a base and a nucleophile?
ii Give, by writing an equation, an example of an amine acting as a base
Ethanoyl chloride, CHTCOCI, reacts with both amines and alcoholsi Give the full structural formula of the compound produced when ethanoyl chloride reacts with
ethylamine, CrHrNHr. tll
ii Name the type of the compound produced in c part i. tU
iii State one of ihe advantages of reacting ethanoyl chloride with ethanol to make an ester rather than
reacting ethanoic acid with ethanol. tUiv Write the full structural formula of the ester made in c part iii tll
(Total 9 marks)
ffanuary 2002 Unit Test 4 question 5 - modified]
tll
tlItUtlItll
P R A C T I C E
3 Compound X is a secondary alcohol which has a molecular formula of CnH,oO.X exhibits optical isomerism.
a Draw the structural formulae of the two optical isomers of X. l2lb Suggest reagents and conditions which would enable its preparation via a reaction involving a
Grignard reagent. (You are not expected to describe how you would prepare a Grignard reagent) t3Ic Compound Y is obtained by oxidising the secondary alcohol X with potassium dichromate(Vl)
acidified with dilute sulphuric acid.i Draw the structural formula of Y.ii To which class of compounds does Y belong?iii Describe the tests you would do on Y, and the results you would expect, to show that your
classification is correct. t4Iiv Both X and Y give a yellow precipitate when treated with iodine in the presence of sodium
hydroxide solution. Write the structural formulae of the organic products of this reaction. l2l(Total 13 marks)
lJwte 2OO1 CH I questiotr 1- modiftefl
At about 1000 oC, when aluminium chloride vapour is heated with solid aluminium, the followingequilibrium is set up:
AlCl.(g) + 2Al(s) -'3AlCl(g)
t1lt1I
a Give the expression for the equilibrium constant, Kr, for this reaction.
b At a particular temperature, a mixture of the above system at equilibrium was found to contain 0.67 g
of AlCl, and 0.63 g of AlCl vapours at an equilibrium pressure of 2.0 atm. Calculate the value of theequilibrium constant, Ko, at this temperature, stating its units. t4l
c The position of equilibrium moves to the right as the temperature is raised. State what this suggestsabout the enthalpy change for the forward reaction. tll
d State, with reasoning, the effect of an increase in the pressure on the system on the value of theequilibrium constant and on the position of the equilibrium. t3I
e Suggest how this reversible reaction could be used in a process to recycle aluminium by extracting itfrom impure aluminium. r2l
(Total LL marks)lJune 2001 CH3 question 3l
S a i Write an equation for the reaction between magnesium oxide and dilute sulphuric acid, including
IU
the state symbols.Describe what you would see during this reaction.
r2lr2lii
b i Write an equation for the reaction between phosphorus(V) oxide and aqueous sodium hydroxide
solution. r2lii With the aid of two equations, show how aluminium hydroxide exhibits amphoteric behaviour.
t3Ic With reference to the reactions in a and b, describe the variation in the metallic character of the
elements across Period 3 of the Periodic Table (sodium to argon) tzld Suggest, with reasoning, the acid-base character of Indium(III) oxide, InrOr. Indium is the fourth
element down Group 3 of the Periodic Table. I2l(Total L3 marks)
llune 2002 Unit Test 4 question 4l
E I " a * r o D l c r r y ,
q u A N T r r A T r v E E e u r L r B R ' A A N D F U N . T T . N A L G R o u p c H E M r s r R y
6a
b iii
c
d
In the context of Brdnsted-Lowry acid-base behaviour, explain the terms:acid weak acid dilute acid.Define pH
t3ltll
Calculate the concentration, in mol dm-t, of a solution of chloric(I) acid, HOCI, which has a pH of4.23. Chloric acid is a weak acid with K^=3.72 x 10* mol dm-t. t4ISketch, on a grid of pH (y-axis) versus volume of alkali added / cm' (x-axis), the titration curve forthe titration of 25 cm' of the chloric(I) acid solution in b part ii with 50 cm' of a solution ofsodium hydroxide of equal concentration. tsI
I
ll
The concentration of hydrogen ions in a 0.100 mol dm-' solution of sulphuric acid is 0.105 moldm-t.
Write equations to show the two successive ionisations of sulphuric acid, H2SO,, in water. tzlSuggest why the concentration of hydrogen ions is not 0.20 mol dm-'in 0.100 mol dm-' sulphuricacid. t1I
t You must revise Topic 1.5 (lntroduction to oxidation and reduction).
* The sign of E' indicates the direction of spontaneous reaction.
t A useful mnemonic is OIL RIG. When a substance is Oxidised It Losesone or more electrons, and if Reduced It Gains one or more electrons.
o A half equation always contains electrons.
o Half equation data are usually given as reduction potentials, i.e. withthe electrons on the left-hand side.
o An oxidising agent becomes reduced when it reacts.
Things to learn and understand
The standard hydrogen electrode
This consists of hydrogen gas at 1 atm pressure bubbling over a platinumelectrode immersed in a 1 mol dm-' solution of H'ions, at a temperatureof 25 "C. By definition the potential of a standard hydrogen electrode iszero (see Figure 5.1).
Standard electrode potential, E'
f This is the electric potential (EMF) of a cell composed of the standardelectrode connected to a standard hydrogen electrode, measured whenthe concentrations of all the ions are 1 mol dm-3, the temperature is25 "C and any gases are at 1 atm pressure.
O for a metal the standard electrode potential is when the metal isimmersed in a solution of its ions at a concentration of 1 mol dm-t.For the reaction:
Zn'.(aq) +2e- - Zn(s)
it is for a zinc electrode immersed in a 1 mol dm-' solution of Zn2'ions at 25 "C.
o for a non-metal the standard electrode potential is when the non-metal at 1 atm pressure (or a solution of it at a concentration of1 mol dm*) is in contact with a platinum electrode immersed in a
z(g)+_atm
H*(aq)1 mol dm-3
Fig 5.1 A stondord hydrogen electrode
E I t * o N s r r r o N M E T A L = , e u A N T r r A T r v E K r N E T r c s A N D A , , p ' - t E D . R G A N t c c H E M I s r R y
solution of the non-metal's ions at a concentration of 1 mol dm-'at 25 "C.
For the reaction:' l rcl ,(g)+e +Cl-(aq)
it is for chlorine gas, at 1 atm pressure, being bubbled over aplatinum electrode dipping into a 1 mol dm-'solution of Cl- ionsat 25 "C.
o for a redox system of two ions of an element the standardelectrode potential is when a platinum electrode is immersed in asolution containing all the ions in the half equation at aconcentration of 1 mol dm-t.
For the reaction:
Fet.(aq)+e +Fe'.(aq)
it is for a platinum electrode immersed in a solution which is 1 moldm-t in both Fet* and Fe'* ions at 25 "C.
O The equations are always given as reduction potentials, i.e. with theelectrons on the left.
O Comparing substances on the left of two half equations, the one withthe larger positive value of E'is the more powerful oxidising agent (itis the most easily reduced).
o Comparing substances on the right of two half equations, the one withthe smaller positive (or more negative) value of E'is the morepowerful reducing agent.
Calculation of -8..".,"" (E""o)
The value can be deduced in one of three ways:L From a cell diagram:
E "tt
= Eright-hand electrode - Et"rt-n"na electrode
If E""u works out to be a negative value, |ou have written the celldiagram backwards, and so the reaction will go right to left.
2 From an overall equation: write both half reactions as reductionpotentials, i.e. with electrons on the left.
E,"".tio. is calculated as:
E,"u.tio, = (E of half equation of the reactant being reduced) - (E ofhalf equation of reactant being oxidised.
Thus for the reaction:
Zn(s) + Cu'*(aq) +Znz*(aq) + Cu(s)
The half equations are:
Cu2.(aq) + 2e- -- Cu(s) E' = +0.34 V
Zn'.(aq) + 2e- =. Zn(s) E' = -0.76 V
In the reaction the Cu2* ions are being reduced (electron gain), and thezinc atoms are being oxidised (electron loss):
Et,""*.n = +0'34 - (-O'76) = +1'10 V
3 From half equations:
To do this one of the half equations has to be reversed (see Workedexample below). This process alters the sign of its Eo.
'lr}l, + e- + Cl- E+ = +1.36 V' lr}f , + e- + Bi E+ = + 1.07 VAs +1.36 > +1.07, chlorine is astronger oxidising agent thanbromine.
Sno* + 2e- =. Sn'- E* = +0.15 VF e t * + e - + F e ' * f = + 0 . 7 7 VAs +0.15 < +0.77, Sn" ions are abetter reducing agent than Fe'- ions.
O R E D o x E e u r L r B R r A
.:r,i"{FF.;
Work^eoL oxiloLph,
Stoichiometry of an overall equation
The total increase in oxidation number of one element must equal the
total decrease in another.
Spontaneous change
O Electrode potential data can be used to predict the feasibility of a
chemical reaction. A reaction is feasible (thermodynamically unstable) if
E.",, is positive.
o However the rate of the reaction may be so slow that the reaction is not
observed (kinetically stable).
Non-stondord cells
O Non-standard conditions may result in a reaction taking place even if
the standard electrode potential is negative.
I The reaction:
ZCu'.(aq) + 4l-(aq) + 2CuI(s) + I,(aq)
should not work because E' (assuming all species are soluble) = -9.39 y.
However copper(I) iodide is precipitated and this makes [Cu'(aq)] verymuch less than 1 mol dm-'. The equilibrium is driven to the right bythe removal of Cu. (aq) ions, so that E,"u.tio. becomes positive and thereaction takes place.
o Acidified potassium manganate(Vll) will quantitatively oxidise manyreducing agents.
r The procedure is to pipette a known volume of the reducing agent into
a conical flask and add an excess of dilute sulphuric acid.
O The potassium manganate(Vlf solution of known concentration is put
in the burette and run in until there is a faint pink colour.
f This shows that there is a minute excess of the manganate(Vll) ions.
o If the stoichiometry of the reaction is known, the concentration of thereducing agent can be calculated. No indicator is needed as themanganate(vll) ions are very intensely coloured.
ri;.Erag'
When a redox half equation isreversed, its sign must be changed.The number of electrons on the left-hand side of one half equation mustequal the number on the r ight-handside of the other half equation.When a redox half equation ismultiplied, its E+ value is notaltered.Always ensure that the reactantsthat are specified in the question(here they are iron(l l) andmanganate(Vll) ions) are 0n the left-hand side of the final overallequation.
Use the following data to deduce the overall equation and the value of E*,,.o.nfor the reaction between acidified potassium manganate(Vll) and iron (ll) ions.
i MnOi(aq) + 8H-(aq) + 5e- -- Mn"(aq) + 4H,0(aq) Es = +1.52 Vii Fe&(aq) + e- + Fe'-(aq) f = +0.77 V
Answer.Reverse equation (ii) and multiply it by 5. Then add it to equation (i):5Fe'-(aq) + 5Fe'-(aq) + 5e- f = -(+0.77 V) = -0.77 VMn0i(aq) + 8H-(aq) + 5e- + Mn"(aq) + 4H,0(l) E+ = +1 .52V
In the reaction above, the oxidationnumber of the manganese decreasesby 5 from +7 lo +2, and as the ironincreases by 1, there must be 5iron(l l) ions to each MnOo- ion in theoverall equation.
T R A N S I T T O N M E T A L S , Q U A N T T T A T T V E K T N E T I C S A N D A P P L T E D O R G A N I C C H E M I S T R Y
Work^ei' exa*np-lo
The rat io of iron(l l) tomanganate(Vll) ions in the equationis 5:1 and so the number of molesof Fe'- is 5 times the number ofmoles of manganate(Vl l).
The ratio of iodine to thiosulphateions is 1:2, so the moles of iodineare'lrthe moles of thiosulphate.
Iodine/thiosulphate titrations - estimation of oxidisingagents
o The procedure is to add a 25.0 cm' sample of an oxidising agent toexcess potassium iodide solution (often in the presence of dilutesulphuric acid).
The oxidising agent liberates iodine, which can then be titrated againststandard sodium thiosulphate solution. When the iodine has faded to apale straw colour, starch indicator is added, and the addition of sodiumthiosulphate continued until the blue colour disappears.
r Iodine reacts with thiosulphate ions according to the equation:
I' + 2S.O,1- -' 2I- + S*On'-
Work^ei, oxil44f-lt,
Disproportionation
A disproportionation reaction can be predicted by using electrodepotentials. Use the data to work out E.",, for the proposeddisproportionation reaction. If it is positive, the reaction will occur.
25.0 cm'of a solution of iron(ll) sulphate was acidified, and titrated against0.0222 mol dm-'potassium manganate(Vll) solution.23.4 cm3 were requiredto give a faint pink colour. Calculate the concentration of the iron(ll) sulphatesolution.Answer: The equation for the reaction is:
MnOi(aq) + 8H.(aq) + SFe'.(aq) - Mn'*(aq) + 4H,0(l)+ SFe3.(aq)Amount of manganate(Vll) =0.0222x23.411000 = 5.195 x 1Q molAmount of ironlill suldnaie = 5.195 x 10* x 5/_t = 2.597 x 10-' mol _?Concentration of iron(ll) sulphate =2.597 x 10
- + 0.0250 = 0.104 mol dm -
25.0 cm'of a solution of hydrogen peroxide, H,0,, was added to an excess ofacidified potassium iodide solution, and the liberated iodine required 23.8 cm'of 0.106 mol dm-'sodium thiosulphate solution. Calculate the concentration ofthe hydrogen peroxide solution.
Answen The equation for the oxidation of iodide ions by hydrogen peroxide is:Hr0r+2H.+21-- l r+2Hr0
Amount of sodium thiosulphate = 0.106 x 23.8/1000 =2.523 x 10-'molAmount of iodine produced = 2.523 x 10-' x 112 = 1 .261 x 10-' molAmount of hydrogen peroxide = 1.261 x 10-3x 1/1 = 1.261 x 10-3 molConcentration of H,0, = 1.261 x 10-3 + 0.0250 = 0.0505 mol dm-3
O R e o o x E e u r L r B R r A
oz(g)
Fig 5.2 Rusting
Stainless steel is iron containing ahigh proport ion of chromium, whichforms a protective layer of Cr,0,over the whole surface of the metal.lf the surface is scratched, a newprotective layer of Cr,O, is formed.
The reactions for charging the cellare the opposite, and only take placeif a potential >2.05 V is applied, withthe anode being connected to thenegative terminal of the chargings0urce.
ttWrkpj, oxiloLPb
Corrosion (rusting)
This is an electrolytic process. When iron is stressed or pitted, some areasbecome anodic and some cathodic. In the presence of water and oxygenthe following reactions take place (see Figure 5.2).a At the anodic areas iron atoms become oxidised and lose 2 electrons:
Fe(s) -* Fe'*(aq) + 2e-
o The electrons travel through the metal and reduce oxygen :'lrOr(aq) + 2e- + HrO --' 2OH-(ag)
o The Fe'. and the OH- ions meet and iron(Il) hydroxide is precipitated:
Fe'.(aq) + 2OH-(a9) - Fe(OH)'(s)
I Finally oxygen oxidises the iron(Il) hydroxide to iron(lll) oxide (rust):
2Fe(OH)r1s; +'/,O,(aq) - Fe,O.(s) + 2H,O(l)
Prevention of corroslon
This can be done by:O placing a physical barrier between the steel and the environment. Such
barriers are paint, tin or chromium plating.
o adding a sacrificial metal. This can be done by coating with zinc(galvanising), or by attaching blocks of magnesium at intervals.
Storage cells
These store electrical energy as chemical energy. The reactions must befully reversible and the chemicals produced in the redox reactions must beinsoluble.
The leod ocid bottery
When electricity is drawn from the cell (dischargitg), the followingreactions take place.At the anode (oxidation) which is negative:
Pb(s) + SO,'-(aq) 'PbSO,(s) + 2e- Eu = +0.36 V
At the cathode (reduction) which is positive:PbO,(s) +2e- + SO,z-(aq) + 4H.(aQ) - PbSO.(s) + 2H,O(t)
Will copper(l) ions disproportionate in aqueous solution to copper andcopper(ll) ions?
Cu-(aq) + e-+ Cu(s) Es = +0.52 VCu'-(aq) + e- + Cu'(aq) Eo = +0.15 V
Answen Reverse the second equation and add it to the first. This gives theequation:20u-(aq) + Cu*(aq) + Cu(s) E*o,, - +0.52 - (+0.15) = +0.37 VBecause F,,,, is positive, the reaction is feasible, and so aqueous copper(l) ionswill disproportionate.
l. I
92 E i t * o N s r r r o N M E T A L ' , q u A N T r r A T r v E K T N E T I c s A N D A p p L t E D o R G A N t c c H E M I S T R '
.rHo -; I Checklist. , -H(H
Before attempting the questions on this topic, check that you can:
iJ Define standard electrode potential.
-J Describe the construction of a standard hydrogen electrode.
iJ Write half equations and use them to deduce overall equations.
,J Predict the feasibility of redox and disproportionation reactions.
J Deduce oxidation numbers and use them to balance redox equations.
rJ Recall the principles of manganate(Vll) and thiosulphate titrations.
ii Recall the corrosion of iron and its prevention.
'J Understand the chemistry of the tead/acid battery.
Testing your knowledge and understanding
For the following set of questions, cover the margin before you answer,
then check to see if you are correct.I Write the equation representing the reaction that takes place in a
standard hydrogen electrode.
f What are the conditions, other than temperature, for this electrode?
* What are the oxidation numbers of manganese in MnOn- and in Mn'.?
What are the oxidation numbers of sulphur in SO-,2- and in SOn'-?
.r, What is the ratio of SOr'- to MnOn- in the reaction between them?
O Why is it not necessary to have an indicator present in potassium
manganate (VII) titrations?
L a Write ionic half equations for the reduction of:
i CrrO rt to Crt* in acidic solution E' = +1.33 V
ii Snn* to Snr* Eu = +0.15 V
ii i Iodate(V) ions, (IOr) to I, in acidic solution E'= +1.19 V
iv I, to I- E'= +0.54 Vb Write overall ionic equations, calculate E,"u.,,o, values and hence
comment on the feasibility of the reactions between:
i potassium dichromate(Vl) and tin(Il) chloride in acid solution
ii potassium iodate(V) and potassium iodide in acid solution.
2 1.32 g of mild steel filings was reacted with excess dilute sulphuric acid,
and the resulting solution made up to a volume of 25O cm'. 25.0 cm'
samples of this were titrated against 0.0200 mol dm-' potassium
manganate(Vll) solution. The mean titre was 23.5 cm'. Calculate thepercentage of iron in the steel.
3 Write the half equations for the redox reactions involved in the
corrosion of iron.
' /, H,(g) + H-(aq) + e-.State symbols are essential.
H, (g) at 1 atm pressure,[H-(aq)] =1mol dm-'+7 and +2
5:2 So that oxidation numbers ofboth change by 1 0
Because the manganate(Vll) ions areintensely coloured.
The answers to the numberedquestions are on pages 136-137.
@ T n a r u s r r r o N M E T A L c H E M r s r R Y
Neither scandium nor zinc is atransition metal although they are inthe d block, because their ions havethe electronic structure [Ar] 3do, 4soand [Ar], 3d'0, 4so respectively.
Note that the number of d electronsincreases left to right, except that Crand Cu have 4s' electron structures,whereas the others have 4s'.This isbecause stability is gained when thed shell is half ful l or full.
Fig 5.3 Electron arrongement in thehexo-aquoiron(ll) ion
4p4s3d| | | | t l
@ Transit ion ITf;t, ir i ; ; ' ;rnistrY
Introduction'..J You should know the colour of the aqua complex ions and of the
hydroxides of the d block elements scandium to zinc.
iJ you must be able to link the reactions in this topic to the theory of
redox equil ibria, Topic 5.1.
,tHg - r o
, &-- Things to learnH
" d block elements are those in which the highest occupied energy
l e v e l i s a d o r b i t a l .
c A transition element is one that has at least one of its ions with a
partly filled d shell.
*' Electron structure of the atoms
Electron structure of the ions. The element first loses its 4s
electrons when forming an ion. Thus the electron structures of iron and
its ions are:
Fe [Ar], 3d6, 4s2
Fe'* [Ar], 3d6, 4so
Fet* [Ar], 3ds, 4sn
Things to understandProperties of transition metals
Complex ion formotion
o Bonding in complex ions. The simplest view is that the ligands formdative covalent bonds by donating a lone pair of electrons into emptyorbitals of the transition metal ion. These could be empty 3d, 4s, 4p or4d orbitals in the ion (e.g. see the arrangement of electrons in boxes inFigure 5.3).
Sc Ti V Cr Mn Fe Co Ni Cu Zn
3d 1 2 3 J 5 6 7 8 10 10
4s 2 z 2 1 2 2 2 2 I 2
\E Fl--T'f
Fe2* electr on / \ ,rrurro electrons: ,i* puir, /
[Cu(H,O)r] ' . is turquoise blue,[Cu(NH.)*(H,0),1'. is deep blue, and[CuCl.]'- is yellow.
Hydrated cations of charge 4+ ormore do not exist because the sumof all the ionisation energies is toolarge to be compensated for by thehydration energy.Copper forms Cu. and Cu'*, andchromium Cf. and Cr'..
T R A N S T T T O N M E T A L S , Q U A N T T T A T T V E K T N E T | C S A N D A P P L T E D O R G A N T C C H E M I S T R Y
Typical ligands are:
neutral molecules: HrO (aqua), NH. (ammine)
anions: F- (fluoro), Cf (chloro), (CN)- (cyano).
o Shapes of complex ions (see pages 14-15). If the complex ion has 6 pairs
of electrons donated by ligands, the complex will be octahedral. Thisis so that the electron pairs in the ligands will be as far apart from eachother as possible.
Coloured complex ions
a If the ion has partially filled d orbitals, it will be coloured. Sc'* and Ti'*have d0 structures, and Cu. and Zn'* have dto and so are not coloured.
o d orbitals point in different directions in space, and so interact todifferent extents with the electrons in the ligands.
O This causes a splitting of the d orbitals into two of higher energy andthree of lower energy.
I When white light is shone into the substance, a d electron is movedfrom the lower energy to the higher energy level.
o The frequency of the light that causes this iump is in the visible range,so that colour at this frequency is removed from the white light.
t The colour depends on the size of the energy gap which varies with themetal ion and with the type of ligand.
Vorioble oxidotion stote
o Successive ionisation energies increase steadily until all the 4s and the3d electrons have been removed, after which there is a large jump inthe value.
o Formation of cations. The increase between successive ionisationenergies is compensated for by a similar increase in hydration energies.Thus cations in different oxidation states are energetically favourable forall transition metals.
O Formation of covalent bonds. Transition metals can make available avariable number of d electrons for covalent bonding. The energyrequired for the promotion of an electron from a 3d to a higher energyorbital is compensated for by the bond energy released.
o Formation of oxoions. In ions such as MnOi, VOr-, VOi and VO2-, theoxygen is covalently bonded to the transition metal which uses avarying number of d electrons.
O Manganese exists:
in the +2 state as Mnt*in the +4 state as MnO,in the +6 state as MnOn'-in the +7 state as MnOn-.
Catalytic octlvity
Tlansition metals and their compounds are often good catalysts.O Vanadium(V) oxide is used in the oxidation of SO, to SO, in the contact
process for the manufacture of sulphuric acid.
o Iron is used in the Haber process for the manufacture of ammonia.
O Nickel is used in the addition of hydrogen to alkenes (hardening ofvegetable oils).
I
Reactions of transition metal compounds
Reoctions with sodium hydroxide ond ommonio solutions
The table gives a summary of these reactions.
" The correct formula for the ions should be the hexa-aqua ion, except for zinc, which forms a tetraqua ion.o The precipitate of Mn(OH), goes brown as it is oxidised by air.' The precipitate of Fe(OH), goes brown on the surface as it is oxidised by the oxvgen in the air.o The precipitate of Cu(OH). goes black as it loses water to form CuO.
Green pptSandy pptbDirty green ppt'Foxy red pptGreen pptBlue pptdWhite ppt
Green solutionPpt remainsPpt remainsPpt remainsPpt remainsPpt remainsColourless solution
Green pptSandy pptoDirty green ppt'Foxy red pptGreen pptPale blue pptWhite ppt
Ppt remainsPpt remainsPpt remainsPpt remainsBlue solnDeep blue solnColourless soln
E l t * o N s r r r o N M E T A L S , e u A N T r r A T r v E K r N E T r c s A N D A p p ' r E D . R G A N r c c H E M r s r R y
Vanadium chemistry
o Vanadium compounds exist in four oxidation states:
+5: VOr-(aq) which is colourless, but in acid solution reacts to form ayellow solution of VOr.(aq) ions:V O ; + 2 H * + V O r * + H r O
+4: VO'.(aq), which is blue.+3: V'.(aq) which is green and is mildly reducing.+2: V'.(aq) which is lavender and is strongly reducing.
o Reduction by zinc.lf zinc and dilute hydrochloric acid are added to asolution of sodium vanadate, NaVO' the colour changes in distinctsequence, as it is steadily reduced from the +5 to the +2 state:
start: yellow caused by VO'.(aq)ions (+5 state)then: green caused by a mixture of yellow VOr.(aq) and blue
VO'z.(aq)then: blue caused by VO'z.(aq) ions (+4 state)then: green caused by V't(aq) ions (+3 state)finally: lavender caused by V'.(aq) ions (+2 state)
t Redox reactions
Vanadium compounds can be oxidised or reduced by suitable reagents. Fora redox reaction to work, the E'.",, value must be positive. A list of E€valuesfor the half reactions of vanadium in its various oxidation states and forsome oxidising and reducing agents is given below. These values show, forinstance, that vanadium(V) will be reduced only to vanadium (IV) by Fe'*ions, whereas Snt* ions will reduce it first to vanadium (lV) and thento vanadium (l l l).
this topic, check that you:
E Can define d block and transition elements.
[,] Cun write the electronic structure of the d block elements and their ions.
-H'?\i t Checklistl / /H
H
Before attempting the questions on
+5 VO,*
VO,'
v'*
+5AV O r * + 2 H * + e + V O ' . + H r O
Fe'* +e - -Fe '*' l rc l r+e- +c l -
v o ' * + 2 H * + e - + v t * + H r oSnn* +2e- +Sn2*Fet* +e +Fet*
V t * + e + V ' *Zr ' t ' * +2e- +ZnH* + e- + ' l rH ,
E€ = +1.0 VE' = +O.77 YE ' = + 1 . 3 6 V
Eu = +0.34 VE '= +0 .15 VEu = +O.77 Y
E' = -0.26 VE' = -0.76VEu = 0.00 V
Fet* CT,
V+4
Sn'*
+4
AFe'*
+3
4.H'
+2
V,*V
+3
'I znlH.I
V+2
8t..,, for the reduction of:VOr* by Fe'. = +1.0 - (+0.77) - +O.23 Y
VO'* by Sn'* = +0.34 - (+0.15) = +0.19 V
v'. by Zn - -0.26 - (4.7 6) = +0.50 V
Et ",,
for the oxidation of:V'* by H. = o.o - (-0.26)
V3* by Fe3* = +O.77 - (+0.3a)
VO'* by Cl, = +1.36 - (+1.0)
- +0.26 Y
= +0.43 V
= +0.36 V
@ T R r N s r r r o N M E T A L c H E M r s r R Y
The answers to the numberedquestions are on pages 137-138.
rJ Can recall the characteristic properties of the transition elements.
J Understand the nature of the bonding in complex ions, the shape ofthe ions and their colour.
lJ Can recall the action of aqueous sodium hydroxide and ammonia onsolutions of their aqua ions.
J Understand deprotonation and ligand exchange reactions.
rJ Can recall the oxidation states of vanadium, and the colours of its ions.
J Know reactions to interconvert the oxidation states of vanadium.
Can recall examples of vanadium, iron and nickel and their compoundsas catalysts.
IHo
," .Testing your knowledge and understanding- o - (
For the following set of questions, cover the margin of the page before vouanswer, then check to see if you are correct.r' Write down the electronic structure of a vanadium atom and a \''- ion.
State the formula and colour of the aqua complex of copper(ll).
* State the formula and colour of the ammine complex of copper(ll).
o Name the types of bonding in the aqua complex of iron(Il).
o What is the shape of the iron(Il) aqua complex ion?
o State the type of reaction that occurs when aqueous sodium hydroxideis added to a solution of the aqua complex of copper(Il).
o State the type of reaction that occurs when excess aqueous ammonia isadded to a solution of the aqua complex of copper(Il).
o What are the oxidation states of vanadium? Give the formula and thecolour of the cation for each oxidation state.
o Give an example of the use of iron as an industrial catalyst.o Give an industrial use of a vanadium compound as a catalyst.
Explain why the [Cu(H,O)u)'. ion is coloured.
Give the equations for the reactions caused by small additions ofsodium hydroxide solution, followed by excess to:
a a solution of the aqua complex of chromium(Ill)b a solution of the aqua complex of iron(Il)c a solution of the aqua complex of zinc(Il).Give the equations for the reactions caused by small additions ofammonia solution, followed by excess to:
a a solution of the aqua complex of iron(III)b a solution of the aqua complex of copper(Il).Write the half equations for the following changes in oxidation state:
a vanadium(V) to vanadium(Il)b vanadium(V) to vanadium (IV)c vanadium(Ill) to vanadium(IV)
1
2
4
v is [Ar], 3d', 4s'. v'- is 1Ar1, 3d'
ICu(H,0),]'- tu rquoise-blue,
[Cu(N H,),(H,0),1' . dark blue
Dative covalent (ligand to ion) andcovalent (within the water molecule)
Vanadium(V) oxide in the Contactprocess to manufacture sulphuricacid
l r l
E l T n n N s t r t o N M E T A L S , e u A N T r T A T r v E K r N E T r c s A N D A p p l - l E D o R G A N l c c H E M I S T R Y
@ Organic chemistry l l lffiI Things to learn and understandlry
The structure of benzene and reactions of aromaticcompounds
o The benzene ring does not consist of alternate double and single bonds.
o All the bond lengths are the same and the molecule is planar.
t There is an overlap of p orbitals above and below the plane of themolecule, forming a continuous or delocalised n system.
o This gives stability to the benzene structure which is why it undergoessubstitution rather than addition reactions.
o Benzene can be represented either by:
Reactions of benzene
Benzene reacts mainly by electrophilic substitution:O Nitration. Benzene reacts with a mixture of concentrated nitric and
sulphuric acids to form nitrobenzene:
CuHu + HNO, - CuHrNO.,+HrO
I Bromination. Benzene reacts with liquid bromine in the presence of acatalyst of anhydrous iron(III) bromide (made in situ from iron andliquid bromine) or of anhydrous aluminium bromide:
CuHu + Brr(l) 'CuHrBr + HBr
o Friedel-Crafts reaction. Benzene will react with halogenoalkanes orwith acid chlorides in the presence of a catalyst of anhydrousaluminium chloride:
CuHu + CrHrCl + CuHrCrH, + HCIethylbenzene
CuHu + CHTCOCI - C'HTCOCH, + HCIphenylethanone (a ketone)
Reactions of compounds with a carbon-containing sidechain
o Compounds such as methyl benzene C6H'CH3 and ethyl benzeneC6HsC2Hs can be oxidised, and the product will contain a COO- groupon the benzene ring, regardless of the number of carbon atoms in thechain. On acidification benzoic acid is produced:
CuHsCzHs + 6[0] + OH-- C'HTCOO- + CO, + 3HrO
conditions: heat under reflux with potassium manganate(Vll) andaqueous sodium hydroxide.
@ o n c e r u r c c H E M r s r R Y l l l
Phenol is a weaker acid thancarbonic acid, and so it will notliberate C0, from sodium hydrogencarbonate, unlike carboxylic acidswhich wil l .
The solution of diazonium ions canreact with phenol. A yellowprecipitate of diazo compound isobtained.
Phenol
OH
Although it forms some hydrogen bonds with water, it is only partiallysoluble and forms two layers when added to water at room temperature.o + alkali. Phenol is a weak acid and it forms a colourless solution with
aqueous sodium hydroxide:
C6HsOH(l) + OH-(a9) - CuH'O-(aq) + H,O(l)
O + bromine water. The presence of the OH group makes substitutioninto the benzene ring much easier. No catalyst is required:
OH OH
A nr--r\',,nr
| ! |J (aq)+3Brr(aq) +
V
(s)+3HBr(aq)
Br
observation: when brown bromine water is added, a white precipitate israpidly formed.
I + acid chlorides. Phenol reacts as an alcohol, and an ester is formed:
C6H'OH + CHTCOCI' CHTCOOC'H, + HCI
Phenylamine
Phenylamine has a NH, group attached to the benzene ring.
Preporotion
Nitrobenzene is reduced by tin and concentrated hydrochloric acid whenheated under reflux. Then sodium hydroxide is added to liberate thephenylamine, which is removed by steam distillation.
Reoctions
O + acid
C6H'NH2 + H.(aq) - CuHTNH'.(aq)
o + nitrous acid
C6H'NH2 +2H* +NOr--CuHrNr. +ZH.O
conditions: add dilute hydrochloric acid to phenylamine and cool thesolution to 5 'C. Sodium nitrite solution is then added, keeping thetemperafure close to 5 "C and above 0'C.
Ensure that the curly arrow startsfrom a bond or an atom with a lonepair of electrons.Ensure that the arrow pointstowards an atom either forming anegative ion or a new covalent bond.
Another way of polymerising etheneis by using titanium tetrachlorideand aluminium triethyl, but this isnot a homolytic mechanism.
With the addition of a hydrogenhalide to an unsymmetrical alkene,the hydrogen atom adds to thecarbon which already has morehydrogen atoms directly bonded toit.This is because the secondarycarbocation,
t t le.o. -C-C-C--
l @ lis more stable than a primary one.
t t le,o. -C-C-O O- t t l
ABr- Br
\H,C - CH,
l T ' f l ]L"r.
- C*HrJ
Br Br
t lH2C- CHz
Heterolytic, electrophilic substitution
Nltratlon of benzene
1 Sulphuric acid is a stronger acid than nitric and so protonates it:HrSOn + HNO, - HrNOr'+ HSO;
then the cation loses water:H2NO3* tH rO+NOr*
E l t * o N s r T r o N M E T A L ' , e u A N T r r A T r v E K r N E T r c s A N D A p p ' t E D . R G A N I c c H E M t s r R y
Mechanisms
A curly arrow (fJ ) represents the movement of a pair of electrons, eitherfrom a bond or from a lone pair. A half-headed arrow (f\) represents themovement of a single electron.
Homolytic, free radical substitution
Reoction befrnreen on olkone ond chlorlne or bromine
O An example is shown of reaction between methane and chlorine.
O There are three stages in this type of reaction:
1 Initiation: light energy causes homolytic fission of chlorine:Clr- 2g,1'
2 Propagation: each propagation step involves a radical reactingwith a molecule to produce a new radical:
CHn + Cl'- CHr' + HCl
then CHr' + Cl, - CH.CI + Cl' etc.
3 Termirtation: involves two radicals ioining with no radicals beingproduced, €.S.
C H r ' + C H r ' t C r H u
Homolytic, free radical addition
Polymertsotlon of ethene
The conditions are a very high pressure (1000 atm) and a trace of oxygen.The conditions produce radicals, R', which attack an ethene molecule:
R' + H,C=CH, - R-CH2-CH''
then R-CHr-CH 2' * H2C=CH, -- R-CH'-CHr-CH,-CHr' etc
Heterolytic, electrophilic addition
Reoctlons of olkenes wlth hologens or hydrogen holldes
The reaction takes place in two steps:
1 The electrophile accepts the n electrons to form a new bond withone of the carbon atoms and at the same time the Br-Br bondbreaks.
2 The intermediate carbocation then bonds with a Bf ion.
O o R c a r u t c c H E M r s r n v i l l
Similar mechanisms occur forbromination, where the catalystreacts with bromine to provide theelectrophile Br-, and in the Friedel-Crafts reaction where the catalystreacts with the halogen compoundto provide electrophiles such asCH3C'0 or C,Hu-.
In the transition state, the new 0-Cbond forms as the C-Br bondbreaks. The transition state has acharge of -1.
Any optical activity is maintained ina Sr2 reaction.lf the mechanism is Sr1, a reactantwhich is an optical isomer wil tproduce a racemic mixture, as thecarbocation can be attacked fromeither side.
The CN- ion is a catalyst, and theconditions must be such that thereis a significant amount of both HCNmolecules and CN- ions present, i .e.a mixture of HCN and KCN.
23
The Nor. ion is the electrophile and attacks the benzene ring.The HSon- ion pulls off a H. and reforms H2son (the catalystf
The loss of H. results in the reforming of the benzene ring and the gain instability associated with the ring.
Heterolytic, nucleophilic substitution
Reoction between hologenoolkonr ond hydroxide or cyonide ions' There are two mechanisms depending on the type of halogenoalkane.
with primary (1") halogenoalkanes, an S*2 mechanism is dominant.' Reaction proceeds via a transition state when the lone pair of electrons
on the nucleophile attacks the halogenoalkane.
t{' t 9H,
-l - ,cH,
Q l l l /Ho:' .:-g ' s' ------+
lno- I - .ur l---------+ Ho-c. :BiH l | / \ | l "H t*u.rr?rro.,
tu*' H
o with tertiary (3") halogenoalkanes the gl mechanism is dominant.This happens in two steps:1 The halogenoalkane ionises in the relatively slow, and hence rate
determining step, to form an intermediate carbocation.
CHt
I r-rc H r - C - B r
ICH,
CH,
t r \CH?- C* t OH-
ICHg
CH,
ICHr-a" + :Br-
ICHt
CHt
IC H e - C - O H
ICHt
2 This then rapidly forms a bond with the nucleophile, oH-.
Heterolytic, nucleophilic addition
Reaction between o corbonyl compound and hydrogen cyonideo Although the reaction is the addition of HCN, the first step is the
nucleophilic attack by the cN- ion on the 6+ carbon atom in thecarbonyl group.
E l t * o N s r T r o N M E T A L = , e u A N T r r A T r v E K r N E T r c s A N D A p p ' r E D . R G A N I c c H E M r s r R y
O The negatively charged oxygen in the anion then removes an H* froman HCN molecule. This produces another CN- ion, which reacts withanother carbonyl group and so on.
A A-o t u-cNI
R _ C _ H +
ICN
o - HI
R - C - H + C N -
ICN
aot* ' /
R - C - +
r-\\ HcN-
o The conditions for this reaction are either HCN and a trace of base orKCN and a small amount of dilute sulphuric acid.
Reactions of aromatic substances (Figure 5.4) and mechanisms (Figure 5.5)are summarised schematically.
conc HNO3 ----------+ C6H'NO2conc H2SO4
Brt CuHrBrFe catalyst
c2Hs clAlClrcatalyst
cH3cocl c6Hscocl{3AlCl3catalyst
Phenolc6l{soH
OH-(aq)-CoHsg
Brr(aq) --> CuHrBr.OH
cH3cocl --------------> cH3cooc6Fls
Phenylamine [- rf(aq) --------------> c6IlNHJ
c6HsNH2 |HNO2- CoI{sNz-a t 5 o C
Fig 5.4 Summary of reactions of oromatic substances
Substitution Addition
Free radical: alkanes + Cb Free radical: polymerisation of alkenes
ffiChecklistBefore attempting the questions on this topic, check that you:
[J Understand the structure of benzene and why it reacts by substitutionrather than by addition.
O o R c l r u r c c H E M r s r R Y l l l
The answers to the numberedquestions are on page 138.
U Can recall the reactions of benzene with nitric acid, bromine andhalogen compounds.
r.J Can recall the product of oxidising side chains.
U Can recall the reactions of phenol with alkali, bromine and acidchlorides.
i.l Can recall the preparation of phenylamine and its reaction with nitrousacid and the coupling of the product with phenol.
iJ Understand the mechanisms of free radical substitution and addition.
tJ Understand the mechanisms of electrophilic substitution and addition.
if Understand the mechanisms of nucleophilic substitution and addition.
-H| I . ; g '
"4. Testing your knowledge and understanding
For the following set of questions, cover the margin before you answer,
then check to see if you are correct.* State the structural formulae of the organic product obtained by the
reaction of:
a propylbenzene with alkaline potassium manganate(Vll)
b phenol with sodium hydroxide solution
c phenol with propanoyl chloride.
o What would you observe if bromine water were added to aqueousphenol?
O State the conditions for the conversion of nitrobenzene tophenylamine.
a Describe the mechanism for the reaction of bromine with benzene.
b Why does benzene undergo a substitution reaction with brominerather than an addition reaction?
In the reaction between propanone and hydrogen cyanide no reaction
occurs unless a small amount of a base such as sodium hydroxide is
added. Explain these observations.
When benzenediazonium chloride is prepared, the reaction is carried
out at about 5 "C.
a Why are these conditions chosen?
b What is the formula of the product obtained by the reaction of thebenzene diazonium chloride solution with phenol in alkaline
solution?
abc
cuHuc00-CuHuO- Na-c,Huc00cuHu
The brown bromine water would gocolourless and a white precipitatewould form.
Heat under reflux with tin andconcentrated hydrochloric acid, cooland then add aqueous sodiumhydroxide.
E l t * o N s r r r o N M E T A L ' , e u A N T r T A T r v E K r N E T r c s A N D A p p L I E D . R G A N r c c H E M I s r R y
Chemical kinet ics l lffib Introduction
You must also revise Topic 2.3 in Chapter 2, page 44 and especially howthe changes in the Maxwell-Boltzmann distribution of energy withincrease in temperature affect the rate of reaction.
daLN Things to learn
The units of k are:I for a zero-order reaction:
mol dm-'s-'t for a first-order reaction: s-'I for a second-order reaction:
mol-t dmt s'
The half-life of a first-order reactionis constant.
f fne rate equation for the reaction:
xM + /N - products, is:
Rate of reaction - k tMlj [N]b where a and b are integers and areexperimentally determined.
i The rate constant, & is the constant of proportionality in the rateequation:
I Its value depends on the activation energy of the reaction and thetemperature.
o Reactions with a large activation energy will have small values of k.
E fne order with respect to one substance is the power to whichthe concentration of that substance is raised in the rate equation. In theexample above the partial order of the chemical M is a.
E fne order of reaction is the sum of the partial orders. In theexample above, the order of the reaction is a + b.
f fne activation enert[y, Eo, is the total kinetic energy that themolecules must have on collision in order for them to be able to react.
E naff-life, tr,r, is the time taken for the concentration to fall from anyselected value to half that value.
ffi Things to understandiry
Determination of rate equation from initial rates
Consider the reaction:A + B + e + p r o d u c t s
The initial rates of reaction with different concentrations of A, B and C arefound, and by taking the experiments two at a time, the order with respectto each substance can be found.Look for two experiments where the concentration of only one substancevaries. If doubling that concentration causes the rate to double, the orderwith respect to that substance is 1.
@ C H E M T . A L K r N E T r c s t l
100
o From experiments l- and 2: [A] doubles and rate doubles. Thereforeorder with respect to A = 1
o From experiments L and 3: [B] doubles and rate x 4. Therefore orderwith respect to B = 2
o From experiments 3 and 4: [C] doubles and rate unaltered. Thereforeorder with respect to C = 0
I Rate of reactiorr = k [A]' [B]' [C]0, and the reaction is 3rd order.
Experimental techniques for following a reaction
1 Titration method. Mix the chemicals and start the clock. At intervalswithdraw a sample, add it to iced water to slow the reaction. Thentitrate one of the substances in the reaction. This method can be used ifan acid, an alkali or iodine is a reactant or a product.
2 Colorimetric method. This can be used when either a reactant or aproduct is coloured (iodine or potassium manganate(Vll) are examples).The colorimeter must first be calibrated using solutions of the colouredsubstance of known concentrations. Then the reactants are mixed andthe clock started. The intensity of the colour is measured as a functionof time. The concentration of the coloured substance is proportional tothe amount of light absorbed.
3 Gas volume method. If a gas is produced in the reaction, its volume canbe measured at intervals of time. The gas can be collected in ahorizontal gas syringe or by bubbling it into an inverted measuringcylinder filled with water.
Graphs of results
If the concentration of a reactant is plotted against time, three shapes ofgraph are likely depending on the order of the reaction.
Mechanisms
A suggested mechanism must be consistent with the order of reaction. Thepartial order of any species which occurs in the mechanism after the ratedetermining step will be zero. The rate-determining step is the sloweststep, e.g.
Nor(g) + Co(g) -' No(g) + Cor(g) has the rate expression:
Ra te=kD{O, l ' [CO] "
A suggested mechanism is:o Step 1 (slow):
NOr+ NO, -- NOr + NO
o Step 2 (fast):
N O r + C O - N O r + C O ,
As the CO enters the mechanism aftter the rate determining step, themechanism is consistent with the fact that the reaction is zero order withrespect to CO.
E 6 00,)
H 4 0-l
20
Experiment tAl tBl tcl Relative rate
1234
1z11
1122
1112
1z44
Note that for the first-order graph,the half-life remains constantthroughout the reaction, whereas forthe second-order the half-lifeincreases as the concentration falls.
* Zero order* Lst order* Znd order
E l t * o N s r r r o N M E T A L S , e u A N T r r A T r v E K r N E T r c s A N D A p p L t E D . R G A N t c c H E M I s r R y
Energy profile diagrams
The first diagram is for a reaction that takes place in a single step, and the
second diagram is for this reaction with a catalyst (see Figure 5.6).
Fig 5.6 Energy profile diagroms
An example of a reaction with a transition state is the S*2 reaction ofbromoethane with hydroxide ions.An example of a catalysed reaction forming an intermediate is Fez* ions as
a catalyst in:SrOrz-(aq) + 2l-(aq) - 2SO n'-(a9) + Ir(aq)
In this mechanism the catalyst first reduces the SrO.'- ions:ZFe'.(aq) + SrOr'-(a9) - 2Fe'.(aq) + 2SO.,'-(aq)
then the Fe"* ions are reduced back to Fe2* ions:2Fe3.(aq) + Zl-(aq)'- ZFe'*(aq) + Ir(aq)
Hc , g
*4- ChecklistBefore attempting the questions on this topic, check that you:
J Can define rate constant, order of reaction and half-life.
iJ Can deduce rate equations from initial rate data.
iJ U.tderstand the concept of activation energy and its relation to the rateconstant.
E Understand that information about mechanisms can be deduced fromthe partial orders of the reactants.
iJ Can recall the energy profiles of reactions with and without catalysts.
ff Ca.r suggest suitable methods for following reactions.
f Can deduce the order of a reaction from concentration/time graphs.
>.bo
r t l
botrq.)
: - - - - - - - - T rans i t ion
Intermediate
Catalysed
O c n e M r c A L K r N E T r c s t l
The answers to the numberedquestions are on page 138.
Testing your knowledge and understandingFor the following set of questions, cover the margin before you answer,then check to see if you are correct.O The following results were obtained from a study of the reaction:
NO,(g) + CO(g) -' NO(8) + CO,(g)
What is the order with respect to NOr?What is the order with respect to CO?What is the order of reaction?State the rate expression.
o Reaction A has a high value of E., and reaction B has a lower E. r'alue.
a Which reaction has the larger rate constant?b Which is the faster reaction?
o A reaction takes place in 3 steps. The E^lkI mol-' for each step are:
step 1: 32, step 2: 51, step 3: 20
Which step determines the rate?
O The decomposition of NrO, is first order. At 200 "C the reaction has ahalf-life of 25 minutes. Calculate how long it will take for theconcentration of NrO, to fall to 6.250/o of its original value.
L The rate of the second order reaction:
2HI(g) -- H,(g) + I,(8)
is 2.0 x 10' mol dm-' s-' when [HI] = 0.050 mol dm-' at 785 K. Calculate
abcd
the value of the rate constant, giving its units.
Describe how you would follow, at 60 'C, the rate of the reaction:
CH3COOH(aq) + CH'OH(aq) - CH,COOCH.(|) + H,O(l).
The decomposition of 3-oxobutanoic acid, CH3COCHTCOOH, wasstudied:
CHTCOCHTCOOH - CH.COCH. + CO,
The results, at 40'C, are tabulated below.
Time/min [3-oxobutanoic acid]/mol dm-'
0265278
t . 60.80.4o.2
a From experiments 1 and 2, orderwith respect to N0, = |
b From experiments 1 and 3, orderwith respect to C0 = 0
c 0rder of reaction is 2 + 0 = 2d R a t e = k [ N 0 , ] '
TI E l t * o N s r r r o N M E T A L ' , e u A N T r r A T r v E K r N E T r c s A N D A p p ' r E D . R G A N t c c H E M t s r R y
@ Organic chemistry lV
Introduction
I Questions on this topic will form part of the synoptic assessment inUnit Test 5.
I This Topic brings together all the organic chemistry that has beencovered in earlier topics, so you will need to revise Topics 2.2, 4.5and 5.3.
Things to learn and understand
Organic analysis
You should know the tests for:s Alkenes
Test: Add bromine in hexane.Observation: The brown bromine becomes colourless.
o llalogenoalkanesTest: Heat under reflux with sodium hydroxide solution, then
acidify with dilute nitric acid. Add silver nitrate solution.Observation: Chlorides give a white precipitate that is soluble in dilute
ammonia.Bromides give a cream precipitate that is insoluble indilute ammonia but dissolves in concentrated ammonia.Iodides give a yellow precipitate that is insoluble inconcentrated ammonia.
O OH group (in alcohols and acids)Test: To the dry compound add phosphorus pentachloride.Observation: Steamy fumes of hydrogen chloride produced.
I AlcoholsTest: Warm with dilute sulphuric acid and potassium
dichromate(Vl) solution.Observation: Primary (1') and secondary (2')alcohols reduce the
i11i#1HT,::'{:il ;"""::? $::i,?i .m:r as,hey arenot oxidised.To distinguish between 1o and 2o rcpeat the experiment,but distil the product into ammoniacal silver nitratesolution:1' alcohols are oxidised to aldehydes, which give a silvermirror.2'alcohols are oxidised to ketones, which do not react.
I Carbonyl, C=O group (aldehyde or ketone)Test: Add a solution of 2,4-dinitrophenylhydrazine.Observation: A red or orange precipitate is seen.
i
II
O o R c l l u c c H E M r s r R Y l v
The solid must be purified byrecrystallisation in order to have asharp and accurate meltingtemperature.
When identifying species that causelines in the spectrum, always give astructural formula for the ion and its+ charge.
Fig 5.7 Mass spectrum of ethonol
The peaks given by 0-H and N-H areusually broad owing to hydrogenbonding.
To distinguish between aldehydes and ketones:Test: Warm with ammoniacal silver nitrate solution.Observation: Aldehydes give a silver mirror, ketones have no reaction.
O Iodoform reactionTest: Gentlv warm with a solution of sodium hydroxide and
iodine.Observation: A pale yellow precipitate of CHI,
This test works with carbonyl compounds containing theCH,C=O and alcohols containing the CH'CH(OH) groups.
I Carboxylic acids, COOH groupTest: Add to a solution of sodium hydrogen carbonate.
Observation: Bubbles of gas, which turn lime water milky.
Interpretation of data
You should be able to deduce structural formulae of organic moleculesgiven data obtained from:r Chemical methods
Carry out tests as above to determine the functional groups in themolecule. Measure the melting point of the substance. Make a solidderivative (such as the 2,4-DNP derivative from carbonyl compounds),purify it, and measure its melting temperature. Check the meltingtemperature values with a data bank to identify the substance.
o Mass spectra (see Figure 5.7)o Observe the fragments obtained and look for the value of the
molecular ion (M)+, if any.o See if there is a peak at (M-15)+. If so the substance probably had a
CH, group which is lost forming the (M-15)+ peak.I If there is a peak at (M-28)+, the substance probably had a CO group.
Other fragments will help to identify the structure.
100
b 8 0(tt
9 6 0
.E 40
&, 20
451 510 20 25 30m/z
35 40
O Infrared spectra (see Figure 5.8)o These are used to identify functional groups in the substance, and
also to compare the spectrum of the unknown with a data bank ofspectra ('finger printing').
O Peaks to look for are at:1650 to 1750 cm-' given bY C=O2500 to 3500 cm-' given by O-H3300 to 3500 cm-' given by N-H1000 to 1300 cm-' given by C-O
E l t * o N s r r r o N M E T A L ' , e u A N T r r A T r v E K r N E T r c s A N D A p p L r E D . R G A N r c c H E M r s r R y
100
Fig 5.8 lnfrored spectrum of ethonol
2000 1s00 1000Wavenumber / cm-r
sOU
g s o(a
(t)
Lrt-
O NMR spectrao The phenomenon of nuclear magnetic resonance occurs when nuclei
such as 'H are placed in a strong magnetic field and then absorbapplied radio frequency radiation.
t The nuclei of hydrogen atoms in different chemical environmentswithin a molecule will show up separately in a NMR spectrum. Thevalues of their chemical shift, 6, are different.
o The hydrogen nuclei in a CHr group will have a different chemicalshift from those in a CH, or in an OH group. The value of d of thepeak due to the hydrogen in OH (or in NH) depends upon thesolvent.
* In low resolution NMR, each group will show as a single peak, andthe area under the peak is proportional to the number of hydrogenatoms in the same environment. Thus ethanol, CH.CHTOH will havethree peaks of relative intensities 3:2:1 and methyl propaneCH.CH(CH3)CH3, will have two peaks with relative intensities of 9:1.
O In high resolution NMR spin coupling is observed. This is causedby the interference of the magnetic fields of neighbouring hydrogennuclei. If an adiacent carbon atom has hydrogen atoms bonded toit, they will cause the peaks to split as follows:
The chemical shift is the differencebetween the absorption frequenciesof the hydrogen nuclei in thecompound and those in thereference compound.
Fig 5.9 High resolution NMR spectrumof ethonol
1 neighbouring H atom2 neighbouring H atomsn neighbouring H atoms
peak splits into 2lines (a doublet)peak splits into 3 lines (a triplet)peak splits into (n + 1) lines.
Thus ethanol gives three peaks (see Figure 5.9):1 peak due to the OH hydrogen, which is a single line
(as it is hydrogen bonded)1 peak due to the CH, hydrogens, which is split into four lines
by the three H atoms on the neighbouring CH, group.1 peak due to the CH, hydrogens, which is split into three lines
by the two H atoms on the neighbouring CHr group.
O oRc l ru rc cHera rs rnY IV
A common error is to think thatKCN or HCN will react withalcohols to form RCN. You mustfirst convert ROH to R-halogen.Many questions on organicsynthesis require, somewhere inthe sequence, the conversion ofthe starting substance to acarbonyl compound or ahalogenoalkane.
Recrystallisation is only suitable forpurifying solids.
When a specific safety precaution isasked for, do not give the use ofsafety glasses, lab coats etc. asexamples. The answer is probably'no naked flames (ether solvent)' or'carry out the experiment in a fumecupboard', in which case the specifichazard such as toxicity must bestated.
a Visible and fIV spectrao z bonds will absorb light energy as an electron is promoted from a
bonding to an anti-bonding n orbital.O The energy gap is in the near UV or visible range, and the group
causing this is called a chromophore.O If the n bonded electrons are linked to a conjugated system (alternate
double and single carbon-carbon bonds), the absorption will shift toa lower frequency, and the substance may be coloured. The colourobserved will be the complementary colour of that absorbed.
Organic synthesis
Reogents ond conditions for the preporatlon oJ substonce B from A
If the number of carbon atoms in the chain is:o increased by one, consider:
a halogenoalkane with KCNb carbonyl with HCNc the Grignard reagent CHrMgBrd a Grignard reagent added to CO, or methanal
o increased by more than one, consider:a a Grignard reagent with more than one carbon atomb Friedel-Crafts reaction for aromatic substances
o decreased by one, consider:a the Hofmann degradationb the iodoform reaction.
Practical techniques
O Recrystallisation. Dissolve the solid in a minim m of hot solvent.Filter the hot solution through a preheated funnel using fluted filterpaper. Allow to cool. Filter under reduced pnsssurle (Buchner funnel),wash with a little cold solvent and allow to dry. The solid should havea sharp melting temperature.
o Heating under reflux is necessary when either the reactant has a lowboiling temperature or the reaction is slow at room temperature.
r Safety precaution. You must use a fume cupboard when a reactant orproduct is toxic, irritant or is carcinogenic.
o Fractional distillation. This is used to separate a mixture of twoliquids. These mixtures can usually be separated into pure samples, butif the boiling points are too close, a good separation will be difficult.
If a liquid of composition X is heated, it will boil at Tr 'C to give a vapourof composition Y (see Figure 5.10). If this is condensed and reboiled, it willboil at a temperature of T, 'C giving a vapour of composition Z. Eventuallypure B will distill off the top and pure A will be left in the flask.
Applied organic chemistry
o Targeting pharrnaceutical compounds. Those which are ionic orhave several groups that can form hydrogen bonds with water, will tendto be retained in aqueous (non-fatty) tissue. Compounds with longhydrocarbon chains and with few hydrogen bonding groups will beretained in fatty tissue. The latter will be stored in the body, whereaswater-soluble compounds will be excreted.
o Nitrogenous fertilisers are of three types:1 quick release, such as those containing NO; and NHn. ions2 slow release, such as urea, NHTCONH,3 nafural, such as slurry, compost and manure.The first two are water-soluble and can be leached out. The last is bulkyand contains little nitrogen.
(ul-r
E T 1HC,.'
t r 17.a)
loOo/o X YA
Z TOOo/oB
Fig 5.lO Boilingte m pe rotu re / com position diag ro m
Liquid
E l t * o N s r r r o N M E T A L ' , e u A N T r r A T r v E K r N E T r c s A N D A p p L r E D . R G A N t c c H E M I s r R y
o Esters, oils and fats. Simple esters are used in food flavourings,perfumes and as solvents for glues, varnishes and spray paints. Animalfats are esters of propan-t,2,3-triol and saturated acids such as stearicacid C,'H3'COOH. When hydrolysed by boiling with aqueous sodiumhydroxide, they produce soap which is the sodium salt of the organicacid.Vegetable oils are liquids which are esters of propan-I,Z,3-triol andunsaturated acids.Margarine is made from polyunsaturated vegetable oils by partialhydrogenation (addition of hydrogen) so that only a few double bondsremain.
O Polymers are of two distinct chemical types:1 Addition. The monomers contain one or more C=C groups.
Polymerisation is the addition caused by breaking the n bonds, andthe polymer and the monomer have the same empirical formula.Examples are poly(ethene), poly(tetrafluoroethene) (PTFE) andpoly(propene).
2 Condensation. Both the monomers have two functional groups,one at each end. Polymerisation involves the loss of a simplemolecule (usually HrO or HCI) as each link forms. Examples arenylon, a polyamide (see Figure 5.11), and terylene, a polyester.
a diacid chloride a diamine
Synthetic polymers are not easily biodegraded and can cause anenvironmental problem of disposal. One answer is for them to be recycled.
dar X,=r Checklist\/'/J-I V
t-------!
Before attempting the questions on this topic, check that you:
c.J Can recall the tests for C=C, C-Hal, COH, C=O, CHO and COOH groups.
iJ Can deduce the functional gtoups present from results of chemical tests.
t.l Can interpret mass, IR, NMR and UV spectra.
*J Can deduce pathways for the synthesis of organic molecules.
*J Rppreciate how the structure of a pharmaceutical affects its solubility.
i-l Can distinguish between types of polymer and understandpolymerisation.
Testing your knowledge and understanding
I How would you distinguish between:
a pentan-3-one and pentanal
-( .n,),- i - T- ( .", ).- ii iO H Ha polyamide
/HCl
cr lil i/ H\ \ / \ / \ / ln i- ( .n,)-- ( \ - (cH,)u- N - ts/ \ r - \ / \ / " \ \ l l
o o H H ' o
Polymers usually soften over arange of temperatures rather thanhave a sharp melting temperature.The main reason for this is that thepolymer is a mixture of molecules ofdifferent chain lengths.
Fig 5.ll A polyamide
The answers to the numberedquestions are on pages 138-139.
O o R G A N r c c H E M r s r R Y r v
b 2-bromopropane and2-chloropropane
c methylpropan-2-ol, methylpropan-l-ol and butan-2-ol?
The mass spectrum of a substance X of molecular formula C.HrO hadpeaks at mass/charge values of tZO, 105, and TT.Identify the peaks andsuggest a structural formula for X.
The IR spectrum of aspirin showed peaks at32OO, I72O and 1150 cm-'.Use the data on page L09 to identify the groups which cause thesepeaks.
Low-resolution NMR spectra of two aromatic isomers with molecularformula C.H,o were obtained. Isomer Y had three peaks of relative areas5:2:3, and isomer Zhad two peaks of relative areas 6:4. Suggeststructural formulae for Y and Z.
E l t * o N s r r r o N M E T A L ' , e u A N T r r A T r v E K r N E T r c s A N D A p p ' r E D . R G A N r c c H E M r s r R y
Practice Test: Unlt 5Time allowed thr 3O min
All questions are taken from parts of previous Edexcel Advanced GCE papers
The answers are on pages I39-14L.1, a The kinetics of the hydrolysis of the halogenoalkane (where R is an alkyl group) RCHTCI with aqueous
sodium hydroxide was studied at 50 "C. The following results were obtained:
i Deduce the order of reaction with respect to the halogenoalkane, RCHrCl, and with respect to thehydroxide ion, OH-, giving reasons for your answers. t4l
ii Hence write the rate equation for the reaction. tUiii Calculate the value of the rate constant with its units for this reaction. tzliv Using your answer to part ii, write the mechanism for this reaction. t3I
(Total LO marks)lJanuary 2001CH6 question 1l
Benzene, CuHu, reacts with ethanoyl chloride, CH3COCI, by an electrophilic substitution reaction in thepresence of aluminium chloride as catalyst.
a Identify the electrophile involved in this reaction and write the equation to show its formation. l2lb Draw the mechanism for the electrophilic substitution of benzene by ethanoyl chloride. t3lc Suggest a reaction scheme, stating reagents and conditions, to convert the product of the reaction in
b into OH
CuH, -COOHI-cI
Ethanoyl chloride can be used to prepare esters t:* as 3-methylbutyl ethanoate,(CH,)TCHCHTCHTOOCCH3, which is a bee alarm pheromone that signals danger to a honey bee. If thiscompound is warmed with aqueous sodium hydroxide, a slow reaction takes place to produce sodiumethanoate and 3-methylbutan-1-ol. The reaction is first order with respect to both 3-methylbutylethanoate and the aqueous hydroxide ion. Explain the term first order and give experimental detailsshowing how this information could be obtained. t8I
(Totat L8 marks)lJune 2001 CH6 questions 2 & 4
tsI
P R A C T I C E
3 Study the reaction scheme beloq and then answer the questions that follow.
step 1 +DCrHrBr --+CrHrMgB --+ C4H,'O
KrCrrOr/HrSOn
oH- (aq)E
step 2
CTH'OH -----+C2H4O
B D
a Give the reagents and conditions necessary for step 1b Compound D reacts with ammoniacal silver nitrate solution to give a silver mirror.
i Give suitable reagents for the conversion of B to D.ii What precautions should be taken to ensure that B was converted mainly to D and did not react
turther? tlliii Give the name or the structural formula of the organic compound formed when D reacts with
l2l
tzl
ammoniacal silver nitrate solution.c Give the structural formula of compound E (C4Hr,O).d Compound F reacts with 2,4-dinitrophenythydrazine solution but has no effect upon ammoniacal
silver nitrate solution. F will also undergo the iodoform reaction.i Explain the significance of the results of the test with ammoniacal silver nitrate and 2,4-
dinitrophenyl hydrazine concerning the functional group in F.ii What structural feature in F is identified by the iodoform reaction? Give the formula of the
products of the reaction.iii Give the structural formula of F.
e The mass spectrum of F showed major peaks at mle values of 72,43 andresponsible for these peaks.
tlltzl
r2l
t3tr2l
29. ldentify the speciest3l
(Total 18 marks)[anuary 2002 CH4 question 1]
4 a Describe how you would measure the standard electrode (reduction) potential of the Fe'.(aq)/ Fe'.(aq)
system.b A standard Fe3.(aq)/Fe'.(aq) electrode is connected to a standard gold electrode, Au'-(aq)/Au(s), at 25
The gold electrode acts as the cathode and the electrons flow, in the external circuit, from theFe3.(aq)/Fe'.(aq) electrode to the gold electrode.i Write the half equations for the reactions that take place at each electrode when an electric current
is drawn from the cell. l2lii Hence write the overall ionic equation for the reaction that takes place. tlliii The potential of this cell is +O.73 V and the standard electrode potential for a Fe3.(aq)/Fe'.(aq)
electrode is +0.77 V. Calculate the standard electrode potential of the gold electrode. tzl(Total LO marks)
ffanuary 2001 CH3 question 3]
tsloc.
E l t * o N s r r r o N M E T A L = , e u A N T r r A T r v E K r N E T r c s A N D A p p . r E D . R G A N r c c H E M I s r R y
EEt_l
Fe: EEt_l
3d
t_lEE
EEE
EEE
4s
Et_lE
c The mixture of metals that results from the reduction in b was reacted with excess dilute sulphuricacid to give a solution of iron(Il) sulphate and chromium(Ill) sulphate. Sodium hydroxide solutionwas added to the mixture until in excess. The mixture was then filtered to give a precipitate Q,containing all the iron, and a filtrate R, containing all the chromium.i What is the formula of the iron-containing ion in the aqueous iron(Il) sulphate solution? tUii Write an ionic equation for the reaction between sodium hydroxide solution and this ion. State
what type of reaction is taking place.iii State the colour of the precipitate Q.iv Give the formula of the chromium-containing ion in R.v State what you would see if solution R was slowly acidified with dilute sulphuric acid until there
was no further change.
Electrode reaction trlv Electrode reaction Elv' l r E r + e + F +2.87
C r t * + e - + C r t * -0.41 t l r c l r+e-+Cl - +1.36'lrcrror'- + 7H* + 3e- + Cr3* + 3tlrHro +L.33 'lrBtr+ e- + Br +1.07
'lrrr+ e- + I- +0.54
Use the standard electrode (reduction) potentials of the chromium ions and the halogens shown belowto answer the questions that follow.
iii List those halogens which will oxidise chromium(Il) to chromium(Ill) but not to chromium(Vl).[1]iv Chromium(Il) in aqueous solution is sky blue whereas aqueous chromium(Il! solution is dark
green. Describe how you would show that your prediction in d part ii actually worked in practice.r2l
chromium that is used in the manufactureto produce iron, chromium and carbon
I2l
tsItlll2l
r2l
tzltlI
Iron and chromium are verya Complete the boxes below
important industrial metals.to show the electronic configurations of:
Cr:
Cr't:
tA4
tArl
lArl t3l
b Chromite, FeCrOn or FeO.CrO' is a mixed oxide of iron andof stainless steel. It can be reduced by heating with carbonmonoxide.Write an equation to represent the reduction of chromite.
i Define the term standard electrode potential.ii List those halogens which will oxidise chromium(Il) to chromium(Ilf .
6 A h,borator/Ma'y
Laboratory chemistry I I
O The specification (syllabus) for the tests on this module includes all theAS and A2 material.
o The Unit Test 68 is a synoptic paper, taken by all candidates, and willalso assess the candidate's quality of written communication.
- , 4 = 1 . .' ,Ki Unit Test 6A: Assessment of practical.t# skills II
E ftris is either internally assessed or a practical exam.
f Notes and books may be used in the tests.
E ffre practical exam will be broadly qualitative in its approach.
[l You should be able to:
i observe and interpret details of the chemistry of the elements andcompounds listed in Units 4 and 5.
ii recognize the results of experiments to identify functional groups inorganic compounds.
iii carry out the techniques described in Topic 5.5 and those used involumetric analysis, kinetics and equilibria.
iv present and interpret quantitative and qualitative results.v devise and plan simple experiments based on the chemistry and
techniques as above.
K+a
All candidates will do this paper
The Unit Test question (exam) paper
I Section A will consist of a compulsory question and will assess acandidate's ability to interpret data from laboratory situations.
O Section B will consist of three questions and the candidate mustanswer two of the three questions.
I These questions will require candidates to make connections betweendifferent areas of chemistry, for example by applying knowledge andunderstanding of principles to situations in contexts new to them.
E i o o r r A N c E D L A B . R A T . R Y c H E M r s r R Y
t Questions will be set on any of the topics in the AS and A2
specification.
o The questions will require much less factual recall than those in earlier
unit tests. Much more emphasis will be placed upon application of
knowledge.
o Questions will test a candidate's ability to analyse information from
several different areas of the specification.
o The words 'suggest' and 'deduce'will occur more often in questions
than 'state' or 'recall ' .
Tackling the paper
* Spend some time looking through Section B and decide which questionsyou are going to attempt. Read all of the question before reiecting it.You may be able to answer all but the first part and so still score goodmarks. If you are getting nowhere in a question, abandon it and tryanother, but do not cross out what you have written, because youmight score more for it than for the other question. The examiner willcount your better mark.
o Do not be put off by unusual compounds or situations. In thesequestions you are not expected to know the answer, but to be able towork it out, using your knowledge and understanding of similarcompounds or situations.
O Synoptic questions will contain material from several topics. This isdone by using the links that exist between different branches ofchemistry.
O Each question should have a thread or link connecting the differentparts. Identification of this thread will help you to focus on the relevantchemistry. So do not treat each part of a question in isolation from theother parts of it.
o For example in the questions in the Unit Test 6B on pages lI9 and LZO.
o Question 2 is based on organic and inorganic nitrogen compounds,with questions on fertilisers, pK", polymers, amine preparation, andK,, for ammonium nitrate linked to thermodynamic and kineticstability.
o Question 3 is a reaction scheme where the carbon chain is increasedleading to a carboxylic acid, then a pH and buffer question about thatacid.
O Question 4 is about the chemistry of iron, linking bonding, transitionmental properties, Bronsted-Lowry pairs, tests for Fe'. and redoxtitrations.
O To do well in this paper you must revise the entire specification(syllabus) and especially Topic I.2. Do not become put off by this loadas many of the Topics in A2 are extensions of those in AS. For example:
o Topic 4.1 with 2.1 - Energetics
o Topic 4.2 with I.4, 1.6 and I.7 - The Periodic Table
o Topic 4.3 and 4.4 with 2.4 and 2.5 - Equilibrium
o Topic 4.5, 5.3 and 5.5 with 2.2 - Organic chemistry
O Topic 5.1 with 1.5 - Redox
O Topic 5.4 with 2.3 - Kinetics
This leaves Topics 1.1 (Atomic stmcture), I.2 (Formulae, equations and moles)and 1.3 (Structure and bonding) all of which are fundamental to chemistry andmuch of these topics will,have been revised by your teacher during the A2 year.
P R A c r r c E T E s r : U t q r r 6 E } ( S Y N o P T t c )
Practice Test: Unlt 6B (Synoptic)
Time allowed L hour 3O min
All questions are taken from previous Edexcel Advanced GCE questions.
The answers are on pages I4I-I42.
Section A1 A fertiliser is known to contain ammonium sulphate, (NHn)rSOn, as the only ammonium salt. This
question concerns methods for the determination of the ammonium and sulphate ions.
a A sample weighing 3.80 g was dissolved in water and the volume made up to 250 cm'. To 25.0 cm'
portions of this solution about 5 cm' (an excess) of aqueous methanal was added. The following
The liberated acid was titrated directly with 0.100 mol dm-'' aqueous sodium hydroxide. The average
volume required was 28.0 cm'. Calculate the percentage of ammonium sulphate in the fertiliser. ts]b In a second determination of the ammonium ion content, the same mass of fertiliser (i.e. 3.80 g) was
treated with excess sodium hydroxide and heated. The ammonia liberated was passed into a known
excess of hydrochloric acid. The unreacted hydrochloric acid was then titrated with standard aqueous
sodium hydroxide.The calculation of the percentage composition of the fertiliser gave a value that was 5olo lower than the
value obtained by the method in a. Suggest reasons for this error other than those arising from the
measurement of volumes. tzlc To determine the sulphate ion concentration of the fertiliser, aqueous barium chloride was added in
excess to the fertiliser solution. The precipitate produced was filtered off and dried by strong heating.
i Give the ionic equation, with state symbols, for the precipitation reaction.ii Suggest why the aqueous barium chloride was added in excess.
d Many carbonates are also insoluble, and can be precipitated, dried and weighed in experiments similar
to c. However the strong heating needed to drive off all the water can cause a problem in determining
the mass of the carbonate precipitated. Suggest what this problem is, and, choosing a suitable example,
write an equation for a reaction that might occur when the precipitate is heated. t4I(Total 14 marks)
[June 2002 Urtit Test 68 question 1]
Section BAnswer TIIVO questions only
2 a The covalent compound urea, (NHr)rC = O, is commonly used as a fertiliser in most of the EuropeanUnion, whereas in the UK the most popular fertiliser is ionic ammonium nitrate, NH4NO'. Apart fromits nitrogen content suggest two advantages of using urea as a fertiliser compared with usingammonium nitrate. I2l
b The ammonium ion in water has an acid dissociation constant K^ = 5.62 x 10-to mol dm-'' Theconjugate acid of urea has K" = O.66 mol dm-'. Use this data to explain which of ammonia or urea is thestronger base. t2l
c Some organic nitrogen compounds are used to manufacture polyamides by condensationpolymerisation. With the aid of diagrams, define the terms condensation polyrnerisation and polyamide.
t4ld Ethanamide, CH3CONH, can be converted to methylamine, CHrNHr.
i State the reagents and conditions for carrying out this conversion. t3lii Suggest the formula of the likely product if urea were used instead of ethanamide in this
conversron.e Ammonium nitrate can explode when heated strongly.
NHnNO.(s) - N,O(g) + ZH,O(g) NI = -23 kJ mol-'
r2ltll
tu
A D V A N C E D L A B O R A T O R Y C H E M I S T R Y
With moderate heating the ammonium nitrate volatilises reversibly
NHnNO,(s) + NH,(g) + HNO.(g) NI = +L77 kJ mol-'
i State why the expression for Ko for the reversible change does not include ammonium nitrate. tl]ii Explain the concepts of thermodynamic and kinetic stability with reference to these two reactions.
tsI(Total 18 marks)
lJune 2001CH6 question 4l
The principal source of the odour in sweat is the organic acid IIA, CTHnOz.It will decolourise brominewater immediately, and it shows geometric isomerism.a IIA can be made by the following sequence of reactions:
cH3cHzcHrcH(cH3)cHroH2-methylpentan-L-ol
IY
cH3cH2cHrcH(cH3)cHo
HCI(aq) conc. HzSOn
cH.cH,cH,cH(cH3) cH(oH)cN
ccTHrno3
D
+ heatCrHrzOz
HA
i Give the reagents and conditions for the oxidation of 2-methylpentan-1-ol to B. t3lii State the conditions and write the mechanism for the reaction of HCN with B to give C. t4liii Give the equation for the hydrolysis of C to D, and hence draw the structural formula of D. t2liv Draw the structure of HA. t2l
b HA has a dissociation constant K" = 4.50 x 1Os mol dm-', and is a monobasic (monoprotic) acid whichdissociates according to the equation:
HA(aq) + H,OQ) + HrO.(aq)+ A(aq)
i Write the expression for K" and calculate the pH of a solution of HA of concentration 0.0100 moldm-'. t3l
ii The pH of skin is almost constant. Explain how in principle, this could be achieved by using HAand one of its salts. (Given its smell, it is perhaps fortunate that skin does not use this compoundas a pH regulator) I4l
(Total L8 marks)fJanuary 2002 CH6 question 3]
Iron(III) chloride exists in two forms, the covalent anhydrous chloride and the ionic hydrated chloride.a Explain why anhydrous iron(III) chloride is covalent whereas sodium chloride is ionic. t4Ib Hydrated iron(III) chloride is soluble in water and its solution is acidic.
i Write an ionic equation to show why its solution is acidic and identify the acid-base conjugatepairs in this reaction. t3I
ii Describe what you would see when a solution of sodium hydroxide is added until in excess to asolution of iron(III) chloride.Write an ionic equation for this reaction and identify the type of reaction. t4l
iii A test for Fe'* ions is to add a solution of thiocyanate ions, SCN-(aq), which react to form a bloodred ion of formula [FeSCN(H'O)J'.. What is the bonding between the SCN- ion and the Fe3. ion,and what is the cause of the colour of the ion? t3l
c Iron also forms the intensely red coloured anion, FeOn'-. In acidic solution this is a powerful oxidisingagent, and is itself reduced to Fe3* ions.Calculate the oxidation number of iron in FeOn'. Suggest, in outline, how you could determine theconcentration of a solution of FeOo'- ions. t4l
(Total 1"8 marks)ffanuary 2001 CH6 question 6l
Arcr,tters
Uni t 1Tonic 1 .1 Atomic structure^ v r 7 v L . 1
t The relative isotopic mass is the mass of a singleisotope of an element (relative to 1172th the massof a carbon 12 atom), whereas the relative atomicmass is the average mass taking into account theisotopes of the element. The relative isotopicmasses of the two chlor ine isotopes are 35 and 37,but the relat ive atomic mass of chlor ine is 35.5
2 There are many more tl i
atoms than "Li atoms innatural l i thium, so the average is closer to 7.
3 A, of magnesium
= (24 .O x 78 .6 + 25 .O x 7O. I + 26 .O x 77 .3) = 24 .3100
4 Let the o/o of "ncabe x,and so the o/o of "Ga = (100 _ x)
69.8 = [69.0x + 71.0 (100 - x)J1 0 0
Therefore 6980 = 69x + 7100 - 77x
Therefore 2x = l2O and x = 60.0
Gallium contains 6O.Oo/o of the ""Ga isotope.
5 Group 5. There is a big iump between the 5th and6th ionisation energies. Therefore the 6th electronis being removed from a lower shel l (nearer to thenucleus) than the first f ive electrons.
8
A N S w E R S
iii Although the element with Z = 11 has 8 moreprotons, it also has 8 more shielding electrons, but asthe 3s electron is further from the nucleus than a 2selectron, so less energy is required to remove it.
The electron in a box notat ion for chlor ine is:
muruuunnun1 s 2 s 2 p , 2 p , 2 p , 3 s 3 p . * 3 p , , 3 p ,
All have one electron in their outer shells, but K isbigger (larger atomic radius) than Na, which isbigger than Li. Therefore the outer electron is heldon less firmly (the extra nuclear charge is balancedby the same extra number of shielding electrons)and so is the easiest to remove from potassium andthe hardest to remove from Li. This makespotassium the most reactive.
9 a il l
iii
+ e --+ Li- (S)+ e -+ Cl- (g)
+ e --+ o- (g)
Li (g)cl (g)o (g)
6 a zsoo
bc
>\bot-<
+)
.A
v)
2000
1500
1000
500
0| 2 t^ti*tt
":-l.r*
e 10 11
A 2p electron, which is in a higher energylevel, is being removed. So less energy isrequired to remove it compared to thatrequired to remove a 2s electron.There are two electrons in the 2p, orbitalwhich repel one another, so less energy isrequired to remove one of the pair than ifthere had only been one electron in thatorbital.
O- (S) + e- - O- (g)For the 1st electron affinity, the (negative)electron is brought closer to the positivenucleus in a neutral atom, and so energy isreleased. For the 2nd electron affinity, thenegative electron is brought towards anegative ion, and so repulsion has to beovercome, which requires energy.
Volume of HrS gas = rlol x molar volume= 0.O206 x 24 = 0.50 dm'
6 Ratio mol of H,(g):mol of CH.(g) = 3:1Volume of Hr(g):volume of CHr(g) = 3:LVolume of Hr(g) = 3 x 33 = 99 dm'
Topic 7.3 Structure and bonding
Melting: on heating, the molecules or ions vibratemore until the intermolecular forces (in molecular)or ion/ion forces (in ionic) substances are overcomeand the lattice breaks down.
Boiling: on heating the average kinetic energy otthe molecules increases. Those with a large enoughkinetic energy and hence speed escape, not onlyfrom the surface but also from the body of theliquid, forming bubbles.
a NH.. There are stronger intermolecular forcesin NH. as it forms H bonds and PH. does not.
b HBr. It has more electrons in its molecule, andso it has stronger dispersion forces (whichoutweigh the difference in dipole/dipoleforces).Propanone. Both have about the same strengthof dispersion forces because they have aboutthe same number of electrons, but propanone ispolar and so has dipole/dipole forces as well.58. It has about twice as many electrons as Pn,and so has stronger dispersion forces.NaCl. It has ion/ion forces that are very muchstronger than the weak intermoleculardispersion forces between CCln molecules.
8 In Na (s) there are loosely held delocalisedelectrons that can move through the lattice. In
a metall icb giant atomice simple molecularc ionicf polymericb giant atomice simple molecularc ionicd H-bonded molecularc ionic
d
o-bond
2 a AlCl, because either there is less difference inelectronegativity between Al and Cl thanbetween Al and For A13. is very polarising and the bigger Cl- ionis more polarisable than the smaller F- ion.BeCl, because either there is less difference inelectronegativity between Be and Cl thanbetween Mg and Clor Be'* is smaller than Mg'. and so is morepolarising.
4 iodineice
e simple moleculard H bonded molecular
Do not say that the molecules or ions start to vibrate.
a Hydrogenbond
b Dispersion c Dipole/dipole
HFr2HBrPH,Ar
YesNoNoNoNo
YesYesYesYesYes
YesNoYesYesNo
Number ofo bond
pairs
Numberof lonepairs
Totalnumber ofelectron
pairs
Shape
siHnBF.
BeCl,PCl3SFuXeFn
NHn'PClu-
43
2364
46
00
0102
00
43
2466
46
TetrahedralTriangular
planarLinear
PyramidalOctahedral
Squareplanar
TetrahedralOctahedral
The answer 'hydrogen bonded ' o r ' i on ic ' canno t app ly toe lements.
NaCl(s) the electrons are localised on the ions, notloosely held and the ions are fixed in theirpositions in the lattice and so are not free to move.
Topic 7.4 The Periodic Table I
1 a They all have the same number of electrons intheir outer shell. For Group 2 this number is 2.
b This period is Period 4, and so all the membershave electrons in four shells.
A N s w E R s 1 2 3
Nuclear charge is equal to the number ofprotons in the nucleus.Screening or shielding occurs when an outerelectron is insulated from the pull of thepositive nucleus by electrons in inner shells.The 1st ionisation energy of sodium is forthe removal of the 3s electron. Sodium has anuclear charge of +11, but the 3s electron isshielded by the 10 electrons in the 1st and2nd shel ls. The second electron that isremoved comes from the Znd shell ( it is a 2pelectron). This electron is shielded from thenucleus onlv bv electrons which are in alower shel l , rvhich in th is case is by the two1s electrons and so i t is held much morestronglv bv the nucleus. I t is a lso nearer tothe nuc leus , rvh ich makes i t even harder toremoVe.Sod ium has a nuc lear charge o f +11 and i t s3s e lec t ron is sh ie lded bv the inner 1s t and2nd she l l e lec t rons . \ lagnes ium has anuc lear charge o f +12 (one n lo re thansodium), but i ts 3s electron is shielded bythe same inner electrons. Thus i ts 3selectrons are pulled more strongly to thenucleus and are harder to remove.
2 abcdef
H, is simple molecularNa is metall icSi is giant atomicS* is simple molecularCl, is simple molecularAr is simple molecular (argon is monatomic).
Sodium and magnesium are metals. Magnesium is2+ and has a smaller ionic radius. therefore it wil lhave stronger metall ic bonds and a higher meltingtemperature. Sil icon forms a giant atomic structurewith each sil icon atom ioined covalently to fourother sil icon atoms. These strong covalent bondshave to be broken. This requires a huge amountof energy and so the melting temperature is veryhigh.
White phosphorus, sulphur and chlorine all formsimple molecular solids. The melting temperaturesdepend upon the strengths of their intermolecularforces which are dispersion (induced dipole/induceddipole) forces. The strength of this force dependsupon the number of electrons in the molecule.
Topic 7.5 Introduction tooxidation and reduction1 a Sn"(aq) -- Sn'.(aq) + 2e
4 a i
l l
b i
Fe ' -1aq) +Sn ' - (aq) +
l l
bc
e --'Fe"-(aq)
2Fe''.(aq) -- Snn.(aq) + 2Fe'.(aq)
Pn has 42 x I 7 =has thehighestweakestmelting
x 15 = 60, S* has 8 x 16 = 128 and Cl, has34 electrons per molecule. Thus sulphurstrongest intermolecular forces and themelting temperature and chlorine has theintermolecular forces and the lowesttemperature.
I t is because of th is ext ra oul l that a ma0nesiumatom is smal ler than a sodium atom.
Carbon, in the form of d iamond, has a s imi lar s t ructure,but as the smal ler carbon atom forms st r0nger bondsthan s i l icon, d iamond's mel t ing temperature is evenhigher .
Elec t rons are 0n r igh t because Sn ' . i s ox id ised : 2e-i s n e e d e d b e c a u s e o x i d a t i o n n u m b e r c h a n g e s b y 2 .
The s imp les t way in wh ich to work th is ou t i s to addtogether the a tomic numbers o f a l l the a toms in them o lec u le .
Equat ion b has to be doub led , so tha t the e lec t ronscance l when 2 t imes b is added to a .
HrOr(aq) + 2H.(aq) + 2e- -. 2HrO(l)
H.0 , i s reduced, so e lec t rons are 0n the le f t .
c PbOr(s) + Pb(s) + 2H,SOo(ag) - 2PbSOo(s) +2H,O(t)
Topic 7.6 Group 1 and Group 2L Dip a hot platinum (or nichrome) wire in clean
concentrated HCI and then into the solid. Place in ahot bunsen flame. The substance that turns theflame carmine red is LiCl, the one which turns theflame li lac is KCI and the one that turns the flameapple green is BaClr.
2 The heat of the flame vaporises the sodiumchloride, producing some Na and Cl atoms.Electrons are promoted into the 4th shell in someof the sodium atoms. These electrons then fall backto the ground state, which is the 3s orbital, andenergy is given out in the form of light. The flame
is yellow because the energy difference between the4th shelt and the 3s orbital is equivalent to theyellow line in the spectrum.
The sodium atom is smaller than the potassiumatom, because sodium has three shells of electronsand potassium has four. Thus sodium's outer selectron is held on more firmly, and more energy isrequired to remove it.
4 a 2Ca + O, - ZCaO (or Ca * ' l rO, -- CaO)b Ca + zH,O - Ca(OH), + H,c 2K + ZH.O -- 2KOH + Hzd Mg(s) + H,O(g) - MgO(s) + H,(g)
The white precipitate is the insoluble magnesiumhydroxide. Barium hydroxide is soluble and so isnot precipitated.
Beryll ium, as BeCO. is the least thermally stable ofthe Group 2 carbonates. It decomposes very easilyon heating because the Be'* ion is very polarising.The other Group 2 ions are less polarising becausethey are larger.
7 a 4LiNO3 -- ZLizO + 4NO, + O,2NaNO. - 2NaNOr* O,2Mg(NO,),- 2MgO + 4NO, + O,
b NarCO. --' no reactionM g C O , - M g O + C O ,BaCO, --+ no reaction
Topic 7.7 Group 7
Covalent HCI ionises when added to water, and thehydration energy given out by the ions bondingwith the water molecules is greater than the energyrequired to break the covalent H-Cl bonds:
HCI(g) + H,O(l) -- H,O.(aq) + Cl-(aq)
First a solution has to be made. The sample can bedissolved either in water or in dilute nitric acid. (Ifwater is chosen, dilute nitric acid must then beadded to destroy the carbonate). To this slightlyacidic solution, add silver nitrate solution. If thereis chloride present, there wil l be a white precipitatethat wil l dissolve in dilute aqueous ammonia.
HrSOn is a stronger acid than HCI (or HI), and so itwil l protonate the Cl- ions in the sodium chlorideproducing HCI gas. HCI is not a strong enoughreducing agent to reduce sulphuric acid, and sono further reaction will take place. With sodium
2e- are needed on the left because S is reduced and i tsox idat ion number goes f rom 0 to -2.
Both H,S and H,0, must be 0n the left because they arethe reactants; equation b is reversed and added to a;note that the H- ions cancel out .
This is the case in spite of the fact that potassium haseight more protons than sodium, because i t a lso haseight more e lect r0ns between the nucleus and the outere lect ron, so the ext ra nuclear charge is balanced by theextra sh ie ld ing.
2e- are needed on the left as PbO, is reduced and i tsox idat ion number changes f rom +4 to +2.
2e- are needed on the lef t because PbS0, is reduced andthe ox ida t ion number o f lead changes f rom +2 to 0 .
Because Pb is a reactant, equation i i has to be reversedand then added to i . This is the reaction that takes placewhen current is drawn from a lead-acid batterv.
iodide, HI is produced by the protonation of theiodide ions, but the HI is a very strong reducingagent, and so it wil l reduce the sulphuric acid anditself become oxidised to iodine.
4 Disproportionation is a reaction in which anelement is simultaneously oxidised and reduced.Chlorine disproportionates when added to alkali:
Cl, + ?OH- - Cl- + CIO- + H,O(0) (-1) (+1)
Also, chlorate(I) disproportionates when heated:
3ClO- -- ZCl- + ClO,,(+1) ( -1 ) (+5)
Answers to Practice Test Unit 1The allocation of marks follows Edexcel markschemes.The marks thatapproximately:
you wil l need for each grade are
A 7Oo/o
B 6lo/o
C 52o/o
D 44o/o
E 360/o
ANsweRs 125
b Graphite has covalent bonds between atoms thathave to be broken [1] ; sodium chlor ide haselectrostat ic forces between the ions [1] . In boththese forces are strong and so a lot of energy isneeded to break them [1] = [31
c i Graphi te has delocal ised electrons above andbelow the plane of he.ragons [1] , which canf l ow (wh i ch a re mob i l e t [ 1 ]
l l
L a i Z C a + O r - Z C a Oii NarO + HrO ---+ 2NaOHiii Na,O + ZHCI ---; 2NaCl + H,O
Species [.], balancing [.]b The thermal stabil ity increases down the group
[f.] as the cation size increases [1] and so thecarbonate ion is distorted (or polarised) less
_ tl lt3l
2 a i Graphite has a giant atomic (or giantcovalent) structure [1]. The bonding betweenatoms is covalent [U.Your diagram must show several interlinkedhexagons [1] in layers tU = [4]
= f2ll l The lat t ice is broken don'n in the l iquid and
so the ions are f ree to nlove = [ u
Atomic number is the number of protons inthe nucleus [1] Mass number is the total (orsum of the) number of protons and neutrons
= [1]= [1]
= l2l
t l l .
3 a i
tu, . g t -i i
b i
l l
i i i
= l2l= [1 ]
I t Sodium chloride has a latticeThe bonding is ionic [l.].Your diagram must show a 3-D arrangementof ions, Na. alternating with Cl-. [1] = [3]
4 aDivide each o/o by their A, values [L] and then divideby the smallest [1]
H L 1 . L * 1 = L 1 . 1 t h e n * 7 . 4 = 1 . 5C 88.9 * 12 = 7 .4 then + 7 .4 = 1Multiply both by 2 to get whole numbers.'. E.F. is CrH. tl l
. l
High energy electrons bombard the Br,molecules [1] and knock out an electron [1]
= [2]They are accelerated by charged plates (oracross an electric f ield) = [1]-"Br- ' 'Br.
= [U
&
a S t r u c t u r e m e a n s g r a n t a t o m i c . i o n i c l a t t i c e ,s i m p l e o r h , d r o g e n b o n d e d m o l e c u l a r o r al a t t i c e o f m e t a l r o n s i n a s e a o f e l e c t r o n s .
c A subs tance , ' , r l l conduct e lec t r i c i t y i f i t hasd e l o c a l r s e d e l e c t r o n s t h r o u g h o u t t h e s t r u c t u r eo r i o n s ' , ' . , h r c h a r e f r e e t o m o v e .
b Do not attempt to answer this type of question interms of size match.
a i i As the part icle has 1 fewer electrons thanprotons, i t wi l l be a posi t ive ion.
a i i i In a mass spectrometer most e lementsproduce ions conta in ing one atom, butd iatomic e lements such as N, , 0 , and thehalogens g ive molecular ions such as B12which may conta in two d i f ferent isotopes.
structure
^f*
= [3]
:1ffi
b
c
Arqswens
HI has more electrons tll .'. induced dipoleor dispersion or London or v.d.Waalsare stronger [1] = I2fi 3 bond pairs and 1 lone pair repel to get
as far apart as possible [1] see diagrambelow [1] = l2l
ii 4 bond pairs around Al repel to get asfar apart as possible [1] see diagrambelow [t] = I2l( i i ) i
T jF-h
Moles of gas = volume + molar volume= 8.0 ,mt/2.4 x 10n cm' mol-' = 3.33 x 10" mol
Itr"o., of molecules = 6.0 x 10" x number ofmoles = z.o x 1o'o 1t1 = fzl
b Do not say because the molar mass of Hl is larger.c Remember that i t is the e lect ron pai rs not the
atoms nor the bonds that repel to g ive maximumseparation.
The oxidation number of sulphur: in HrSOn is+6 [1], in H,S is +2 [1], in SO, is +a [1] = [3]The I- ion is the stronger reducing agentbecause it reduces sulphur from +6 to +2 fAl,whereas Br- only reduces it to +4 [1] = l2fZCl- - Cl, + 2e- or Cl- -
'lrcl, * e- = [1]
[L] for the electrons around S and [1] for theelectrons around both Cl atoms. = l2l
a t t a
S** *. .* ..
CI CI+ + + ++ + + +
The 2 bond pairs and the 2 lone pairs [L]repel each other to get as far apart aspossible [.] = I2lCl is more electronegative than S, so thebonds are polar [1], but as the molecule isbent, the polarities do not cancel [1] = fzl
Unit 2
Topic 2.1- Energetics I
Gases at a pressure of t atmosphete, a statedtemperature (298 K) and all solutions ofconcentration 1".00 mol dm-'. Substances in theirmost stable state.
i zC(s) + ZH,(g) + O,(8) - CHaCOOH(I)ii CH,COOH(I) + 2O,(g) - 2CO,(8) + 2H,O(l)i i i CH3COOH(aq) + NaOH(aQ) - CH,COONa(aq)
A big jump in ionisation energy means that anelectron has been removed from a new shell [1]The first big jump is after 6 electrons and sothere are six electrons in the outer shell [1]The second big jump is after 8 electrons and so
MI,= +90.3 kJ + (-57.1) kJ = +33.2 kJ mol-'
Note that ethanoic acid is a weak acid, and so i tsformula must be wr i t ten in f u l l and not as H' .
i & i i For an oxidation reaction, the electrons areon the r ight, and for reduction they are on the left .i i i You would not be penal ised for fa i l ing to cancel
one of the Cl- ions, i .e . hav ing the equat ion.OCl- + 2H. + 2Cl -Cl , + H,0 + Cl -
No(g) +tboz(g)
t/, Nr(g) + oz(g)
Note that the + s ign is g iven in the answer in orderto emphasise that the react ion is endothermic.
where A-FI, = 12 x enthalpy of combustion of carbon(graphite) + 12 x enthalpy of combustion of hydrogenandLH, = minus the enthalpy of combustion of lauric acid.LH, = 12x(-394) +12x(-286) -(-7377) = -783 kJ mol-'.
6 HCI + NaOH -- NaCl + HrO
Amount of HCI = corC(mol dm-') x vol(dm')
= 1.00 x 1001000
= 0 .100 mol
Rise in temperatur€ = +6.80'CHeat produced when 0.100 mol reacts= total mass x specific heat capacity x rise in
Total = -1120 kJNI = net result of bonds broken and bonds made= +1075 - 1720 = -45 kJ mol- ' .
Topic 2.2 Organic chemistry IL a
B r F Fl l l
B r - C - C - C - Hl l l
C I C I H
Nf,2NOr(g) ---------------+ NzOn(g)
\ /\ /
* ' \
, / * ,Nz(g) + zOzG)
H Ht l
H - C - C - O - Hl l
H H
H\ /\ /
r - - r -\ - - \ - f
/ \/ \H
H
H
Break (endo)1 x C=C = +6"12 kJ1 x H-O = +463 kJ
Total = +1075 kJ
bC I O H H Ht t t l
H - C - C - C - C - Ht l l l
H H H H
2 a CH, ,CH.CH,OH and CH,CH(OH)CH, .
c
3 a
b
4 a
b
5 a
b
6 a
b
c
d
cH.cHrcH.cH.cH,, cH3c(cHr)rcH, andcH.cH=cH(cH,)cH.CH,CH=CHCH, (both the c is and transisomers), (CH,).C=CH, and CH.CHTCH=CH,
A free radical is a species with a single unpairede lec t ron , e .g . C l . .Homolytic f ission is when a bond breaks withone electron going to each atom.
An electrophile is a species which seeks outnegative centres and accepts a lone pair ofelectrons to form a new covalent bond.Heterolytic f ission is when a bond breaks withthe two electrons going to one atom.
C.Hn + Cl, -* CrH.Cl + HCIthen: CrH.Cl + Cl, - CrH*Cl, + HCI etcconditions: ultraviolet l ightc.H,, + 3 ' l ror- ZCo, + 3Hroconditions: f lame or a spark
CH,CH=CH, + H, - CH.CHTCH,conditions: heat and Pt catalystCH.CH=CH, + Br, - ' CH3CHBTCHTBTconditions: solution in hexaneobservation: goes from brown to colourlessCH.,CH=CH, + HI - CH..CHICH.conditions: mix gases at room temperatureCH,CH=CH, + [O] + H,O -- CH.,CH(OH)CH,OHcondition: aqueousobservation: goes from purple to brownprecipitate
LH, = minus 2 t imes the enthalpy of format ion ofN0,(g) , because the react ion represents theformat ion of N0,(g) reversed and LH2= enthalpy offormat ion of N,0, (g) .
Th is t ype o f unambigu0us f 0 rmu la i s a l l owed un less afu l l s t ructura l formula is asked for .
Because the mixture became hotter, the reaction isexothermic. The mass used is the mass of the wholeso lu t ion ,100 cm '+ 100 cm3 = 200 cm '=200 g , no t themass of the HCl .
r rsrr l i i t* f i
128 ANswERs
7 a CH.CHICH. + NaOH' CH.CH(OH)CH, + NaIpropan-2-ol
b CH3CHICH. + KOH--'CH.CH=CHz + KI + HrOpropene
c CH3CHICH, + ZNH, - CH.CH(NHr)CH3 + NH4I2-aminopropane
8 CH.CH(OH)CH, - H,O -'CH.CH-CH,
conditions: concentrated acid at l7O "C
9 aH H ol t / -
H - C - C - Cl l \
H H O - HPropanoic acid
observation: solution goes from orange to greenb No reaction
observation: solution stays orangec CH3CHCICH3
2-chloropropaneobservation: steamy fumes
1O The n carbon-carbon bond in ethene is weakerthan the o carbon-carbon bond in ethane. Thismakes the activation energy less, and so thereaction is faster.
1,1 The C-Br bond is stronger than the C-I bond. Thismakes the activation energy more, and so thereaction is slower.
L2 C6H1o + Br, - CuH,oBr,5.67 g of CuH,o = 5.67182 = 0.0691 moIBecause the reaction is a L:L reaction, the amountof product is also 0.0691 mol.Therefore the theoretical yield of CuH,oBr,= 0.0691 x 242 = 16.7 gThe o/o yield = 15.4 g x 100 = 92o/o.
1 6 . 7 g13 They are stable in air owing to the strong C-Cl
and C-F bonds.They then diffuse into the stratosphere wherethey form radicals and destroy the ozone layer.
Tonic 2.3 Kinetics II
1 i The activation energy is the minimum energythat the reacting molecules must have whenthey collide in order for them to react. It ismeasured in kJ mol-'.
ii A catalyst is a substance that speeds up achemical reaction without being used up. Itworks by providing an alternatlve route with asmaller activation enetgy.
Pressure. As the pressure is increased, the numberof molecules in a given volume increases, and thefrequency of collision increases and hence the rateof reaction, which is dependent upon thisfrequency of coll ision, increases too.
Temperature. If the temperature is increased, theaverage kinetic energy of the molecules is alsoincreased. This means that a greater proportion ofthe colliding molecules will have, between them,energy greater than or equal to the activationenergy. There will be a larger number of successfulcollisions and a faster rate of reaction.
Catalyst. This increases the rate of reaction byproviding an alternative path with a loweractivation energy, thus a greater proportion of thecolliding molecules have energy greater than orequal to the lower catalysed activation energy.
Energy Ea
Bacterial and enzyme reactions are also sloweddown by a decrease in temperature for exactly thesame reason as ordinary chemical reactions, whichis that fewer colliding molecules will have thenecessary activation energy to react on collision. Sothe low temperature in the refrigerator will slowdown the biological decay reactions.
i i i
ttt(u
Uq.)
(|{
(J
>.@
OJ
tll lv
boq)
a r - l
Never say that a catalyst lowers the activation energy.Light is not a catalyst, because i t is not a substanceand i t is used up. l t is just a specif ic form of energy.
product.R is the reactant and P is the
Topic 2.4 Chemical equilibria Ii trueii falseiii truei Position moves R-L, because that is the
exothermic direction.i i Pressure increases, therefore position moves
R-L as fewer gas moles on leftiii None, catalyst speeds up the forward and
reverse reactions equally, reducing the timetaken to reach equil ibrium.
3 i Alkali removes ethanoic acid, therefore positionmoves R--L.
i i No change because AH is zero4 Solid dissolves because high [Cll drives equil ibrium
L-R.
5 Too high a temperature produces too low a yield.Too low a temperature produces too slow a rateCatalyst allows reaction to proceed at a fast rate ata temperature where the yield is high.
Topic 2.5 Industrial inorganicchemistry
L Reaction is slow at moderate temperatures.Therefore a catalyst is used which allows a fast rateand reasonable yield at 400 "C, that is, an economicrate and yield. A very high pressure (200 atm) isused, even though this is very expensive becausethe yield at ordinary pressures is very low. A veryhigh pressure drives the equilibrium L'R to theside with fewer gas moles. After NH, is separated,the unreacted gases are recycled.
2 The ammonia l iquefies under the high pressure andis removed from the unreacted nitrogen andhydrogen, which are then put back through thereaction chamber.
Answers to PracticeTest Unit 2Edexcel markThe allocation of marks follows
schemes.The marks that you will need each grade areapproximately:
AusweRs 129
It is a series of compounds with the same generalformula [1] which differ by CH, [L], and havethe same functional group [1] = [3]-(CH(CH,)CHr).- : correct carbon chain [1] wi thextension bonds at both ends [.] = l2li See Fig 2.4 on page 38. [1] for each
isomer = I2li i Rotation around the z bond is restricted = [1]
See Fig 2.9 on page 45: [1] for having the axescorrect (y-axis labelled fraction of molecules,x-axis energy), [1] for having the curves with thecorrect shape, and [1] for having the T, curveflatter with the peak moved to the right = [3]Draw the activation energy on the graph to ther ight of both peaks [1, ] , explanat ion that the areaunder the curve to the r ight of the E. l inemeasures the number of molecules wi th E > E.,[1] , which is larger at T, than at T, [1] , so thereare a greater proport ion of successful col l is ions
L a
b
c
2 a
3 a
b
tu
b
d
= [4]
i The reagent is HBr [1] and the conditions arein the gas phase [l.] = I2f
i i S is CH,CHBTCH, = [1]i P is CH.CH(OH)CH, = [1]i i The type of reaction in step 1 is electrophil ic
lr,l addition [L] = l2lin the conversion of S to P the type isnucleophil ic [1] substitution [1] = [21
i For step 2, the reagent is conc sulphuric (orphosphoric) acid or aluminium oxide [1] andthe condition is heat [L] = f?l
i i For step 3, the reagents are potassiumdichromate(Vl) [1] and sulphuric acid [1] andthe conditions are heat under reflux [1] = [3]Q is CH,CH(OH)CH,OH [r.] = [1]
b Do not have the 3 carbon atoms in a straightcha i n .
c Do not draw the st ructure wi th a bond angles of90 'a round the C=C group .
a The tota l area under the two curves should be thesame.
b Ment ion must be made of co l l is ions.
l * rtul l"Iitri
a l f the fu l l s t ructura l formula had been asked, youmust then show each atom separately, with al lbonds shown.
b As there are 2 marks for each, you must make2 points .
c i i Potass ium manganate(Vl l ) is an a l ternat ive topotass ium dichromate(Vl ) . A lways g ive e i ther thefu l l names or the fu l l formulae, e.g. not'acidif ied dichromate' n0r 'Crr0, '- lHr' . Also 'ref lux'
by i tse l f does not imply heat .
130 Auswens
Hess's law states that the enthalpy change for areaction [1] is independent of the route [1] = [2]The definition is the enthalpy (or heat but notenergy) change when 1 mole [1] of a substance iscompletely burnt in oxygen (or burnt in excessoxygen) [.] under standard conditions (at L atm
A lower temperature will move the positionto the right [1], because that is theexothermic direction (or this produces heat)
tll = fzlDecreasing the volume increases the pressureand so moves the equilibrium to the right [1]as the right has fewer gas molecules [1] = [2]
ZOt- - O, + 4e- (or Ot- -- ' lrO, + 2e-)A l ' . + 3e- 'A lThe reaction at the anode. = [1]The oxygen produced at the anode reacts withthe carbon anode and eats it away = [1]Aluminium oxide (pure bauxite) has anextremely high melting temperature [.], and sothe process would be too expensive [1] = l2l
Dynamic means that the reaction is occurringin both directions [1] and equil ibrium meansthat there is no change in concentrations
tlI = l2f
= [1]= [1]
tl
C I
ll
4 a
b
5 a i
ll
pressure) [1]i Bonds broken
1 x C - C = + 3 4 85 x C - H = + 2 0 6 O1 x C - O = + 3 6 01 x O - H = + 4 6 33
" Q=O=l-1-a88Total = + 47L9 kJ tU
L,H = + 47t9 - 5750 = -103t kJ mol-' [t] = [3]i i See Fig. 2.1 on page 33.
Labelling of reactants and products [1],with energy of reactants higher thanproducts and AFI shown [L] = f2l
between rate and yield [1]The catalvst used is iron
= [3]bonds made4 x C = O = - 2 9 7 26 x O - H = - 2 7 7 8Total = -5750 kJ tll
= [3]= [1]
6 abcd
e
tll
lv
Your answer for the temperature should bebetween 350 and 500 "C (623 and 773K) = [1]A high temperature favours a high rate ofreaction [1], but gives a low yield [1], so atemperature is chosen which gives a balance
Answers to Practice Test Unit 38The marks that you will need for each grade areapproximately:
Hydrogen: burns with a squeaky pop [1]Oxygen: relights splint [1]Carbon dioxide turns lime-water [f.] milky [1]Sulphur dioxide: the solution turns green [1]The gas which turns moist litmus red then bleachesit is chlorine [1] = [6]
2 a i There is a loss in mass because CO, gas [1] islost from the flask [1] = I2l
ii At first the slope of the graph is steep, so the. reaction is fast, then as the slope gets less it
A catalyst provides an alternative pathway
[1], which has a lower activation energy [1]= I2l
b i Homogeneous means that all the substancesare in the same phase (here all are in thegaseous state) = [U
a i i In indus t r ia l p rocesses , the cho ice o ftempera ture depends on ra te (h ightempera ture good) and pos i t ion o fequ i l ib r ium (h igh tempera ture bad fo rexothermic react ions). A catalyst a l lows afast rate at a lower temperature.
a Remember that the ox ide ion has a charge of 2- .b In both e lect ro lys is and cel ls , ox idat ion occurs at
the anode (o and a are both vowels) .
c i Remember that bond breaking is a lwaysendothemic (+) and bond making exothermic(-)
i i l f the reaction profi le had been asked for,then you must show an energy barr ier ( theactivation energy) in your enthalpy leveld iagram.
slows down [1], when the l ine is horizontal(after 6 or 7 minutes) the reaction hasstopped (all the acid has been used up) [1]
= f2fb For experiment 2: your graph should init ially be
steeper than in 1 but is horizontal earlier, butwith the same mass loss (1 B) as in 1. [1]For experiment 3: your graph should init ially beless steep. It should become horizontal later, butat the same mass loss as in 1. [1]For experiment 4: your graph should be steeperthan in 1 and become horizontal at twice themass loss (2 8) as in 1. [f.]
c i Amount of HCI = 1.00 mol dm-"x 50/1000 dm'= 0.0500 mol [L] ,amount o f CaCO3 = '1 , ,0 .0500 mol
= [3]
= 0.0250 mol [L] : mass of CaCO, =0.0250 mol x 100 g mol-' = 2.5O g tl l = [3]
i i As the amount of acid in experiment 4is twice that of the other experiments[1], the mass of CaCO. must be at least2 x 2 . 5 0 = 5 . 0 0 9 t l ] = [ 2 ]
3 a Any two of:use a pipette to measure out the CuSO, solution[1] as the volume is measured more accurately;[1] use smaller pieces of iron [1] as there wil l beless heat loss in a faster reaction [1];use a polystyrene (or plastic) cup [f.] as thisreduces heat loss [1];measure temperature for several minutes beforeaddition of iron [f.] as this allows for a moreaccurate determination of the init ialtemperature [1,];
A r u s w e R s 1 3 1
4 a i Any one of:use a larger beaker [1] as this wil l preventacid spray escaping [1];powder the l imestone [1] as this speeds upthe reaction [1,];use more acid [L] as this gives larger andhence more accurate titres [,]. = [21Choose titres 1 and 3 [1];mean = '1, (14.9O + 14.85) = 14.87 5 cm' 1L1-- r2lAmount of NaOH in t i t re = 0.100 mol dm-' ' x14 .87517000 dm'= 0 .0014875 mol = [1 ]Amount of HCI in port ion = 0.001488mo1
= [1]i i i Total amount of HCI remaining =
10 x 0.001488 = 0.0L488 mol = [11iv Original amount of HCI = 2.0 mol dm-' x
50/1000 dm" = 0.100 mol = [1]v Amount of HCI used = 0.100 - 0.01488
= 0.08512 mol = [1]v i Amount of CaCO. = '1r r 0.085 L2 = O.O4256
mol [1] = 0.04256 mol x 100 g mol- ' = 4.256gtll = l2l
v i i P u r i t y = 1 0 0 x 4 . 2 5 6 9 1 5 . 2 4 9 = 8 t . 2 o / o = [ 1 ]i C which is between I4.75 and 14.95 = [1]ii One of:
not reading from bottom of meniscus, notreading meniscus at eye level, burette notvertical, air bubble below tap, funnel left intop of burette = [U
5 a Place the chemicals in a round bottomed flask[1], f itted with a reflux condenser [1]. Heat forseveral minutes [1]. Disti l off the propan-2-ol[1]. Collect the fraction that boils off around82 "C [1] = [5]
b Amount = 6.15 g + 123 g mol- ' = 0.0500 mol [1]vol of NaOH = 0.0500 mol * 2.O mol dm-' =0.025 dm' = 25 cm' [L] = f}l
c 100o/o yield would give 0.0500 mol of propan-2-ol = 0.0500 mol x 60 g mol- '= 3.00 g t l l .80o/o yield = 3.00 x 80/1,00 = 2.4 g [1] = I2-l
d One of: the reaction did not go to completion;it is an equil ibrium reaction;some propene was formed by elimination. = [1]
ll
b i
l l
measure temperature more frequently [1] asmakes the extrapolation more accurate [1];use a thermometer which reads to 0.5 or 0.1[1] as this gives a more accurate temperaturestir the reaction mixture gently [1] as thisensures an even temperature in the liquid [1].
= [4]
this
O C
[1];
b i Heat given out = 50.09 x 4.18Jg= 3L80 J o r 3 .18 kJ
ii Amount of CuSOn = 0.500 moldm'= 0 .025 mol
ii i -3.18 kJ + 0.025 mol = -L27 kJ(Lmark for the negative sign andnumber)
- 'd.g- ' x 15.2 'C
= [1]dm- 'x 50 /1000
= [1]mol-' = [zl
1. for the
b The tempera ture g0es L . rp , s0 heat i s p roduced, andthe reac t ion is exo thermic ; . ' . AH is negat ive .
a To answer th is you have to work out i f thereact ion is faster or s lower than in 1 and whetherthe mass of C0, lost is the same, more or lesst h a n i n 1 .
c i You have to halve the moles of HCI becausethere are hal f as manv moles of CaC0, asmo les o f HCl .
a i l f poss ib le t i t ra t ion exper iments shou ld bedes igned to g ive t i t res o f about 25 cm' .
i i You shou ld on ly choose t i t ra t ion va lues tha tare less than 0 .2 cm3 apar t .
b v i Don' t forget that the rat io of HCI to CaCO, is2 ' . 1 .
:1" L l l l ,uEr r !
(a) The reaction is slow, so i t must be heated.1-bromopr0pane is vo lat i le , so a ref lux condensermust be used. F inal ly the product must be d is t i l ledoff to separate i t from the reaction mixure.
132 AnsweRs
Unit 44.7 Energetics II
Ca*(g) + Cl(g)
ii AH, of CaCl(s)= +193 +590 +LZt + (-36a) + (-650) = -110 kJmol-t
iii AH, of (ZCaCl(s) -' CaCl,(s) + Ca(s))= -790 -2 x (-L10) = -57O kJ mol-'Therefore products are morethermodynamically stable than reactants.Thus CaCl will disproportionate.
2 NaF: Experimental lattice energy= theoretical lattice energy.Therefore it is nearly t0oo/o ionic.MgIr: Experimental >> theoretical.Therefore it is considerably covalent, because thedoubly charged Mg"'ion is much more polarisingthan the singly charged Na- ion.Also the much bigger I- ion is more polarisable thanthe smaller F- ion.
Topic 4.2 The Periodic Table IINaCl(s) + aq - Na'(aq) + Cl-(aq)S i c l n + Z H r O - S i O z + 4 H C l
PCl, + H.,O -' HrPOn + 5HC1.
NaCl is ionic and so iust dissolvesSicln and PCl, are covalent and so react to giveHCI.A lone pair of electrons on the oxygen atom ofHrO forms a dative bond into the empty 3dorbital in the sil icon atom. This releasesenough energy to break the Si-Cl bond.In CCln, the carbon atom has no 2d orbital andso a dative bond cannot form before the C-Clbond breaks, and the four large chlorine atomsaround the small carbon atom sterically hinderthe approach of a water molecule.
As a base:Al(OH).(s) + 3H.(aq) - 'Al'.(aq) +3H,O(|)and as an acid:
At(OH),(s) + 3OH-(aq) ' [AI(OH), ] ' - (aq)
Topic1 a i
l+ntca(g) +lrcb1sy
| +resCa(s) +'lrclr(g)
ca'*(g) + 2cl(g)
l
| +seoCa(g) + 2Cl(g)
+l + I 2 I
x 2Ca(g) + Clr(g)
-364
Ca*(g) + CI(g)
CaCl(s)
-364 x 2
| . 'noCa(g) + Cl(g)
J .,,,0Ca-(g) + 2cl(g)
+| +1e3
Ca(s) + Clr(g)
l a ilt
lll
It
Note that the enthalpy of atomisat ion of chlor ine isfor the react ion ' / ,C1,(g) -- Cl(g).
b i NIr of CaCl,(s)= +L93 +590 +1L50 + 2 x I2 '1. + 2 x (-364) +
(-2237)= -79O kJ mol-'
Do not be tempted to g ive the tota l ly wrong equat ion:NaCl + H ,0 - - NaOH + HCl .
ca'*(g) + zcl-(g)
CaClr(s)
For the reactions of oxides and hydroxides, ionicequations are usually easier. Basic oxides/hydroxidesreact wi th H- ions to g ive s imple posi t ive ions, and ac id icoxides/hydroxides react with 0H- ions to give negativeions.
1 mole o f S0, g ives 1 mole o f S0, and ' / , mo le o f 0 ,
3 a i Mg(OH),(s) + 2H.(ae) * Mg'.(aq) + 2H,O(l)i i PbO(s) + 2H.(aq) -- 'Pb'.(aq) + H,O(l)i i i SO, + acid: no reaction.
b i Mg(OH), + alkali: no reactionii PbO(s) + ZOH-(aq) + H,O(l) - Pb(OH),'-(aq)iii SO,(g) + ZOH-(aq) * SO,'-(aq) + H,O(l)
b i Because reaction L*R is exothermic, adecrease in temperature causes K. toincrease, and so the position of equil ibriummoves to the right.
i i As the reaction going from right to leftcauses an increase in the number of moles ofgas, a decrease in pressure wil l cause theposition of equil ibrium to shift from right toleft. K. is unaltered.
iii There will be no effect on either. A catalystspeeds up the rate of both forward and backreactions equally, and so equilibrium isreached sooner.
Half way to end point is 12.5 cm',which is when [HA] = [A-].pH a t th is po in t i s 4 .95 .Therefore pK = 4.95.
Phenolphthalein.A buffer is a solution of known pH, which wil l
3 . :
25
3 a ISOr],"no,'
ffi= (q.2\,
(o .2 ) " x 0 .1= 10 mol-t dm'
which is not equal to K., so it is not at equilibrium.As the quotient <K., the reaction moves left toright, unti l equil ibrium is reached.
point is at 25 cm'NaOH.
End point
Half way
Th is i s a su i tab le ind ica tor because i t s range o f pHfor i t s en t i re co lour change is w i th in the ver t i ca l par to f the graph.The [ ] term only equals K when the system is in
equ i l i b r ium.
1 3 4 A N s w E R S
resist change in pH when contaminated withsmall amounts of either acid or base. It consistsof any weak acid and its coniugate base, such asethanoic acid and sodium ethanoate.
The salt is fully ionised:CH'COONa(aq) - CH,.COO-(aq) + Na-(aq)
The weak acid is partially ionised:CH.COOH(aq) + H.(aq) + CH.COO-(aq)
The CHTCOO- ions from the salt suppress mostof the ionisation of the acid.If H- ions are added to the solution, almost allof them are removed by reaction with the largereservoir of CH.COO- ions from the salt, thusthe pH hardly alters.
H.(aq) + CH.COO-(aq) -* CH.COOH(aq)If OH- ions are added, almost all of them areremoved by reaction with the large reservoirof CHTCOOH molecules from the weak acid,thus the pH hardly alters.OH-(aq) + CH,COOH -' CH.COO-(aq) + HrO(l)
[H] = K" x [weak acid]/[salt][weak acid] = 0.44 mol dm-'Amount of sal t = 4.4182 = 0.0537 mol.Therefore [sal t ] = 0.0537/0.L = 0.537 mol
[ H ] = I . 7 4 x l O - s x 0 . 4 40 . 5 3 7
= 1.43 x 10-' mol dm-'Therefore pH = -log (1.43 x 1,0*; = 4.85.
3cH(cH3)cooH
3 a Add water to each: ethanoyl chloride gives offsteamy fumes of HCI; ethanoic acid shows novisible reaction.Add PCl, to each: ethanoic acid gives off steamyfumes of HCI; ethanoyl chloride shows novisible reaction.
b Add to a solution of iodine and sodiumhydroxide: propanone gives a pale yellowprecipitate; propanal gives no precipitate.Add ammoniacal silver nitrate solution andwarm: propanal gives a silver mirror;propanone gives no mirror.
Answers to Practice Test Unit 4The allocation of marks follows Edexcel markschemes.The marks that you will need for each grade areapproximately:
1 - 3om
Topic 4.5 Organic chemistry II
CI
CHs
H CH"\ /
C - C/ \
cHs ct
(a) H\ /
C - C/ \
CHs
Do not say that the pH is constant. 'Nearly constant 'is acceotable.
l f there is a double bond, look for geometr ic isomers.Draw them with the correct bond angles, not 90".Look for four d i f ferent groups around a carbon atomfor opt ica l isomer ism.
Four equat ions are necessary for a good answer,w i th one o f them as an eou i l ib r ium reac t ion .
An increase in the number o f carbon a toms in a cha inind ica tes the use o f a Gr ignard reagent o r a cyan ide .
The concentrat ion not the number of moles should beused in buf fer ca lcu lat ions.
1 a
1 . a
2 a
b
c
1 e 5A r u s w e R s t v t
iii One of: the reaction is faster, or no catalyst isrequired, or with ethanol the reaction is anequil ibrium
Mg'.(g) zcl- (g
r tMs"(g) zcl(s)
r tMg(s) + Cl,(g)
I \ r H H
H l lI ro-?-c-H
H - C - f | |
i \ H H
H o-* MgClr(s)
CtH.t
C .I
' . .
t 'oH
H
= [1]
= [4]= IZf
= [1]
b
Correct cycle with all state symbols [2]Mfo,.ution = SU[l Of all Othef ChangeS- 6 4 2 = + L 5 0 + 7 3 6 + 1 4 5 0 + 2 x 1 2 1+ 2 x ( - 3 6 4 ) + L E [ 2 ]LE = -2492 kJ mol-' [1,] = [5]i The ionic radius of Sr'* is < that of Ba'' [1],
and so there is a stronger force of attractionbetween cat ion and anion in Sr(OH). than inBa(OH), [r.] = l2'l
i i Energy is released when the d-oxygen atom inwater [1] is attracted to the positive cation [.]
= I2liii A.Ff,oru,,on = -L?ttice energy + A-F1n,.,,o,inn of cation +
Mhyd,u,ion of the anions [.]AI1,,, Ba(OH), = +2228 - 1360 - 92O = -52kJ
tllAI1,., Sr(OH), = +2354 - 1480 - 920 = -46kJ
tllso A-F/,o,u,,u. of Ba(OH), is 6 kJ more exothermicand so it is more soluble [,] = l4l
b Reagents: C,H.CHO[l] and CH.MgBr [1] orCH*CHO [1] and C,H,MgBr [1]Conditions; dry ether solvent [1] = [3]
c i Y i S C H , C H T C O C H , = [ 1 ]i i Y i s a k e t o n e = [ 1 ]iii Add 2,4-dinitrophenylhydrazine [1] which
gives an orange ppt. [1]Add Fehling's solution (or ammoniacal silvernitrate) [1] which stays blue (does not give a
CHs
silver mirror) [L]iv CH[,, [f.] and C,H.COONa [1]
= [1]= [1]= [1]
= [1]= [1]
3 a
4 a
C.Ht
I. . .c_
Ho I -.n,
H
i cH.cHrcHrc=Nii LiAlHn or NaBHniii cH3cH2cHzcHzNH2i The lone pair of electrons on the N atom
make it a baseii RNH, + H* - RNH3.I
HI
H - C -I
H
b Amount of AtCl = 0.63g162.5 g mol-' = 0.01 molAmount of Alcl, = 0.67 9/133.5 $ rnol-' =0.005mo1 [1] total moles = 0.015mole fractions: AlCl = O.667, AlCl, = 0.333 [1]
a As a fu l l s t ruc tu ra l fo rmula was no t asked, v0u needn o t s h o w a l l t h e a t o m s .
b A n y a m i n e r , v o u l d d o .c F u l l s t r u c t u r a l f o r m u l a e v r e r e a s k e d . s o a l l a t o m s
a n d b o n d s m u s t b e s h o u r n .
a Remember that la t t ice energ ies are negat ive.b i i i l f you forgot to mul t ip ly -460 by 2 ( there
are 2 0H- ions) you would lose 1 mark.Another way to score the 3rd and 4thmarks is : f rom Sr-- Ba, the la t t ice energydecreases by 126 kJ [1], but AHnro,,,,o. of thecation only decreases by 120 kJ [1]
a You must make an at tempt at a 3-D drawing andthe two isomers must be mir ror images of eachother .
b i i i You must show that i t is a carbonylcomoound and then that i t is not analde hyd e.
b iv C,H.C00H would be accepted as the otherorganic product .
, --
et a l
\ -
ii It forms a substituted amide
136 AxsweRs
c An increase in temperature shifts the equilibriumin the endothermic direction. . ' .AH ispositive = [1]
d Ko stays constant [1] The position of equilibriummoves to the side with fewer gas molecules [1],which is to the left [1] = [3]Mix scrap aluminium metal with AlCl, and heatto form AlCl [U, remove from excess Al, thencool (or increase pressure) and pure Al and AlCl3are formed. The latter is then used again [1] = [2]
5 a i MgO(s) + H,SOn(aQ) -- MgSO.,(aq) + H,O(l)Correct equation [1] state symbols [L] = lz'J
ii A white solid [1] reacts to form a colourless
25
Vol. NaOH/cmE
Shape of curve correct [L], start at pH = 4 llf,vertical at 25 cm' [1] and at a pH of between 7 and11 [f.], curve flattening off at pH 13 [1] = [5]i HrSOl --' H* + HSO.,- [L]
HSO.*- + H* + SOn^ [1]ii The second ionisation is incomplete and is
suppressed by the H. from the first ionisation,and so produces very few H. ions
ii For having 2 equations, one with H' (or HCI)and the other with OH- (or NaOH) [1,] and:AI(OH)3 + 3H* -- Al''* + 3HrO orAI(OH)3 + 3HCl -- ' AICIr + 3H,O [1.] plusAI(OH), + 3OH- -r Al(OH)u'- orAI(OH), + 3NaOH -'Na,Al(OH). [1] = [3]
MgO is basic, AI(OH). is amphoteric and PnO,o isacidic [L] showing the metall ic character of theelement decreasing across the period tl l = f2lAs a Group is descended the oxides get morebasic [1]. Indium is in the same Group asamphoteric aluminium, so indium oxide wil l bebasic [1] = I2l
= [2]
= [1]
6 a
Uni t 5Topic 5 .7 Redox equilibria
1 a i CrrOr'- (aq) + 14H'(aq) + 6e- -- 2Cr"(aq) +THrO
ii Snn.(aq) + 2e- -- Sn'.(aq)i i i IO, (aq) + 6H.(aq) + 5e- - '1,1,+ 3HrOiv l, + 2e- -- Zl-(ag)
b
An acid is a proton donor [,],Weak means that it is incompletely(ionised) [L], but dilute means thatconcentration is small [1]i pH = -log,u [H.]ii Ku = 3.72 x lO-" = [H.]tOCl-l
a Sol ids do not appear in Ko expressi0ns.b Part ial pressure is mole fract ion x total pressure.c K is only a l tered by temperaure.
a Don't conf use weak (how ionised) with dilute(how many moles) .
b To get f rom pH to [H- ] , use the 10" but ton onyour ca lcu lator .
d Sulphur ic ac id is only s t rong in i ts 1stion isat ion, so th is so lut ion has a pH of 0.98 not0.699.
Many students are very bad at writ ing equations.You must make sure that you learn the equat ions inTopic 4.2 o l the speci f icat ion.a Whenever you are asked to state what you would
see, you must descr ibe the appearance before andafter reaction.
Al l are reduct ions and so have e lect rons on the le f t .The number of e lect rons equals the tota l change inoxidat ion number.
b i
Equat ion a i i is mul t ip l ied by 3, reversed andadded to equation a i .
t The water ligands split the d orbitals into two ofhigher energy and three of lower energy. Whenwhite light shines on the solution, an electronabsorbs visible light energy and moves (jumps)from the lower to the higher energy level,absorbing some of the red/green light and leavingblue light.
2 a [Cr(H,O). ]"(aq) + 3OH-(ag)-Cr(OH),(s) + 6H,O
This react ion is deprotonat ion.
then Cr (OH) , (s ) + 3OH-(aQ) + Cr (OH) . ' - (aq)
,/I
T h i s o c c u r s b e c a u s e C r ( l l l , y i s a m p h o t e r i c
then no further reaction.[Cu(H,O)J'.(aq) + 2NH.(ae) - Cu(OH),(s) +2NH; (aq) + 4H"O
then Cu(OH),(s) + 4NH,(aq) + 2H,O(l) -- '
[Cu(NH,)o(H,O)J'.(ag) + 20H-(aq)
The overal l react ion is l igand exchange.
4 a VOI(aq) + 4H.(aq) + 3e--- V'.(aq) + ZH,O
c
3 a
b
taI--ta
Equat ion a iv is mul t ip l ied by 512, reversed andadded to equat ion a i i i .
This reaction is also deprotonation.
Vanadium(V) is reduced, so electrons are on the left . Theoxidat ion number changes by 3, so there must be threeelectrons.
Topic 5.3 Organic chemistry III1 a The bromine reacts with the iron catalyst to
form FeBr.:ZFe + 3Br, --' 2FeBr,
The FeBr. then reacts with more bromine toform Br'(the electrophile) and FeBrn.
The intermediate cation loses H. to the FeBr".
H t - \ s t l -
GyB'l,,rl.-,, 1v / l t lL B r I
O-u'* nu, + FeBr,
It is energetically favourable for the intermediatecation to lose an H* and gain the stability of thebenzene ring, rather than add Br- as happens withalkenes.
The first step is the addition of a CN- ion. HCN istoo weak an acid to produce a significant amountof CN- ions. NaOH will deprotonate HCNmolecules, producing the necessary CN- ions.
3 a If the temperature is too low (<5 'C), the rate is
too slow.
Check that:,s. the arrow starts on the delocal ised r ing and goes
towards the Br- (and not to the + of Br-). ' f f i the intermediate has a broken delocal ised r ing
w i t h a + i n s i d e i t
.r the arrow starts from the o bond of the r ing/Hatom and goes ins ide the hexagon (but notdirect ly to the +).
138 ANsweRs
b VOI(aq) + ZH.(aq) 1 e - VO"(aq) + H,Oc V"(aq) + HrO -- VO'.(aq) + 2H.(ae) + e-d VO"(aq) + H,O -- VOi (aq) + 2H.(aq) + e-
If the temperature is too high (>5 'C), the
benzene diazonium chloride wil l decompose.
b C.H.N=NC6H*O- but C.H'N=NC.H.,OH isacceptable.
Tonic 5.4 Chemical kinetics IIL v r P v v . ^
I Rate = klHll '
k = rate.tHI]"
= 2.0 x 10'mol dm-' s- '
@= 0.080 s' mol' dm'
2 Cany out the following procedure:s* Place equal volumes of solutiorl, e.8. 50 cm'of 0.1.0
mol dm-'ethanoic acid and methanol in flasks in a
thermostatically controlled tank at 60 "C.
,,,:' Mix, start the clock and replace in the tank.'ii:i At intervals of time, pipette out 10 cmt portions
and add to 25 cm'iced water in a conical f lask.
+ Rapidly titrate with standard sodium hydroxidesolution using phenolphthalein as the indicator.
.# Repeat several times.llr Plot a graph of the titre (which is proportional to
the amount of ethanoic acid left) against time.
3 Time taken for the concentration to halve from 1.6to 0.8 mol dm-' = 26 minutes.Time taken to halve again to 0.4 mol dm-'= 26minutes.Time taken to halve again to O.2 mol dm-'= 26minutes.tr,ris a constant, therefore reaction is 1st order.
Topic 5.5 Organic chemistry IV1 a Add 2,4-dinitrophenylhydrazine. Both give an
orange precipitate.Add ammoniacal silver nitrate solution. Onlypentanal will give a silver mirror on warming.
Heat under reflux with aqueous sodiumhydroxide.
Cool and acidify with dilute nitric acid, thenadd silver nitrate solution.
Fehl ing 's so lut ion, which g ives a red prec ip i ta te,could be used in place of ammoniacal si lver nitrate.
Halogenoalkanes are covalent and so must f i rst behydro lysed to produce hal ide ions.
The 2-bromo compound gives a creamprecipitate insoluble in dilute ammonia butsoluble in concentrated ammonia.The 2-chloro compound gives a whiteprecipitate which dissolves in dilute ammonia.
Warm with dilute sulphuric acid and potassiumdichromate(VI) solution and disti l l off anyproduct into ammoniacal silver nitratesolut ion.Methylpropan-Z-ol does not change the colourof the potassium dichromate(VI).Methylpropan-1-ol turns it from orange togreen and the disti l late gives a silver mirror.Butan-2-ol turns it from orange to green andthe disti l late has no effect on the silver nitratesolution. To confirm, add a few drops of thebutan-2-ol to iodine and aqueous sodiumhydroxide and warm gently. A yellowprecipitate of iodoform will be produced.
The substance is probably aromatic (> 6 carbonatoms and about the same number of hydrogenatoms).120 is the molecular ion.105 is 15 less than 12O and is probably caused byloss of CH,..77 is probably caused by the (CuH,). group (77 is 43less than 720 and is probably caused by loss ofcocH.).X is probably C.H.COCH..The 120 peak is caused by (C.H.COCH3).,the 105 peak by (C.H.CO)',and the 77 peak by (CuH,)..
32OO cm-' is due to O-H, 1720 cm-' is due to C=O,and 1L50 cm-' is due to C-O.
Six of the carbon atoms are in a benzene ring. Theremainder are either CrH, group or 2 x CH, groups.
Y is C.H,CHrCH..
[5 H's in CoH., 2 H's in CH, and 3 H's in CH..]
Z is CoH,(CH.)r.
[6 H's in 2 x CH, groups, 4 H's in CuHr.]
Answers to Practice Test Unit 5
A N s w E R s 1 3 9
approximately:
a i Consider exper iments 1 & 2: when [RCHrCl] isincreased 3 t imes, rate also increases 3 t imes[U, . ' . 1st order wi th respect to [RCHrCl] . t l lCons ider exper iments 1 & 3 : whenconcentrat ions of both are doubled, the rate isincreased - l t imes [1] , . ' .1st order wi th respectto [OH ] as n 'e l l [1]
i i Rate = k [RCH.Cl l x [oH ]i i i k = ra te / ( [RCH-Cl l x [OH- ] )
= - 1 . 0 x l 0 ' m o l d m ' s
'
0 . 0 5 0 m o l d m x 0 . 1 0 r n o l d m - '= 0.080 [r] mol
' clm s
' [U = l2' l
iv As i t is ls t order in both, i t is a S,,2 mechanism
tu.
= [4]= [1]
R\
/--\ ^H - ()'
,(:- Cl ----------------
H lH
_ R
I
I
H - () ----C---- Cl ----------------D r
/ \ :- H H - J
R//
H O - C ^ C Ir \l HH
Curlv arrows from O of OH- to C and from C-Cl obond to Cl [1], correct transition state [L] = [3]
2 a
b
The electrophi le is CH.CO* [1] .The equat ion is:CH,COCI + AlCl,, --+ CH,,CO. + AICI*- [t] = I2l
H ii At ^v\
c+ _-_+ l ( * . I t . - - . - } l ( ) l cH,\ v / \ vCH,
- CH,
The allocation of marksschemes.The marks that you wil l
follows Edexcel mark
need for each grade are
+ H *
Curly arrow from ring to C of CO' [1], correctintermediate [,], curly arrow from C-H o bond toring [L] = [3]React the phenylethanone, C.HTCOCH3, withHCN [1], with a trace of base (or in a solutionbuffered anywhere between pH 5 to 9 or add amixture of HCN and KCN) tl l . This produces
1 2
T h e 1 ' a l c o h o l i s p a r t i a l l y o x i d i s e d t o a n a l d e h y d e .The 2" a lcoho l i s ox id ised to a ke tone.The 3" a lcoho l i s no t ox id ised .Butan-2-o l con ta ins the CH,CH(0H) group and sog ives a pos i t i ve iodoform tes t .
i You must make i t c lear what data you areus ing and how you ar r i ve a t each order .
iv Make sure that al l your cur ly arows starte i ther a t a bond and go to an a tom 0r a t anatom and go to a bond.
i
140 ANswERs
CHs
ICsHs - C -CN
IoH tll
Now add aqueous sulphuric acid (or anynamed acid) [L] and heat under reflux [1]
_
d Mix known amounts of ester and oH- [1]
= [51
remove sample [t]at known time [1.]quench the reaction by adding ice cold water [1,]titrate unreacted OH- with acid [t]Calculate rate = change in [OH-] + time [1]Repeat with double [ester] and the same tOHl tllRepeat with double [OHl and the same[ester]-tl]
= [8]
3 a The reagent for step 1 is magnesium [1], and theconditions are dry ether [1] = l2l
b i Potassium dichromate(Vl) [1] and sulphuricacid [1] = f2l
ii Distil off D as it is formed [1] = [1]iti Ethanoic acid (CH.COOH) or ethanoate ions
(cH3coo) tll = [1]E is CHTCH,CH(OH)CH. [f.] for any secondaryalcohol and [.] for the correct formula = f2li F is a carbonyl compound as it reacts with
Z,4-dinitrophenyl hydrazine [U, but it mustbe a ketone and not an aldehyde as it doesnot react with ammoniacal silver nitrate [1]
= I2lii In order to do the iodoform reaction F must
have the CH.CO group [1] and the products
4 a Dip a platinum elec-trode [1] into a solutionwhich is L mol dm-'in both Fe'* and Fe'. ions [1].This is connected via a salt bridge (containingpotassium chloride solution) [U to a standardhydrogen electrode [1]. The potential is measuredwith a high resistance voltmeter (or apotentiometer) [L] = [5]
b i Au''* + 3e- - Au [1]Fe" -t Fe"* + e- [1] =
ii Au" + 3Fett - Au + 3Fet' =
iii Au'. + 3e- - Au E (gold)3Fe'* - - 3Fet '+ 3e- E'= -0.77 vadding the two half-equations gives E cellE (gold) + (-0.77) = E' (cell) = + 0.73 [UE (gold) = * 0.73 + O.77 = + 1.50V [1] = [21
or [Fe(H,O)u]'. + ZOH- -' [Fe(OH),(H,O)J + ZH.Ocorrect iron species in product [1] balance
tll = fzfiii The precipitate's colour is pale green = [1]iv [Cr(OH)J'- = [1]v The solution of R would first give a Sreen
precipitate [1], which forms a green solutionwith excess acid [1] = I2l
i The standard electrode potential is thepotential difference between a standardhydrogen electrode and the half-cell [1] whereall concentrations are 1- mol dm-'and thepressure of all gases is 1 atm. [L] = I2l
ii All four halogens will oxidise Cr'* = [l]iii Both bromine and iodine will not oxidise Cr''*
' further = [1]
l2ltll
4s
m
d
b
are CHI,. [1] and CH.CH,COO- [U =
iii F is CH,CH,COCH. [2] =
e The species are: (CH.CHTCOCH..). for m/e = 72(COCH.). for 43 [1] and (CH3CH,)' for 29 tll
_
t3lr2l
[1],
t3l
d
b The curly arrow must go towards the C of the C0 nol theC of the CH, group. The in termediate must have a brokenr ing going across a l l but the C atom which has formedthe bond wi th the CH,C0 group.
d The pH would hard ly change, so a method involv ingmeasur ing the pH change over t ime would score amaximum of 6.
b i Reduction (gain of electrons) takes place at thegold electrode because i t acts as the cathode.Thus Au' . ions gain e lect rons that f lowed in theexternal c i rcu i t f rom Fe" ions which lost them.
i i i Mul t ip ly ing the Fe ' . ha l f -equat ion by 3 does nota l ter i ts F va lue.Another way of doing the calcu lat ion is to usethe standard reduction potentials thus:F of ox id is ing agent (gold) - f o f reducingagent ( i ron) = P of the cel l .
b i i i The a ldehyde D is ox id ised to an ac id by thes i l ver /ammonia complex ions .
c Grignard reagents react wi th an aldehyde to give asecondary a lcoho l .
d i i The only substances to do the iodoform react ion area lcoho ls w i th a CH,CH(0H) group, ke tones w i th aCH, ,C0 group and e thana l .
e Don' t forget the charge on the species in a mass spectrum.
a Remember tha t 3d ' , 4s ' i s more s tab le than 3do,4s ' (and s imi la r ly fo r Cu wh ich is 3d '0 , 4s ' ) . A lsothe 4s electrons are lost f i rst when cat ions aref o r m e d .
c i A l l t rans i t ion meta l ions are hydra ted inso lu t ion , w i th 6 water mo lecu les coord ina te lyb o n d e d o n .
iv Chromium is amphoter ic . ICr (0H) , ] - wou ld bean acceptable answer.
v Add i t ion o f ac id w i l l f i r s t p rec ip i ta te thehydrox ide wh ich , as i t i s bas ic , w i l l then reac tw i th excess ac id to fo rm a so lu t ion o f thehydrated metal ions.
Mix the blue Cr'* solution separately withbromine and iodine [1], the solution goesgreen and stays green with excess halogentll = [2]
Answers to Practice Test Unit 68The allocation of marks follow Edexcel mark schemes.The marks that you wil l need for each grade areapproximately:
1 a Amount of NaOH = 0.100 mol dm-' x 0.0280 dm'= 0.00280 mol [f.] = moles of H.Ratio H. to NHn* = 1 : 1".Amount of NH.,. in 25 cm' = 0.00280 mol.Amount in 250 cm' = 0.0280 mol [L]Amount of (NHr)rSOn - '1, , O.o280 = 0.0140 mol[1] ,mass of (NHn)rSOn - 0.0140 mol x 132 g mol-' =1.848 I [1] ,o/o (NH*)'SO.* in ferti l iser = 1".848 x 100/3.80 =48.60/o faf = [5]Any two of: not all ammonia driven off [1];ammonia incompletely absorbed by the HCI [f,];some ammonia gas escapes [1] = I2li Ba'.(aq) + SOn'- (aq) - BaSO.(s)
Species [1], state symbols [1.] = f2lii To ensure that all the sulphate ions were
precipitated = [UMany carbonates wil l decompose on heating [L],and the CO, gas evolved will result in a lowermass being recorded [1]. An example is CaCO. [1]CaCO, - CaO + CO, [1] = [4]
A ,NSwERS 141
Answer any two of the section B questions
Any two of : i t is considerably cheaper [1] ;less r isk of leaching tUit does not effect the pH of the soil [1]it releases the nitrogen slowly [L] = [21The smaller the K" of the conjugate acid, thestronger the base [1]. Therefore ammonia is thestronger base [.] = I2lPolymerisation occurs when many molecules iointo form a long chain [1]; condensation is withthe elimination of water or a small inorganicmolecule such as HCI [L].A polyamide contains the -CONH- l ink [1] as in:
lv
2 a
b
O O H H/ l l , \ l l | , \ l \
+. -( cH,)--i- *-(. n,),-*+ tu\ l
d i
l l
e l
The reagents are : l iquid bromine [1] andconcentrated sodium hydroxide [1]. Thecondition is that it must be heated [1] = [3]NH.NH, = [uSolids are not included in Ko expressions andammonium nitrate is a solid. = [1]
ii As the reaction is endothermic, the energylevel of the products is higher than that of thereactants [1], therefore the reaction is said to
b
d
a Don ' t fo rge t to mu l t ip ly by 10 to ge t the to ta la m o u n t o f t h e N H , ' i o n s , a n d t h e n h a l v e i t a s t h e r ei s I a m o l e o f a m m o n i u m s u l p h a t e p e r 1 m o l o fN H . i o n s .Forget t ing to d iv ide by 2 g ives the answer 97 .3Yo(obta ined by about ' / , o f the cand ida tes) wh ichscores 4 marks .
c i In a p rec ip i ta t ion reac t ion there are jus t two ionso n t h e l e f t a n d t h e f u l l f o r m u l a o f t h e s o l i d o nt h e r i g h t o f t h e e q u a t i o n
d Don ' t g ive a g r0up 1 carbonate , nor BaCO, , as i td o e s n t d e c o m p o s e 0 n h e a t i n g i n t h e l a b , n o rB e C 0 b e c a u s e i t d e c o m p o s e s b e l o w r o o mte m oe rat u re.
a The d isadvantages of us ing urea are that some ureaevaporates especial ly i f there is no rain within 4 days ofappl icat ion. l t cannot be b lended wi th phosphate norpotass ium fer t i l isers, and i t is less su i tab le on chalkysoi ls . l t can cause damage to seedl ings. l t does not worki f the soi l is too cold.
c The stronger the base, the weaker i ts conjugate acid.d Thermodynamic s tabi l i ty is to do wi th AH and k inet ic
stabi l i tv is to do with the rate of the reaction.
142 ANswERs
be thermodynamically stable [1] and theequilibrium lies to the left [U.As the reaction proceeds on moderate heating,the activation energy is fairly small [1] and sothe reaction is kinetically unstable [1]
3 a i The reagents are potassium dichromate(Vl) [1]and sulphuric acid [1]. The conditions arecareful heating and distil off the aldehyde as itis formed [.]. = [3]
ii The conditions for the addition of HCN to acarbonyl compound are KCN in aqueousethanol at a pH of 8 (or a mixture of KCN andHCN) [1]The mechanism is:
[1] for a curly arrow going from the C in CN-
[1] for a curly arrow going from the C:O z bondto the oxygen atom
[1] for the intermediate with its - charge = [4]i i i CH,CH,CH,CH(CH.)CH(OH)CN + HCI + zH,O
-- CH.CH,CH,CH (CH3) CH(OH) COOH+ NH'CI [.],correct formula of organic product [1] = l2l
cH, H\.-./
o/ \ /
cHs - cHz- cHz c.
H A O - H
= [1]
K. = [H,O.] x [A-] or [H] instead of [H,O.] [1]IHA]
lH .o l= [A l= . / - ( r r l=@= 6.71x L0" mol dm-' [1]
4 a The Fe ion is 3+ whereas the Na ion is only 1+ [1]The Fe'* ion is much more polarising than the Na.ion [L], and so it draws the electrons [.] from thelarge Cl- ion towards itself and the bond becomes
iii The bonding is dative covalent (coordinate)[1]The ligands split the d-orbitals into two levels
tllThe light is absorbed and a d electronpromoted from the lower to the higher level.tll = [3]
c Each oxygen is -2, :. 4 oxyg€rrs = -8. The ion is2-, and so the Fe is +6 (6+ and 8- make 2-) tllThere are two ways. The first is: add excessacidified potassium iodide solution to an aliquotof the FeOn'- solution [1], then titrate theliberated iodine against standard sodiumthiosulphate solution [1], adding starch when thesolution becomes pale yellow and stopping whenit is colourlesss [1]The second method is: add excess acid to analiquot [1], then titrate against standard FeSOosolution [1] until the solution becomes very palepink [1] = [4]
H/
R- i {-'cN=---------- n-/ \\.--o
H
IR- C -CN
IOH
H
IC - C N €l nI t t
o- H-cN\-/
iv
b i
ll
pH = -log (6.7L x LO') = 3.L7 l l lThe acid is partially ionised: HA =. H* + A-The salt is totally ionised: NaA - Na. + A-This suppresses the ionisation of the acidso both [HA] and [Al are large comparedany H* or OH- that may be added. [1]When H'is added, it is removed by thereaction:H. + A--- HA [1]If OH- is added, it is removed by:OH- + HA -- A- + H,O [1.]
= [3]
tllandto
a i Care has to be taken i f you wish to stop the oxidationof a pr imary a lcohol at the a ldehyde.
i i As always make sure that your curly arows start onan atom and go to form a bond or s tar t on a bondand go to an atom.
i i i the react ion is s imply R-CN - RCO0H.iv the conc sulphur ic dehydrates the a lcohol to a C=C.
You were to ld in the stem that HA decolour isedbromine water and so i t is unsaturated.
b i i This is the standard answer to the mode of act ion ofa buffer and must be learnt.
a A compar ison wi th NaCl must be made.b i Make sure that you identi fy both acid-base pairs and
label them c lear ly .i i l ron is not amphoter ic and so the ppt s tays in excess.i i i After the electrons have been excited they gradually
fal l back to the lower level giving out heat, and arethen ready to absorb photons.
c FeO., '- is very l ike Mn0.,- in colour and in oxidising power.So you can use either the standard method of est imatingoxid is ing agents ( iod ine t i t ra t ion) or the manganate(Vl l )method which is d i rect t i t ra t ion wi th i ron( l l ) ions.
= [4]
A P P E N D I X
Affi organic reactions
1. AS and A2 (synoptic)Alkanes, e.g. ethane CIITCII,. Ethane ---+ carbon dioxide and water
Halogenoalkanes, e.g. l-bromopropaneo l-bromopropane ---+ propan-l-ol
Reactant: sodium (or potassium) hydroxideEquation: CH3CHTCHTBT + NaOH --- CHTCHTCHTOHConditions: heat under reflux in aqueous solutionClassification: nucleophilic substitution
o l,-bromopropane ---' propeneReactant: potassium hydroxideEquation: CH,CH,CHTBT + KOH --r CHrCH:CH, + KBr + HrOConditions: heat under reflux in ethanolic solutionClassification: elimination
o L-bromopropane --+ butanenitrileReactant: potassium cyanideEquation: CH3CH,CH,BT + KCN --+ CH.CH2CH2CN + KBrConditions: heat under reflux in a solution of ethanol and waterClassification: nucleophilic substitution
o L-bromopropane ---; 1-aminopropaneReactant: ammoniaEquation: CH.CHTCHTBT + 2NH, + CH.CH2CH2NH2 + NHnBrConditions: heat a solution of ammonia in ethanol in a sealed tubeClassification: nucleophilic substitution
A P P E N D I x
. 1-bromopropane ---+ Grignard reagent (A2 only)Reactant: MagnesiumEquation: CH,CHTCH,BT + Mg ---' CH,CH,CH,MgBTConditions: warm (in water bath) under reflux in dry ether
Alcohols, €.S. ethanolo ethanol ---; ethanal
Reactant: potassium dichromate(Vl) + dilute sulphuric acidEquation: C,H,OH + [O] ----r CH.CHO + HrOCondit ions: heat careful ly and dist i l out the aldehyde as i t is formedClassif icat ion: oxidation
o ethanol ---+ bromoethaneReactant: hydrogen bromideEquation: C,H.OH + HBr - CrH.Br + H,OCondit ions: HBr made in sit t t from 50(% sulphuric acid and sol id potassium
bromideo ethanol ----r iodoethane
Reactant: hydrogen iodideEquation: CTH.OH + HI ---+ C,H.l + H,OCondit ions: HI made in situ from iodine and moist red phosphorus
o Ethylmagnesium bromide ---' a secondary alcoholReactant: an aldehvde such as ethanalEquation: CrH,MgBr + CH.CHO --+ CH,CH(OH)C,H.Conditions: dry ether solution, then hydrolyse with dilute acidClassification: nucleophilic addition to the aldehyde
o Ethylmagnesium bromide ---+ a tertiary alcoholReactant: a ketone such as propanoneEquation: CrHrMgBr + CH,COCH. - (CH.),C(OH)C,H,Conditions: dry ether solution, then hydrolyse with dilute acidClassification: nucleophilic addition to the ketone
o Ethylmagnesium bromide ---; a carboxylic acidReactant: (solid) carbon dioxideEquation: CrHrMgBr + CO, ---t CTHTCOOHConditions: dry ether solution, then hydrolyse with dilute acid
Carboxylic acids, e.g. ethanoic acido ethanoic acid ---+ an ester
Reactant: ethanolEquation: CH,COOH + C,H,OH:CH.COOC,H, + H,OConditions: heat under reflux with a few drops of concentrated sulphuric acidClassification: esterification
Condit ions: heat
A P P E N D I X
o ethanoic acid ----r an alcoholReactant: l i thium aluminium hydride ( l i thium tetrahydridoaluminate(II I))
Equation: CH,COOH + a[Hl - CH,CH'OH + H'O
Conditions: dry ether solution then hydrolyse with dilute acid
Classification: reduction. ethanoic acid ---+ an acid chloride
Reactant: phosphorus pentachloride (or PCl, or SOCIr)
Reactant: tin and concentrated hydrochloric acidEquation: C'H,NO, + 6[Hl - C"H'NHr+ 2H.OConditions: heat under reflux, then add sodium hydroxideClassification: reduction
PhenylaminePhenylamine --r diazonium ion + azo dye
Reactant: step 1: nitrous acid at 5 'C; step 2: phenolEquation: CuHsNH2 + 2H' + NO, - C.H.N,' + 2H.O
CoHrN2'+ C"H.OH - HOC.H.-N:N-C"H. + H.OCondit ions: mix phenylamine with sodium nitr i te and hydrochloric acid at
5 "C, then add phenol in sodium hvdroxide solut ion.
