Answers to SAQsadvancedsciences.cambridge.org/media/COAS_C2_assess_bkSAQan… · 216 Chapter 1 1 a b Alternatively: CH 3C(CH 3)(OH)CH 2CH 3 c 2 C 4H 9Br 3 a 3-chloro-4,5-dimethylheptane

216

Chapter 1 a1

b

Alternatively: CH3C(CH3)(OH)CH2CH3c

2 C4H9Br

a3 3-chloro-4,5-dimethylheptane. b 1-methyl-3-nitrobenzene. c Bromoethanoic acid.

a4 NH2CH2CO2Hb

c

5

Answers to SAQs 6

a7 Homolyticfissioniswhenacovalentbondisbroken toformtwo(free)radicals.

b Thetwopropagationstepsconstituteachain reaction.Inthesestepschlorineradicals,Cl•(g), areconsumedandregenerated.Aslongasthereis asufficientsupplyofchlorine,Cl2, and ethane, the reactionwillcontinue.

8 Twomolesofbromine,Br2.

a9 Anelectrophileisareagentwhichacceptsan electron-pairasitattacksanelectron-richcentre, leadingtotheformationofanewcovalentbond.

b The CH2BrCH2+ carbocation, formed in the initial

attackbyBr2,reactswithawatermolecule(oran OH–ionfromwater)toform2-bromoethanol.

10 Anucleophileisareagentwhichdonatesalone-pairof electronsasitattacksacentre(suchasacarbonatom) whichhasapartialpositivecharge,δ+.Theattack resultsintheformationofanewcovalentbond.

Chapter 2 a1 6electrons(onefromeachcarbonatominthe

benzenering). b 2porbitals(onefromeachofthesixcarbonatoms). c Onceinvolvedinπbonding,theseelectronsarenolongerassociatedwithanyoneparticularcarbonatom.

2 Allthebondsarethesamelength. BenzenedoesnotbehavelikeanalkenewithC=C

doublebonds.Forexample,itdoesnoteasilyundergo additionreactions.

Theenthalpyofhydrogenationisnotasexothermicas predictedfromthetheoreticalvaluecalculatedusing Kekulé’smodel.

a3

C

H H H

H

C

O

H

H C

H

H

C

H

H

C H

H

CH3 CH2CH3

CH3

OH

C

HO

FIGURE NUMBER: 0521798825c01awf027JOB No: 8369CLIENT: C.U.PJOB TITLE: A2 Core ChemistrySAVED AS: 0521798825c01awf027_pf2

OHCOOH

Br BrOH

Br O

(butanoic acid)OH

CH3 CH2 CH2 C

O

(methyl propanoate)O

CH2

CH3CH3

C

O

(ethyl ethanoate)

OCH3

CH2

CH3

C

(2-methylpropanoic acid)

CH3

CH3

C

O

CH OH

(propyl methanoate)

O

OH

CH2

CH3CH2

C

(1-methylethyl methanoate)

O

OH

CHCH3

CH3C

O

(butanoic acid)OH

CH3 CH2 CH2 C

O


CH2

CH3CH3

C

O

(ethyl ethanoate)

OCH3

CH2

CH3

C


CH3

CH3

C

O

CH OH

(propyl methanoate)

O

OH

CH2

CH3CH2

C


O

OH

CHCH3

CH3C

O

(butanoic acid)OH

CH3 CH2 CH2 C

O


CH2

CH3CH3

C

O

(ethyl ethanoate)

OCH3

CH2

CH3

C


CH3

CH3

C

O

CH OH

(propyl methanoate)

O

OH

CH2

CH3CH2

C


O

OH

CHCH3

CH3C

O

(butanoic acid)OH

CH3 CH2 CH2 C

O


CH2

CH3CH3

C

O

(ethyl ethanoate)

OCH3

CH2

CH3

C


CH3

CH3

C

O

CH OH

(propyl methanoate)

O

OH

CH2

CH3CH2

C


O

OH

CHCH3

CH3C

O

(butanoic acid)OH

CH3 CH2 CH2 C

O


CH2

CH3CH3

C

O

(ethyl ethanoate)

OCH3

CH2

CH3

C


CH3

CH3

C

O

CH OH

(propyl methanoate)

O

OH

CH2

CH3CH2

C


O

OH

CHCH3

CH3C

O

(butanoic acid)OH

CH3 CH2 CH2 C

O


CH2

CH3CH3

C

O

(ethyl ethanoate)

OCH3

CH2

CH3

C


CH3

CH3

C

O

CH OH

(propyl methanoate)

O

OH

CH2

CH3CH2

C


O

OH

CHCH3

CH3C

COOH

HH

HOOCC C

H

COOHH

HOOCC C

CH3

NO2

CH3

1-methyl-2-nitrobenzene 1-methyl-3-nitrobenzene

1-methyl-4-nitrobenzene

NO2

CH3

NO2

Answers to self-assessment questions

217

2 a Butan-1-olshouldbeheatedgentlywithacidifieddichromate(VI);thebutanalshouldbedistilled off immediately.

b CH3CH2CH2CH2OH + [O] CH3CH2CH2CHO + H2O

c Butan-1-olshouldberefluxedwithacidifieddichromate(VI);thebutanoicacidshouldbedistilledoffafteratleast15minutes’refluxing.

d CH3CH2CH2CH2OH + 2[O] CH3CH2CH2COOH + H2O

Alternatively: CH3CH2CH2CH2OH + [O] CH3CH2CH2CHO + H2O followedby CH3CH2CH2CHO + [O] CH3CH2CH2COOH

a3

b

a4 Nucleophilicaddition. b Hydride ion, H−. c

5 Withpentanalasilvermirrorisobservedformingontheinsidesurfaceofthetesttubebutnochangeisseenwithpentan-3-one.Withpentanal,thealdehydeisoxidisedtothecarboxylicacid(pentanoicacid)byAg+ionsintheTollens’reagent,whicharereducedthemselvestoAgmetalintheredoxreaction.Ontheotherhand,ketones,suchaspentan-3-one,cannotbeoxidisedtocarboxylicacidsbymildoxidisingagentssonochangeisseenwithTollens’reagent.

Chapter 4 a1 Ester.

b Carboxylicacid. c Ester.

a2 Zn + 2CH3CH2COOH (CH3CH2COO)2Zn + H2 b Na2CO3 + 2HCOOH 2HCOONa + CO2 + H2O c MgO+2CH3COOH (CH3COO)2Mg+H2O

d

b

4

a5 Theattackingspeciesinvolvedintheadditiontobenzeneisa(free)radical.

b Theattackingspeciesinvolvedintheadditionofchlorinetoanalkeneisanelectrophile.

a, b6

a7 Potassiumphenoxideandwater. b

8 Thebrominemoleculeispolarisedbythedelocalisedπelectronsonphenol.(Theenhancedreactivityofthebenzenering,causedbythe–OHgroup,isalsorequired.Notethataqueousbrominewillnotreactwithbenzene–ahalogencarrier,suchasFeBr3isneeded.)

Chapter 3 1 Thehydrocarbonpartsofthemoleculesareonlyattractedtoothermoleculesbyweak,instantaneousdipole–induceddipoleforces(vanderWaals’forces).Thecarbonylgrouphasapermanentdipoleandwillhydrogenbondtowatermoleculesviaalone-pairontheC=Ogroup.Aldehydesandketoneswithlessthanfourcarbonatomsaremiscible(theymixfreely)withwaterbecausetheintermolecularforcesinthemixturearesimilarinstrengthtothoseintheseparateliquids.Asthelengthofthecarbonchainisincreased,thevanderWaals’forcesincreaseandstarttodominatethephysicalpropertiesofthealdehydesandketones.Ifthelargermoleculesweretodissolve,theywoulddisruptthehydrogenbondingbetweenwatermolecules,sothesecompoundsremainimmisciblewithwater.

CH3

NO2

NO2

O2N

Cl Cl AlCl3

δ+ δ–

H Cl

+

Cl

+ [AlCl4]–

+ HCl + AlCl3

OH

CO

ether

aldehyde

OCH3

H

phenolic–OH group

+ H2OOH + KOH O–K+

CH2 CH3CH

OH

H3C

CH2 CH2 CH2OHH3C

CH3CH2C

H H

H–

OH + OH–

Hpropan-1-ol

HO

HO

HCH3CH2C

CH3CH2C

O–

H

H

CH3CH2CO–

H

δ+

δ–¨

¨

+COOH NaOH

+COONa H2O


218

6

a7

b CH3CH2COOCH3(l) + H2O(l) H+(aq) CH3CH2COOH(aq) + CH3OH(aq) propanoic acid methanol

a8 Water. b Esterification. c 3 d

a9

b

c (9Z,12Z)-octadec-9,12-dienoic acid.

a10 9% b 86% c 69% d 17% e TherearenoC=Cbondssoyoudon’tneedtoindicatetheirpositions.

11

12 C18,3(9,12,15)

13 Advantages of biofuels:conserveourdiminishingsuppliesoffossilfuels•plantsabsorbCO• 2astheygrow,reducingthegreenhouseeffectbiodegradableifspilled.•

Disadvantages of biofuels:not‘carbon-neutral’asyet•deforestationtogainmorelandtogrowthecrops,•destroyinghabitatsandreducingrainforestcropsusedforfuelinsteadoffood,affectingprices•andcausinghardshipindevelopingcountries.

Needabalancedargumentwithawell-reasonedconclusion.

a3

b

Pentanoic acid, CH3CH2CH2CH2COOH, and other isomericcarboxylicacids.

Methoxybutan-2-one,CH3OCH2COCH2CH3, and otherisomerswithanetherplusaketoneoranaldehydegroup.

Cyclicisomersarealsopossible. Forexample:

a4

b Propanoic acid and propan-2-ol. c

a5 Butylmethanoate. b

O

OOCH3

O

O

O

Hmethyl butanoate propyl ethanoate butyl methanoate

O

OOCH3

O

O

O

Hmethyl butanoate propyl ethanoate butyl methanoate

O

O

OCH3

O

O

OH

O

OH

O

OH

1-methylpropylmethanoate

1-methylethylethanoate

methyl 2-methylpropanoate2-methylpropylmethanoate

1,1-dimethylethylmethanoate

O

O

OCH3

O

O

OH

O

OH

O

OH





O

O

OCH3

O

O

OH

O

OH

O

OH





O

O

OCH3

O

O

OH

O

OH

O

OH





OH

O

O

OCH3 CH3

CH3

CH2

CHC

O

OH

CH3 CH2 C

O

O

CH3 CH2 CCHHO+ +CH3

CH3

CH3

CH3

H2O

CH

O

OH

CH3 CH2 C

O

O

CH3 CH2 CCHHO+ +CH3

CH3

CH3

CH3

H2O

CH

O

OH

H C CH2CH2CH2CH3

CH2CH2CH2CH3

HO+

O

O

H C H2O+

O CCOOH O

CH3 CH3COOH+

COOCH3(l) + NaOH(aq) COONa(aq) + CH3OH(aq)

sodium benzoate methanolCOOCH3(l) + NaOH(aq) COONa(aq) + CH3OH(aq)

sodium benzoate methanol

O

O

CH2OH

CHOH HC

H2C

H2C

CH2OH

+ 3CH3(CH2)16COOH + 3H2O

O

C (CH2)16CH3

O

O

C (CH2)16CH3

O

C (CH2)16CH3

COOH

COOH

COOH

O

OH

H C CH2CH2CH2CH3

CH2CH2CH2CH3

HO+

O

O

H C H2O+


219

a5

b

6 OnlycompoundAhascis–transisomers.The isomersare:

a7

b

c

8

a9

b

Chapter 5 a1 Butylamine.

b Amine C: C2H5NHCH3.

a2 Propylamine, CH3CH2CH2NH2 b 4-aminophenol,

a3 CH3CH2CH2CH2NH2 + HNO3 CH3CH2CH2CH2NH3

+ NO3–

b

c

a4

b

c

Chapter 6 a1 HOOCCH2NH3

+Cl–(aq) b NH2CH2COO–Na+(aq)

a2 HOOCCH2NH2(aq) + HCl(aq) HOOCCH2NH3

+Cl–(aq) b NH2CH2COOH(aq) + NaOH(aq)

NH2CH2COO–Na+(aq) + H2O(l)

3 TheconcentrationofthelysinezwitterionisatamaximumataboutpH9.5.Thesidegroup(R)on theα-aminoacidisbasicinitsnature.

a4

b Water. c Condensationreaction.

H2N OH

H2N –ClH3N+OH + HCl OH

H2N H2NOH + NaOH O–Na+ + H2O

NOH

HN O

NH2 HNO2HO HCl+

++

+ +

HO OH

N2+Cl–

N2+Cl–

2H2OHO

HO NN OH HCl

O

H

H

N

H

O

O H orH

O

C

HHN C C

C C

C

H

H

H

H H

H H

C

HH

C

O

H

H

N

H

O

OH

C

C

H

N C C

H C

O

HH

H

H

H

H

H

C

CH

HH

C

HO

O

OO OCH3

H2N

N

H

ester

amide bonds that may bebroken by hydrolysis

CHO

trans (E)CHO

cis (Z )

CH3CH2CHBrCH3

*NH2

*

OH

HO

HO

*N

a

b

cH

COOH HOOC

C

CH3

H2N

H

mirror plane

C

CH3

NH2

H

COOH

CH2

C

CH(CH3)2

H2NH

HOOC

CH2

C

CH(CH3)2

NH2

Hmirror plane

H2N C*

*

H

CHCH2CH3

CH3

COOH

H2N –ClH3N+OH + HCl OH

H2N H2NOH + NaOH O–Na+ + H2O

NOH

HN O

NH2 HNO2HO HCl+

++

+ +

HO OH

N2+Cl–

N2+Cl–

2H2OHO

HO NN OH HCl

NOH

HN O

NH2 HNO2HO HCl+

++

+ +

HO OH

N2+Cl–

N2+Cl–

2H2OHO

HO NN OH HCl

NOH

HN O

NH2 HNO2HO HCl+

++

+ +

HO OH

N2+Cl–

N2+Cl–

2H2OHO

HO NN OH HCl

NOH

HN O

NH2 HNO2HO HCl+

++

+ +

HO OH

N2+Cl–

N2+Cl–

2H2OHO

HO NN OH HCl

HO

O

OO OCH3

H2N

N+ 2H2O

H

HO

O

O O

H2N H2NOH

+ CH3OH +OHmethanol

2-aminobutanedioic acid(aspartic acid)

2-amino-3-phenylpropanoic acid (phenylalanine)

HO

O

OO OCH3

H2N

N+ 2H2O

H

HO

O

O O

H2N H2NOH

+ CH3OH +OHmethanol

2-aminobutanedioic acid(aspartic acid)

2-amino-3-phenylpropanoic acid (phenylalanine)


220

a4

b

c

Chapter 7 a1

b Repeatunit:

Monomer:

a2

b

a3

b

C

H

nCH2 = CHCN

H

C

C

N

n

H

C

H H H

H

C

C

H

n

H

poly(propene)

CH

H

propene

H

HH

C

HC

Cn

n

nO

O

H

C

O H

O

C

H HH

C

H

O

+

+ (2n – 1) H2O

C

O

O

C

O CH2 CH2 CH2 CH2

H

O

H H

CC O H

H

H

Cn

n

nO

O

H

C

O H

O

C

H HH

C

H

O

+

+ (2n – 1) H2O

C

O

O

C

O CH2 CH2 CH2 CH2

H

O

H H

CC O H

H

H

Cn

n

nO

O

H

C

O H

O

C

H HH

C

H

O

+

+ (2n – 1) H2O

C

O

O

C

O CH2 CH2 CH2 CH2

H

O

H H

CC O H

H

H

Cn

n

nO

O

H

C

O H

O

C

H HH

C

H

O

+

+ (2n – 1) H2O

C

O

O

C

O CH2 CH2 CH2 CH2

H

O

H H

CC O H

H

H

or

O OH COOH

OH OH

O

O nor

O

O n

Nn n

n

H

H

N C C

O

O H

OH

O

H

H

+

+ (2n – 1) H2ON

H

N

H

C C

O

O

amide link

n

N

H

N

H

C C

O

O

amide link

Nn n

n

H

H

N C C

O

O H

OH

O

H

H

+

+ (2n – 1) H2ON

H

N

H

C C

O

O

amide link

n

N

H

N

H

C C

O

O

amide link

Nn n

n

H

H

N C C

O

O H

OH

O

H

H

+

+ (2n – 1) H2ON

H

N

H

C C

O

O

amide link

n

N

H

N

H

C C

O

O

amide link

Nn n

n

H

H

N C C

O

O H

OH

O

H

H

+

+ (2n – 1) H2ON

H

N

H

C C

O

O

amide link

n

N

H

N

H

C C

O

O

amide link

Nn n

n

H

H

N C C

O

O H

OH

O

H

H

+

+ (2n – 1) H2ON

H

N

H

C C

O

O

amide link

n

N

H

N

H

C C

O

O

amide link

d Onlyonetypeofmonomerisusedtomakepoly(ethene),whereastherearetwodifferentmonomersusedtomakeKevlar.

Poly(ethene)ismadeinanadditionpolymerisationreactioninwhichthepolymeristheonlyproduct.However,Kevlarismadeinacondensationpolymerisationreactioninwhichwaterisproducedaswellasthepolymer.

Kevlarandpoly(ethene)arebothpolymersconsistingoflong-chainmoleculesinvolvingmanyrepetitionsofarelativelysimplerepeatunit.

5 Somepointstolookfor: Advantages:biodegradable;madefromrenewable

resources;grease/oilresistant;doesnotcontaminatefood;transparent;machinable.

Disadvantages:PLAcontaminatesotherplasticscollectedforrecycling;slightlymoreexpensivethanplasticssuchaspoly(chloroethene)–althoughtraditionalplasticswillbecomemoreexpensiveascrudeoilrunsout;candevelopa‘haze’overtime;shrinksifheat-sealedtoonearitsmeltingtemperature;fieldsusedforcropsusedasrawmaterialforPLAcouldbeusedforgrowingfood.


221

d

a5 Lessdosagerequired;reducesriskofside-effects astheunwantedenantiomermightpresenta health hazard.

b Reducesthechancesoflitigationagainstthedrugcompanyasaresultofside-effectscausedbytheunwantedenantiomer.

6 Thalidomidewasprescribedtopregnantwomenasasedativeduringtheearly1960s.Foratimeitwasthepreferredsedativeduringpregnancyasthealternatives,suchasvalium,wereaddictive.Unfortunately,oneofthestereoisomersofthalidomideprovedtohavedisastrousside-effects,causingbabiestobebornwithcongenitaldeformities(teratogenicity).Notsurprisingly,thalidomidewasquicklywithdrawnfromuseandlawsuitswerefiledagainstthemanufacturerstocompensatethoseaffectedandtohelpfinancetheircare.(Note,however,thatthalidomideracemisesinthebody,soevengivingthedrugasapureenantiomerwouldnothavepreventedtheproblems.)

7 Pointstoinclude:Racemicmixtureproducedintraditionalsynthetic•routes.Thisresultsintheneedtoseparatethemixtureofenantiomers.Thiscanuselargevolumesoforganicsolventswhichhavetobedisposedof,alongwiththeunwantedenantiomer.Theprocesswillalsousemorechemicals,whichrequire naturalresources.Enzymesarestereo-specific.•Wholeorganismscanbeused(avoidingtheneedto•isolateenzymes).Fewerstepsinprocess,resultinginmoreefficiency.•

Chapter 9 a1 Calciumcarbonate.

b Ethanol.

2 Any twoofthefollowing:Thin-layerchromatographyisfasterthanpaper•chromatography.Thethinlayermaybemadefromdifferentsolids.•Soawidevarietyofmixturescanbeseparatedbycarefulchoiceofthemobileandstationaryphases.

Chapter 81

2 CompoundC,asithasthesamestructureasthepartsmarkedwithablueboxintheanswertoSAQ1.However,compoundAhasthesamenumberofelectronsinthesameorbitalsasaspirinandthereforehasasimilarshapetoo.

3 CompoundB.

a4

b

c

Oaspirin

O

OH

O

2-hydroxybenzoic acid(salicylic acid)

O

OH

OH

salicin

OO

HO

HO

OH

OH

OH

H2C CH2

HCl(aq)

heat with alcoholic ammonia under pressure

CH3CH2Cl

CH3CH2NH2

heat with H2SO4(aq) andK2Cr2O7(aq)

heat with ethanoland acid catalyst

CHO

CO2H

CO2CH2CH3

CH3CH2CH2CH2Brreflux withNaOH(aq)

reflux with H2SO4(aq) and excess K2Cr2O7(aq)

CH3CH2CH2CH2OH

CH3CH2CH2COOH

CH3CH2COCH3

warm with NaBH4 in water

distil from mixturewith conc. H2SO4 and KBr(s)

CH3CH2CHOHCH3

CH3CH2CHBrCH3

heat with alcoholic NH3

CH3CH2CHNH2CH3


222

b Fromthesplittingpatternsgiven,thefollowingcanbededuced.

Toproduceatriplet,thethreeprotonsat•chemicalshift1.2mustbeadjacenttoa –CH2–group.Toproduceaquartet,thetwoprotonsat•chemicalshift4.1mustbeadjacenttoa –CH3group.Toproduceasinglet,thethreeprotonsat•chemicalshift2.0musthavenoprotonsontheadjacentcarbonatom.

c Thetypesofprotonsareasfollows:

Chemical shift, δ/ppm Type of proton

1.2 CH3–R

2.0C

O

CH3

4.1 –O–CH2–R

ThepresenceoftheC=Ogroupissupportedbytheinfrareddata,as1750cm–1istheabsorptionfrequencyforthisbond.

d Thestructureofthecompoundis

4 Methylpropanoatehasthefollowingstructure:

Thechemicalshiftsandsplittingpatternsare asfollows:

Type of proton

Chemical shift, δ/ppm

Relative number of protons

Splitting pattern

CH3–O 3.3–4.3 3 singlet

CH3–R 0.7–1.6 3 triplet

C

O

RCH2

2.0–2.9 2 quartet

5 Thepeakat5.4ppmwoulddisappear.

6 Thecarbonatominthebenzeneringthatisbondeddirectlytotheethylgroup.

7 Asinglepeak,somewherebetween110and165ppm.

a8 3peaks,onefortheCatomin–CH3, one for –CH2– and one for –CH2OH.

b 2peaks,onefortheCatomsinthetwo–CH3 groupsandonefor–CHOH.

Thin-layerchromatographycanbeusedforquickly•selectingthebestconditionsforlarger-scaleseparations.Thin-layerchromatographyworkswithvery •smallsamples.

3 Compound1:Rf=1.512.5=0.12

Compound2:Rf=9.112.5=0.73

Compound1hasagreateraffinityforthethinlayerthandoescompound2.Asthethinlayerissilicagel,compound1ismorepolarthancompound2.

a4 Bymeasuringthedifferenceintimebetweentheinjectionofthesampleandthecentreofthepeakfor a component.

b Therelativeareasunderthepeaksrepresenttherelativeamountsofthecomponentsinthemixture.

5 Therelativeheightofitspeakcomparedtothesumoftheheightsofallthepeaks.

a6 A methanol, CH3OH B ethanol, C2H5OH C butan-1-ol,CH3CH2CH2CH2OH D 2-methylbutan-1-ol

b Usingpeakheights:

Peaks A B C D

% of total 6.7 66.7 13.3 13.3

a7 9minutes. b About180–190. c Bycomparingitsmassspectrumtoadatabaseofknownspectra.

Chapter 10 1 EthanolcontainsHatomssoitproduces(3)peaksofitsownontheprotonNMRspectrum,makingitmorecomplicated to interpret.

a2 3peaks,correspondingtothe3differenttypesof1H atom in propanal: CH3CH2CHO

b The CH3peakwillbesplitintoatriplet(astheadjacentCatomisbondedtotwo1Hatoms, son=2andn+1=3).

3 a Chemical shift, δ/ppm

Relative number of protons

Splitting pattern

1.2 3 triplet

2.0 3 singlet

4.1 2 quartet

CH2 CH2 CH2OH

CH3

C

O

OCH3

CH2 CH3

C

O

OCH2

CH3

CH3 C


223

reagent.Oneapproachistoensurealargeexcessofmethanol.RelativetotheconcentrationofHCl,themethanolconcentrationcouldthenbeassumedtobeconstant.Thiswouldallowtheconcentrationoftheacidtobemonitored.ThedataobtainedwouldenabletheorderwithrespecttoHCltobededuced.

7 Separateexperimentsneedtobeconducted.Ineachexperiment,theconcentrationofjustonereactantisallowedtochange,withotherreactantspresentinexcess.Theeffectofchangingtheconcentrationof H+(aq)canbeinvestigatedbytheadditionofastrongacid(suchassulfuricacid).Similarly,theeffectofchangingtheconcentrationofCl–(aq) can be investigatedbytheadditionofsodiumchloride.

a8 Thereactionrate=k [N2O5],sotheorderofthereactionis1.

b k=1.05×10–5s–1

a9 reactionrate=k [H+] [CH3COCH3] [I2(aq)]0 (ineffect,reactionrate=k [H+] [CH3COCH3]as

[I2(aq)]0=1) b Substitutingdatafromexperiment1intherateequation:

10.9×10–6 mol dm–3s–1 =k × 1.25 mol dm–3 ×0.5×10–3 mol dm–3

k= 10.9×10–6 mol dm–3s–1

1.25 mol dm–3 ×0.5×10–3 mol–1 dm3s–1

=1.74×10–2 mol–1 dm3s–1

10

Chapter 12 1 Experiment1,Kc=0.276;experiment4,Kc=0.272.

a2 CH3COOH + CH3CH2OH CH3COOCH2CH3 + H2O

b 0.1728mol c 0.5000–0.1728=0.3272mol

d Kc=[CH3COOCH2CH3][H2O][CH3COOH][CH3CH2OH]

e Kc=(0.3272)×(0.3272)(0.1728)×(0.1728)

=3.59

3 IfwecallthenumberofmolesofproductsatequilibriumM1andM2,andthenumberofmolesofreactantsatequilibriumM3andM4,andthetotalvolumeisV,thentheconcentrationsofthesubstancesat

equilibriumareM1V ,

M2V ,

M3V and

M4V .

a9 • MassspectrometrytofindMr from the molecularionpeakandapossiblestructure forthemolecule.IRtoidentifythefunctionalgroupsC=Oand•O–Hintheacid.OruseNMR.

b • MassspectrometrytofindMrandapossiblestructure.IRtoidentifytheC=Ogroupintheester. •OruseNMR.

c • MassspectrometrytofindMrandapossiblestructure.NMRtoidentifythechemicalgroupscontaining•protonsandfindtheirarrangementinthemolecule.IRtoidentifythefunctionalgroups–OHand•C=O.

Chapter 11 a1 Infrared.

b 2.00×10–13s

2 Experimentalevidenceshowsthattherateofreactionisdirectlyproportionaltotheconcentrationofcyclopropane.Iftheconcentrationofcyclopropaneishalved,therateofreactionishalved.

3 Table11.2hasmoreaccuratedataasthetangentsdrawnwerelonger,givinglargertriangles.

4 Thereactionisfirstorderwithrespecttocyclopropaneandfirstorderoverall.

5 Bytitratingsmallsamplesofthereactionmixturewithstandardisedbase,forexample1.00moldm–3 aqueoussodiumhydroxide,atsettimesduringthechemicalreactionyoucouldfindtheconcentrationofhydrochloricacidasthereactionprogressed.YoucouldalsomonitorthisconcentrationusingeitherapHmeteroraconductivitymeter.Bothdevicesrespondtochangesinhydrogenionconcentration,whichisitselfan indication of the concentration of hydrochloric acid over time.

6 Rememberthatthetemperatureofthereactionmixturemustbeconstantthroughout. a Wecannottell.Thisdatacanbeexplainedifthereactionisfirstorderwithrespecttobothreactants,inwhichcasebothreactantsaffecttherate.Alternativelythedatacanbeexplainedifthereactionissecondorderwithrespecttoonereactantandzeroorderwithrespecttotheother,inwhichcaseonlyonereactantaffectstherate.

b Severalapproachesarepossible.Toprovideafairtest,theexperimentshouldbedesignedtostudytheeffectofchangingtheconcentrationofonlyone

H H + Cl–

H

CH+

O

H

H

H Cl + O

H

HH

C


224

b Theequilibriumpositionshiftstotheright.Thisisbecausetheforwardreactioncausesadecreaseinthenumberofgasmoleculessohigherpressurefavourstheforwardreaction,therebydecreasingthepressureaccordingtoLeChatelier’sprinciple.Kcisunchanged.(Thehigherpressurewillalsocauseanincreasedrateofreaction.)

c Theequilibriumpositionisunchanged.Kcis unchanged.(Thecatalystwill,however,cause anincreasedrateofreaction.)

Chapter 13 a1 H2SO4(l) + H2O(l) H3O+(aq) + HSO4

–(aq) B–Lacid B–Lbase B–Lacid B–Lbase

b CH3COOH(aq) + H2O(l) CH3COO–(aq) + H3O+(aq) B–Lacid B–Lbase B–Lbase B–Lacid

c CH3NH2(aq) + H2O(l) CH3NH3+

(aq) + OH–(aq) B–Lbase B–Lacid B–Lacid B–Lbase

d NH3(g)+HCl(g) NH4+(s)+Cl–(s)

B–Lbase B–Lacid B–Lacid B–Lbase

2 Purewaterisapoorconductorofelectricity,which showsthatitcontainsveryfewionsthatcancarrya directcurrent.

3 Kc=[H+][OH–]

[H2O]

Kcisverysmall,sotheconcentrationsoftheproducts mustbeverymuchsmallerthantheconcentrationof wateritself.Thisindicatesthatonlyatinyproportion ofpurewaterexistsatanyonetimeasprotonsand hydroxideions,adeductionbackedbytheevidence thatwaterisapoorconductorofelectricity.

a4 H2SO4+CuO CuSO4 + H2O and 2H+ + O2– H2O

b H2SO4 + Na2CO3 Na2SO4 + H2O + CO2 and 2H+ + CO3

2– H2O + CO2 c H2SO4+Mg MgSO4 + H2

and 2H++Mg Mg2+ + H2 d H2SO4 + 2KOH K2SO4 + 2H2O

and H+ + OH– H2O

a5 Energy=200g×4.2Jg–1 °C–1 ×13.5°C=11340J b 100cm3of2.00moldm–3acidcontain0.200moles

of H+ions.100cm3of2.00moldm–3alkalicontain 0.200molesofOH–ions.Thiswillproduce 0.200molesofwater.

c Enthalpychangeofneutralisation=–113400.200

=−56700Jmol–1or–56.7kJmol–1

a6 3.5 b 2.0 c 7.4

TheexpressionforKcis

M1V ×

M2V

M3V ×

M4V

The‘Vs’allcancel,soKc=[M1] × [M2][M3] × [M4]

andsothetotalvolumeisirrelevant.

4 The10molesofhydrogeniodidebegintodissociate,forminghydrogenandiodine:

2HI(g) H2(g)+I2(g)

Foreverymoleculeofiodineformed,twomoleculesofhydrogeniodidehavetosplitup.Toform0.68molesofiodinemolecules,2×0.68molesofhydrogeniodidemustdissociate.Thismeansatotallossof1.36molesofhydrogeniodide,from10downto8.64moles.

a5 Kc=[N2O4(g)][NO2(g)]2

unitsaredm3 mol–1

b Kc=[NO2(g)]2

[NO(g)]2[O2(g)]unitsaredm3 mol–1

c Kc=[NH3(g)]2

[N2(g)][H2(g)]3unitsaredm6 mol−2

6

Kc=[CO] [H2O][H2][CO2]

= 9.47 × 9.470.53×80.53

=2.10 (Nounitsasinthishomogeneousreaction,i.e.all

reactantsandproductsareinthesamestate,thetotalmolesofreactants=totalmolesofproducts,sotheunitscancelout.)

7 Theforwardreactionisexothermic,soanincreaseintemperaturecausestheequilibriumtoshifttotheleft.Therearelessproductsandmorereactantsatequilibrium.Kcdecreases.

a8 Theequilibriumpositionshiftstotheleft.Thisisbecausetheforwardreactionisexothermicandhighertemperaturefavourstheendothermicbackwardreaction,therebyloweringthetemperatureaccordingtoLeChatelier’sprinciple.Kc decreases.(Thehighertemperaturewill,however,causeanincreasedrateofreaction.)

H2(g) CO2(g) CO(g) H2O(g)

Initial concentration/mol dm–3

10.0 90.0 0 0

Equilibrium concentration/ mol dm–3

10.0–9.47 =0.53

90.0–9.47 =80.53

9.47 9.47


225

8 Acid Base

HNO3 NO3–

H2SO3 HSO3–

[Fe(H2O)6]3+ [Fe(H2O)5(OH)]2+

HF F–

HNO2 NO2–

HCOOH HCOO–

C6H5COOH C6H5COO–

CH3COOH CH3COO–

CH3CH2COOH CH3CH2COO–

[Al(H2O)6]3+ [Al(H2O)5(OH)]2+

CO2 + H2O HCO3–

SiO2 + H2O HSiO3–

HCO3– CO3

2–

HSiO3– SiO3

2–

H2O OH–

a9 Ka=[H+]2

[benzoic acid]

Therefore [H+]=6.3×10–5 ×0.020 pH=–log10[H+]=2.95

b pH=3.5.Aqueoussolutionsofaluminiumsaltsaresurprisinglyacidic.AnaccidentaltippingofaluminiumsaltsintoareservoirinCornwallcreatedtapwateracidicenoughtodissolvecopperfrompipesandtoworrylargenumbersofpeopleaboutthepossibilityofbeingpoisoned.

c pH=2.4

a10 1.26 ×10–3 mol dm–3 b 2.00×10–4 mol dm–3 c 6.31 ×10–12 mol dm–3

a11 [H+]is5.01×10–5 mol dm–3. Kais1.26×10–7 mol dm–3.

b 0.0500moldm–3propanoicacidhasapHof3.1.[H+]is7.94×10–4 mol dm–3. Kais1.26×10–5.

c 0.0100moldm–3methanoicacidhasapHof2.9.[H+]is1.26×10–3 mol dm–3. Kais1.58×10–4.

12 Strongacid–strongbase:theslopeofthegraphissteepovertherangepH3.5topH10.Anyindicatorwithacolourchangerangewithintheselimitsissuitable:bromocresolgreen,methylred,bromothymolblueorphenolphthalein.Theothersinthetablearenotsuitable.

a7 pH=0.0 b pH=0.3 c Theaqueoussolutioncontains3.00gofhydrogen

chloride, HCl, per dm3.TofindthepHweneedthehydrogenionconcentrationinmoldm–3.

TherelativemolecularmassofHCl=(1.0+35.5) =36.5. Thustheconcentrationofhydrogenchloride

=3.0036.5 mol dm–3=0.0822moldm–3.

Becausethehydrogenchloridedissociatescompletelytoformhydrogenionsandchlorideions,theconcentrationofhydrogenionsis0.0822moldm–3.

ThepHofthisacid=–log10[H+]=–log10[0.0822] =1.09.

d Potassiumhydroxidedissociatescompletely insolution:

water KOH(s) K+(aq) + OH–(aq) 0.001mol 0.001mol 0.001mol

Theconcentrationofhydroxideionsisthesameastheconcentrationofthepotassiumhydroxide.

Kw=[H+][OH–]=1.00×10–14 mol2 dm–6

so

[H+] =1.00×10–14 mol2 dm–6

[OH–]

=1.00×10–14 mol2 dm–6

0.001moldm–3

=1.00×10–11 mol dm–3

ThepHofthisacid=–log10[H+]=–log10[10–11] =11.0

e Sodiumhydroxideionisescompletelyinaqueoussolution:

water NaOH(s) Na+(aq) + OH–(aq)

TherelativemolecularmassofNaOH =(23.0+16.0+1.0)=40.0 Anaqueoussolutioncontaining0.200gofNaOH

per dm3contains0.20040.0 mol NaOH, i.e.

5 ×10–3 mol dm–3.Theconcentrationofhydroxideionsistherefore5.00×10–3 mol dm–3.

Kw=[H+][OH–]=1.00×10–14 mol2 dm–6 Therefore [H+] ×5.00×10–3=1.00×10–14

[H+]=1 ×10–14

5 ×10–3 =2.00×10–12 mol dm–3

ThereforepH=–log10[H+]=–log10(2.00×10–12) =11.7


226

Chapter 14 a1 Theenergychangeassociatedwithachemical

reaction. b Achemicalchangeinwhichenergyisreleasedtothesurroundings;ΔH isnegative.

c Achemicalchangeinwhichenergyistakeninfromthesurroundings;ΔH ispositive.

2 Thetotalenthalpychangeforachemicalreactionisindependentoftheroutebywhichthereactiontakesplace,providedinitialandfinalconditionsarethesame.

a3 ½O2(g) O(g) b Cs(g) Cs+(g)+e–

c K(s)+12 Cl2(g) KCl(s)

d I(g)+e– I–(g) e Ba(s) Ba(g)

a4

b –787kJmol–1

a5

13 Strongacid–weakbase:theslopeissteepovertherange7.0–2.0.Theindicatorswecouldusearemethylyellow,methylorange,bromophenolblue,bromocresolgreenormethylred.Wemightgetawaywithusingbromothymolblue,butalltheothersinthetableareunsuitable.

14 Weakacid–strongbase:methylorangestartschangingcolourwhenthepHis3.2andstopsatpH4.4,soyouwouldseenocolourchangewiththisindicator.

a15 Strongacid–weakbase:methylorangeorbromophenolblue.

b Strongacid–strongbase:bromocresolgreen,methylred,bromothymolblueorphenolphthalein.

c Theequilibriumconstantforaspirinissimilartothatofmethanoicacid,soaspirinisaweakacid.Potassiumhydroxideisastrongbase,sothesensitiveregionfortheindicatorwouldbeintherangepH7–11.Phenolphthaleinwouldbethebestchoice of indicator.

a16 TheconjugateacidisNH4+andtheconjugatebase

isNH3. b Whendilutehydrochloricacidisadded,theadditionalhydrogenionsareacceptedbytheammoniamolecules.Youcanthinkoftheadditional H+ionsreactingwiththeOH–ionsintheequilibriummixtureshownbelow(formingH2O):

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

Thepositionofequilibriumthenshiftstotherightto replace OH–ionsremovedfromthemixture.

Whendilutesodiumhydroxideisadded,thepositionofequilibriumshiftstotheleft,toremovethe additional OH–(aq)ions.

c BecauseNH3isaweakbasetherewillnotbeenoughNH4

+ionsintheequilibriummixtureofanammoniasolution(astheequilibriumlieswellovertotheleft)to remove additional OH–ions(alkali)added.

a17 Theequationfortheequilibriumreactionis

HCOOH(aq) H+(aq) + HCOO–(aq)

fromwhichwecanwritetheequilibriumconstantexpression:

Ka=[H+][HCOO–]

[HCOOH] Rearrangingthisequation:

[H+]=Ka × [HCOOH][HCOO–] mol dm–3

Substitutingthedatagivenproduces:

[H+]=1.6×10–4 × 0.05000.100 mol dm–3

=8.00×10–5 mol dm–3 sopH=–log10(8.00×10–5) =–(–4.096)=4.10

b Usingthemethodinparta,pH=4.8

Na(s) + Cl2(g)

Na(g) + Cl2(g)

∆H f (NaCl)

∆H at (Na)

∆H i1 (Na)

∆H at (Cl)

∆H latt (NaCl)

∆HEA (Cl)Na+(g) + Cl2(g)

Na+(g) + Cl(g)

Na+(g) + Cl–(g)

NaCl(s)

∆H latt

MgO(s)

∆H i1

∆H ea2

∆H i2

Mg(s) + O2(g)

Mg+(g) + O2(g)

Mg(g) + O2(g)

Mg2+(g) + O2(g)

Mg2+(g) + O(g)

∆H at

∆H at

Mg2+(g) + O–(g)

Mg2+(g) + O2–(g)

∆H ea1

∆H f


227

Chapter 15 1 Theentropyofasubstanceisthedegreeofdisorderofthatsubstance.

2 Startingwithmostdisordered,thereforetheonewiththehighestentropy:oxygen,carbonmonoxide,water,copper.

a3 386–(393+192)=–199JK–1 mol–1. The entropy changeisnegative.Thereisadecreaseinthedisorderofthesystem.

b 396–(131+152)=+113JK–1 mol–1. The entropy changeispositive.Thereisanincreaseinthedisorderofthesystem.

a4 Theentropychangeisnegativeandthereisadecreaseinthedisorderofthesystembecausethereisadecreaseinthenumberofgasmoleculesinvolved.

b Theentropychangeispositiveandthereisanincreaseinthedisorderofthesystembecausethereisanincreaseinthenumberofgasmoleculesinvolved.

a5 72°C+273=345K b 246K–273=–27°C c 0K–273=–273°C d 273 K and 373 K

a6 ΔG=ΔH – TΔS b ΔG=–190000–(360×–121)=–146440Jmol–1 or–146kJmol–1

c ΔGhasalargenegativevalue.Thereactionisspontaneousandgoestocompletion.

d Thereactioncausesareductioninthenumberofgasmolecules,sothereisadecreaseinthedisorderofthesystem.

a7 ΔG=ΔH – TΔS ΔG=+84000–(360×83.0)=+54120Jmol–1 or+54.1kJmol–1

b ΔGhasapositivesignanditsmagnitudeiswellover+20kJmol–1.Thisreactionwillnotoccurspontaneouslyat360K.

c IfthereactiontakesplaceatasufficientlyhightemperaturesothatΔH – TΔSbecomesnegativeitwilloccurspontaneously.

d ΔG=ΔH – TΔS –20000=+84000–(T ×83.0)

T=104000

83 =1253K

e Thereactioncausesanincreaseinthenumberofgasmolecules,sothereisanincreaseinthedisorderofthesystem.

b

a6 CaO b K2O c SrI2

7 LiF,Li2O,MgO. LiFiscomposedofsinglychargedionssohastheleast

attractionbetweenions,Li2Ohasonedoublychargedion,andMgOhastwodoublychargedionssohasthemostattractionbetweenions.

8 Sodiumbromidehasanegativeenthalpychangeofsolution.Itissolubleinwater.Sodiumchloridehasasmallpositiveenthalpychangeofsolution.Itissolubleinwater.Silverchlorideandsilverbromidehavelargerpositiveenthalpychangesofsolution.Theyareinsolubleinwater.

a9 Na+(g)+(aq) Na+(aq) b Cl–(g)+(aq) Cl–(aq) c NaCl(s)+(aq) NaCl(aq)

a10 AsensiblepredictionfortheΔH hyd of Na+wouldbearound–420kJmol–1(theactualvalueis–406kJmol–1).Thisislargerinmagnitude(ormore exothermic)thanthevalueforK+asNa+hasasmallerionicradiusthanK+andthereforeahigherchargedensity.ItissmallerthanthevalueforLi+asNa+hasalargerionicradiusthanLi+ and therefore alowerchargedensity.

b Mg2+ismorehighlychargedandhasasmallerionicradius(duetoonelessoccupiedshellofelectrons)thanK+sohasamuchhigherchargedensity,thereforetheΔH hydofMg2+ismuchmoreexothermicthantheΔH hyd of K+.

11 ΔH latt+ΔH sol=ΔH hyd –2592kJ+–55kJ=ΔH hyd(Mg2+) + 2 ×ΔH hyd(Cl–) –2647kJ=–1920kJ+2×ΔH hyd(Cl–) –2647kJ+1920kJ=2×ΔH hyd(Cl–) ThereforeΔH hyd(Cl–)is–363.5kJmol–1.

∆H latt

NaO2(s)

2 × ∆H i1

∆H ea2

2Na(s) + O2(g)

2Na(g) + O2(g)

2Na+(g) + O2(g)

2Na+(g) + O(g)

∆H at

2 × ∆H at

2Na+(g) + O–(g)

2Na+(g) + O2–(g)

∆H ea1

∆H f


228

9

salt bridge

VO2+, 1 mol dm–3

V3+, 1 mol dm–3

H+ , 1 mol dm–3

H+, 1 mol dm–3

voltmeter

H2(g),1 atmosphere

298 K

platinum

platinum

a10 S + 2e– S2–

b E =–0.51V

salt bridge

I– , 1 mol dm–3 H+, 1 mol dm–3

voltmeter

+ –

H2(g),1 atmosphere

298 K

platinum

platinum

iodine (s)

0.54 V11

a 12 Cr2+

b Ag

a13

chromiumrod salt bridge

Cr3+, 1 mol dm–3 Cl–, 1 mol dm–3

– +

Cl2(g),1 atmosphere

298 K

voltmeter2.10 V

platinum

b 1.36–(–0.74)=2.10V c Chlorine half-cell.

Chapter 16 a1 Cu2+

b Fe c Cl– d Cu2+

e Fe

a2 +6 b +2 c +3 d +6

a3 +4 b +6 c 0 d +7 e +3 f +2

4 First reaction: Fe:0 +2;reducingagent

H: +1 0;oxidisingagent Cl: –1 and –1 Second reaction: Fe:0 +3;reducingagent H: +1 0;oxidisingagent Cl: –1 and –1

5 For the Fe2+/Fe half-cell: a Fe2+ + 2e– Fe b –0.44V c Fe2+:1.00moldm–3

For the Cr2+/Cr half-cell: a Cr2+ + 2e– Cr b –0.91V c Cr2+:1.00moldm–3

For the Ag+/Ag half-cell: a Ag+ + e– Ag b +0.80V c Ag+:1.00moldm–3

Inallthreecellsthetemperaturemustbe298KandinthestandardhydrogenelectrodestheH+(aq) concentrationmustbe1.00moldm–3, the H2pressuremustbe1atmosphere(101kPa)andelectricalcontactmustbemadebyplatinum.

6 +1.52 V

7 298K,allgasesatapressureof1atmosphere(101kPa),allrelevantconcentrationsat1.00moldm–3.

8 Platinumdoesnottakepartinreactions.Itisan inert electrode.


229

d Increasingtheconcentrationsofreactantsforcesequilibriumtoshifttotherightinordertoreducetheseconcentrations.Therefore,Egoesupand the Cr2O7

2–/H+solutionbecomesastrongeroxidisingagent.

19 E -basedpredictionsrefertostandardconditions,butlabconditionsarenotusuallystandard.(However,ifthe E valuesforthetwohalf-equationsdifferby morethan0.30volts,E -basedpredictionsare usuallycorrect.)

E valuesmaypredictthatareactionwilloccur,eventhoughinrealitythereactionmayhavesuchaslowratethatitisnotobserved.

20 Addacatalyst;increasetemperature;increaseconcentrationofdissolvedreactants;increasepressureofgaseousreactants;increasesurfaceareaofsolidreactants.

a i21 0to+2 ii +4 to +2

b Leadisoxidisedattheleadplate.Itsoxidationnumberisincreasing;itislosingelectrons.Leadisreducedattheleadoxideplate.Itsoxidationnumberisdecreasing;itisgainingelectrons.

a 22 NiO2 + 2H2O + 2e– Ni(OH)2 + 2OH–hasamorepositiveE soitproceedsforwards.

Cd(OH)2 + 2e–→Cd+2OH–hasalesspositive E soitproceedsbackwards.

b 1.30V c Cd + NiO2 + 2H2O Cd(OH)2 + Ni(OH)2 d Cd(OH)2 + Ni(OH)2 Cd + NiO2 + 2H2O

23 Fuelcellvehicles.

a24 HydrogenloseselectronsformingH+ions(oxidation).

b OxygengainselectronsandcombineswithH+ionsformingwater(reduction).

c Bydiffusingthroughamembrane. d Viatheexternalcircuit. e 2H2 + O2 2H2O

a25 MethaneisCH4. Onemoleofmethanehasamassof16.0g,outof

which4.0gishydrogen.

4.016.0 ×100%=25.0%

MethanolisCH3OH. Onemoleofmethanolhasamassof32.0g,outof

which4.0gishydrogen.

4.032.0 ×100%=12.5%

b CH3OH + 1½O2 CO2 + 2H2O (or 2CH3OH + 3O2 2CO2 + 4H2O)

a14

b (–0.13)–(–1.18)=1.05V c Lead half-cell.

15 Fe3+ + I– Fe2+ + I2

a16 Yes. MnO4

– + 5Cl– + 8H+ Mn2+ + 52 Cl2 + 4H2O

MnO4– + 8H+ + 5e– Mn2+ + 4H2O,withits

morepositiveE value,willproceedinaforwarddirectionwhileCl2 + 2e– 2Cl–proceedsinabackwarddirection.

b No. MnO4

– + 8H+ + 5e– Mn2+ + 4H2O,withitslesspositiveE value,cannotproceedinaforwarddirectionwhileF2 + 2e– 2F–proceedsinabackwarddirection.

c Yes. V2+ + H+ 1

2 H2 + V3+

2H+ + 2e– H2,withitsmorepositiveE value,willproceedinaforwarddirectionwhile V3+ + e– V2+proceedsinabackwarddirection.

d No. 2H+ + 2e– H2,withitslesspositiveE value,

cannotproceedinaforwarddirectionwhile Fe3+ + e– Fe2+proceedsinabackwarddirection.

a17 Cellvoltage=1.52–1.36=+0.16V,thereforeyes. b Cellvoltage=1.52–2.87=–1.35V,thereforeno. c Cellvoltage=0.00–(–0.26)=+0.26V,thereforeyes.

d Cellvoltage=0.00–0.77=–0.77V,thereforeno.

a i18 E=morethan1.33V ii E=lessthan1.33V iii E=lessthan1.33V

ib Strongeroxidisingagent. ii Weakeroxidisingagent. iii Weakeroxidisingagent.

c HighconcentrationCr2O72–,highconcentrationH+,

lowconcentrationCr3+.

manganeserod

lead rod

salt bridge

Mn2+, 1 mol dm–3 Pb2+, 1 mol dm–3

voltmeter

– +1.05 V

298 K


230

a2 +6 b +3

a3 Ni2+(aq) + 2OH–(aq) Ni(OH)2(s) b Ti3+(aq) + 3OH–(aq) Ti(OH)3(s)

a4 Acentralpositiveionwithoneormoreligandspeciesdativelybondedtoit.

b Anatom,neutralmoleculeornegativeionwhichisabletouseoneormorelone-pairstobonddativelytoapositiveion.

c Aligandwithtwolone-pairswhichcaneachformadativebondtoapositiveion.

a5 Thenumberofdativebondsformedbetween theligandsandthecentralpositiveionina complexion.

ib 6 ii 4 iii 6 iv 4

ic 3+ ii 2+ iii 3+ iv 3+

a6

b

c

c Ahydrogen–oxygenfuelcellproduceswaterasitsonlyproduct.Amethanol–oxygenfuelcellalsoproducescarbondioxide,whichmayberesponsibleforglobalwarming.

a26 40× 5 ×107=2×109 J b 2 ×109 J ×0.4=8×108 J

c 8 ×108

1 ×106 =800km

a27 400×143000=5.72×107 J b 5.72 ×107 J ×0.6=3.43×107 J

c 3.43 ×107

1 ×106 =34.3km

d TherangeoftheFCVbetweenrefuellingisfartoosmallwhencomparedwiththepetrol-enginedcar inSAQ26.

28 Ifamaterialadsorbshydrogenthenthehydrogenisstoredonthesurfaceofthematerial.

Ifamaterialabsorbshydrogenthenthehydrogenisstoredinthebulkofthematerial.

29 Theproblemsinclude:highproductioncostsofhydrogen–oxygenfuel•cellstoxicchemicalsusedintheproductionof•hydrogen–oxygenfuelcellslimitedlifetimeofhydrogen–oxygenfuelcells•thecostofreplacinganddisposingofspent•hydrogen–oxygenfuelcellsthedevelopmentofnewengineeringskills•associatedwiththemaintenanceofhydrogen–oxygenfuelcellsthedevelopmentofnewengineeringskills•associatedwiththesupplyofhydrogenproblemsassociatedwithstoringandtransporting•hydrogen,includingdifficultyinstoringalargeamountofagas,energycostsofliquefaction,safetyissues,adsorptionorabsorptionmethodsarestillindevelopment,adsorbersandabsorbershavealimitedworkinglifetimepublicreluctancetoaccepthydrogenasanewfuel•politicalreluctancetocommittoexpensivelong-•termplanningtheneedforalarge-scalesourceofelectricalenergy•tosupplythehydrogen.

Chapter 17 a1 [Ar]3d54s1

b [Ar]3d3

c [Ar]3d104s1

d [Ar]3d9

e [Ar]3d54s2

f [Ar]3d5

Fe

H3N

H3N

NH3

NH3

NH3

NH3

3+a

b c

FeCl

Cl

Cl

Cl

2–

Au

Cl

Cl Cl

Cl

–

Fe

H3N

H3N

NH3

NH3

NH3

NH3

3+a

b c

FeCl

Cl

Cl

Cl

2–

Au

Cl

Cl Cl

Cl

–

Fe

H3N

H3N

NH3

NH3

NH3

NH3

3+a

b c

FeCl

Cl

Cl

Cl

2–

Au

Cl

Cl Cl

Cl

–


231

a11 [Zn(CN)4]2–sinceitsKstabislarger. b CN–ligandswillsubstituteforNH3ligands

in [Zn(NH3)4]2+,forming[Zn(CN)4]2–.Thishappensbecause[Zn(CN)4]2–ismorestablethan[Zn(NH3)4]2+.

equation:

4CN– + [Zn(NH3)4]2+ 4NH3 + [Zn(CN)4]2–

c Nothinghappens(exceptatveryhighconcentrationsofammonia).

Nothinghappensbecause[Zn(CN)4]2–ismorestablethan[Zn(NH3)4]2+.

a12 AsaligandtoanFe2+ioninahaemoglobinmolecule.

b Haemoglobincarriesoxygenfromthelungsbecauseoxygenconcentrationinthelungsishigh,soligandsubstitutionoccursattheFe2+ions.Haemoglobincarriescarbondioxidefromrespiringtissuesbecausecarbondioxideconcentrationintherespiringtissuesishigh,soligandsubstitutionoccursattheFe2+ions.

c CarbonmonoxidegasformsamorestablecomplexwithFe2+ionsthanoxygendoes.ThecarbonmonoxidemoleculesremainboundtotheFe2+ionseveninsituationswheretheoxygenconcentrationsarehigh.Thehaemoglobinthereforefailstotransportoxygentotherespiringtissuesthatneedit.

13 6Fe2+ + Cr2O72– + 14H+ 6Fe3+ + 2Cr3+ + 7H2O

a14 23.50cm3=0.02350dm3

n=V × c son=0.02350×0.0400=0.000940moles

b Theequationforthereactioninthetitrationis:

5Fe2+ + MnO4– + 8H+ 5Fe3+ + Mn2+ + 4H2O

ThereforethenumberofmolesofFe2+ =5×0.000940=0.00470moles

c molesofFe=molesofFe2+=0.00470moles massofFe=n×Ar=0.00470×55.8=0.262g

d percentagemassofiron=0.2620.420 ×100%=62.4%

a15 47.81000 ×0.200=0.00956mol

b 0.00956mol c 0.00956×63.5=0.60706g

d0.60706g

1g ×100%=60.7%

Thefinalanswerhasbeenroundedto3significantfiguresasthatistheaccuracyofthedata.

a7

b

a8 The transisomerisisomer3. b Opticalisomerism. c

d Co2+

a9 Theheatofthebenchmatmakesthewaterevaporate. [CoCl4]2–ionsareformedandthepinkcircleturnsblueliketherestofthepaper.

b Ligandsubstitution. c [Co(H2O)6]2+ + 4Cl– [CoCl4]2– + 6H2O

or [Co(H2O)6]2+ + 4Cl– [CoCl4]2– + 6H2O

a10 [FeSCN2+][Fe3+][SCN–]

[Hg(CN)42–][Hg2+][CN–]4

b [Hg(CN)4]2–asitsKstabis(much)larger.

+

H3N

H3N

Co

Cl

Cl

NH3

NH3

a

b

+

H3N

H3N

Co

Cl

Cl

NH3

NH3

+

H3N

H3N

Co

Cl

Cl

NH3

NH3

+

H3N

Co

Cl

Cl

NH3

NH3

NH3

+

H3N

H3N

Co

Cl

Cl

NH3

NH3

a

b

+

H3N

H3N

Co

Cl

Cl

NH3

NH3

+

H3N

H3N

Co

Cl

Cl

NH3

NH3

+

H3N

Co

Cl

Cl

NH3

NH3

NH3

Co

mirror

redrawingisomer 2,with a90˚ rotationanticlockwise

leaveisomer 1as it is

H2N

H2N

NH2

NH2

H2C

CH2

CH2

H2C

Co

Cl

ClCl

Cl NH2

H2N

NH2

H2N CH2

H2C

CH2

H2C

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Answers to SAQsadvancedsciences.cambridge.org/media/COAS_C2_assess_bkSAQan… · 216 Chapter 1 1 a b Alternatively: CH 3C(CH 3)(OH)CH 2CH 3 c 2 C 4H 9Br 3 a 3-chloro-4,5-dimethylheptane