2023 HCI Prelim P4 Mark Scheme

HWACHONGINSTITUTION2023C2H2CHEMISTRYPRELIMINARYEXAMINATIONMARKINGSCHEME Paper 4 1 (a) (i) Sampletable: massofcappedcontainer+FA1/g 8.50massofcappedcontainer+remainingFA1/g 5.49massof FA1used/g 3.01 initialtemperatureof FA2,Ti / o C 29.1lowesttemperatureofmixture,Tf / o C 25.4changeintemperature/ o C ‒3.7 ● correct headers+units+massrange(3.00g±0.10g) [1]neednotshow‘changeintemperature’ ● correct precision[1]o massesto2d.p.o temperaturesto1d.p. (ii) ∆T=Tf ‒Ti =25.4–29.1=‒3.7 o C(1d.p.) [1]q=mc∆T=50.0×1.00×4.18×3.7=773.3J=773J(3s.f.) [1] (iii) ηcitricacid used= =0.01568mol=0.0157mol(3s.f.) 3.01192 ηNaHCO3 used= ×1.00=0.0500mol(3s.f.)(> ) [1] 501000 3×η𝑐𝑖𝑡𝑟𝑖𝑐 𝑎𝑐𝑖𝑑 = 0.0471Citricacidisthelimitingreagent(shown) (iv) ∆H1 =+ =+49307Jmol –1 =+49.3kJmol –1 [1] 7730.01568 (v) q=mc∆TΔ𝐻 = 𝑞𝑛𝑙𝑖𝑚𝑖𝑡𝑖𝑛𝑔 𝑟𝑒𝑎𝑔𝑒𝑛𝑡 As FA 2isinexcess, doublingitsvolumedoesnot changetheamountofheatabsorbed,q,since∆H×narebothunchanged,butitdoublesthemassofsolutionthat thesameamount ofheatabsorbedmustcooldown.Hencethetemperaturechangewillbehalfthatmeasuredintheexperiment.[0.5]sameamountofheatabsorbed+mass/volumeofsolutiondoubled[0.5]temperaturechangehalved (b) (i) Sampletable: VFA2 /cm 3 VsolutionA /cm 3 10.00 40.015.00 35.0 2023HCIC2H2ChemistryPreliminaryExaminationMarkingScheme/Paper4 1

20.00 30.025.00 25.030.00 20.040.00 10.0[1]● 6setsofVFA2 +VsolutionA ● Vtotal =50cm 3 ● chosenvaluesofVFA2 shouldbewell-spaced Procedure1.FilltheburettewithFA2.2. Placetwopolystyrenecupstogetherandputthemina250cm 3 glassbeaker.Transfer40cm 3 of FA2intothepolystyrenecupusingtheburette.3.Measuretheinitialtemperatureof FA2usingthethermometer.RecordtheinitialtemperatureasTFA2 .4.Washanddrythethermometer.5.Useameasuringcylindertomeasure10cm 3 of solution A.Measuretheinitialtemperatureof solution Ausingthethermometer.RecordtheinitialtemperatureasTsolutionA .6.Add solution Ato FA2inthepolystyrenecup.Stirthemixturegentlyusingthethermometerandrecordtheminimumtemperature,Tmin ,reached.7.Washanddrythepolystyrenecup.8.Repeatsteps2to6usingthevolumesof FA2and solution Aasgiveninthetableabove.9. The weighted initial temperature should be calculated for each experimentaccordingtotheformulagivenandrecordedasTweighted,initial :10. Calculate the temperature change of each experiment by taking (ΔT =Tmin −Tweighted,initial ),andrecorditasΔTasshowninthetableabove. [0.5]correct apparatus(burettefor FA 2&measuringcylinderfor solution A+polystyrenecupsascalorimeter.AlsoacceptuseofmeasuringcylindersforbothFA2andsolutionA)[0.5]wash+drythermometer&polystyrenecup[0.5]measures&recordsTFA2 &TsolutionA [0.5]measures&recordsTmin &calculatesTweightedinitial [0.5]repeatsexperiments[0.5]calculates&recordsΔTforeachexperiment (ii) 2023HCIC2H2ChemistryPreliminaryExaminationMarkingScheme/Paper4 2

● correctaxeslabels+shapeofgraph● markingofΔTmin &Vmin ● graphmustpassthroughorigin+x-axisat50cm 3 mark (iii) ηNaHCO3 (FA2)used= ×1.00mol 𝑉𝑚𝑖𝑛 1000 ηcitricacid (solutionA)presentinsolution= mol [0.5] 𝑉𝑚𝑖𝑛 1000 × 1.00 × 13 [citricacid]= moldm –3 [0.5] 𝑉𝑚𝑖𝑛 1000 × 1.00 × 1350 − 𝑉𝑚𝑖𝑛 1000 (iv) q=mc∆T=(50×1.00)×4.18×∆Tmin [0.5] ∆H2 = ×3[0.5] 𝑞η𝑁𝑎𝐻𝐶𝑂3 = 50 × 1.00( ) × 4.18 × ∆𝑇𝑚𝑖𝑛𝑉𝑚𝑖𝑛 1000 × 1.00 OR ∆H2 = ×3 𝑞η𝑐𝑖𝑡𝑟𝑖𝑐 𝑎𝑐𝑖𝑑 = 50 × 1.00( ) × 4.18 × ∆𝑇𝑚𝑖𝑛𝑉𝑚𝑖𝑛 1000 × 1.00 × 13 = 50 × 1.00( ) × 4.18 × ∆𝑇𝑚𝑖𝑛𝑉𝑚𝑖𝑛 1000 × 1.00 PenaliseonceforM13orM14ifstudentdoesnotmultiplybystoichiometriccoefficientin1(b)(iii)and1(b)(iv)) (c) (molar)enthalpychangeofsolutionofcitricacid[1] (d) Citricacidisaweakacid,whilesodiumhydrogencarbonateisaweakbase.AsthereisnorapidpHat theequivalencepointsforaweakacid–weakbasetitration, thereisnosuitableindicatorthatcanbeusedtosignaltheendpoint.Hencethetitrimetricmethodcannotbeusedtodeterminetheconcentrationofcitricacidinthiscase. [0.5]identificationofweakacid&weakbase[0.5]norapidpHchange+nosuitableindicator 2023HCIC2H2ChemistryPreliminaryExaminationMarkingScheme/Paper4 3

2 (a) t td /min finalburettereading/cm 3 initialburettereading/cm 3 VFA5 /cm 3 Vt /cm 3 min s ● correct headersandunits[1]o transfertime(minands)o decimaltimetd (min)o finalandinitialburettereadings,VFA5 andVt (cm 3 )Accept‘seconds’,‘minutes’.Reject‘sec’.Acceptifdataisrecordedin2tables,aslongasitisclearwhichtitreislinkedtowhichtransfertime.Penalisethismarkifrawtransfertimeisnotrecordedorfinalandinitialburettereadingsareinverted. ● correct precision[1]o transfertimesinminands(tonearests)o td to0.1mino burettereadingsandvolumesto0.05cm 3 Fortransfertimes,e.g.for2min30s,reject2’30’’or2.30minIgnorefirstpointiftransfertimeinminandsarenotrecorded. [1]:● calculatestd correctly 1–2s=0.0min 21–26s=0.4min 45–50s=0.8min3–8s=0.1min 27–32s=0.5min 51–56s=0.9min9–14s=0.2min 33–38s=0.6min 57–59s=1.0min15–20s=0.3min 39–44s=0.7minandcalculatesVFA5 andVt correctly [1]:● recordsfivesetsofdatawithin16minutes(acceptmax16.4min)● firstdatatakenwithin0to2.4minNopenaltyiftimeintervalsarenotwell-spaced (b) (i) [1]:● plotsVt (y-axis)vstd (x-axis)withcorrectaxeslabelsandunits and● sensiblelinearscalechosensothatplottedpointsoccupyatleast½ofgraphgridinbothxandydirections. 2023HCIC2H2ChemistryPreliminaryExaminationMarkingScheme/Paper4 4

[1]:● plotsallpointscorrectlywithin½smallsquare● includesVmax =20.00cm 3 pointingraph [1]:● drawsbest-fitsmoothcurveDonotallowthismarkifanomalouspointsareincludedinthecurve.Donotallowmarkifshapeofgraphiswrong.Awardmarkbasedonwhat thestudent hasplotted(donot re-plot forwronglyplottedpoints). [1]:● Atleast4outof5pointsarenomorethan1smallsquareawayfromthebest-fitcurveinbothxandydirections.Donotallowthismarkifmorethan1pointisplottedwrongly.Donotallowmarkifshapeofgraphiswrong.Awardmarkbasedonwhat thestudent hasplotted(donot re-plot forwronglyplottedpoints). (ii) Totalη(I2 )formed=η(S2 O8 2− )used=0.010× =3.00×10 –4 mol 301000 η(I2 )in10cm 3 =3.0×10 –4 × =3.00×10 –5 mol [1] 10100 η(S2 O3 2− )needed=2×η(I2 )=2×3.00×10 –5 =6.00×10 –5 molVolumeof FA3= ×1000=20.0cm 3 [1] 6.00 × 10 −5 0.0030 Ignoremissingworkingatthispoint (iii) ● quotesatleasttwot½ valuesorlabelongraph(acceptnon-consecutivet½ )● givesconclusionbasedonwhethert½ is(approximately)constantornote.g.firstorderwithrespectto[S2 O8 2− ]ast½ isapproximatelyconstant [0.5]ate.g.½(5.3+5.7)≈5.5min[0.5]averaget½ canbeshownin(b)(iv)● rate=k’[S2 O8 2− ] (iv) valueofk’= = =0.1260=0.126(3s.f.) (units=min −1 ) 𝑙𝑛 2𝑡½ 𝑙𝑛 25.5 [0.5]usingcorrecteqn[0.5]usingaveraget½ Ignoremissingworkingands.f.errorsatthispoint (v) [I − ]=( ×0.400)÷( )=0.280moldm −3 (3s.f.) [1] 701000 1001000 valueofk= = =0.450(3s.f.)ecffrom(iv) [1] (units=mol −1 dm 3 min −1 ) 𝑘'[𝐼 − ] 0.1260.280 Ignoremissingworking,incorrect/missingunitsors.f.errorsatthispoint. [1]:Showsworkinginallcalculationsin1(a)(ii)-(iv); 2(b)(ii)and2(b)(iv)-(v).Allcalculationsmustberelevantalthoughtheymaynotbecompleteorcorrect.Anycalculationnot attemptedlosesthismark. [1]:Showsappropriatesignificantfigures(3or4s.f.)ordecimalplacesin allfinalanswersin1(a)(ii),(iv)and2(b)(iv)-(v). 2023HCIC2H2ChemistryPreliminaryExaminationMarkingScheme/Paper4 5

Finalanswersrefertothosewrittenonthelineprovided.Ifthisisabsent,considerfinalanswerinworking.Anycalculationnot attemptedlosesthismark. [1]:Showsappropriateunitsin allanswersin 1(a)(ii)(°CorK;JorkJ), 1(a)(iv)(Jmol ─1 orkJmol ─1 )and2(b)(v)[I − ]only(moldm −3 )Anycalculationnot attemptedlosesthismark. (c) t½ =ln2/k’=ln2/(k[I − ])ORrateishalvedORrate [0.5]∝ 1𝑡1/2 When[I − ]ishalved,t½ doubles. [0.5]nomarkfor‘t½ increases’ (d) [1]:Ironisabletoexhibitvariable(accept‘multiple’,‘afew’,‘many’)oxidationstates.[1]:Step1:S2 O8 2– +2Fe 2+ →2SO4 2– +2Fe 3+ [1]:Step2:2I – +2Fe 3+ →I2 +2Fe 2+ 3 (a) Table3.1tests observations obs1 Placeabout1cmdepthofdilutesulfuricacidintoatest-tube.Addaspatulaof FA6tothistesttube,followedbyanother2cmdepthof FA3. Filterthemixtureintoacleantest-tubeandretainthefiltrate. Tothefiltrate,addaqueoussodiumthiosulfatedropwise,withshaking,untilthesolutionfirstbecomescolourless.Retainthissolutionforuseintest2. ● Blackresidue.● Yellowish-brown/yellow-orangefiltrate. 1[0.5]2[0.5] 2 Tothesolutionfromtest1,addaqueoussodiumhydrox