Scholarship Chemistry (93102) 2011

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931020

SSUPERVISOR’S USE ONLY

Scholarship 2011Chemistry

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Ch�ck �h�� �h� ���i�n�l �����n� ������ (���) �n ���� ���issi�n slip is �h� s��� �s �h� n����� �� �h� ��p �f �his p�g�.

Y�� sh��l� �nsw�� ALL �h� q��s�i�ns in �his ���kl��.

A periodic table is provided on page 2.

If ��� n��� ���� ���� f�� �n� �nsw��, �s� �h� �x��� sp�c� p���i��� �� �h� ��ck �f �his ���kl��.

Ch�ck �h�� �his ���kl�� h�s p�g�s � – �6 in �h� c����c� ����� �n� �h�� n�n� �f �h�s� p�g�s is �l�nk.

Y�� ��� ���is�� �� sp�n� �pp��xi����l� 35 �in���s �n ��ch q��s�i�n.

YOU MUST HAND THIS BOOKLET TO THE SUPERVISOR AT THE END OF THE EXAMINATION.

9 3 1 0 2

© New Zealand Qualifications Authority, 2011. All rights reserved.No part of this publication may be reproduced by any means without the prior permission of the New Zealand Qualifications Authority.

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Y�� h��� �h��� h���s �� c��pl��� �his �x��in��i�n.

QUESTION ONE

(a) (i) Discuss how the charges on subatomic particles contribute to the size of atoms and their ions.

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(ii) Explain the trends in the atom and ion sizes (in pm) in Figure 1 below.

Relate your answer to your discussion in part (i).

Li+ Be2+

60 31 152 111

Na+ Mg2+ Al3+

95 65 50 186 160 143

K+ Ca2+ Ga3+

133 99 62 231 197 122

Rb+ Sr2+ In3+

148 113 81 244 215 162

N3– O2– F–

171 140 136 70 66 64

S2– Cl–

184 181 104 99

Se2– Br–

198 185 117 114

Te2– I–

221 216 137 133

Figure 1: Atomic and ionic radii (pm)

Note: • Thedarkcirclesrepresenttheions.

• Theboldednumbersaretheionicradii.

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(b) TheBoudouardreactionisthenamegiventotheoxidation-reductionreactioninvolvingamixtureofcarbon(graphite),carbondioxideandcarbonmonoxideatequilibrium,atagiventemperature.Thereactionisanimportantprocessinsideablastfurnaceintheproductionofmetals from metal oxides.

Thegraphbelowshowshowthereactionmixturecompositionofthegasespresentchangeswithchangingtemperatureatatmosphericpressure(101kPa).

00 500

CO2 CO

1 000

0.10.20.30.40.50.60.70.80.9

1

Temperature / °C

Frac

tion

of C

O an

d CO

2

Boudouard Equilibrium

(i) Identify,withjustification,theproductoftheexothermicprocessintheBoudouardreaction.

Use your answer to discuss the reaction products when the hot gases from a blast furnace reach the cooler air at the top of the chimney.

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(ii) Alarge,heat-proofsyringewithafreelymoveableairtightpistoninitiallycontained50mLofcarbondioxideat101kPaand25°C,plus1gofgranularcarbon.Theapparatus was heated and maintained at a certain constant temperature until equilibrium wasreached.Thepressureofthesystemremainedconstantat101kPathroughout.Theapparatuswasthencooledrapidlyto25°C.(Atthistemperatureanyfurtherchangeincomposition was negligible.) A total of 60 mL of gas was then present.

Determine the temperature at which the reaction was carried out.

Note:• Thesameconditionsapplytothisexperimentandtothatrepresentedbythegraph

onthepreviouspage.

• Thevolumeofasubstanceinitsgasphaseataparticulartemperatureandpressureis directly proportional to the amount in moles of the substance present.

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QUESTION TWO

(a) Account for the differences in the properties of the compounds in the table below.

Name Structural formula

Melting Point / °C

Solubility in water at 25°C

CyclohexaneC6H12

7 Insoluble

CyclohexanolC6H11OH

OH25 Sparingly soluble

cis-cyclohexane-1,2-diolC6H10(OH)2

OH

OH98 Very soluble

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(b) A year 11 student carried out an experiment to compare the energy released when octane and ethanol were burned as fuels. Octane was found to release considerably more energy than ethanol.Thestudentconcludedthatthiswasbecauseoctanewasalargermolecule,andsothereweremorechemicalbondstobreak,andhencemoreenergyreleasedinthereaction.

Discuss the misconceptions in the student’s explanation and use appropriate data from the tablebelowtoprovideyourownanswerfortheobservation.

M(octane) = 114.0 g mol–1 M(ethanol) = 46.1 g mol–1

Bond enthalpy data / kJ mol–1

C–H 414 O–H 463C–O 358 C≡O 1076C=O 804 H–H 436C–C 346 O=O 498

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QUESTION THREE

(a) A mixture contains oxalic acid, H2C2O4, sodium oxalate, Na2C2O4, and a water soluble impurity that does not react with solutions of sodium hydroxide or potassium permanganate.

Todeterminethecompositionofthemixture,2.496gofthemixturewasdissolvedinwatertogive100.00mLofsolution.

In one test, 5.00 mL of the solution was titrated with 0.01803 mol L–1acidifiedpotassiumpermanganatesolutionandneeded23.35mLtoreachtheequivalencepoint.

In another test, 10.00 mL of the solution was titrated with 0.1040 mol L–1 sodium hydroxide solutionandneeded17.30mLtoreachtheequivalencepoint.

Determine the mass fractions of oxalic acid, oxalate ion and the impurity.

M(H2C2O4) = 90.04 g mol–1 M(Na2C2O4) = 134.0 g mol–1

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(b) Manganeseisanelementthatexhibitsanumberofdifferentoxidationstates.Half-cellreactionsandpotentialsforthedifferentoxidationstatesvarydependingontheconditions.Thetablegivesthestandardreductionpotentialsformanganesespeciesinaqueoussolutionranging from Mn(II) to Mn(VII) at pH = 0 and at pH = 14.

Conditions Reduction half-equation E° / VpH = 0 Mn2+(aq)+2e–→Mn(s) –1.19

MnO4–(aq)+H+(aq)+e–→HMnO4

–(aq) +0.90MnO2(s)+4H

+(aq)+2e–→Mn2+(aq)+2H2O(ℓ) +1.23MnO4

–(aq)+8H+(aq)+5e–→Mn2+(aq)+4H2O(ℓ) +1.51Mn3+(aq)+e–→Mn2+(aq) +1.54MnO4

–(aq)+4H+(aq)+3e–→MnO2(s)+2H2O(ℓ) +1.69HMnO4

–(aq)+3H+(aq)+2e–→MnO2(s)+2H2O(ℓ) +2.10MnO2(s)+4H

+(aq)+e–→Mn3+(aq)+2H2O(ℓ) +0.95pH = 14 2MnO2(s)+H2O(ℓ)+2e–→Mn2O3(s)+2OH

–(aq) +0.15Mn2O3(s)+3H2O(ℓ)+2e–→2Mn(OH)2(s)+2OH

–(aq) –0.23Mn(OH)2(s)+2e

–→Mn(s)+2OH–(aq) –1.19

(i) Explain why Mn(II) is not oxidised by O2 in solutions at pH = 0, but is oxidised by O2 solutions in which [OH–] is 1 mol L–1.

O2(g)+2H2O(ℓ)+4e–→4OH–(aq) E°=+0.40V(pH=14)

O2(g)+4H+(aq)+4e–→2H2O(ℓ) E°=+1.23V(pH=0)

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(ii) KMnO4 is a common oxidant, and in acidic conditions may be reduced to MnO2 or Mn2+.

Discuss why the intermediate species HMnO4

– does not accumulate during the reduction of MnO4

– to MnO2.

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(iii) Discuss the pH dependence of the stability of Mn(III) in aqueous solution.

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QUESTION FOUR

(a) (i) ThenormalpHinbloodplasmais7.40.ThepHofbodyfluidsisregulatedbythepresenceofCO3

2–,HCO3–andCO2dissolvedinthesefluids.

At25°C,Ka(H2CO3)=4.2× 10–7 and Ka(HCO3–) = 4.7 × 10–11.

IdentifythecomponentsfromthelistabovethatwouldformthebestbufferatthispHand calculate their ratio.

Determinewhetherthisbufferismoreeffectiveagainstaddedacidorbase.

(ii) Lactic acid (HLac) is often said to be produced by the body during rapid exercise.

Show by calculation that lactic acid is mainly present as the lactate ion (Lac–) at pH = 7.40.

pKa(lactic acid) = 3.86

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(b) Predispositionfortheconditionknownasgoutoccurswhentheconcentrationofuricacid(HUr) and urate ions (Ur–)inthebloodbecomestoohigh.Uricacidisaweakacid.pKa = 5.4 at37°C.

At37°C,thesolubilityofsodiumurateis8.00× 10–3 mol L–1 and the solubility of uric acid is 5.00 × 10–4 mol L–1.

In blood serum, [Na+] = 0.130 mol L–1(ieatpH=7.4andtemperatureof37°C).

(i) Calculatethemaximumurateconcentrationatwhichsodiumuratewillnotprecipitate,and show by calculation that uric acid will not precipitate at this concentration either.

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(ii) Oneformofkidneystoneismadeofcrystalsofuricacid.Thesecanformwhentheconcentration of uric acid and urate is high, and the pH of urine drops to around 5 to 6.

CalculatethepHatwhichthesestonescanform.

Assumethatthetotalconcentrationofuricacidandurateis2.00× 10–3 mol L–1.

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QUESTION FIVE

(a) CompoundAhasthemolecularformulaC4H4O4.CompoundB is a stereoisomer of CompoundA.ThemeltingpointsofCompoundAandCompoundBare135°Cand287°Crespectively,duetotheabilityofCompoundA to form an intramolecular hydrogen bond.CompoundA reacts, on heating, with dilute H2SO4toformCompoundC,C4H6O5. CompoundCisabletoexistasenantiomers.CompoundA will react with potassium permanganatetoformCompoundD,C4H6O6.WhenCompoundD is oxidised by periodate ions (IO4

–),CompoundEisproduced.CompoundEhasthemolecularformulaC2H2O3 andgivesapositivetestwithTollens’reagent(ammoniacalsilvernitrate).EachmoleculeofCompoundDproduces2moleculesofCompoundE.

(i) IdentifythestructuresofCompoundsA–E. Justify your choice of the stereoisomers A and BandwriteabalancedequationforthereactionofCompoundEwithTollens’reagent.

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(ii) DrawtwootherisomersofCompoundA: one that has different functional groups from CompoundA,andanotherthathasthesamefunctionalgroupsasCompoundA.

ExplainwhyneitheroftheseisomersmeettherequirementstobeCompoundA as described in part (a).

Question Five continues on the following page.

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(b) DiscussthepotentialforCompoundC to form polymers.

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QUESTION NUMBER

Extra space if required.Write the question number(s) if applicable.

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QUESTION NUMBER

Extra space if required.Write the question number(s) if applicable.

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QUESTION NUMBER

Extra space if required.Write the question number(s) if applicable.

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QUESTION NUMBER

Extra space if required.Write the question number(s) if applicable.

93

10

2

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Q��s�i�n M��k

ONE

(8)

TWO

(8)

THREE

(8)

FOUR

(8)

FIVE

(8)

TOTAL(40)

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