COAS C2 assess bkeocans - Cambridge University...

COAS Chemistry 2 1

Answers to End-of-chapter questions

Chapter 2 1 a

(or multiple substitution of –NO2 groups in the benzene ring) [1] b 1 mole of phenol → 1 mole of 4-nitrophenol relative molecular mass of phenol (C6H5OH) = 72.0 + 5.0 + 16.0 + 1.0 = 94.0 [1]

number of moles of phenol = 0.94

100 [1]

0.94

100 moles of 4-nitrophenol (C6H4OH(NO2)) has a mass of 0.94

100 × 139.0 g [1]

The yield is 27% = 0.94

100 × 139.0 g × 10027 = 39.9 g [1]

2 Discussion of the π bonding p orbitals overlap [1] above and below the ring [1] (to form) π bonds / orbitals [1] any of the first three marks are available from a labelled diagram, for example

π bonds / electrons are delocalised [1] Other valid points – any two of: • ring is planar • C–C bonds are equal length / have intermediate length/strength between C=C and C–C • σ bonds are between C–C and/or C–H • bond angles are 120° (max 2 out of 4 marks) [2] Quality of written communication two or more sentences with correct spelling, punctuation and grammar [1]

COAS Biology 2

3 a

b greater reactivity of phenol/the ring is activated because … lone-pair from oxygen atom is delocalised into the ring [1] so electron density (of the ring) is increased [1] so electrophiles are more attracted (to the ring)/dipole in electrophile more easily

induced [1] (NOT just more easily ‘attacked’ or ‘susceptible’)

Chapter 3 1 a at least one correct skeletal formula [1] correct cis (Z) and trans (E) isomers of but-2-enal [1] b i heat with Tollens’ reagent/ammoniacal silver nitrate [1] to give silver mirror/precipitate [1]

ii Aldehydes can be oxidised to a carboxylic acid (or words to that effect)/ aldehydes can reduce Ag+ to Ag [1]

c i But-2-ene-1-ol or CH3CH=CHCH2OH (either stereoisomer) [1] ii reduction/redox/addition (not hydrogenation) [1]

d C4H6O + 5O2 → 4CO2 + 3H2O [1]


COAS Chemistry 2 3

2 a

curly arrow from O of OH– to C [1] dipole on C=O and curly arrow breaking C=O [1] structure of the intermediate [1]

curly arrow from lone pair on O– (of the correct intermediate) to H of H2O (allow O to H+ ion here) [1]

curly arrow breaking the H–O bond in H2O [1] b 1 mark for the correct answer to each step below with ‘error carried forward’

throughout. The steps may come in any order.

one week’s supply = 21 × dose = 5.25 g or 0.0317 mol [1]

If 100% yield, mass of trichloroethanal = 5.1655.147

× mass of chloral hydrate = 4.68 g

(223 mg if done first) [1] or mass of trichloroethanal = 0.0317 × 146.75 = 4.68 g

60% yield, so need 60

100 times as much

60

100 × 4.68 = 7.8(0) g [1]

Common errors for 2 marks are: 9.82 g (mass ratio upside down) 8.75 g (mass ratio not done) 2.60 g (3× not done) 1.11 g (7× not done) 0.371 g (21× not done) 7798 g (mg to g not done), etc. CCl3CH(OH)2 + [O] → CCl3COOH + H2O [1]

Chapter 4 1 a

propanoate and ester group [1] 2-methylpropyl group [1]

b propanoic acid [1] 2-methylpropan-1-ol [1] heat [1] conc. H2SO4 [1] (allow ‘error carried forward’ from part a for the equation) CH3CH2COOH + (CH3)2CHCH2OH → CH3CH2COOCH2CH(CH3)2 + H2O reactants [1] products [1]


COAS Chemistry 2 4

2 a i

glycerol [1] sodium salt of carboxylic acid [1] balanced equation (only with correct formulae) [1]

ii moles of tristearin = 890

1000 = 1.12(4) [1]

molar mass of soap = (12 × 18) + (16 × 2) + 35 + 23 = 306 g mol–1 [1] mass of soap = 3 × 1.12(4) × 306 = 1032 g [1] accept answers in range 1028 to 1032 g If wrong in part i, they can score the moles of tristearin [1 mark] and then follow

their Mr of soap and their balancing number from their part i to give a method [1 mark] mass of soap = 1.12(4) × their balancing number × their Mr [max 2 out of 3]

b Non-polar (long) hydrocarbon chains on tristearin [1] Dissolve in hexane by van der Waals’ attraction (not hydrophobic) [1] Idea of instantaneous/fluctuating/temporary dipoles (or words to that effect) [1] c i Saturated formula would be C21H43COOH (or words to that effect; there must

be a quantitative approach showing that there are 2 hydrogens too few for saturated) [1]

ii Avoid using lubricants from fossil fuels/renewable/thermally stable at high temperatures/biodegradable (or similar sensible remark) [1]

3 a i NaOH/Na [1] ii C6H5OH + NaOH → C6H5ONa + H2O

C6H5OH + Na → C6H5ONa + 21 H2 or 2C6H5OH + 2Na → 2C6H5ONa + H2 [1]

b i allow a dipole on just one C=O bond [1] ii

iii lone/electron pair from oxygen is delocalised into the ring/interacts with π electrons [1]

increases π electron density/negative charge (around the ring) [1] attracts electrophiles more [1] c Mr salicylic acid = 138 [1]

moles (in 1:1 reaction) = 138

103500 6× = 2.536 × 107 [1]

mass of phenol needed = 2.536 × 107 × 94 = 2384 tonnes [1]

allowing for 45% yield = 2384 × 45

100 = 5298 tonnes

(allow 5297.5 to 5300) [1] (allow for errors carried forward throughout)


COAS Chemistry 2 5

Chapter 5 1 a nitrous acid/nitric(III) acid/HNO2 [1] b [1] c diazonium (ion/salt) [1] d To prevent decomposition/it reacting or diazonium ion is unstable

(or words to that effect) [1] e structure showing the amine coupled to the phenol or its salt, e.g.

–N=N– [1] rest of structure (joined by two nitrogens) [1] 2 C6H5NH2 (or structural formula) [1]

3 [1] 4 a ‘diamino’ refers to two amine/–NH2 groups [1] ‘1,4’ refers to the positions of the amine groups on the benzene ring, with the

numbering starting on a carbon atom bonded to an amine group [1] b i reduction [1]

ii tin/Sn and conc. hydrochloric acid/HCl [1] reflux/boil [1]

iii 12 [H] [1] + 4H2O [1]

c i Lone-pair (of electrons) on the nitrogen (N) atom [1] donated to H+ ion (from acid) [1] ii [2]

Chapter 6 1 a i RCH(NH2)COOH allow groups R, CH, NH2 and COOH in any order

ii any unambiguous structure, e.g.


COAS Chemistry 2 6

b i molecule/ion/‘it’ has both + and – charges [1] ii description or diagram to show proton/H+ transfer from COOH to NH2 [1]

not just ‘hydrogen’ transfer

c i heat/warm/reflux [1] named strong acid/base (not conc. HNO3 or conc. H2SO4)/an enzyme (which need not be named) [1]

ii hydrolysis [1] d i ethanolic ammonia [1]

ii any mention of chiral/optical isomers [1] leucine synthesised in the laboratory contains a mixture of (two optical) isomers [1] leucine from meat/natural source contains only one (optical) isomer [1] 2 a b structure with correct use of at least two 3D bonds allow error carried forward if lactic acid is labelled in a e.g.

not if all four bond angles at 90° [1] 3 a i carboxylic acid [1]

ii NH2CH(CH3)COONa or NH2CH(CH3)COO–Na+ [2] allow drawing of structure of the sodium salt

iii water [1] b [2]

c The amine end of one amino acid reacts with the carboxylic acid end of another [1] The two amino acids bond via a peptide link [1] in a condensation reaction/with a water molecule produced [1] [1] [1]


COAS Chemistry 2 7

Chapter 7 1 a second starting material: H2N–(CH2)6–NH2 [1] products:

peptide bond displayed [1] correct repeat [1] + HCl [1] allow error carried forward on the carbon skeleton of the diamine b i [1]

[1] ii any valid suggestion to explain or describe stronger intermolecular forces –

e.g. Nomex® is planar so packs together more easily/greater H-bonding/ van der Waals’/forces between molecules (ignore arguments based on Mr) [1]

2 a i

[2] one mark for reactants and one mark for product

ii Addition polymerisation involves monomers with C=C double bonds in which the polymer is the only product formed [1] whereas condensation polymerisation involves monomers with reactive groups at either end of the molecule, and produces small molecules as well as the polymer in the reaction. [1]

b [1] c i

[2] ii H reacts with NaOH / poly(propene) does not [1]

H is an ester / is polar ... [1] will be hydrolysed by NaOH [1] poly(propene) is non-polar [1] any 3 out of 4 marks ‘hydrolysed by NaOH’ gets the ‘reacts with NaOH’ mark as well


COAS Chemistry 2 8

3 Monomer for polymer L: [1] Monomers for polymer M: [1] [1]

Chapter 8 1 a i Sodium hydroxide solution/NaOH(aq) [1]

ii Nucleophilic [1] substitution [1]

iii C6H5CH2Cl + NaOH → C6H5CH2OH + NaCl [1] b i

[1] ii C6H5CH2OH + CH3COOH → C6H5CH2OOCCH3 + H2O [1] iii flavourings/perfumes/solvents [1] iv

curly arrow from CH3COO– to Cδ+ [1] curly arrow from C–Cl bond to Clδ– [1] δ+ and δ– correctly positioned [1] 2 a i Hot/reflux/boil [1] with named strong acid, e.g. dil. / e.g. 6 mol dm–3 / conc. hydrochloric acid [1] (do not accept conc. sulfuric acid or conc. nitric acid) or Hot / reflux/boil [1] with named strong alkali, e.g. sodium hydroxide solution [1]

ii (alpha) amino acids [1]


COAS Chemistry 2 9

iii If acid was chosen as reagent in a i

[2] allow dipeptides, allow salts/cations with positive charge on N atom or If alkali was chosen as reagent in a i

[2] iv Breakdown of a compound by water/acid/alkali [1]

Show peptide link (–NHCO–) being broken by water to give –NH2 + –COOH or by acid to give amino acid or cation or by alkali to give carboxylates [1] b i • There has to be at least one chiral centre

• a C atom with 4 different atoms/groups of atoms bonded to it/ asymmetric carbon atom • the compound will then exist as ‘mirror-image’ isomers/enantiomers • which rotate the plane of polarised light in opposite directions/are optically active

maximum of 3 marks awarded [3] ii There are two chiral centres [1]

so they can be arranged as L-L, L-D, D-L or D-D/rotate plane of polarised light left-left, left-right, right-left or right-right [1]

iii Use chiral catalyst/enzymes/enzymes in an organism such as bacteria/ use amino acids as starting materials that have the same required optical activity as the drug [1] iv • Adds time/cost in separating the mixture

• the separation requires extra reagents/uses more resources • extra reagents and waste stereoisomers must be disposed of • possible side-effects of unwanted stereoisomers

maximum of 2 marks awarded [2] 3 a RCH(NH2)COOH [1] b 1 mark for each correct charge on –+NH3 and –O– [2]


COAS Chemistry 2 10

c i Optical isomerism [1] ii e.g.

1 mark for each correct stereoisomer [2] d Compound D has one more C atom than phenylalanine [1] the amine (NH2) and carboxylic acid (COOH) groups are separated by an extra C atom in D but in phenylalanine they are on adjacent C atoms [1] compound D has an –OH/hydroxyl/alcohol group but phenylalanine does not [1] e i

1 mark for each chiral centre [2] ii A mixture, as there is an equal chance of each of the two chiral centres being in either of their optically active orientations depending on how reactants collide. [1]

f

1 mark for the N–H, 1 mark for the C=O [2]


COAS Chemistry 2 11

Chapter 9 1 a The diagram below could score 3 marks if appropriately annotated

Sample is injected into spectrometer Carrier gas carries it through the column Column is heated Samples are separated by their attraction for the column/partition Different components have different times of emergence Samples may be analysed by mass spectrometry or by using standards 1 mark for each point, to a total of 5 marks [5] Correct terminology, e.g. carrier gas, mobile phase, stationary phase, gives the 1 mark for quality of written communication [1] b i Rf value is the distance moved by a component divided by the distance moved by the solvent front. [1]

ii Retention time is the time between injection and the emergence of a component. [1] 2 a i Adsorption [1]

ii Any two of the following: Variety of adsorbents can be used Better/faster separation Components can be extracted Smaller samples can be used [2] b i Theobromine [1]

ii Caffeine peak area = 2

631× = 93

Theobromine peak area = 2

446× = 92 [1]

9392

92+

= 49.7 ± 1%

marked consequentially [1] iii Dyes are not usually volatile [1]

3 a i The gas that carries the solutes through the column [1] ii The solid (or liquid on a solid support) that is packed inside the column [1]


COAS Chemistry 2 12

b percentage A = 6030 = 50% [1]

percentage B = 6010 = 16.7% [1]

percentage C = 6020 = 33.3% [1]

Chapter 10 1 a

splitting: doublet [1] quartet [1] ignore any other peaks position: doublet peak is at δ = ~1.4 and quartet peak is at δ = ~4.3 [1] allow error carried forward from one incorrect splitting pattern areas: 1 and 3 on the correct peaks (or either way round as an error carried forward from above) [1] b 4 peaks in total would appear on the NMR spectrum using an inert solvent [1]

OH/labile protons now visible (or words to that effect) [1] 2 a Mass spectrum/spectrometry [1] molecular ion peak/m/e or mass of the peak furthest right (or words to that effect) [1]


COAS Chemistry 2 13

b δ value/chemical shift gives the ‘type’ of proton/chemical environment (or words to that effect) [1] example quoted from data sheet [1] number of peaks gives the number of different types of proton/ chemical environments [1] relative number/ratio of [1] peak areas / height of steps in integration trace gives the number of protons (of each type) [1] splitting gives number of neighbouring/adjacent protons [1] description of n + 1 rule/example of doublet, triplet or quadruplet showing 1, 2 and 3 protons on neighbouring (carbon) atom (or words to that effect) [1] D2O can be used to identify OH groups [1] any 7 for 7 marks

Quality of written communication mark for correct use and organisation of at least two of the following technical terms:

proton, environment, singlet (doublet, etc.), ppm, equivalent, chemical shift, splitting, labile, integration [1] 3 a The peak is due to the CH3CO– group [1] It is not split, so next to a C with no protons/has no neighbouring proton/ δ value is in the range 2.0–2.9 [1] b Adjacent to a C with three protons/to a –CH3 group [1] c/d

relative peak areas 2 : 3 : 3 [1] triplet [1] at 0.7–1.6 [1] mark any additional incorrect peaks first, taking 1 mark off for each error

Chapter 11 1 a i time for concentration (of a reactant) to fall to half the original value [1]

ii t1/2 = 460 ± 10 s [1] constant half-life [1] evidence on graph to support constant half-life (at least two half-lives shown) [1]

iii no change [1]


COAS Chemistry 2 14

b k = 2/1

693.0t =

460693.0 = 1.51 × 10−3 [1]

s−1 [1]

for consequential marking: answer should be: iiatoanswer

693.0

c rate = k[C6H5N2Cl(aq)] [1] d i after 800 s, [C6H5N2Cl(aq)] = 1.75 × 10−4 mol dm−3 [1] allow any value from 1.7 × 10−4 to 1.8 × 10−4

ii rate = k[C6H5N2Cl(aq)] = (1.51 × 10−3) × (1.75 × 10−4) = 2.6 × 10−7 [1] mol dm−3 s−1 [1]

iii measure gradient of line at t = 800 s [1] 2 a (change in) concentration/mass/volume with time [1] b i O2: experiment 2 has 4 times [O2] as experiment 1: rate increases by 4 [1] so order = 1 with respect to O2 [1] NO: experiment 3 has 3 times [NO] as experiment 2: rate increases by 9 [1] so order = 2 with respect to NO [1]

ii rate = k[O2][NO]2 [1]

iii k = 22 NO][]O[rate

= 2)00100.0(00100.010.7

× = 7.10 × 109 [1]

units are dm6 mol–2 s–1 [1] c i the slowest step [1]

ii 2NO2 → NO + NO3 [1] NO3 + CO → NO2 + CO2 [1] (or similar stage involving intermediates) 3 a i rate at start (of reaction)/t = 0 [1]

ii 0.048 (mol dm−3 s−1) [1] b i C12H22O11(aq): experiment 2 has 2 times [C12H22O11(aq)] as experiment 1 and rate increases by 2 [1] so order = 1 with respect to C12H22O11 [1] HCl(aq): experiment 3 has 1.5 times [HCl] as experiment 1 and rate increases by 1.5 [1] so order = 1 with respect to HCl(aq) [1] order has to be correct to get reason mark

ii 2 or second order [1] this will be dependent on answer to part i

iii rate = k[C12H22O11] [HCl] or rate = 2.4 × [C12H22O11] [HCl] [2] rate = k[C12H22O11] [H2O] scores 1 mark rate = [C12H22O11] [HCl] scores 1 mark k[C12H22O11] [HCl] scores 1 mark k = [C12H22O11] [HCl] scores zero check for error carried forward from part b i c increases [1]


COAS Chemistry 2 15

d C12H22O11: 0.05 mol dm−3 [1]

in one half-life, [C12H22O11] concentration halves / 210.0 [1]

HCl: 0.1 mol dm−3 (since the acid is a catalyst) [1] assume units are mol dm−3 unless told otherwise assume mol dm3 means mol dm−3 but penalise wrong unit once only 4 a i Br−(aq): 1st order [1] as [Br−(aq)] triples the rate triples [1] H+(aq): 2nd order [1] as [H+(aq)] doubles the rate quadruples [1] BrO3

−(aq): 1st order [1] as [BrO3

−(aq)] doubles the rate doubles [1] ii rate = k[Br−(aq)] [H+(aq)]2 [BrO3

−(aq)] [1] (state symbols not needed)

iii k = aq)](BrO[aq)]([Haq)]([Br

rate–

32– +

= 1.01.01.0

102.12

3–

××× [1]

rate constant, k = 12 [1] units: dm9 mol−3 s−1 [1] answer of 0.0833 would score 1 mark b i slowest step [1]

ii rate equation shows reaction is 1st order with respect to HBr and 1st order with respect to O2 [1] which corresponds to molecules in step 1 [1] iii 4HBr + O2 2Br2 + 2H2O [1]

5 From graph, constant half-life [1] therefore first order with respect to [CH3COCH3] [1] From table, rate doubles when [H+] doubles [1] therefore first order with respect to [H+] [1] From table, rate stays the same when [I2] doubles [1] therefore zero order with respect to [I2] [1] order with no justification does not score rate = k[H+][CH3COCH3] (from all three pieces of evidence) [1]

k = ]COCHCH[][H

rate

33+ or

3–

9–

105.102.0101.2××

×

[1]

= 7.0 × 10–5 [1] dm3 mol–1 s–1 [1] accept 7 × 10–5 rate-determining step involves species in rate equation [1] two steps that add up to give the overall equation [1] The left-hand side of a step that contains the species in the rate-determining step [1] i.e. CH3COCH3 + H+ → [CH3COHCH3]+ [CH3COHCH3]+ + I2 → CH3COCH2I + HI + H+ Quality of written communication Student organises relevant information clearly and coherently, using specialist vocabulary where appropriate, using the following four words/phrases: constant, half-life, order, doubles/×2 [1]


COAS Chemistry 2 16

6 a i H2: experiment 2 has 2.5 times more [H2] as experiment 1 and rate increases by 2.5 [1] so order = 1 with respect to H2 [1] NO: experiment 3 has 3 times [NO] as experiment 2 and rate has increased by 9 (= 32) [1] so order = 2 with respect to NO [1] Quality of written communication At least two complete sentences where the meaning is clear [1]

ii rate = k[NO]2[H2] [1]

iii k ]H[[NO]rate

22 = or

20.0)10.0(6.2

2 × [1]

= 1300 [1] units are dm6 mol–2 s–1 [1] allow 1 mark for 7.69 × 10–4 or 1.3 × 10x (where x not 3) b i 1.5O2(g) → O3(g) or O2(g) + 0.5O2(g) → O3(g) [1] NO is a catalyst [1] as it is (used up in step 1 and) regenerated in step 2 or not used up in the overall reaction [1] allow 1 mark for ‘O/NO2 with explanation of regeneration’

ii rate = k[NO][O3] [1] species in rate equation match the reactants in the slow step/ rate-determining step [1] 7 a i constant half-life [1]

ii rate = k[N2O5] [1] a common error will be to use ‘2’ from equation

iii curve downwards getting less steep [1] curve goes through (1200, 0.30), (2400, 0.15), (3600, 0.075) [1]

iv tangent shown on graph at t = 1200 s [1] v rate = 6.2 × 10–4 × 0.60 = 3.7(2) × 10–4 [1]

mol dm–3 s–1 [1] error carried forward possible from part ii using [N2O5]x second-order answer = 2.2(3) ×10–4 b i slow step [1]

ii (CH3)2C=CH2 + H2O (CH3)3COH [1] iii H+ is a catalyst [1]

H+ used in first step and formed in second step/regenerated/not used up [1] iv rate = k[(CH3)2C=CH2] [H+] [1]

common error will be the use of H2O instead of H+

Chapter 12

1 a Kc = OH]H[CCOOH]CH[O]H[]HCOOCCH[

523

2523

[2]

award 1 mark if upside down b i

CH3COOH C2H5OH CH3COOC2H5 H2O 6.0 12.5 0 0 1 7.5 5 5

reactant concentrations [1] product concentrations [1]


COAS Chemistry 2 17

ii Kc = 5.71

55××

= 3.3 [1]

no units [1] or error carried forward based on answers to part a and/or part b i c leave experiment longer [1] monitor compositions and repeat until constant value [1] d i more CH3COOC2H5 and H2O/less CH3COOH and C2H5OH [1] equilibrium shifts to the right to reduce the concentration of ethanol [1]

ii Kc stays the same [1] e stays the same/catalyst does not shift equilibrium position [1] forward and reverse reactions altered by same amount/equilibrium achieved in less time [1] f i equilibrium has shifted to the left [1] more reactants/less products [1]

ii forward reaction is exothermic [1] 2 a forward and reverse reactions at same rate [1] achievable from either direction [1] requires closed system [1] concentrations of reactants and products are constant [1] [maximum 2]

b i Kc = 22

3

(g)][HCO(g)][OH(g)]CH[

1 mark for top, 1 mark for bottom [2]

ii Kc = 22–3–

5–

)104.2)(101.3()106.2(

×××

[1]

= 14.6 (dm6 mol–2) [1] c i fewer molecules on right [1] reaction relieves increase in pressure [1]

ii Kc stays the same [1] iii rate increases [1]

increased collisions/more concentrated [1] rates initially forward faster than reverse [1] at equilibrium, rates same [1] d i Kc decreases so products decrease/reactants increase [1] therefore equilibrium moves to the left/to endothermic side [1] second mark dependent on the first ii ΔH is negative because of equilibrium change in part i [1] mark is consequential on answer to part d i

3 a i Kc = ]O[]N[

NO][

22

2

[2]

award 1 mark if upside down Kp expression is worth 1 mark

ii equilibrium lies on the left because Kc is very small [1]


COAS Chemistry 2 18

iii [O2(g)] = c2

2

][NNO][

K× =

31–

216–

108.41.1)100.4(

×××

[1]

= 0.30 mol dm–3 (calculator value is 0.303030303) [1] answer given to 2 sig figs [1] award 2 marks for 3.3 (upside down, calculator value is 3.3) award 2 marks for 7.6 × 1014 (missing out the squared value, calculator value is 7.5757…) award 2 marks for 0.37 (1.1 on top of equation, calculator value is 0.3666…) award 2 marks for 5.2 × 10–46 (‘4’ values swapped, calculator value is 5.236363… × 10–46) b i ΔH is positive [1] equilibrium moves to the right to compensate for increase in temperature/ to lower the temperature/to minimise the change [1]

ii increase in proportion of NO [1] because Kc increases (can be linked to either increased proportion of NO or enthalpy change) [1]

iii 2NO + O2 → 2NO2 [2] 1 mark for species correct 1 mark for ‘simplest’ balanced equation NO + ½ O2 → NO2 also gets 2 marks 1 mark if N2O4 given as product c optimum pressure is low pressure [1] fewer gaseous moles on left [1] optimum temperature is low temperature [1] forward reaction is exothermic [1] the ‘reason’ mark can only be awarded if the ‘condition’ mark is correct 1000 °C used to increase rate with more energetic collisions or so that a greater proportion of molecules exceed activation energy [1] 10 atm used to increase rate by increasing concentration or increasing collisions [1] catalyst used to increase rate by lowering the activation energy/providing a lower energy route [1] NOT increase equilibrium yield Quality of written communication: recognition of a compromise between rate and equilibrium amount [1]

4 a Kc = ]PCl[]Cl[]PCl[

5

23

[1]

b i PCl5 > 0.3 mol dm–3; PCl3 and Cl2 < 0.3 mol dm–3 [1] ii at start, system is out of equilibrium with too much PCl3 and Cl2 and not

enough PCl5/ 3.0

3.03.0 ×

= 0.3 is greater than Kc = 0.245 mol dm–3 [1]

c i Kc does not change as temperature is the same [1] ii fewer moles on left-hand side [1]

system moves to the left to compensate for increase in pressure by producing fewer molecules [1] d i Kc decreases (as more reactants than products) [1] ii forward reaction is exothermic/reverse reaction is endothermic [1] equilibrium moves to the left to oppose increase in energy/because Kc decreases [1]


COAS Chemistry 2 19

e i 4PCl5 + 10MgO → P4O10 + 10MgCl2 [1]

ii 100 g P4O10 = 284100 = 0.35(2) mol [1]

moles PCl5 needed = 4 × 0.352 = 1.408/1.4 mol [1] mass PCl5 = 1.4(08) × 208.5 = 293.568/294/291.9 g [1] use of 284 for P4O10 and 208.5 for PCl5 [1] 3 marks for 73.4/72.975/72.3 g (no use of the ‘4’ factor) 3 marks for 18.35 g (from dividing by 4)

Chapter 13 1 a i ionic product of water [1]

ii Kw = [H+(aq)][OH–(aq)] state symbols not needed [1]

b moles of HCl = 1000

35.21105 3– ×× = 1.067 × 10–4 mol [1]

moles of Ca(OH)2 = 2

10067.1 4–× = 5.34 × 10–5 mol [1]

concentration of Ca(OH)2 = 25

1000 × 5.34 × 10–5

= 2.136 × 10–3 mol dm–3 [1] 2 marks for 4.27 × 10–3/8.54 × 10–3 mol dm–3 (no factor of 4) c [OH–] = 2 × 2.7 × 10–3 = 5.4 × 10–3 mol dm–3 [1]

[H+(aq)] = aq)]([OH –

wK

=

3–

14–

104.5100.1×× = 1.85 × 10–12 mol dm–3 [1]

pH = –log10(1.85 × 10–12 mol dm–3) = 11.73/11.7 [1] allow for errors carried forward for the pH mark, providing that the [H+] value

has been derived from ]OH[

K–

w

If pOH method has been used, pOH = 2.27 would get the first mark pH = 14 – 2.27 = 11.73 gets the second mark The commonest mistake will be to not double OH– and to use 2.7 × 10–3 This gives an ‘error carried forward’ mark of 11.43/11.4, worth 2 marks pH = 11.13 from dividing by 2: worth 2 marks d 8 [1] 2 a acid is a proton/H+ donor [1] base is a proton/H+ acceptor [1] conjugate acid has H+ more than conjugate base [1] equation showing acid–base pairs [1] 2 acid–base pairs labelled correctly [1] dilute acid has small number of moles dissolved per volume [1] weak acid has partial dissociation [1] Quality of written communication: At least two complete sentences that are legible and where the spelling, punctuation and grammar allow the meaning to be clear. At least one equation shown. [1]


COAS Chemistry 2 20

b i Ka = [HCN(aq)]

]aq)(CN[]aq)(H[ –+

[1]

ii Ka = [HCN(aq)]

aq)](H[ 2+

, therefore 4.9 × 10–10 = 0.010

aq)](H[ 2+

[1]

[H+(aq)] = )010.0()109.4( 10– ×× = 2.2 × 10–6 mol dm–3 [1] pH = –log10[H+(aq)] = –log10(2.2 × 10–6) = 5.65/5.66/5.7 accept calculator value [1] 3 a i strong acid: completely dissociates/ionised [1]

Brønsted–Lowry acid: proton/H+ donor [1] ii NO3

– [1] b i pH = –log10[H+]/–log10(0.015) [1] = 1.82/1.8 (not 2) [1]

ii [H+] = 0.0075 mol dm–3 pH = –log10(0.0075) = 2.12/2.1 [1] c i Kw = [H+(aq)][OH–(aq)] state symbols not needed [1] ii [H+(aq)] = 10–pH = 10–13.54 = 2.88/2.9 × 10–14 mol dm–3 [1]

[NaOH]/[OH–(aq)] = aq)]([Hw

+

K =

14–

14–

1088.2100.1×× = 0.347/0.35 mol dm–3 [1]

d i a solution that minimises/resists/opposes pH changes [1] ii the buffer must contain both CH3COOH and CH3COONa/CH3COO–/weak acid and conjugate base [1]

Solution A is a mixture of CH3COOH and CH3COONa/has an excess of acid/ is acidic [1] Solution B contains only CH3COONa/only CH3COO–/only the salt/is neutral [1] CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)/acid/alkali has been neutralised/CH3COOH(aq) and NaOH react together [1] e [H+] increases [1] H2O ionises more/for H2O ⇌ H+ + OH–, equilibrium moves to the right [1] exo/endo is irrelevant 4 a i pH = –log10[H+(aq)] state symbol not needed [1]

ii HBr is stronger than CH3COOH because pH is lower [1] HBr dissociates more/more H+ ions … for the same concentration [1]

iii diluting by a factor of 10/10-fold dilution [1] pH = 3 [1] credit a calculated pH for error carried forward from a wrong dilution with working shown b i Kw = [H+(aq)][OH–(aq)] state symbols not needed [1]


COAS Chemistry 2 21

ii [H+(aq)] = aq)]([OH –

wK

=

020.0100.1 14–×

= 5 × 10–13 mol dm–3 [1]

pH = –log10(5 × 10–13) = 12.30 [1] accept the calculator value of 12.30103 error carried forward is possible for the pH mark providing that the [H+]

value has been derived from ]OH[

K–

w

If the pOH method is used, pOH = 1.7 would get the first mark, pH = 14 – 1.7 = 12.3 gets the second mark c i start at pH = 3.4 (approx. halfway up 0–7 rise) [1] sharp rise at 20 cm3 (must have a vertical part) [1] finish higher above pH 7 than starting pH, with line continued to 50 cm3, but finish pH is less than 14 [1] ii indicator that has a pH range coinciding with steepest part of titration curve in i, likely to be thymol blue or brilliant yellow [1] pH range coincides with … pH change during sharp rise/equivalence point [1]

5 a partial dissociation: HCOOH ⇌ H+ + HCOO– [1] b i pH = –log10(1.55 × 10–3) = 2.81/2.8 [1]

[H+] deals with negative indices over a very wide range/pH makes numbers manageable/removes very small numbers [1]

ii Ka = ][HCOOH(aq)

aq)]([HCOOaq)](H[ –+

[1]

(state symbols not needed)

iii Ka = ][HCOOH(aq)

aq)](H[ 2+

=

015.0)1055.1( 23–× [1]

= 1.60 × 10–4 (mol dm–3) [1] pKa = –log10Ka = –log10(1.60 × 10–4) = 3.80 [1]

iv percentage dissociating = 015.0

100)1055.1( 3– ×× = 10.3%/10% [1]

working not required c i HCOOH + NaOH → HCOONa + H2O [1] state symbols not needed

ii n(HCOOH) = 1.50 × 10–2 × 1000

00.25 = 3.75 × 10–4 [1]

volume of NaOH(aq) that reacts is 30.0 cm3 [1]

so [NaOH] = 3.75 × 10–4 × 30

1000 = 0.0125 mol dm–3 [1]

iii Kw = [H+(aq)][OH–(aq)] [1]

pH = –log10 ⎥⎦

⎤⎢⎣

⎡ ×0125.0101 14–

= 12.10/12.1 [1]

(calculator value is 12.09691001) iv metacresol purple [1]

pH range coincides with pH change during sharp rise or pH 6–10/coincides with equivalence point/end-point [1]


COAS Chemistry 2 22

Chapter 14 1 a 1 = B and 3 = C [1] 2 = D and 4 = E [1] 5 = A and 6 = F [1] b –443 = +76 + (+122) + (+376) + (–349) + LE [1] LE = –668 kJ mol–1 allow error carried forward if one mistake only in part a [1] c Na+ smaller than Cs+ don’t accept ‘sodium smaller’ first time [1] Na+ has a larger charge density [1] attracts the anion/Cl– more strongly/sodium chloride has the stronger bonding [1] d dissolves/no reaction do not accept ‘nothing’ [1] colourless/neutral/pH 7 [1] e add aqueous AgNO3 [1] chloride gives a white precipitate [1] iodide gives a pale yellow precipitate [1] 2 a i ΔH1 = –796 ΔH2 = +178 ΔH3 = +590 ΔH4 = +1150 [1] ΔH5 = +244 [1] ΔH6 = –696 [1] allow 178, 590, 1150, 244 (without plus sign)

ii –796 – (+178) – (+590) – (+1150) – (+244) – (–696) [1] = –2262 kJ mol–1 [1] 2 marks for –2262 with no working shown allow error carried forward from the wrong figures on the Born–Haber cycle 1 mark if 1 error, 0 marks if 2 or more errors

iii magnesium fluoride more exothermic than calcium chloride (or reverse argument) because ionic radius of Mg2+ is less than that of Ca2+/charge density of magnesium ion is greater than that of calcium ion (or reverse argument) [1]

ionic radius of F– is less than that of Cl–/charge density of fluoride ion is greater than that of chloride ion (or reverse argument) [1] stronger (electrostatic) attraction between cation and anion in MgF2 than in CaCl2/stronger ionic bonds in MgF2 [1] Answer must refer to the correct particle. Not ‘Mg/magnesium has a smaller radius’ or ‘F/fluorine has a smaller radius’. Allow ‘magnesium/fluorine has a smaller ionic radius’. b Any two from: for second ionisation energy the electron lost is closer to the nucleus [1] for second ionisation energy the electron is lost from a particle that is already positive [1] for second ionisation energy there is one more proton than electron [1] for second ionisation energy the electron is more firmly attracted to the nucleus [1] allow reverse argument


COAS Chemistry 2 23

3 Definition – maximum 3 marks Mg2+(g) + 2Cl–(g) MgCl2(s) [1] the enthalpy change that accompanies the formation of 1 mole of a solid (compound) [1] from its constituent gaseous ions [1] Born–Haber cycle – maximum 5 marks correct formulae on cycle [1] correct state symbols [1] use of 2 moles of Cl(g), i.e. 246 [1] use of 2 moles of Cl–(g), i.e. 698 [1] –2526 kJ mol–1 [1] Comparison – maximum 3 marks (any three from): Na+ has a larger radius than Mg2+ (or reverse argument) [1] Br– has a larger radius than Cl– (or reverse argument) [1] Na+ has a lower charge than Mg2+ (or reverse argument) [1] strongest attraction is between Mg2+ and Cl–/MgCl2 has the strongest attraction between its ions (or reverse argument) [1] or Na+ has a lower charge density than Mg2+ (or reverse argument) [1] Br– has a lower charge density than Cl– (or reverse argument) [1] strongest attraction between ions which have the highest charge density/MgCl2 has the strongest attraction between its ions (or reverse argument) [1] Quality of written communication correct spelling, punctuation and grammar in at least two sentences [1]

Chapter 15 1 a 525 kJ mol–1 [1] b 193.6 J K–1 mol–1 [1] c uses ∆G = ∆H – T∆S [1] to be feasible, ∆G = 0 or ∆G < 0 (has a negative value) [1]

minimum T = SHΔΔ [1]

converts ∆S from J to kJ/divides by 1000 or converts ∆H from kJ to J [1] 2712 K or 2438 °C or 2439 °C [1] (units essential) 2 a i ∆G = ∆H – T∆S. ∆G = +489 000 – (500 × 541), so ∆G = +218.5 kJ mol–1. [1]

ii ∆G = ∆H – T∆S. 0 = +489 000 – (T × 541), so T = 541

000489 = 904 K [1]

iii Reaction 1 will only be spontaneous at temperatures over 904 K (631 °C). [1] b i Reaction 2: ∆G = ∆H – T∆S. ∆G = –394 000 – (500 × 7), so ∆G = –397.5 kJ mol–1 [1] Reaction 3: ∆G = ∆H – T∆S. ∆G = +172 000 – (500 × 180), so ∆G = +82.0 kJ mol–1 [1] Reaction 4: ∆G = ∆H – T∆S. ∆G = –27 000 – (500 × 12), so ∆G = –33.0 kJ mol–1 [1]

ii Reaction 3 is not spontaneous at 500 K. [1]

iii ∆G = ∆H – T∆S; 0 = +172 000 – (T × 180), so T = 180

000172 = 956 K [1]

c The temperature in a blast furnace must be at least 904 K. (In practice temperatures of over 1200 K are used. ) [1]


COAS Chemistry 2 24

d The single-step mechanism involves a reaction between two solids. This is likely to be very slow due to lack of collisions between reactant particles. [1]

Each step in the three-step mechanism involves a reaction between a gas and a solid. These reactions are likely to be much quicker. [1]

Chapter 16 1 a emf/voltage/potential difference [1] half-cell combined with standard hydrogen electrode [1] standard conditions of 298 K, 1.00 mol dm–1, 1 atm/101 kPa all 3 conditions required for 1 mark [1] b i diagram shows: voltmeter and salt bridge and complete circuit [1] solution labelled Cu2+ and electrode labelled Ag [1]

ii direction from Cu(s) to Ag(s) (must be in/close to wire) [1] iii 0.80 V – 0.34 V = 0.46 V [1] iv Cu + 2Ag+ Cu2+ + 2Ag [1]

c Standard electrode potential for chlorine is more positive than for Fe3+, therefore it is a better oxidising agent than Fe3+ [1] (do not accept Eo is larger or smaller) Standard electrode potential for iodine is less positive than for Fe3+, therefore it is a poorer oxidising agent than Fe3+ [1]

(accept release of electrons/equilibrium arguments) 2 a labels on diagram to show: Ni(s) and Ni2+(aq) [1] salt bridge and suitable circuit [1] platinum electrode [1] I2 and I– [1] concentration of 1 mol dm–3 for at least one solution/298 K [1] b i 0.79 V [1]

ii Ni → Ni2+ + 2e– [1] I2 + 2e– → 2I– [1]

iii Ni + I2 → Ni2+ + 2I– [1] iv from nickel towards iodine since nickel half-cell standard electrode potential is more negative [1]


COAS Chemistry 2 25

3 a A = platinum (electrode) B = H+(aq)/HCl(aq)/other suitable acid C = voltmeter D = Cl2(g) [2] state symbols needed for B and D all correct for 2 marks, 3 correct for 1 mark b i from hydrogen half-cell to chlorine half-cell (left to right) [1] electrons flow to half-cell with more positive standard electrode potential [1]

ii pressure = 1 atm/100 kPa/101 kPa temperature = 298 K/25 °C concentration = 1/1.00 mol dm–3 [2] all correct for 2 marks, 2 correct for 1 mark c the standard electrode potential for ClO3

–/½Cl2 is more positive than that of ½Cl2/Cl– [1] ClO3

– has a greater tendency to gain electrons than Cl2/ClO3– is a better oxidising

agent than Cl2 [1] alternative:

because Eo is positive, the reaction will go from left to right, therefore ClO3– is

reduced so it must be a better oxidising agent than chlorine

Chapter 17 1 a are often catalysts [1] allow compounds are often paramagnetic, do not allow mention of metallic properties b i tetrahedral/or a clear drawing of a tetrahedral ion [1] bond angle of 109.5 ± 0.5° [1] allow square planar (1 mark) with bond angle of 90° (1 mark) tetrahedral structure must have at least one wedge bond

ii Cl– [1] iii (concentrated) hydrochloric acid/(concentrated) solution of an ionic chloride [1] iv suitable equation

e.g. [Cu(H2O)6]2+ + 4Cl– → [CuCl4]2– + 6H2O or [Cu(H2O)6]2+ + 4NH3 → [Cu(H2O)2(NH3)4]2+ + 4H2O [1] reaction in which a ligand is swapped or displaced by another ligand/can be shown by artwork [1] not ‘ligand is substituted’ 2 Ligand substitution

Suitable example, e.g. reaction of thiocyanate ions with hexaaquairon(III) to give [Fe(H2O)5(CNS)]2+ [1]

Observations, e.g. formation of a blood-red colour [1] Suitable equation, e.g. [Fe(H2O)6]3+ + CNS– → [Fe(H2O)5(CNS)]2+ + H2O [1] Suitable example can be awarded from equation. Equations do not need state symbols Precipitation Suitable example, e.g. reaction between (aqueous) iron(II) chloride with (aqueous) sodium hydroxide [1] Observations, e.g. formation of a green precipitate/formation of a green solid [1] Suitable equation, e.g. Fe2+(aq) + 2OH–(aq) → Fe(OH)2(s) [1] Precipitate mark can be awarded from state symbol in equation


COAS Chemistry 2 26

Redox Suitable example, e.g. oxidation of iron(II) chloride by chlorine to make iron(III) chloride [1] Observation, e.g. green solution becomes yellow/rust solution [1] Suitable equation, e.g. 2FeCl2 + Cl2 2FeCl3 [1] Other examples could include iron and chlorine to make iron(III) chloride/iron and HCl to make FeCl2/MnO4

– and Fe2+ to make Fe3+ Quality of written communication: correct spelling, punctuation and grammar in at least two sentences [1] must refer to answers that address the question 3 a Zn2+ is 1s2 2s2 2p6 3s2 3p6 3d10 and Cu2+ is 1s2 2s2 2p6 3s2 3p6 3d9 [1] copper has at least one ion with an incomplete filled d subshell (zinc does not)/ copper(II) ion has an incomplete set of d electrons (zinc ion does not)/copper(II) ion has an incomplete d subshell (zinc ion does not) (or reverse argument) [1] b Cu2+ compounds are coloured but Zn2+ compounds are not [1] Cu2+ compounds may be catalytic but Zn2+ compounds are not [1] allow Cu2+ forms complexes but Zn2+ does not, allow correct chemistry of Cu2+ compared to Zn2+, e.g. Cu2+ and NaOH gives blue precipitate but Zn2+ gives white precipitate (that redissolves in excess) c moles of hydrogen = 3.17 × 10–3/moles of zinc = 3.17 × 10–3 [1] not 3 × 10–3 mass of zinc = 0.207 g/mole of zinc × 65.4 [1] not 0.2 percentage of copper = 83.2 [1] allow errors carried forward, final answer must be to 3 or 4 sig figs, penalise

significant figures just once, allow values between 82.9–83.2

COAS C2 assess bkeocans - Cambridge University...

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