Chemistry Paper 3 November 2020

1. A$NOCl_{(g)} \rightleftharpoons NO_{(g)} + Cl_{(g)}$
2. B$2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$
3. C$2NOCl_{(g)} \rightleftharpoons N_{2(g)} + O_{2(g)} + Cl_{2(g)}$
4. D$NOCl_{(g)} \rightleftharpoons NO_{2(g)} + Cl^-_{(g)}$

1. A0.36 atm
2. B0.43 atm
3. C0.72 atm
4. D1.08 atm

1. A$3.16 \times 10^{-2}$ moldm⁻³
2. B$6.25 \times 10^{-2}$ moldm⁻³
3. C$3.16 \times 10^{-6}$ moldm⁻³
4. D$1.98 \times 10^{-2}$ moldm⁻³

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1. AIt reacts with oxygen in the air to release ozone, which is what actually kills bacteria
2. BIt is a weak acid whose acidic, low-pH character helps kill bacteria and other microorganisms
3. CIt is a strong base that neutralises the natural acidity of most household surfaces on contact
4. DIt is completely insoluble in water, so it forms a long-lasting protective film over surfaces

1. ADissociation is essentially complete at equilibrium, since Kp values below 1 are typically taken to signal a completed reaction
2. BDissociation is only partial at equilibrium, since Kp is small and NOCl remains the dominant species
3. CNo dissociation occurs at all, since Kp is not exactly equal to 1
4. DThe reaction proceeds entirely in the reverse direction, reforming only NOCl at equilibrium

1. Athe energy released when one mole of the bond is stretched
2. Bthe energy required to convert one mole of a liquid into a gas
3. Cthe energy required to remove one mole of electrons from a bond
4. Dthe energy required to break one mole of the bond in the gaseous state

1. A606 J
2. B1 212 J
3. C1 316 J
4. D2 424 J

1. Athe neutralisation reaction is incomplete
2. Bthe acid and the alkali were not of exactly equal concentration
3. Cwater is formed as a liquid and not as a gas
4. Dheat is lost to the surroundings

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1. ABoth solutions contribute equally to the final mixture, so their average gives a fairer starting point for the rise
2. BThe two solutions must be at exactly the same temperature before mixing, so no average is really needed
3. COnly the HCl's initial temperature actually matters in this calculation, since the NaOH is the one being added to it
4. DUsing an average is thought to help reduce the effect of any heat lost to the surroundings during mixing and reaction

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1. A4 050 C
2. B6 750 C
3. C243 000 C
4. D67.5 C

1. A125.9 cm³
2. B251.8 cm³
3. C1 007 cm³
4. D62.9 cm³

1. Aallow the conditions that give an economical rate to be selected
2. Bgive the enthalpy change of the reaction
3. Cdetermine the position of equilibrium
4. Dshow which products are formed

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1. AThe reaction is second order with respect to [Y], since a constant half-life is sometimes mistaken for a second order signature
2. BNo conclusion about order can be drawn from the half-life of a reaction under any circumstances
3. CThe reaction is first order with respect to [Y], since a constant half-life is characteristic of first order kinetics
4. DThe reaction is zero order with respect to [Y], since the half-life should stay constant only for a zero order process

1. AFirst order, since halving [XY] halves the rate, showing rate is directly proportional to [XY]
2. BZero order, since the rate does not depend on [XY] at all under these conditions
3. CSecond order, since halving [XY] should quarter the rate for a second order dependence
4. DThird order, since the relationship between rate and [XY] cannot be determined from two experiments

1. AZero order overall, since the two first order dependences cancel each other out
2. BFourth order overall, since the individual orders (1 x 1 x 1 x 1) are multiplied together
3. CFirst order overall, since only the higher of the two individual orders is counted
4. DSecond order overall, since the individual orders (1 + 1) are added together

1. A$Mg(HCO_3)_2 + H_2SO_4 \rightarrow MgSO_4 + 2H_2O + 2CO_2$
2. B$Mg(HCO_3)_2 + 2H_2SO_4 \rightarrow Mg(HSO_4)_2 + 2H_2O + 2CO_2$
3. C$Mg(HCO_3)_2 + H_2SO_4 \rightarrow MgSO_4 + 2CO_3^{2-} + 2H^+$
4. D$Mg(HCO_3)_2 + H_2SO_4 \rightarrow MgSO_4 + H_2CO_3$

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1. Asimple molecular
2. Bmetallic
3. Cgiant ionic
4. Dgiant covalent

1. A1
2. B4
3. C7
4. D10

1. AMg(HCO3)2 is generally considered far too expensive to be used on any significant scale for large-scale paper conservation work
2. BMg(HCO3)2 reacts explosively with cellulose fibres in the paper on contact
3. CExcess Mg(HCO3)2 can leave the paper alkaline, or the CO2 released may itself cause further damage
4. DMg(HCO3)2 makes the paper permanently radioactive after repeated treatment

1. ANitric acid and carbonic acid, formed from atmospheric nitrogen oxides and carbon dioxide dissolving in moisture
2. BHydrochloric acid and hydrofluoric acid, both released directly into the air from the burning of certain fossil fuels
3. CPhosphoric acid and boric acid, commonly found dissolved in rainwater
4. DAcetic acid and formic acid, produced by plants growing near paper archives

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1. AYCl4 simply dissolves in water to give a neutral solution, while sodium chloride instead hydrolyses vigorously to give an acidic solution
2. BBoth YCl4 and NaCl hydrolyse vigorously with water, each giving an equally acidic solution
3. CNeither YCl4 nor NaCl reacts with water in any way under normal laboratory conditions
4. DYCl4 hydrolyses vigorously, giving HCl fumes and an acidic solution, while NaCl simply dissolves to give a neutral solution

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1. A$OH^-$ on the carbonyl carbon
2. BHCN on the carbon to oxygen double bond
3. C$H^+$ on the carbonyl oxygen
4. D$CN^-$ on the carbonyl carbon

1. A$(CH_3)_2C(OH)CN$
2. B$(CH_3)_2C(OH)CONH_2$
3. C$HO-C(CH_3)_2-CO_2CH_3$
4. D$CH_2=C(CH_3)CO_2CH_3$

1. Aoxidative coupling
2. Baddition polymerisation
3. Ccondensation polymerisation
4. Dsubstitution polymerisation

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1. A2 CH2=C(CH3)CO2CH3 -> CH3-C(CH3)(CO2CH3)-CH2-CH(CO2CH3)(CH3)
2. Bn CH2=C(CH3)CO2CH3 -> [-CH2-C(CH3)(CO2CH3)-]n
3. Cn CH2=C(CH3)CO2CH3 -> [-CH2-CH(CO2CH3)-]n
4. Dn CH2=C(CH3)CO2CH3 -> [-CH2-C(CH3)(CO2CH3)-]n + n H2O

1. AA nitrile group and a sulphonic acid group
2. BAn amide group and an ester group
3. CAn alkene group and a halogenoalkane group
4. DA nitro group and an anhydride group

1. AOptical isomerism, since Taxol contains chiral carbon atoms bonded to four different groups
2. BTaxol shows no isomerism at all, since its structure is completely symmetrical
3. CGeometric (cis-trans) isomerism, since Taxol contains a restricted carbon-carbon double bond
4. DPositional isomerism, since the functional groups can occupy different positions on the same skeleton

1. ABenzoic acid, an amino-alcohol/amino-acid fragment, and an alcohol, from hydrolysis of its amide and ester links
2. BA nitrile group and a sulphonic acid group, both introduced during a separate late-stage synthetic step of the molecule
3. COnly carbon dioxide and water, since hot acid completely combusts the molecule
4. DOnly benzene, since hot acid removes every substituent from the aromatic ring

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1. Aa carbon to carbon double bond and a ketone group
2. Ba carboxylic acid group and an aldehyde group
3. Ca carbon to carbon double bond and an aldehyde group
4. Dan aldehyde group and a hydroxyl group

1. Aelimination
2. Boxidation
3. Chydrolysis
4. Dreduction

1. Agenerates $CN^-$, which is a stronger nucleophile than HCN
2. Bremoves the water formed in the reaction
3. Cincreases the concentration of HCN in the mixture
4. Dprotonates the carbonyl oxygen

1. AIt contains a hydroxyl group that condenses with another molecule's hydroxyl group, releasing water
2. BIt contains a benzene ring that opens up under heat to link separate monomer units together
3. CIt contains a reactive carbon-carbon double bond that can open up so monomer units join directly together
4. DGeometric (cis-trans) isomerism, since Taxol is said to contain a carbon-carbon double bond with restricted rotation

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1. AThe C=C bond adds hydrogen directly, giving a saturated alcohol with no oxidation occurring
2. BOnly the aldehyde group is reduced to a primary alcohol in this reaction, leaving the carbon-carbon double bond completely unchanged
3. CThe C=C bond is oxidatively cleaved and the aldehyde group is oxidised, giving carboxylic acid products
4. DThe molecule is completely unaffected, since KMnO4 does not react with aldehydes or alkenes

1. AFree-radical addition, since CN- is generated as a radical under basic conditions
2. BElectrophilic substitution, since CN- replaces a hydrogen atom on the carbon chain
3. CNucleophilic substitution, since CN- displaces the oxygen atom of the carbonyl group entirely
4. DNucleophilic addition, since CN- adds across the polar C=O double bond of the carbonyl group

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1. Aester and amide
2. Baldehyde and primary amine
3. Ccarboxylic acid and phenol
4. Dphenol and primary amine

1. Aa benzene ring which can be attacked by acids and bases
2. Ba side chain that can be both oxidised and reduced
3. Can acidic phenolic $-OH$ group and a basic $-NH_2$ group
4. Dtwo acidic $-OH$ groups on the same ring

1. A$NH_3$ in ethanol, then $LiAlH_4$ in dry ether
2. Baqueous NaOH, then $NH_3$ in ethanol
3. CKCN in ethanol, then $LiAlH_4$ in dry ether
4. DHCN, then hot dilute $H_2SO_4$

1. AIt reacts chemically with water to form an entirely new, unrelated compound
2. BIts polar -OH and -NH2 groups form hydrogen bonds with water molecules
3. CIts non-polar benzene ring dissolves readily in water due to van der Waals forces alone
4. DIt has no polar groups at all, so it dissolves purely by simple mechanical mixing

1. AElectrophilic substitution occurs on the activated benzene ring (ortho/para to the -OH groups), with no catalyst needed
2. BIts non-polar benzene ring is sometimes thought to dissolve readily in water through van der Waals forces alone, without any hydrogen bonding
3. CNo reaction occurs at all, since dopamine's ring is deactivated by its -OH substituents
4. DNucleophilic substitution occurs, with bromide ions displacing the -NH2 group from the side chain

1. AThe primary amine, -NH2, group on the side chain
2. BThe carbon-carbon bonds within the benzene ring itself
3. CThe -CH2CH2- chain linking the ring to the amine group
4. DThe phenolic -OH groups on the aromatic ring

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