Danho
ZIMSEC A Level · N2014

Chemistry Paper 2 November 2014

Questions
35
Total marks
48
Time allowed
75 min

Sit this paper online

Questions
35
Pass mark
21
Sit this paper

Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[1 marks]chemical equilibrium
For the equilibrium N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), the expression for KpK_p is
  1. ApN2 pH23pNH32\frac{p_{N_2}\,p_{H_2}^3}{p_{NH_3}^2}
  2. BpNH3pN2 pH2\frac{p_{NH_3}}{p_{N_2}\,p_{H_2}}
  3. C2pNH3pN2+3pH2\frac{2p_{NH_3}}{p_{N_2} + 3p_{H_2}}
  4. DpNH32pN2 pH23\frac{p_{NH_3}^2}{p_{N_2}\,p_{H_2}^3}

Question 102

[1 marks]chemical equilibrium
2.0 mol of nitrogen and 2.0 mol of hydrogen were allowed to reach equilibrium and 0.20 mol of nitrogen reacted. The amounts of N2N_2, H2H_2 and NH3NH_3 at equilibrium are
  1. A0.20, 0.60 and 0.40 mol
  2. B1.8, 1.4 and 0.20 mol
  3. C1.8, 1.4 and 0.40 mol
  4. D1.8, 1.8 and 0.40 mol

Question 103

[1 marks]chemical equilibrium
At equilibrium the mixture contains 1.8 mol N2N_2, 1.4 mol H2H_2 and 0.40 mol NH3NH_3. The mole fraction of ammonia is
  1. A0.111
  2. B0.200
  3. C0.389
  4. D0.500

Question 104

[1 marks]chemical equilibrium
At equilibrium the mixture contains 1.8 mol N2N_2, 1.4 mol H2H_2 and 0.40 mol NH3NH_3. The total number of moles present at equilibrium is

Answer this when you sit the paper.

Question 105

[3 marks]chemical equilibrium
Given the equilibrium mole fractions xN2=0.500x_{N_2}=0.500, xH2=0.389x_{H_2}=0.389 and xNH3=0.111x_{NH_3}=0.111, and Kp=40.7 atm−2K_p = 40.7\ \text{atm}^{-2} at 400 K, the total pressure of the equilibrium mixture is closest to
  1. A9.7 atm
  2. B0.10 atm
  3. C0.42 atm
  4. D2.4 atm

Question 201

[1 marks]electrochemistry
The standard conditions used when measuring a standard electrode potential are
  1. A298 K, 10 atm and 0.1 mol dm−30.1\ mol\,dm^{-3} solutions
  2. B373 K, 1 atm and saturated solutions
  3. C273 K, 1 atm and 1 mol dm−31\ mol\,dm^{-3} solutions
  4. D298 K, 1 atm and 1 mol dm−31\ mol\,dm^{-3} solutions

Question 202

[1 marks]electrochemistry
Given S2O82−+2e−⇌2SO42−S_2O_8^{2-} + 2e^- \rightleftharpoons 2SO_4^{2-}, Eθ=+2.01 VE^\theta = +2.01\ V and Ag++e−⇌AgAg^+ + e^- \rightleftharpoons Ag, Eθ=+0.80 VE^\theta = +0.80\ V, the standard cell potential is
  1. A+0.61 V+0.61\ V
  2. B+1.21 V+1.21\ V
  3. C+2.01 V+2.01\ V
  4. D+2.81 V+2.81\ V

Question 203

[1 marks]electrochemistry

The standard electrode potentials for two half cells are

S2O82−+2e−⇌2SO42−S_2O_8^{2-} + 2e^- \rightleftharpoons 2SO_4^{2-}, E⊖=+2.01E^\ominus = +2.01 V

Ag++e−⇌AgAg^+ + e^- \rightleftharpoons Ag, E⊖=+0.80E^\ominus = +0.80 V

Concentrated ammonia is added to the Ag+/AgAg^+/Ag half cell. The magnitude of the cell potential, Ecell⊖E^\ominus_{cell}

  1. Adecreases, because silver is oxidised
  2. Bis unchanged, because ammonia is not an electrolyte
  3. Cincreases, because [Ag+][Ag^+] falls as the complex ion forms
  4. Dfalls to zero, because the electrode dissolves

Question 204

[1 marks]electrochemistry
Standard electrode potential is defined as the electromotive force of a half-cell, measured under standard conditions, relative to
  1. Athe concentration of ions present at equilibrium
  2. Bthe standard hydrogen electrode
  3. Cany reference metal chosen by the experimenter
  4. Dthe electrode of highest atomic number in the cell

Question 205

[2 marks]electrochemistry
For the half-cells S2O82−+2e−⇌2SO42−S_2O_8^{2-} + 2e^- \rightleftharpoons 2SO_4^{2-} (Eθ=+2.01 VE^\theta = +2.01\ V) and Ag++e−⇌AgAg^+ + e^- \rightleftharpoons Ag (Eθ=+0.80 VE^\theta = +0.80\ V), the overall equation for the cell reaction is
  1. A2SO42−+2Ag+→S2O82−+2Ag2SO_4^{2-} + 2Ag^+ \rightarrow S_2O_8^{2-} + 2Ag
  2. BS2O82−+2Ag+→2SO42−+2AgS_2O_8^{2-} + 2Ag^+ \rightarrow 2SO_4^{2-} + 2Ag
  3. CS2O82−+Ag→2SO42−+Ag+S_2O_8^{2-} + Ag \rightarrow 2SO_4^{2-} + Ag^+
  4. DS2O82−+2Ag→2SO42−+2Ag+S_2O_8^{2-} + 2Ag \rightarrow 2SO_4^{2-} + 2Ag^+

Question 206

[1 marks]electrochemistry
In the reaction between the S2O82−/SO42−S_2O_8^{2-}/SO_4^{2-} and Ag+/AgAg^+/Ag half cells, the species that acts as the oxidising agent is

Answer this when you sit the paper.

Question 207

[2 marks]electrochemistry
The S2O82−/SO42−S_2O_8^{2-}/SO_4^{2-} half cell acts as the cathode of the combined cell because
  1. Ait produces a colourless solution, unlike the silver half cell
  2. Bit contains a significantly larger number of atoms and electrons per formula unit
  3. Cit has the more positive standard electrode potential, so it is reduced
  4. Dsilver is a solid metal, so it cannot be reduced further

Question 208

[1 marks]electrochemistry
The half-equation for the process occurring at the anode of the combined S2O82−/SO42−S_2O_8^{2-}/SO_4^{2-} and Ag+/AgAg^+/Ag cell is

Answer this when you sit the paper.

Question 301

[1 marks]industrial chemistry / equilibrium
A major source of sulphur dioxide in the atmosphere is
  1. Athe decay of nitrate fertilisers
  2. Bthe evaporation of sea water
  3. Cthe burning of fossil fuels containing sulphur
  4. Dthe electrolysis of brine

Question 302

[1 marks]industrial chemistry / equilibrium
In the Contact process 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) is carried out at a low pressure because
  1. Aa high pressure would destroy the catalyst
  2. Bthe reaction is endothermic
  3. Ca low pressure gives a higher yield
  4. Dthe yield is already high, so high pressure plant is not worth its cost

Question 303

[1 marks]industrial chemistry / equilibrium
Vanadium(V) oxide is used in the Contact process because it
  1. Aabsorbs the sulphur trioxide formed
  2. Bprovides an alternative path of lower activation energy
  3. Cshifts the equilibrium to the right
  4. Dincreases the yield of sulphur trioxide

Question 304

[1 marks]industrial chemistry / equilibrium
Which of the following is an adverse effect of sulphur dioxide on the environment?
  1. Ait raises the salinity of rivers and lakes
  2. Bit displaces oxygen from the atmosphere, causing suffocation
  3. Cit dissolves in atmospheric moisture and falls as acid rain
  4. Dit depletes the ozone layer by catalytic chain reactions

Question 305

[1 marks]industrial chemistry / equilibrium
Sulphur dioxide preserves food mainly because it
  1. Ableaches away the food's natural colourings
  2. Bkills microorganisms and prevents oxidation of the food
  3. Craises the internal pH of the food to alkaline levels, slowing enzyme action
  4. Dneutralises stomach acid once the food is eaten

Question 306

[2 marks]industrial chemistry / equilibrium
In the Contact process, 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g), ΔHθ=−96 kJ mol−1\Delta H^\theta = -96\ kJ\ mol^{-1}, the reaction mixture is cooled because
  1. Acooling speeds up the vanadium(V) oxide catalyst, so it works faster
  2. Bcooling condenses the SO3SO_3 formed into a liquid, continuously removing it from the gaseous equilibrium mixture as more forms
  3. Cthe reaction is endothermic and needs heat removed to keep proceeding
  4. Dthe reaction is exothermic, so lowering the temperature shifts the equilibrium towards SO3SO_3, raising the yield

Question 307

[1 marks]industrial chemistry / equilibrium
In the Contact process, an excess of air is pumped into the reactor because it
  1. Adilutes the SO2SO_2 so the reaction proceeds more slowly
  2. Bincreases the proportion of oxygen present, shifting the equilibrium further towards SO3SO_3
  3. Cremoves the SO3SO_3 formed from the mixture continuously, preventing it decomposing back into SO2SO_2 and O2O_2
  4. Dacts as an additional catalyst for the oxidation of SO2SO_2

Question 401

[1 marks]organic chemistry / amines
In the conversion CH3Cl→A→CH3CH2NH2CH_3Cl \rightarrow A \rightarrow CH_3CH_2NH_2, the reagent and conditions for step I are
  1. Aethanolic KCN, heat under reflux
  2. Bammonia in ethanol, sealed tube
  3. Cconcentrated H2SO4H_2SO_4 at 170 °C
  4. Daqueous NaOH, warm

Question 402

[1 marks]organic chemistry / amines
Intermediate A, CH3CNCH_3CN, is converted into ethylamine by
  1. Aheating with concentrated HCl
  2. Bwarming with aqueous NaOH
  3. Crefluxing with acidified K2Cr2O7K_2Cr_2O_7
  4. Dreduction with LiAlH4LiAlH_4 in dry ether

Question 403

[1 marks]organic chemistry / amines
Dimethylamine is a stronger base than ethylamine because
  1. Atwo alkyl groups release electrons to nitrogen, making the lone pair more available
  2. Bit has a higher relative molecular mass
  3. Cit can form more hydrogen bonds with water
  4. Dits nitrogen atom has no lone pair

Question 404

[1 marks]organic chemistry / amines
The functional group present in intermediate A, CH3CNCH_3CN, is the

Answer this when you sit the paper.

Question 405

[1 marks]organic chemistry / amines
Dimethylamine and ethylamine are isomers. The condensed structural formula of dimethylamine is

Answer this when you sit the paper.

Question 406

[1 marks]organic chemistry / amines
Which of the following is an industrial use of halogenated hydrocarbons?
  1. Aas refrigerants and aerosol propellants
  2. Bas reducing agents in the blast furnace
  3. Cas catalysts in the Haber process
  4. Das fertilisers supplying nitrogen to crops

Question 407

[3 marks]organic chemistry / amines
Uncontrolled use of halogenated hydrocarbons such as CFCs harms the environment mainly because they
  1. Aare chemically stable and persist in the atmosphere, allowing them to catalytically break down stratospheric ozone
  2. Bincrease soil salinity, which reduces crop yields near industrial sites
  3. Care highly reactive and burn readily in air, releasing acidic combustion gases that fall as acid rain over wide areas
  4. Dbiodegrade rapidly in soil, releasing toxic breakdown products into groundwater

Question 501

[1 marks]organic chemistry / polymers
Adding sulphuric acid to nylon-6,6 gives hexanedioic acid and 1,6-diaminohexane because the acid
  1. Ahydrolyses the amide links in the chain
  2. Bdehydrates the polymer
  3. Creduces the carbonyl groups
  4. Doxidises the polymer chain

Question 502

[1 marks]organic chemistry / polymers
Urea, NH2CONH2NH_2CONH_2, is used in commercial fertilisers because it
  1. Aneutralises acidic soils
  2. Bhas a high nitrogen content
  3. Ckills weeds in the soil
  4. Dis insoluble in water

Question 503

[1 marks]organic chemistry / polymers
Bromine in the presence of iron(III) bromide reacts with a benzene ring by
  1. Afree radical substitution
  2. Belectrophilic addition
  3. Celectrophilic substitution
  4. Dnucleophilic addition

Question 504

[2 marks]organic chemistry / polymers
Which pair of functional groups is present in fluorescamine?
  1. Aa lactone (ester) group and a ketone (carbonyl) group
  2. Ban aldehyde group and a primary amine group, both capable of forming hydrogen bonds
  3. Ca carboxylic acid group and an amide group
  4. Dan alcohol group and a nitrile group

Question 505

[1 marks]organic chemistry / polymers
When fluorescamine is refluxed with aqueous NaOH, the reaction that occurs is
  1. Aelectrophilic substitution on its aromatic ring
  2. Boxidation of its aromatic side chain to a carboxylic acid under vigorous acidic conditions
  3. Chydrolysis of its lactone (ester) group, forming a carboxylate salt and an alcohol
  4. Dreduction of its ketone group to a secondary alcohol

Question 506

[1 marks]organic chemistry / polymers
NaBH4NaBH_4 reduces the ketone (carbonyl) group of fluorescamine to a

Answer this when you sit the paper.

Question 507

[1 marks]organic chemistry / polymers
Hot concentrated KMnO4KMnO_4 reacts with the aromatic ring system of fluorescamine mainly by
  1. Aoxidising any alkyl side chain on the ring to a carboxylic acid
  2. Bhydrolysing the ester group attached to the ring
  3. Creducing the aromatic ring to a cyclohexane ring
  4. Dnitrating the ring to give a nitro-substituted product under fuming nitric acid conditions

Question 508

[2 marks]organic chemistry / polymers
Terylene is a better sweat absorber than nylon-6,6 mainly because
  1. Aterylene has a much lower relative molecular mass than nylon-6,6, so more of it dissolves directly into the sweat layer during use
  2. Bterylene fibres are woven more loosely, letting sweat pass straight through
  3. Cnylon's amide groups are far more polar than terylene's ester groups
  4. Dits polar ester (C=O) groups form more hydrogen bonds with water than nylon's amide groups, which are largely tied up in intramolecular bonding

The answers, and why they are the answers

Sit the paper here to see which ones you got right. Danho explains every question, keeps your score, and works without a connection.