Danho
ZIMSEC A Level · N2002

Chemistry Paper 2 November 2002

Questions
29
Total marks
38
Time allowed
75 min

Sit this paper online

Questions
29
Pass mark
18
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]thermochemistry / Born-Haber cycle
The standard enthalpy change of combustion of hydrogen is −286 kJ mol−1-286\ kJ\,mol^{-1}. The thermochemical equation for this change is
  1. AH2(g)+12O2(g)→H2O(g)H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)
  2. BH2(g)+12O2(g)→H2O(l)H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)
  3. C2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l)
  4. DH2(g)+O2(g)→H2O(l)H_2(g) + O_2(g) \rightarrow H_2O(l)

Question 102

[1 marks]thermochemistry / Born-Haber cycle
In the Born-Haber cycle for aluminium oxide shown, the enthalpy change ΔH6=−15300.1 kJ mol−1\Delta H_6 = -15300.1\ kJ\,mol^{-1} is
  1. Athe lattice enthalpy of Al2O3Al_2O_3
  2. Bthe first electron affinity of oxygen
  3. Cthe enthalpy change of atomisation of aluminium
  4. Dthe enthalpy change of formation of Al2O3Al_2O_3

Question 103

[1 marks]thermochemistry / Born-Haber cycle
The lattice enthalpy of Al2O3Al_2O_3 (−15300 kJ mol−1-15300\ kJ\,mol^{-1}) is far more exothermic than that of MgO (−3889 kJ mol−1-3889\ kJ\,mol^{-1}) because
  1. AAl2O3Al_2O_3 is amphoteric whereas MgO is basic
  2. BAl3+Al^{3+} is smaller and more highly charged than Mg2+Mg^{2+}
  3. CAl2O3Al_2O_3 contains more oxide ions per formula unit
  4. Daluminium is a better conductor of electricity than magnesium

Question 104

[2 marks]thermochemistry / Born-Haber cycle
In the Born-Haber cycle for aluminium oxide shown in Fig. 1.1, which row correctly identifies both ΔH1\Delta H_1 (the step from 2Al(s)+32O2(g)2Al(s) + \frac{3}{2}O_2(g) to 2Al(g)+32O2(g)2Al(g) + \frac{3}{2}O_2(g)) and ΔH4\Delta H_4 (the step from 2Al3+(g)+3O(g)2Al^{3+}(g) + 3O(g) to 2Al3+(g)+3O−(g)2Al^{3+}(g) + 3O^-(g))?
  1. AΔH1 = lattice enthalpy of Al2O3; ΔH4 = atomisation of oxygen
  2. BΔH1 = atomisation of aluminium; ΔH4 = first electron affinity of oxygen
  3. CΔH1 = atomisation of aluminium; ΔH4 = second electron affinity of oxygen
  4. DΔH1 = first ionisation energy of aluminium; ΔH4 = first electron affinity of oxygen

Question 105

[2 marks]thermochemistry / Born-Haber cycle
Fig. 1.1's cycle contains three further enthalpy changes: ΔH2\Delta H_2 (from 2Al(g)+32O2(g)2Al(g)+\frac{3}{2}O_2(g) to 2Al3+(g)+32O2(g)2Al^{3+}(g)+\frac{3}{2}O_2(g)), ΔH3\Delta H_3 (from 2Al3+(g)+32O2(g)2Al^{3+}(g)+\frac{3}{2}O_2(g) to 2Al3+(g)+3O(g)2Al^{3+}(g)+3O(g)), and ΔH5\Delta H_5 (from 2Al3+(g)+3O−(g)2Al^{3+}(g)+3O^-(g) to 2Al3+(g)+3O2−(g)2Al^{3+}(g)+3O^{2-}(g)). Which row correctly identifies all three?
  1. AΔH2 = lattice enthalpy of Al2O3; ΔH3 = ionisation energies of aluminium; ΔH5 = atomisation of oxygen
  2. BΔH2 = ionisation energies of aluminium; ΔH3 = second electron affinity of oxygen; ΔH5 = atomisation of oxygen
  3. CΔH2 = ionisation energies of aluminium; ΔH3 = atomisation of oxygen; ΔH5 = second electron affinity of oxygen
  4. DΔH2 = atomisation of oxygen; ΔH3 = ionisation energies of aluminium; ΔH5 = first electron affinity of oxygen

Question 106

[1 marks]thermochemistry / Born-Haber cycle
Standard conditions, as used when defining a standard enthalpy change, are a temperature of ___ K (25 degrees C) and a pressure of 1 atmosphere.

Answer this when you sit the paper.

Question 107

[2 marks]thermochemistry / Born-Haber cycle
By Hess's Law applied to Fig. 1.1, the enthalpy change of formation, ΔHf\Delta H_f, of Al2O3(s)Al_2O_3(s), going the long way round the cycle, is equal to
  1. AΔH6 minus the sum of ΔH1, ΔH2, ΔH3, ΔH4 and ΔH5
  2. BΔH2 + ΔH3 + ΔH5 only, since ΔH1, ΔH4 and ΔH6 were already identified separately
  3. CΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5 + ΔH6
  4. DΔH1 + ΔH4 + ΔH6 only, since these are the only three enthalpy changes given a numerical value

Question 201

[1 marks]transition elements / catalysis
A transition element is best defined as an element which
  1. Aforms at least one stable ion with a partially filled d subshell
  2. Blies in the d block of the Periodic Table
  3. Cforms coloured compounds
  4. Dhas a completely filled d subshell

Question 202

[1 marks]transition elements / catalysis
The electronic configuration of the iron(II) ion is
  1. A[Ar]3d6[Ar]3d^6
  2. B[Ar]3d64s2[Ar]3d^64s^2
  3. C[Ar]3d44s2[Ar]3d^44s^2
  4. D[Ar]3d54s1[Ar]3d^54s^1

Question 203

[1 marks]transition elements / catalysis
In the iron-catalysed reaction between peroxodisulphate ions and iodide ions, the equation for the step which restores the pale green colour is
  1. A2Fe2++S2O82−→2Fe3++2SO42−2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}
  2. B2Fe3++S2O82−→2Fe2++2SO42−2Fe^{3+} + S_2O_8^{2-} \rightarrow 2Fe^{2+} + 2SO_4^{2-}
  3. C2Fe3++2I−→2Fe2++I22Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2
  4. D2Fe2++I2→2Fe3++2I−2Fe^{2+} + I_2 \rightarrow 2Fe^{3+} + 2I^-

Question 204

[2 marks]transition elements / catalysis
Which pair of properties below is characteristic of transition elements, such as iron, but not of s-block elements, such as sodium or calcium?
  1. Alow melting point and high volatility
  2. Bhigh reactivity with cold water and strongly basic oxides
  3. Cvery large atomic radii and very low ionisation energies
  4. Dvariable oxidation states and the formation of coloured ions/complexes

Question 205

[2 marks]transition elements / catalysis
In the reaction between peroxodisulphate and iodide ions catalysed by iron, which row correctly matches each observed colour to the iron species responsible?
  1. Apale green = Fe2+Fe^{2+}; yellow-brown = Cr3+Cr^{3+}
  2. Bpale green = Fe2+Fe^{2+}; yellow-brown = Fe3+Fe^{3+}
  3. Cpale green = SO42−SO_4^{2-}; yellow-brown = Fe3+Fe^{3+}
  4. Dpale green = Fe3+Fe^{3+}; yellow-brown = Fe2+Fe^{2+}

Question 206

[1 marks]transition elements / catalysis
State the general property of transition metal ions, illustrated by iron in this reaction, that allows them to act as catalysts.

Answer this when you sit the paper.

Question 301

[1 marks]electrolysis / Avogadro constant
A current of 2.0 A is passed through aqueous silver nitrate for 180 minutes. The quantity of charge passed is
  1. A21 600 C
  2. B43 200 C
  3. C360 C
  4. D6 000 C

Question 302

[1 marks]electrolysis / Avogadro constant
During the electrolysis of aqueous silver nitrate between inert electrodes, the reaction at the anode is
  1. AAg++e−→AgAg^+ + e^- \rightarrow Ag
  2. B2H++2e−→H22H^+ + 2e^- \rightarrow H_2
  3. C2NO3−→2NO2+O2+2e−2NO_3^- \rightarrow 2NO_2 + O_2 + 2e^-
  4. D2H2O→O2+4H++4e−2H_2O \rightarrow O_2 + 4H^+ + 4e^-

Question 303

[1 marks]electrolysis / Avogadro constant
Passing 21 600 C through aqueous silver nitrate deposits 24.17 g of silver (Ar=108A_r = 108). Taking the electronic charge as 1.6×10−19 C1.6 \times 10^{-19}\ C, the Avogadro constant, in mol−1mol^{-1}, is
  1. A9.65×10239.65 \times 10^{23}
  2. B1.35×10241.35 \times 10^{24}
  3. C3.01×10233.01 \times 10^{23}
  4. D6.03×10236.03 \times 10^{23}

Question 401

[1 marks]organic chemistry / hydrolysis
Aspirin is 2-ethanoyloxybenzoic acid. The functional group in aspirin which is readily hydrolysed by refluxing with dilute hydrochloric acid is the
  1. Aketone
  2. Bphenol
  3. Camide
  4. Dester

Question 402

[1 marks]organic chemistry / hydrolysis
Aspirin liberates carbon dioxide from aqueous sodium carbonate because it contains
  1. Aan alcohol group
  2. Ba carboxylic acid group
  3. Ca phenol group
  4. Dan ester group

Question 403

[1 marks]organic chemistry / hydrolysis
The calcium salt of aspirin is far more soluble in water than aspirin itself because
  1. Aaspirin is a gas at room temperature
  2. Bthe salt is hydrolysed completely in water
  3. Cthe salt is ionic and is readily hydrated by water
  4. Dthe salt has a much lower molar mass

Question 404

[1 marks]organic chemistry / hydrolysis
Hydrolysis of X (C6H4(OH)(CONHCH2CH3)C_6H_4(OH)(CONHCH_2CH_3)) with dilute HCl under reflux gives 2-hydroxybenzoic acid and ___.

Answer this when you sit the paper.

Question 405

[2 marks]organic chemistry / hydrolysis
Write a balanced equation for the reaction between aspirin, C9H8O4C_9H_8O_4, and aqueous sodium carbonate, which releases carbon dioxide.

Answer this when you sit the paper.

Question 406

[1 marks]organic chemistry / hydrolysis
In compound X, which contains a −CONHCH2CH3-CONHCH_2CH_3 group and a phenolic −OH-OH group, the functional group that is hydrolysed on reflux with dilute hydrochloric acid is the
  1. Aketone
  2. Bphenol
  3. Camide
  4. Dester

Question 501

[1 marks]organic chemistry / amino acids and polymers
Tyrosine and glycine react together to form a polymer. The linkage joining the monomer units is the
  1. Aester
  2. Bether
  3. Cglycosidic
  4. Damide

Question 502

[1 marks]organic chemistry / amino acids and polymers
The polymerisation of amino acids to form proteins is
  1. Afree radical
  2. Bionic
  3. Caddition
  4. Dcondensation

Question 503

[1 marks]organic chemistry / amino acids and polymers
In the pure solid state glycine exists as the dipolar ion
  1. AH2NCH2COO−H_2NCH_2COO^-
  2. BH2NCH2CO2HH_2NCH_2CO_2H
  3. CH3N+CH2COO−H_3N^+CH_2COO^-
  4. DH3N+CH2COOHH_3N^+CH_2COOH

Question 504

[1 marks]organic chemistry / amino acids and polymers
The polymer formed by the condensation of many amino acid molecules, such as tyrosine and glycine, joined by amide (peptide) links is classified as a
  1. Apolyamide (polypeptide/protein)
  2. Bpolyalkene
  3. Cpolyether
  4. Dpolyester

Question 505

[2 marks]organic chemistry / amino acids and polymers
Of tyrosine and glycine, the compound that shows an isomerism not found in the other, and the type of isomerism involved, is
  1. Aglycine; optical isomerism
  2. Btyrosine; optical isomerism
  3. Cglycine; structural isomerism
  4. Dtyrosine; cis-trans isomerism

Question 506

[2 marks]organic chemistry / amino acids and polymers
In alkaline solution, the dipolar (zwitterion) form of glycine, H3N+CH2COO−H_3N^+CH_2COO^-, loses a proton from its −NH3+-NH_3^+ group to give the ion

Answer this when you sit the paper.

Question 507

[2 marks]organic chemistry / amino acids and polymers
Tyrosine shows optical isomerism but glycine does not because
  1. Atyrosine's alpha carbon is bonded to four different groups, while glycine's alpha carbon is bonded to two identical hydrogen atoms
  2. Btyrosine has a C=C double bond in its side chain, allowing cis-trans isomerism instead
  3. Ctyrosine contains a benzene ring, and only ring-containing molecules can be chiral
  4. Dglycine is a much smaller molecule and so cannot rotate plane-polarised light

More sittings of this paper

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.

Get it on Google PlayDownload on the App Store