Questions 1
(a)Topic-26.1 Simple rate equations, orders of reaction and rate constants
(b)Topic-8.3 Homogeneous and heterogeneous catalysts
(c)Topic-6.1 Redox processes: electron transfer and changes in oxidation number (oxidation state)
(a) The equation for reaction 1 is shown.
reaction 1 \(X \to 2Y\)
Reaction 1 is first order with respect to the concentration of X. The half-life of the reaction, t12, is 900 s at 20°C.
(i) A solution of X with a concentration of 0.180 moldm⁻³ is prepared at 20°C. Calculate the average rate of reaction 1 over the first 1800s.
(ii) Complete the rate equation for reaction 1.
(iii) Show that the rate constant, k, is \(7.70 × 10^{–4} s^{–1}\) at 20°C.
(iv) Calculate the initial rate of reaction 1 when the concentration of X is 0.150 moldm⁻³. Include units.
(b) Catalysts may be homogeneous or heterogeneous.
(i) Platinum is a transition element. Explain why transition elements behave as catalysts.
(ii) Name the metal catalyst in the Haber process and explain why it is a heterogeneous catalyst.
(iii) Platinum acts as a heterogeneous catalyst in the removal of nitrogen dioxide, NO₂, from the exhaust gases of car engines. Describe the mode of action of a platinum catalyst in this process.
(iv) NO₂ acts as a homogeneous catalyst in the oxidation of atmospheric sulfur dioxide, SO₂. Write equations for the two reactions that occur.
(c) SO₂ dissolves in water, forming H₂SO₃. H₂SO₃ can be oxidised under acidic conditions. The relevant electrode reaction and its \(E^o\) value are shown.
Four more half-equations for reactions occurring under acidic conditions, and their \(E^o\) values, are shown. 
Select the oxidising agent that could oxidise H₂SO₃ to SO₄²⁻ ions under acidic conditions. Write an equation, and give the \(E^o\) cell value, for the reaction that occurs.
▶️Answer/Explanation
Ans
(a)(i) Change in concentration = 0.135 [1]
0.135 / 1800 = 7.5 × 10⁻⁵
(ii) Rate = k[X]
(iii) k = 0.693 / t½ / k = 0.693 / 900 / k = ln2 / t½ / k = ln2 / 900 / t½ = ln2 / k
(iv) 1.16 × 10⁻⁴
mol dm⁻³ s⁻¹
(b)(i) Any two from:
• variable oxidation state
• vacant / empty / unfilled d orbitals
• can form dative bonds / can accept electrons
(ii) Iron AND
iron is solid, reactants are gases OR
catalyst and reactants are in different phases / states
(iii) Two for one mark, three for two marks:
• adsorption of reactants by Pt
• bonds of reactants weaken
• desorption of products
(iv) \(NO_2 + SO_2 \to NO + SO_3\) AND
\(2NO + O_2 \to 2NO_2\)
(c) \(BiO^+\)
3H₂SO₃ + 2BiO⁺ + H₂O → 3SO₄²⁻ + 8H⁺ + 2Bi
0.11 V
Questions 2
(a)Topic-23.2 Enthalpies of solution and hydration
(b)Topic-5.2 Hess’s law
(c)Topic-23.3 Entropy change, ΔS
(d) Topic-23.3 Entropy change, ΔS
(e)Topic-23.3 Entropy change, ΔS
(a) Predict and explain the variation in enthalpy change of hydration for the ions \(Na^+\), \(Mg^{2+}\) and \(Al^{3+}\).
(b) Fig. 2.1 shows an incomplete energy cycle.
(i) Complete line C on Fig. 2.1. Include state symbols.
(ii) Use both words and symbols to identify change 2 on Fig. 2.1. Use changes 1 and 3 as examples of how this should be done.
(iii) Calculate a value for the lattice energy of magnesium chloride, \(\Delta H_{latt}\) MgCl₂(s), by selecting and using appropriate data from Table 2.1.
(c) Define entropy.
(d) At 25°C the enthalpy change of solution of compound Z is +26 kJ mol⁻¹. The entropy change of solution of Z at the same temperature is \(+52JK^{–1}mol^{–1}\). Calculate the value of the Gibbs free energy change, ΔG, for the solution of Z at 25°C.
(e) (i) Use your answer to (d) to predict whether or not Z is soluble in water at 25°C. Explain your answer.
(ii) Predict whether Z becomes more or less soluble as the water is heated from 25°C to 95°C. Explain your answer.
▶️Answer/Explanation
Ans
(a) \(\Delta H_{hyd}\) increases from left to right due to increase in charge
ionic radius decreases from left to right
causing increased attractive force to water molecules
(b)(i) \(Mg^{2+} (aq) + 2Cl^–(aq)\)
(ii) Enthalpy change of hydration of magnesium ions and chloride ions
\(\Delta H_{hyd} Mg^{2+} + 2\Delta H_{hyd}Cl^–\)
(iii) Selects 155, 1920 and 364 only
2 × 364
answer –2493
(c) The number of arrangements of the particles and of the energy in the system
(d) ΔG = ΔH – TΔS
answer +10.5
(e)(i) No, ΔG is positive
(ii) Becomes more / soluble because ΔG becomes more negative / less positive / smaller / closer to zero
Questions 3
(a)Topic-7.2 Brønsted–Lowry theory of acids and bases
(b)Topic-7.2 Brønsted–Lowry theory of acids and bases
(a) The pH of a saturated solution of calcium hydroxide is 12.35 at 298K.
(i) Show that the concentration of hydroxide ions in a saturated solution of calcium hydroxide is 0.0224 moldm⁻³ at 298K.
(ii) Use data given in (i) to calculate the solubility product, \(K_{sp}\), of calcium hydroxide at 298K. Include the units of \(K_{sp}\) in your answer.
(iii) A spatula measure of solid calcium chloride is stirred into a sample of saturated calcium hydroxide solution. All of the calcium chloride dissolves. Describe one other observation that would be made and give an estimated value of the pH of the solution obtained. Explain both your answers.
observation …………………………………………………………………………………………………………
pH of solution ……………………………
explanation …………………………………………………………………………………………………………
(iv) Calcium hydroxide reacts with dilute sulfuric acid to form calcium sulfate. Barium hydroxide behaves in a similar way, forming barium sulfate. Explain why calcium sulfate is more soluble in water than barium sulfate.
(b) Some solid calcium is added to an excess of aqueous ethanoic acid, CH₃COOH, and left until all the calcium has reacted. The resulting mixture, mixture D, contains no undissolved solids.
(i) Write an equation for the reaction of calcium with CH₃COOH.
(ii) Use formulae of molecules and ions to identify two conjugate acid–base pairs present in mixture D.
Pair 1 should consist of organic species.
(iii) Write the expression for the \(K_a\) of CH₃COOH.
(iv) The concentration of calcium ethanoate, (CH₃COO)₂Ca, in mixture D is 0.394 moldm⁻³. The concentration of CH₃COOH in mixture D is 0.270 moldm⁻³. The \(K_a\) of CH₃COOH is 1.74 × 10⁻⁵ moldm⁻³ at 298K. Calculate the pH of mixture D.
(v) Write two equations to show how mixture D can act as a buffer solution.
equation 1 ………………………………………………………………………………………………………….
equation 2 ………………………………………………………………………………………………………….
▶️Answer/Explanation
Ans
(a)(i) \([H^+] = 10^{–12.35}\) OR \([H^+] = 4.47 \times 10^{–13}\)
\(K_w / 4.47 \times 10^{–13}\)
OR
pOH = 1.65
[OH⁻] = 10⁻¹.⁶⁵
(ii) \(K_{sp} = (0.0112)(0.0224)^2\) OR \(K_{sp} = [Ca^{2+}][OH^{–}]^2\)
answer 5.62 × 10⁻⁶
\(mol^3 dm^{–9}\)
(iii) white solid / white ppt AND pH between 7.01 and 12.34
common ion effect
hydroxide removed by precipitation OR ppt is Ca(OH)₂
(iv) lattice energy and hydration energy greater for CaSO₄ OR
lattice energy and hydration energy decrease down group
hydration energy decreases more / is dominant factor
enthalpy of solution is more endothermic for BaSO₄ OR is more endothermic down grp
(b)(i) Ca + 2CH₃COOH → Ca(CH₃COO)₂ + H₂
(ii) Pair 1: CH₃COOH and CH₃COO⁻
Pair 2: \(H_3O^+\) and \(H_2O\) OR \(H_2O\) and \(OH^–\)
(iii) \(\frac{[H^+][CH_3COO^-]}{[CH_3COOH]}\)
(iv) [CH₃COO⁻] = 0.788, [CH₃COOH] = 0.270
[\(H^+\)] = 5.96 × 10⁻⁶
pH = 5.22
(v) \(CH_3COO^– + H^+ \to CH_3COOH\)
OR \(Ca(CH_3COO)_2 + 2H^+ → 2CH_3COOH + Ca^{2+}\)
\(CH_3COOH + OH^– \to CH_3COO^– + H_2O\)
Questions 4
(a)Topic- 28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(b)Topic-6.1 Redox processes: electron transfer and changes in oxidation number (oxidation state)
(c)Topic-6.1 Redox processes: electron transfer and changes in oxidation number (oxidation state)
Transition metal atoms and transition metal ions form complexes by combining with species called ligands.
(a) When NaOH(aq) is added to an aqueous solution containing \([Co(H_2O)^6]^{2+}\) a precipitation reaction occurs accompanied by a colour change. In this reaction, two of the water ligands each lose one H+ ion. The H⁺ ions are gained by OH⁻ ions from the NaOH (aq).
(i) State the colour change seen in this precipitation reaction.
from ……………………………………………………. to ……………………………………………………
(ii) Complete the ionic equation for this precipitation reaction
(iii) This precipitation reaction can also be described as a different type of reaction. Name this type of reaction.
(b) L is an uncharged tridentate ligand. L donates three lone pairs to a metal atom or ion. Cobalt forms an octahedral complex ion, E, with L. Complex ion E has a 2+ charge.
(i) Give the formula of E.
(ii) Identify the oxidation state of cobalt in E.
(iii) The d-orbitals of the cobalt atom or ion present in E are split in energy. State the number of d-orbitals that are at a higher energy level and the number of d-orbitals that are at a lower energy level.
(iv) Define the term non-degenerate d-orbitals.
(c) The mineral chromite contains a compound which has the formula \(FeCrnO_4\). The oxidation state of iron in \(FeCrnO_4\) is +2. A sample of 4.18g of FeCrnO₄ is dissolved in an excess of sulfuric acid. The resulting solution is made up to 250 cm³. This is solution F. All the Fe²⁺ ions in 25.0 cm³ of solution F are oxidised to Fe³⁺ ions by exactly 18.7 cm³ of 0.0200 mol/dm³ KMnO₄. One \(MnO_4^–\) ion reacts with five Fe\(^{2+}\) ions. Assume no other oxidation reaction occurs.
(i) Write an equation for the reaction of Fe\(^{2+}\) ions with \(MnO_4^–\) ions in acid solution.
(ii) Calculate the number of moles of Fe\(^{2+}\) ions in 25.0 cm³ of solution F
(iii) Calculate the Mr of \(FeCrnO_4\) and use your answer to deduce the value of n.
▶️Answer/Explanation
Ans
(a)(i) red/pink blue
(ii) \([Co(H_2O)_6]^{2+} + 2OH^– \to Co(H_2O)_4(OH)_2 + 2H_2O\)
(iii) acid-base
(b)(i) \([CoL_2]^{2+}\)
(ii) +2
(iii) 2
3
(iv) Not of the same energy / have different energy
(c)(i) \(5Fe^{2+} + MnO_4^– + 8H^+ \to 5Fe^{3+} + Mn^{2+} + 4H_2O\)
(ii) 0.02 × 18.7/1000 = 3.74 × 10⁻⁴
3.74 × 10⁻⁴ × 5 = 1.87 × 10⁻³
(iii) 1.87 × 10⁻³ × 10 = 1.87 × 10⁻²
4.18/1.87 × 10⁻² = 224 / 223.5(294)
n = 2
Questions 5
(a)Topic- 28.1 General physical and chemical properties of the first row of transition elements, titanium to copper
(b)Topic-6.1 Redox processes: electron transfer and changes in oxidation number (oxidation state)
\(Ni^{2+}\) ions form a number of different complex ions, including \([Ni(H_2O)_6]^{2+}\), \([Ni(NH_3)_6]^{2+}\) and \([Ni(en)_3]^{2+}\). The abbreviation en represents 1,2-diaminoethane. The numerical values of two stability constants, \(K_{stab}\), are given in Table 5.1.
(a) Complete the expression for the \(K_{stab}\) of [Ni(en)₃]²⁺.
(b) A solution of [Ni(H₂O)₆]²⁺ is added to a solution that contains 0.10 mol/dm³ NH₃ and 0.10 mol/dm³ en.
(i) Predict which complex ion, \([Ni(NH_3)_6 ]^{2+}\) or \([Ni(en)_3]^{2+}\), is present in the resulting mixture in the highest concentration. Explain your answer.
complex ion present in largest concentration = ……………………………………..
explanation ………………………………………………………………………………………
(ii) Complete the equation for the ligand exchange reaction occurring in (i).
Complete Fig. 5.1 to show the three-dimensional structures of the two isomers of [Ni(en)3]2+.
Use 
to represent the en ligand. Name the type of isomerism shown.
type of isomerism shown …………………………………………………………………………………………….
▶️Answer/Explanation
Ans
(a) \(K_{stab}=\frac{[[Ni(en)_3 ]^{2+} ]}{[Ni^{2+}][en]^3}\) OR \(\frac{[[Ni(en)_3 ]^{2+} ]}{[Ni(H_2O)_6^{2+} ][en]^3}\)
(b)(i) \([Ni(en)_3]^{2+}\) AND
larger \(K_{stab}\) OR more stable
(ii) \([Ni(H_2O)_6]^{2+} + 3en → [Nien_3]^{2+} + 6H_2O\)
Questions 6
(a)Topic- 21.1 Organic synthesis
(b)Topic- 21.1 Organic synthesis
(c)Topic- 21.1 Organic synthesis
(d)Topic- 21.1 Organic synthesis
Fig. 6.1 shows two reactions of ethanedioic acid, HOOCCOOH.
(a) (i) Draw the organic product G in the box in Fig. 6.1.
(ii) In Fig. 6.1, SOCl₂ is given as the reagent that reacts with HOOCCOOH to produce G. Identify a different reagent that also reacts with HOOCCOOH to produce G.
(b) Identify two different reagents that oxidise HOOCCOOH to form carbon dioxide and water.
(c) HOOCCOOH ionises as shown.
\(HOOCCOOH \rightleftharpoons HOOCCOO^– + H^+\)
HOOCCOOH is a much stronger acid than methanoic acid, HCOOH. Suggest an explanation for this difference in acidity.
(d) Benzene-1,4-dicarboxylic acid, \(HOOCC_6H_4COOH\), can be made from benzene, \(C_6H_6\), in two steps as shown in Fig. 6.2.
(i) Suggest the identity of J by drawing its structure in the box in Fig. 6.2.
(ii) Identify the reagents and conditions for step 1 and step 2.
(iii) Draw the structure of exactly one repeat unit of the polymer formed when benzene-1,4-dicarboxylic acid reacts with ethane-1,2-diol, HOCH₂CH₂OH. The linkage formed between the monomers should be shown fully displayed.
(iv) State the type of polymerisation that occurs when benzene-1,4-dicarboxylic acid reacts with ethane-1,2-diol and name the linkage formed between the monomers.
type of polymerisation …………………………………….
linkage …………………………………………………………..
▶️Answer/Explanation
Ans
(a)(i) ClOCCOCl
(ii) PCl₃ / PCl₅
(b) oxygen
acidified KMnO₄
(c) C=O are electron withdrawing / electronegative
weakens O–H bond OR stabilises anion
(ii) step 1: CH₃Cl + AlCl₃
step 2: hot alkaline KMnO₄
(iv) condensation AND ester
Questions 7
(a)Topic- 29.2 Characteristic organic reactions
(b)Topic- 21.1 Organic synthesis
(c)Topic- 21.1 Organic synthesis
(d)Topic- 21.1 Organic synthesis
(e)Topic- 21.1 Organic synthesis
(f)Topic- 21.1 Organic synthesis
Benzene reacts with chlorine gas to form chlorobenzene. This reaction can be described as the reaction between benzene molecules and Cl⁺ ions. The Cl⁺ ions are formed by adding a suitable catalyst to the chlorine gas.
(a) Give the name or formula of a catalyst that can be used for this reaction.
(b) The mechanism for this reaction is shown.
(i) The movement of a pair of electrons is represented by x in diagram 1.
• State where this pair of electrons is before step 1 takes place.
• State where this pair of electrons is after step 1 has taken place.
(ii) The movement of another pair of electrons is represented by y in diagram 2.
• State where this pair of electrons is before step 2 takes place.
• State where this pair of electrons is after step 2 has taken place.
There are six carbon atoms in diagram 2.
State how many of these carbon atoms are sp hybridised, \(sp^2\) hybridised, and \(sp^3\) hybridised.
sp hybridised ……………………………
\(sp^2\) hybridised ……………………………
\(sp^3\) hybridised ……………………………
(d) Complete the equation for this reaction between benzene and chlorine.
(e) The mechanism for this reaction is electrophilic substitution. Complete the following sentence. Write formulae in the gaps provided. During this reaction, the electrophile is …………………………… and a …………………………… atom in benzene is substituted by a …………………………… atom.
(f) Chloroethane reacts with NaOH(aq). Chlorobenzene does not.
(i) Name the mechanism of the reaction that chloroethane undergoes with NaOH(aq), and identify the major organic product that is formed.
mechanism …………………………………………………………………………………………………………
major organic product ……………………………………………………………………
(ii) Explain the difference in reactivity of chloroethane and chlorobenzene when treated with NaOH(aq).
▶️Answer/Explanation
Ans
(a) aluminium chloride OR AlCl₃
b)(i) delocalised system / delocalised ring / pi system / pi ring
C–Cl bond
(ii) C–H bond
delocalised system / delocalised ring / pi system / pi ring
(c) 0, 5, 1
(d) \(C_6H_6 + Cl_2 \to C_6H_5Cl + HCl\)
(e) \(Cl^+\)
hydrogen / H
chlorine / Cl
(f)(i) nucleophilic substitution AND ethanol
(ii) delocalisation of LP of Cl with π system
C–Cl bond is stronger in chlorobenzene / partly double
Questions 8
(a)Topic-34.4 Amino acids
(b)Topic- 21.1 Organic synthesis
(c)Topic- 21.1 Organic synthesis
(d)Topic- 21.1 Organic synthesis
(e)Topic- 21.1 Organic synthesis
(f)Topic- 21.1 Organic synthesis
(g)Topic- 21.1 Organic synthesis
(h)Topic- 21.1 Organic synthesis
The amino acid serine, \(HOCH_2CH(NH_2)COOH\), exists in two optically active forms. These optical isomers, isomer P and isomer Q, are shown in Fig. 8.1.
(a) Isomer P and isomer Q have identical physical and chemical properties, with the exception of two specific properties. One of these two properties is their differing effect on plane polarised light. State the other property by which they differ.
(b) A solution of pure isomer P of a particular concentration rotates plane polarised light by 5.0° in a clockwise direction. Describe how a solution of pure isomer Q of the same concentration affects plane polarised light.
(c) State another term, in addition to stereoisomers, optical isomers and non-superimposable mirror images, which can be used to describe this pair of chiral compounds, isomer P and
isomer Q.
(d) Give the term used to describe a mixture containing equal amounts of isomer P and isomer Q.
(e) Describe one way in which a single pure optical isomer of serine can be produced, instead of making a mixture of isomer P and isomer Q.
(f) Complete Table 8.1 to describe the peaks seen in the proton \((^1H)\) NMR spectrum of HOCH₂CH(NH₂)COOH dissolved in D₂O. Use as many rows in Table 8.1 as you need to, leaving the other rows blank
(g) Proline is a naturally occurring amino acid. The skeletal formula of proline is shown. 
State the number of peaks in the carbon-13 \((^{13}C)\) NMR spectrum of proline.
(h) Glutamic acid is a naturally occurring amino acid. The skeletal formula of glutamic acid is shown.
The isoelectric point of glutamic acid is pH 3. A sample of glutamic acid is dissolved in a solution of pH 1. A strong alkali is then added until the pH of the mixture reaches pH 14. During this process all possible ionised forms of glutamic acid are present at different times, depending on the pH of the solution. Complete the boxes below to show four different ionised forms of glutamic acid that are present at the stated pH values. 
▶️Answer/Explanation
Ans
(a) biological activity
(b) rotates plane polarised light 5.0° in anticlockwise direction
(c) enantiomers
(d) racemic
(e) use of chiral catalyst
(f) CH₂ and CH only in column one
CH₂ gives a doublet, CH gives a triplet
doublet due to 1 proton on neighbouring carbons
triplet due to 2 protons on neighbouring carbons
(g) 5
