Cambridge IAS-Level · Thinka-original Practice Paper

2025 Cambridge IAS-Level Chemistry (9701) Practice Paper with Answers

Thinka Nov 2025 (V4) Cambridge International A Level-Style Mock — Chemistry (9701)

140 marks270 mins2025
AnalysisSample paper
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V4) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.

Paper 1 (Multiple Choice)

Answer all 40 questions by choosing the correct option A, B, C or D.
40 Question · 40 marks
Question 1 · multiple_choice
1 marks
The standard enthalpy changes of combustion, \(\Delta H_c^{\ominus}\), for cyclopropane\((g)\) and propene\((g)\) are given below:

* cyclopropane\((g)\): \(-2091\text{ kJ mol}^{-1}\)
* propene\((g)\): \(-2058\text{ kJ mol}^{-1}\)

What is the standard enthalpy change, \(\Delta H^{\ominus}\), for the isomerisation of cyclopropane to propene?

\(\text{cyclopropane}(g) \rightarrow \text{propene}(g)\)
  1. A.\(-33\text{ kJ mol}^{-1}\)
  2. B.\(+33\text{ kJ mol}^{-1}\)
  3. C.\(-4149\text{ kJ mol}^{-1}\)
  4. D.\(+4149\text{ kJ mol}^{-1}\)
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Worked solution

By Hess's law, we can construct an enthalpy cycle using the enthalpy changes of combustion.

\(\Delta H^{\ominus} = \sum \Delta H_c^{\ominus}(\text{reactants}) - \sum \Delta H_c^{\ominus}(\text{products})\)

\(\Delta H^{\ominus} = \Delta H_c^{\ominus}(\text{cyclopropane}(g)) - \Delta H_c^{\ominus}(\text{propene}(g))\)

\(\Delta H^{\ominus} = -2091 - (-2058) = -33\text{ kJ mol}^{-1}\)

Marking scheme

Award 1 mark for the correct calculation of \(-33\text{ kJ mol}^{-1}\) (Option A).
Question 2 · multiple_choice
1 marks
When solid sodium bromide is heated with concentrated sulfuric acid, a gas is evolved that turns acidified potassium dichromate(VI) paper from orange to green.

What is the oxidation state of the sulfur atom in this gas?
  1. A.\(-2\)
  2. B.\(0\)
  3. C.\(+4\)
  4. D.\(+6\)
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Worked solution

When solid sodium bromide is heated with concentrated sulfuric acid, the bromide ions initially form hydrogen bromide, but because bromide is a moderately strong reducing agent, it reduces the sulfuric acid (where sulfur is in the \(+6\) oxidation state) to sulfur dioxide (\(\text{SO}_2\)). Sulfur dioxide is a reducing gas that reduces the orange dichromate(VI) ions to green chromium(III) ions. The oxidation state of sulfur in \(\text{SO}_2\) is \(+4\).

Marking scheme

Award 1 mark for identifying the oxidation state of sulfur in sulfur dioxide as \(+4\) (Option C).
Question 3 · multiple_choice
1 marks
A sample of \(20\text{ cm}^3\) of a hydrocarbon gas, \(\text{C}_x\text{H}_y\), is completely burned in \(150\text{ cm}^3\) of oxygen (an excess). After cooling to room temperature, the total volume of gas remaining is \(110\text{ cm}^3\).

This remaining gas mixture is passed through aqueous sodium hydroxide, and the total gas volume decreases to \(50\text{ cm}^3\).

What is the molecular formula of the hydrocarbon?
  1. A.\(\text{C}_3\text{H}_4\)
  2. B.\(\text{C}_3\text{H}_6\)
  3. C.\(\text{C}_3\text{H}_8\)
  4. D.\(\text{C}_4\text{H}_{10}\)
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Worked solution

1. The decrease in volume when passed through aqueous sodium hydroxide is due to the absorption of carbon dioxide: \(V(\text{CO}_2) = 110 - 50 = 60\text{ cm}^3\).
2. Carbon atoms per molecule, \(x = \frac{60}{20} = 3\).
3. The remaining \(50\text{ cm}^3\) of gas is the unreacted excess oxygen. Therefore, the volume of oxygen reacted is \(150 - 50 = 100\text{ cm}^3\).
4. The stoichiometry of combustion is given by: \(\text{C}_3\text{H}_y + (3 + \frac{y}{4})\text{O}_2 \rightarrow 3\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}(l)\).
5. The ratio of volumes is: \(\frac{V(\text{O}_2)}{V(\text{C}_3\text{H}_y)} = \frac{100}{20} = 5 = 3 + \frac{y}{4}\).
6. Solving for \(y\): \(\frac{y}{4} = 2 \implies y = 8\). Thus, the formula is \(\text{C}_3\text{H}_8\).

Marking scheme

Award 1 mark for calculating the formula of the hydrocarbon as \(\text{C}_3\text{H}_8\) (Option C).
Question 4 · multiple_choice
1 marks
An element \(X\) in Period 3 forms a chloride that is a liquid at room temperature and reacts violently with water to form a solution with pH 2. Element \(X\) also forms an oxide that is insoluble in water but reacts with hot concentrated aqueous sodium hydroxide.

What is element \(X\)?
  1. A.Aluminium
  2. B.Silicon
  3. C.Phosphorus
  4. D.Sulfur
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Worked solution

1. Silicon tetrachloride, \(\text{SiCl}_4\), is a liquid at room temperature and undergoes violent hydrolysis in water to yield \(\text{SiO}_2\) and white fumes of \(\text{HCl}\) gas, resulting in a strongly acidic solution (pH \(\approx 2\)).
2. Silicon dioxide, \(\text{SiO}_2\), is a giant covalent oxide that is insoluble in water but shows weakly acidic character by dissolving in hot concentrated aqueous sodium hydroxide to form sodium silicate, \(\text{Na}_2\text{SiO}_3\).

Marking scheme

Award 1 mark for identifying element \(X\) as Silicon (Option B).
Question 5 · multiple_choice
1 marks
An organic compound \(Y\) with molecular formula \(\text{C}_4\text{H}_9\text{Br}\) reacts with hot aqueous sodium hydroxide to produce an alcohol that cannot be oxidized by acidified potassium dichromate(VI).

What is the structural formula of \(Y\)?
  1. A.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}\)
  2. B.\((\text{CH}_3)_2\text{CHCH}_2\text{Br}\)
  3. C.\(\text{CH}_3\text{CH}_2\text{CH(Br)CH}_3\)
  4. D.\((\text{CH}_3)_3\text{CBr}\)
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Worked solution

1. Reaction with hot aqueous \(\text{NaOH}\) is a nucleophilic substitution that converts the halogenoalkane to an alcohol.
2. Since the resulting alcohol cannot be oxidized by acidified potassium dichromate(VI), it must be a tertiary alcohol.
3. The only tertiary alcohol with four carbon atoms is 2-methylpropan-2-ol, \((\text{CH}_3)_3\text{COH}\).
4. Therefore, the starting halogenoalkane \(Y\) must be 2-bromo-2-methylpropane, \((\text{CH}_3)_3\text{CBr}\).

Marking scheme

Award 1 mark for identifying \((\text{CH}_3)_3\text{CBr}\) as the correct structure (Option D).
Question 6 · multiple_choice
1 marks
How many stereoisomers exist for 4-chlorohex-2-ene?
  1. A.2
  2. B.3
  3. C.4
  4. D.8
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Worked solution

1. First, analyze the structure of 4-chlorohex-2-ene: \(\text{CH}_3-\text{CH}=\text{CH}-\text{CH(Cl)}-\text{CH}_2-\text{CH}_3\).
2. The double bond between C2 and C3 can exhibit cis-trans (E/Z) isomerism because C2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\) (two different groups), and C3 is bonded to \(-\text{H}\) and \(-\text{CH(Cl)CH}_2\text{CH}_3\) (two different groups).
3. C4 is a chiral centre because it is bonded to four different groups: \(-\text{H}\), \(-\text{Cl}\), \(-\text{CH}_2\text{CH}_3\), and \(-\text{CH}=\text{CHCH}_3\).
4. Since the molecule is unsymmetrical, the stereocentres act independently, giving \(2^n = 2^2 = 4\) unique stereoisomers: (E, R), (E, S), (Z, R), and (Z, S).

Marking scheme

Award 1 mark for concluding there are 4 stereoisomers (Option C).
Question 7 · multiple_choice
1 marks
Nitrogen dioxide, \(\text{NO}_2\), dimerises to form dinitrogen tetroxide, \(\text{N}_2\text{O}_4\), according to the following equilibrium:

\(2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)\)

At a certain temperature, 2.00 mol of \(\text{NO}_2\) is placed in a closed vessel. When equilibrium is established, the vessel contains 0.60 mol of \(\text{N}_2\text{O}_4\) and the total pressure in the vessel is \(1.50 \times 10^5\text{ Pa}\).

What is the value of the equilibrium constant, \(K_p\), at this temperature?
  1. A.\(8.75 \times 10^{-6}\text{ Pa}^{-1}\)
  2. B.\(1.14 \times 10^5\text{ Pa}^{-1}\)
  3. C.\(4.08 \times 10^{-6}\text{ Pa}^{-1}\)
  4. D.\(7.50 \times 10^{-1}\text{ Pa}^{-1}\)
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Worked solution

1. Determine equilibrium amounts:
- Initial: \(n(\text{NO}_2) = 2.00\text{ mol}\), \(n(\text{N}_2\text{O}_4) = 0\text{ mol}\).
- At equilibrium: \(n(\text{N}_2\text{O}_4) = 0.60\text{ mol}\).
- Therefore, \(\text{NO}_2\) reacted = \(2 \times 0.60 = 1.20\text{ mol}\).
- Equilibrium \(n(\text{NO}_2) = 2.00 - 1.20 = 0.80\text{ mol}\).
2. Total moles at equilibrium: \(n_{\text{total}} = 0.80 + 0.60 = 1.40\text{ mol}\).
3. Partial pressures:
- \(p(\text{NO}_2) = \frac{0.80}{1.40} \times (1.50 \times 10^5\text{ Pa}) = 8.571 \times 10^4\text{ Pa}\).
- \(p(\text{N}_2\text{O}_4) = \frac{0.60}{1.40} \times (1.50 \times 10^5\text{ Pa}) = 6.429 \times 10^4\text{ Pa}\).
4. Calculate \(K_p\):
- \(K_p = \frac{p(\text{N}_2\text{O}_4)}{[p(\text{NO}_2)]^2} = \frac{6.429 \times 10^4\text{ Pa}}{(8.571 \times 10^4\text{ Pa})^2} \approx 8.75 \times 10^{-6}\text{ Pa}^{-1}\).

Marking scheme

Award 1 mark for the correct calculation of \(K_p\) (Option A).
Question 8 · multiple_choice
1 marks
An organic compound \(Z\) has the structural formula \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{OH}\).

Which reagent reacts with \(Z\) to produce an organic product that contains both a carboxylic acid group and a ketone group?
  1. A.Hot acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) under reflux
  2. B.Sodium metal at room temperature
  3. C.Phosphorus pentachloride at room temperature
  4. D.Concentrated sulfuric acid under reflux
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Worked solution

Compound \(Z\) is butane-1,3-diol, which has a primary alcohol group (\(-\text{CH}_2\text{OH}\)) at carbon-1 and a secondary alcohol group (\(-\text{CH(OH)}-\)) at carbon-3.

* When reacted with hot acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) under reflux, the secondary alcohol group is oxidized to a ketone group (\(-\text{CO}-\)) and the primary alcohol group is fully oxidized to a carboxylic acid group (\(-\text{COOH}\)).
* This results in the formation of 3-oxobutanoic acid, \(\text{CH}_3\text{COCH}_2\text{COOH}\).
* Other reagents do not produce this combination of functional groups: sodium metal forms an alkoxide salt, phosphorus pentachloride replaces the hydroxyl groups with chlorine atoms, and concentrated sulfuric acid dehydrates the diol to form alkenes.

Marking scheme

Award 1 mark for selecting the correct oxidizing agent (Option A).
Question 9 · Multiple Choice
1 marks
A gaseous hydrocarbon, \( \text{C}_x\text{H}_y \), has a volume of \( 0.10\text{ mol} \) and is completely combusted in \( 0.90\text{ mol} \) of oxygen (which is in excess). After cooling to room temperature and pressure, the remaining gas volume is \( 0.65\text{ mol} \). When this remaining gas is passed through concentrated aqueous sodium hydroxide, the volume of gas decreases to \( 0.25\text{ mol} \). What is the molecular formula of the hydrocarbon?
  1. A.\( \text{C}_4\text{H}_6 \)
  2. B.\( \text{C}_4\text{H}_8 \)
  3. C.\( \text{C}_4\text{H}_{10} \)
  4. D.\( \text{C}_3\text{H}_8 \)
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Worked solution

Let the reaction be: \( \text{C}_x\text{H}_y + (x + \frac{y}{4})\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O} \).

1. At room temperature and pressure, water is a liquid, so the remaining gas consists only of carbon dioxide, \( \text{CO}_2 \), and unreacted excess oxygen, \( \text{O}_2 \). Total volume of remaining gas = \( 0.65\text{ mol} \).
2. Concentrated NaOH absorbs carbon dioxide. The gas remaining after NaOH treatment is unreacted \( \text{O}_2 \) = \( 0.25\text{ mol} \).
3. This means the amount of \( \text{CO}_2 \) produced is \( 0.65\text{ mol} - 0.25\text{ mol} = 0.40\text{ mol} \).
4. Since \( 0.10\text{ mol} \) of \( \text{C}_x\text{H}_y \) produced \( 0.40\text{ mol} \) of \( \text{CO}_2 \), we have \( x = \frac{0.40}{0.10} = 4 \).
5. The oxygen reacted is \( 0.90\text{ mol} - 0.25\text{ mol} = 0.65\text{ mol} \).
6. Using the ratio of hydrocarbon to oxygen: \( 1 : (4 + \frac{y}{4}) = 0.10 : 0.65 \), which gives \( 4 + \frac{y}{4} = 6.5 \).
7. Solving for \( y \): \( \frac{y}{4} = 2.5 \Rightarrow y = 10 \).

Therefore, the molecular formula of the hydrocarbon is \( \text{C}_4\text{H}_{10} \).

Marking scheme

Award 1 mark for the correct option (C). Correct calculation of the moles of CO2 produced (0.40 mol) and oxygen reacted (0.65 mol), followed by the correct determination of the carbon (x = 4) and hydrogen (y = 10) subscripts.
Question 10 · Multiple Choice
1 marks
Solid potassium halides are reacted separately with concentrated sulfuric acid. Which observation is correct?
  1. A.With solid KF, a purple vapor is observed because the halide ion is easily oxidized.
  2. B.With solid KCl, a green gas is produced because chloride ions reduce sulfuric acid.
  3. C.With solid KBr, a reddish-brown vapor is produced containing bromine gas and sulfur dioxide.
  4. D.With solid KI, the only gaseous product is hydrogen iodide because iodide is a weak reducing agent.
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Worked solution

A: With solid KF, a misty vapor of HF gas is observed, but fluoride ions are not oxidized by concentrated sulfuric acid, so no purple vapor (which would be fluorine or iodine) is formed.
B: Chloride ions are not strong enough reducing agents to be oxidized by concentrated sulfuric acid; only misty fumes of HCl are produced, not green chlorine gas.
C: Bromide ions are oxidized by concentrated sulfuric acid to produce bromine gas (a reddish-brown vapor), while the sulfuric acid is reduced to sulfur dioxide gas. This is a correct observation.
D: Iodide ions are very strong reducing agents and easily reduce concentrated sulfuric acid to a mixture of products including iodine, sulfur dioxide, sulfur, and hydrogen sulfide.

Marking scheme

Award 1 mark for the correct option (C). Correct identification that bromide ions are oxidized to bromine (red-brown vapor) while concentrated sulfuric acid is reduced to sulfur dioxide.
Question 11 · Multiple Choice
1 marks
Calculate the standard enthalpy change of formation, \( \Delta H_f^\ominus \), of liquid ethanol, \( \text{C}_2\text{H}_5\text{OH}(l) \), using the standard enthalpy changes of combustion, \( \Delta H_c^\ominus \), given below:

\( \Delta H_c^\ominus[\text{C}(s)] = -394\text{ kJ mol}^{-1} \)
\( \Delta H_c^\ominus[\text{H}_2(g)] = -286\text{ kJ mol}^{-1} \)
\( \Delta H_c^\ominus[\text{C}_2\text{H}_5\text{OH}(l)] = -1367\text{ kJ mol}^{-1} \)
  1. A.\( -279\text{ kJ mol}^{-1} \)
  2. B.\( -687\text{ kJ mol}^{-1} \)
  3. C.\( +279\text{ kJ mol}^{-1} \)
  4. D.\( -1646\text{ kJ mol}^{-1} \)
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Worked solution

The equation for the standard enthalpy change of formation of liquid ethanol is:
\( 2\text{C}(s) + 3\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l) \)

Using Hess's law and enthalpies of combustion:
\( \Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products}) \)
\( \Delta H_f^\ominus = [2 \times \Delta H_c^\ominus(\text{C}(s)) + 3 \times \Delta H_c^\ominus(\text{H}_2(g))] - [\Delta H_c^\ominus(\text{C}_2\text{H}_5\text{OH}(l))] \)
\( \Delta H_f^\ominus = [2 \times (-394) + 3 \times (-286)] - [-1367] \)
\( \Delta H_f^\ominus = [-788 - 858] + 1367 \)
\( \Delta H_f^\ominus = -1646 + 1367 = -279\text{ kJ mol}^{-1} \)

Marking scheme

Award 1 mark for the correct option (A). Correct setup of Hess's cycle using 2 times the carbon value and 3 times the hydrogen value, subtracted by the ethanol combustion value.
Question 12 · Multiple Choice
1 marks
An oxide of a Period 3 element, X, is a white solid with a high melting point. It is completely insoluble in water, but it reacts with both hot concentrated aqueous sodium hydroxide and hot concentrated hydrochloric acid. What is the identity of element X?
  1. A.magnesium
  2. B.aluminium
  3. C.silicon
  4. D.phosphorus
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Worked solution

1. Silicon dioxide (SiO2) is a macromolecular white solid with a high melting point, but it only reacts with concentrated alkalis (like hot NaOH) and not with hydrochloric acid (as it is purely acidic).
2. Magnesium oxide (MgO) is basic and reacts with acids but not with alkalis.
3. Aluminium oxide (Al2O3) is amphoteric. It has a high melting point, is insoluble in water, and reacts with both acids (HCl) and bases (NaOH) to form salts.
Therefore, X must be aluminium.

Marking scheme

Award 1 mark for the correct option (B). Correctly identifying aluminium oxide as the only Period 3 oxide that is amphoteric, insoluble in water, and has a high melting point.
Question 13 · Multiple Choice
1 marks
Equal amounts of four different halogenoalkanes are added to separate test-tubes containing aqueous silver nitrate in ethanol at \( 50^\circ\text{C} \). The time taken for a silver halide precipitate to appear in each test-tube is recorded. Which compound will produce a precipitate in the shortest time?
  1. A.1-chlorobutane
  2. B.2-iodo-2-methylpropane
  3. C.1-iodobutane
  4. D.2-chloro-2-methylpropane
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Worked solution

The rate of hydrolysis of halogenoalkanes depends on two factors:
1. The strength of the C-Halogen bond: C-I is weaker than C-Cl, so iodoalkanes hydrolyze much faster than chloroalkanes.
2. The classification of the halogenoalkane: tertiary halogenoalkanes react via the \( \text{S}_{\text{N}}1 \) mechanism, which is much faster than the \( \text{S}_{\text{N}}2 \) mechanism used by primary halogenoalkanes under these conditions.

Therefore, the tertiary iodoalkane, 2-iodo-2-methylpropane, has both the weakest bond (C-I) and the fastest reacting tertiary structure, making it hydrolyze the fastest.

Marking scheme

Award 1 mark for the correct option (B). Correct deduction based on C-I being the weakest bond and tertiary halogenoalkanes hydrolyzing faster than primary ones.
Question 14 · Multiple Choice
1 marks
Which of the following compounds exhibits both geometrical (cis-trans) isomerism and optical isomerism?
  1. A.3-methylhex-3-ene
  2. B.4-methylhex-1-ene
  3. C.4-methylhex-2-ene
  4. D.2,3-dimethylpent-2-ene
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Worked solution

To exhibit geometrical (cis-trans) isomerism, a molecule must contain a C=C double bond where each carbon atom of the double bond is attached to two different groups.
To exhibit optical isomerism, a molecule must contain at least one chiral carbon atom (attached to four different groups).

Let's analyze 4-methylhex-2-ene: \( \text{CH}_3\text{CH}=\text{CHCH}(\text{CH}_3)\text{CH}_2\text{CH}_3 \):
- The C=C double bond has: C2 bonded to \( -\text{H} \) and \( -\text{CH}_3 \) (different); C3 bonded to \( -\text{H} \) and \( -\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3 \) (different). Thus, cis-trans isomerism is possible.
- C4 is bonded to \( -\text{H} \), \( -\text{CH}_3 \), \( -\text{CH}_2\text{CH}_3 \), and \( -\text{CH}=\text{CHCH}_3 \). All four groups are different, so C4 is a chiral center, allowing optical isomerism.
None of the other options meet both criteria.

Marking scheme

Award 1 mark for the correct option (C). Correct evaluation of both geometrical and optical criteria for the given structures.
Question 15 · Multiple Choice
1 marks
Which of the following species has a non-linear shape and a bond angle of approximately \( 104.5^\circ \)?
  1. A.\( \text{NH}_2^- \)
  2. B.\( \text{NH}_4^+ \)
  3. C.\( \text{CO}_2 \)
  4. D.\( \text{NO}_2^+ \)
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Worked solution

Let's determine the shape and bond angle of each species:
- \( \text{NH}_2^- \): Nitrogen has 5 valence electrons + 1 from the negative charge = 6 valence electrons. It forms 2 single covalent bonds with hydrogen atoms, leaving 2 lone pairs on the nitrogen atom. The 4 electron pairs arrange tetrahedrally, giving a non-linear (bent) shape. The two lone pairs repel more strongly than the bonding pairs, reducing the standard tetrahedral bond angle from \( 109.5^\circ \) to approximately \( 104.5^\circ \).
- \( \text{NH}_4^+ \): 4 bonding pairs, 0 lone pairs; tetrahedral shape, bond angle is \( 109.5^\circ \).
- \( \text{CO}_2 \): 2 double bonds, 0 lone pairs; linear shape, bond angle is \( 180^\circ \).
- \( \text{NO}_2^+ \): Nitrogen has 5 valence electrons - 1 from the positive charge = 4. It forms 2 double bonds with oxygen; linear shape, bond angle is \( 180^\circ \).

Marking scheme

Award 1 mark for the correct option (A). Correct deduction of the number of bonding pairs and lone pairs in the amide ion and mapping to the bent shape with a 104.5 degree bond angle.
Question 16 · Multiple Choice
1 marks
In the redox reaction between acidified potassium dichromate(VI) and tin(II) ions, the unbalanced equation is:

\( a\,\text{Cr}_2\text{O}_7^{2-} + b\,\text{H}^+ + c\,\text{Sn}^{2+} \rightarrow d\,\text{Cr}^{3+} + e\,\text{H}_2\text{O} + f\,\text{Sn}^{4+} \)

What is the ratio of the coefficient of \( \text{Cr}_2\text{O}_7^{2-} \) to the coefficient of \( \text{Sn}^{2+} \) (\( a:c \)) in the fully balanced equation, and which species acts as the oxidizing agent?
  1. A.Ratio is \( 1:3 \); oxidizing agent is \( \text{Cr}_2\text{O}_7^{2-} \)
  2. B.Ratio is \( 1:3 \); oxidizing agent is \( \text{Sn}^{2+} \)
  3. C.Ratio is \( 3:1 \); oxidizing agent is \( \text{Cr}_2\text{O}_7^{2-} \)
  4. D.Ratio is \( 3:1 \); oxidizing agent is \( \text{Sn}^{2+} \)
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Worked solution

1. Write the half-equations:
- Reduction: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)
- Oxidation: \( \text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2\text{e}^- \)

2. Multiply the oxidation half-reaction by 3 to balance the electrons transferred:
\( 3\text{Sn}^{2+} \rightarrow 3\text{Sn}^{4+} + 6\text{e}^- \)

3. Combine the two half-equations:
\( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 3\text{Sn}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 3\text{Sn}^{4+} \)

So, \( a = 1 \) and \( c = 3 \). The ratio \( a:c \) is \( 1:3 \).

4. The oxidizing agent is the species that gets reduced (gains electrons), which is the dichromate(VI) ion, \( \text{Cr}_2\text{O}_7^{2-} \).

Marking scheme

Award 1 mark for the correct option (A). Correctly balancing the electrons between the two half-equations and correctly identifying the dichromate ion as the oxidizing agent.
Question 17 · multiple_choice
1 marks
A mixture of 10 cm³ of a gaseous hydrocarbon \( \text{C}_x\text{H}_y \) and 70 cm³ of oxygen (an excess) is exploded in a sealed vessel at high temperature. After cooling to room temperature, the total volume of gas remaining is 55 cm³. Passing this gas through aqueous sodium hydroxide reduces the volume to 25 cm³. What is the molecular formula of the hydrocarbon?
  1. A.\( \text{C}_3\text{H}_4 \)
  2. B.\( \text{C}_3\text{H}_6 \)
  3. C.\( \text{C}_3\text{H}_8 \)
  4. D.\( \text{C}_4\text{H}_{10} \)
Show answer & marking scheme

Worked solution

1. The decrease in volume upon passing through aqueous sodium hydroxide is due to the absorption of carbon dioxide, \( \text{CO}_2(g) \).
Volume of \( \text{CO}_2 \) produced = \( 55 - 25 = 30\text{ cm}^3 \).
Since 10 cm³ of \( \text{C}_x\text{H}_y \) produced 30 cm³ of \( \text{CO}_2 \), \( x = \frac{30}{10} = 3 \).

2. The remaining 25 cm³ of gas is the unreacted excess oxygen. Therefore, the volume of oxygen that reacted = \( 70 - 25 = 45\text{ cm}^3 \).

3. The equation for the complete combustion of a hydrocarbon is:
\( \text{C}_x\text{H}_y(g) + \left(x + \frac{y}{4}\right)\text{O}_2(g) \rightarrow x\text{CO}_2(g) + \frac{y}{2}\text{H}_2\text{O}(l) \)

From the volume ratio of reacted gases:
\( \frac{\text{Volume of } \text{O}_2 \text{ reacted}}{\text{Volume of hydrocarbon reacted}} = x + \frac{y}{4} = \frac{45}{10} = 4.5 \)

Substituting \( x = 3 \):
\( 3 + \frac{y}{4} = 4.5 \Rightarrow \frac{y}{4} = 1.5 \Rightarrow y = 6 \).

Thus, the molecular formula of the hydrocarbon is \( \text{C}_3\text{H}_6 \).

Marking scheme

1 mark for the correct option B.
- Award 1 mark for calculating the correct volume of CO2 (30 cm³) to find x = 3, determining the volume of reacted O2 (45 cm³), and using the stoichiometric ratio to solve for y = 6.
Question 18 · multiple_choice
1 marks
Use the following standard enthalpy change data to calculate the standard enthalpy change, \( \Delta H^\ominus \), for the reaction shown below.

\( \text{C}(s) + \text{H}_2\text{O}(g) \rightarrow \text{CO}(g) + \text{H}_2(g) \)

Data:
- \( \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\ominus = -393.5\text{ kJ mol}^{-1} \)
- \( \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\ominus = -283.0\text{ kJ mol}^{-1} \)
- \( \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\ominus = -285.8\text{ kJ mol}^{-1} \)
- \( \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(g) \quad \Delta H^\ominus = +44.0\text{ kJ mol}^{-1} \)
  1. A.\( -131.3\text{ kJ mol}^{-1} \)
  2. B.\( -175.3\text{ kJ mol}^{-1} \)
  3. C.\( +131.3\text{ kJ mol}^{-1} \)
  4. D.\( +175.3\text{ kJ mol}^{-1} \)
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Worked solution

By Hess's law, we can construct the target equation by combining the given steps:

1. For \( \text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g) \):
\( \Delta H_1 = \Delta H_{\text{c}}^\ominus[\text{C}(s)] - \Delta H_{\text{c}}^\ominus[\text{CO}(g)] = -393.5 - (-283.0) = -110.5\text{ kJ mol}^{-1} \)

2. For the decomposition of gaseous water, \( \text{H}_2\text{O}(g) \rightarrow \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \):
This is the reverse of the formation of \( \text{H}_2\text{O}(g) \).
\( \Delta H_{\text{f}}^\ominus[\text{H}_2\text{O}(g)] = \Delta H_{\text{f}}^\ominus[\text{H}_2\text{O}(l)] + \Delta H_{\text{vap}}^\ominus[\text{H}_2\text{O}(l)] = -285.8 + 44.0 = -241.8\text{ kJ mol}^{-1} \)
Therefore, for the reverse reaction:
\( \Delta H_2 = +241.8\text{ kJ mol}^{-1} \)

3. Summing these two steps:
\( \Delta H^\ominus = \Delta H_1 + \Delta H_2 = -110.5 + 241.8 = +131.3\text{ kJ mol}^{-1} \)

Marking scheme

1 mark for the correct option C.
- Award 1 mark for finding the standard enthalpy of combustion of carbon to CO (-110.5 kJ/mol), finding the standard enthalpy of formation of gaseous water (-241.8 kJ/mol), and correctly applying Hess's law to obtain +131.3 kJ/mol.
Question 19 · multiple_choice
1 marks
When a solid sample of sodium halide \( \text{NaX} \) is heated with concentrated sulfuric acid, a gas is evolved that turns acidified potassium dichromate(VI) paper green, and a purple vapor is also observed. What is the identity of \( \text{NaX} \) and what is the role of the sulfuric acid in this reaction?
  1. A.\( \text{NaBr} \); sulfuric acid acts only as an acid.
  2. B.\( \text{NaBr} \); sulfuric acid acts as both an acid and an oxidizing agent.
  3. C.\( \text{NaI} \); sulfuric acid acts only as an acid.
  4. D.\( \text{NaI} \); sulfuric acid acts as both an acid and an oxidizing agent.
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Worked solution

1. The purple vapor observed is iodine, \( \text{I}_2(g) \), which is formed when iodide ions, \( \text{I}^- \), are oxidized. Thus, \( \text{NaX} \) must be sodium iodide, \( \text{NaI} \).
2. The gas that turns acidified potassium dichromate(VI) paper from orange to green is sulfur dioxide, \( \text{SO}_2 \). This is formed when concentrated sulfuric acid oxidizes the iodide ions and is itself reduced from the +6 oxidation state to the +4 state.
3. Sulfuric acid also undergoes an acid-base reaction with \( \text{NaI} \) to produce hydrogen iodide, \( \text{HI} \). Thus, sulfuric acid acts as both an acid and an oxidizing agent in this reaction.

Marking scheme

1 mark for the correct option D.
- Award 1 mark for correctly identifying that the purple vapor represents I2 (hence NaI) and that sulfuric acid behaves as both an acid (forming HI) and an oxidizing agent (forming SO2 and I2).
Question 20 · multiple_choice
1 marks
Equal amounts in moles of the four Period 3 oxides, \( \text{Na}_2\text{O} \), \( \text{MgO} \), \( \text{SiO}_2 \), and \( \text{P}_4\text{O}_{10} \), are added separately to equal volumes of water and stirred thoroughly. Which row correctly identifies the oxide that produces the most acidic solution and the oxide that produces the most alkaline solution?
  1. A.Most acidic: \( \text{P}_4\text{O}_{10} \) | Most alkaline: \( \text{Na}_2\text{O} \)
  2. B.Most acidic: \( \text{P}_4\text{O}_{10} \) | Most alkaline: \( \text{MgO} \)
  3. C.Most acidic: \( \text{SiO}_2 \) | Most alkaline: \( \text{Na}_2\text{O} \)
  4. D.Most acidic: \( \text{SiO}_2 \) | Most alkaline: \( \text{MgO} \)
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Worked solution

- \( \text{P}_4\text{O}_{10} \) reacts vigorously with water to form phosphoric(V) acid, \( \text{H}_3\text{PO}_4 \), which is a moderately strong acid and yields a highly acidic solution with a very low pH.
- \( \text{Na}_2\text{O} \) is a basic oxide that reacts completely with water to form sodium hydroxide, \( \text{NaOH} \), which is highly soluble and a strong alkali, giving the most alkaline solution with a very high pH.
- \( \text{SiO}_2 \) is giant covalent and insoluble, having no effect on pH.
- \( \text{MgO} \) is only sparingly soluble, forming a weakly alkaline solution of \( \text{Mg(OH)}_2 \) (pH \( \approx 9 \)).

Marking scheme

1 mark for the correct option A.
- Award 1 mark for identifying P4O10 as the oxide that forms the most acidic solution, and Na2O as the oxide that forms the most alkaline solution.
Question 21 · multiple_choice
1 marks
Three different primary halogenoalkanes, 1-chlorobutane, 1-bromobutane, and 1-iodobutane, are reacted separately with aqueous silver nitrate in ethanol at \( 50\text{ }^\circ\text{C} \). Which statement correctly explains the relative rates of these nucleophilic substitution reactions?
  1. A.1-chlorobutane reacts fastest because the \( \text{C}-\text{Cl} \) bond is the most polar.
  2. B.1-iodobutane reacts fastest because the \( \text{C}-\text{I} \) bond is the weakest.
  3. C.1-chlorobutane reacts fastest because chlorine is the most electronegative halogen of the three.
  4. D.1-iodobutane reacts slowest because the activation energy for the cleavage of the \( \text{C}-\text{I} \) bond is the highest.
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Worked solution

The rate of hydrolysis (nucleophilic substitution) of halogenoalkanes depends on the strength of the carbon-halogen bond rather than its polarity.
- The \( \text{C}-\text{I} \) bond is the longest and weakest bond of the three (lowest bond enthalpy) because iodine has the largest atomic radius.
- Therefore, the \( \text{C}-\text{I} \) bond breaks most easily, resulting in the lowest activation energy for the reaction. Thus, 1-iodobutane reacts the fastest.

Marking scheme

1 mark for the correct option B.
- Award 1 mark for recognizing that bond enthalpy (not bond polarity) determines the rate of hydrolysis of halogenoalkanes, and that C-I is the weakest bond, making 1-iodobutane the fastest to react.
Question 22 · multiple_choice
1 marks
How many distinct stereoisomers exist for the compound pent-3-en-2-ol, \( \text{CH}_3-\text{CH}=\text{CH}-\text{CH(OH)}-\text{CH}_3 \)?
  1. A.2
  2. B.3
  3. C.4
  4. D.8
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Worked solution

1. **Chiral Center**: Carbon-2 is bonded to four different groups: \( -\text{H} \), \( -\text{OH} \), \( -\text{CH}_3 \), and \( -\text{CH}=\text{CH}-\text{CH}_3 \). It is therefore chiral and can exist in two optical configurations (R and S).
2. **Geometrical Isomerism**: The \( \text{C}=\text{C} \) double bond between Carbon-3 and Carbon-4 has two different groups on both carbons (Carbon-3 is bonded to \( -\text{H} \) and \( -\text{CH}_3 \); Carbon-4 is bonded to \( -\text{H} \) and \( -\text{CH(OH)}-\text{CH}_3 \)). It can therefore exist in E and Z configurations.
3. The total number of stereoisomers is \( 2^2 = 4 \) (E-2R, E-2S, Z-2R, Z-2S).

Marking scheme

1 mark for the correct option C.
- Award 1 mark for identifying the single chiral center and the single C=C double bond capable of E/Z isomerism, and calculating 2^2 = 4 stereoisomers.
Question 23 · multiple_choice
1 marks
When 2-methylbut-2-ene reacts with hydrogen bromide, \( \text{HBr} \), a mixture of two halogenoalkanes is formed. Which statement correctly identifies and explains the formation of the major product?
  1. A.The major product is 2-bromo-3-methylbutane because the secondary carbocation intermediate is more stable.
  2. B.The major product is 2-bromo-2-methylbutane because the tertiary carbocation intermediate is more stable.
  3. C.The major product is 2-bromo-3-methylbutane because the secondary carbocation intermediate has less steric hindrance.
  4. D.The major product is 2-bromo-2-methylbutane because the primary carbocation intermediate is more stable.
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Worked solution

The addition of \( \text{H}^+ \) to the unsymmetrical alkene 2-methylbut-2-ene, \( (\text{CH}_3)_2\text{C}=\text{CH}-\text{CH}_3 \), can form two different carbocations:
- Addition of \( \text{H}^+ \) to Carbon-3 yields a tertiary carbocation, \( (\text{CH}_3)_2\text{C}^+-\text{CH}_2-\text{CH}_3 \).
- Addition of \( \text{H}^+ \) to Carbon-2 yields a secondary carbocation, \( (\text{CH}_3)_2\text{CH}-\text{CH}^+-\text{CH}_3 \).

A tertiary carbocation is more stable than a secondary carbocation due to the greater electron-releasing inductive effect of the three alkyl groups. Thus, the reaction proceeds predominantly via the tertiary carbocation intermediate, which is then attacked by \( \text{Br}^- \) to yield the major product, 2-bromo-2-methylbutane.

Marking scheme

1 mark for the correct option B.
- Award 1 mark for identifying that 2-bromo-2-methylbutane is the major product and explaining its formation via the more stable tertiary carbocation intermediate.
Question 24 · multiple_choice
1 marks
In an experiment, 1.0 mol of gaseous \( \text{A} \) and 2.0 mol of gaseous \( \text{B} \) are mixed in a sealed vessel of volume 2.0 dm³ at a constant temperature. The system is allowed to reach equilibrium according to the equation:

\( \text{A}(g) + 2\text{B}(g) \rightleftharpoons 2\text{C}(g) \)

At equilibrium, the vessel is found to contain 0.8 mol of \( \text{C} \). What is the value of the equilibrium constant, \( K_c \), under these conditions?
  1. A.\( 0.74\text{ dm}^3\text{ mol}^{-1} \)
  2. B.\( 0.89\text{ dm}^3\text{ mol}^{-1} \)
  3. C.\( 1.48\text{ dm}^3\text{ mol}^{-1} \)
  4. D.\( 2.96\text{ dm}^3\text{ mol}^{-1} \)
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Worked solution

Using an ICE table to find the equilibrium moles:
- **Initial moles**: \( n(\text{A}) = 1.0 \), \( n(\text{B}) = 2.0 \), \( n(\text{C}) = 0 \)
- **Change in moles**: Since \( n(\text{C})_{\text{eq}} = 0.8\text{ mol} \), the change for \( \text{C} \) is \( +0.8 \). By stoichiometry:
- Change in \( \text{B} = -0.8\text{ mol} \)
- Change in \( \text{A} = -0.4\text{ mol} \)
- **Equilibrium moles**:
- \( n(\text{A}) = 1.0 - 0.4 = 0.6\text{ mol} \)
- \( n(\text{B}) = 2.0 - 0.8 = 1.2\text{ mol} \)
- \( n(\text{C}) = 0.8\text{ mol} \)
- **Equilibrium concentrations** (dividing by volume = 2.0 dm³):
- \( [\text{A}] = 0.6 / 2.0 = 0.3\text{ mol dm}^{-3} \)
- \( [\text{B}] = 1.2 / 2.0 = 0.6\text{ mol dm}^{-3} \)
- \( [\text{C}] = 0.8 / 2.0 = 0.4\text{ mol dm}^{-3} \)
- **Calculate \( K_c \)**:
\( K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^2} = \frac{(0.4)^2}{(0.3) \times (0.6)^2} = \frac{0.16}{0.3 \times 0.36} = \frac{0.16}{0.108} \approx 1.48\text{ dm}^3\text{ mol}^{-1} \)

Marking scheme

1 mark for the correct option C.
- Award 1 mark for finding the correct equilibrium moles of A (0.6 mol) and B (1.2 mol), converting moles to concentrations by dividing by the volume (2.0 dm³), and substituting these values into the Kc expression to obtain 1.48.
Question 25 · multiple-choice
1 marks
The standard enthalpy changes of combustion of ethene, \(\text{C}_2\text{H}_4(\text{g})\), and ethanol, \(\text{C}_2\text{H}_5\text{OH}(\text{l})\), are \(-1411\text{ kJ mol}^{-1}\) and \(-1367\text{ kJ mol}^{-1}\) respectively. What is the standard enthalpy change for the hydration of ethene to form ethanol? \[\text{C}_2\text{H}_4(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{C}_2\text{H}_5\text{OH}(\text{l})\]
  1. A.\(-44\text{ kJ mol}^{-1}\)
  2. B.\(+44\text{ kJ mol}^{-1}\)
  3. C.\(-2778\text{ kJ mol}^{-1}\)
  4. D.\(+2778\text{ kJ mol}^{-1}\)
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Worked solution

According to Hess's Law, the enthalpy change of the reaction can be calculated from the enthalpies of combustion of the reactants and products: \(\Delta H^\theta_r = \sum \Delta H^\theta_c(\text{reactants}) - \sum \Delta H^\theta_c(\text{products})\). Since water is already fully oxidized, its enthalpy of combustion is zero. Therefore, \(\Delta H^\theta_r = \Delta H^\theta_c[\text{C}_2\text{H}_4(\text{g})] - \Delta H^\theta_c[\text{C}_2\text{H}_5\text{OH}(\text{l})] = -1411 - (-1367) = -44\text{ kJ mol}^{-1}\).

Marking scheme

Award 1 mark for the correct calculation showing the correct use of Hess's Law with enthalpies of combustion to arrive at \(-44\text{ kJ mol}^{-1}\).
Question 26 · multiple-choice
1 marks
When a solid sodium halide, \(\text{NaX}\), is reacted with concentrated sulfuric acid, a purple vapor is observed along with a gas that has a characteristic smell of rotten eggs. Which row correctly identifies the halide \(\text{NaX}\) and the product responsible for the rotten egg smell?
  1. A.\(\text{NaX} = \text{NaBr}\); gas = \(\text{SO}_2\)
  2. B.\(\text{NaX} = \text{NaI}\); gas = \(\text{H}_2\text{S}\)
  3. C.\(\text{NaX} = \text{NaI}\); gas = \(\text{SO}_2\)
  4. D.\(\text{NaX} = \text{NaBr}\); gas = \(\text{H}_2\text{S}\)
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Worked solution

Sodium iodide, \(\text{NaI}\), is a strong reducing agent and reduces concentrated sulfuric acid, \(\text{H}_2\text{SO}_4\), to several products. The iodide ions are oxidized to iodine, \(\text{I}_2\), which is observed as a purple vapor. The sulfur in sulfuric acid is reduced from oxidation state +6 to -2, forming hydrogen sulfide, \(\text{H}_2\text{S}\), which has a distinctive rotten egg smell. Sodium bromide is not strong enough to reduce sulfur to this extent, forming only bromine and sulfur dioxide.

Marking scheme

Award 1 mark for identifying sodium iodide as the correct halide and hydrogen sulfide as the reduction product responsible for the smell of rotten eggs.
Question 27 · multiple-choice
1 marks
A sample of \(0.1215\text{ g}\) of magnesium ribbon is reacted with \(25.0\text{ cm}^3\) of \(0.500\text{ mol dm}^{-3}\) hydrochloric acid. What is the volume of hydrogen gas, in \(\text{cm}^3\), collected at room temperature and pressure (r.t.p.)? [Molar volume of gas at r.t.p. = \(24.0\text{ dm}^3\text{ mol}^{-1}\); \(A_r(\text{Mg}) = 24.3\)]
  1. A.\(60\text{ cm}^3\)
  2. B.\(120\text{ cm}^3\)
  3. C.\(150\text{ cm}^3\)
  4. D.\(300\text{ cm}^3\)
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Worked solution

First, write the balanced equation: \(\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\). Next, calculate the moles of each reactant: Moles of \(\text{Mg} = 0.1215 / 24.3 = 0.00500\text{ mol}\). Moles of \(\text{HCl} = (25.0/1000) \times 0.500 = 0.0125\text{ mol}\). Since 1 mole of \(\text{Mg}\) requires 2 moles of \(\text{HCl}\), \(0.00500\text{ mol}\) of \(\text{Mg}\) requires \(0.0100\text{ mol}\) of \(\text{HCl}\). Since \(0.0125\text{ mol}\) of \(\text{HCl}\) is available, \(\text{HCl}\) is in excess and \(\text{Mg}\) is the limiting reactant. Moles of \(\text{H}_2\) produced = \(0.00500\text{ mol}\). Volume of \(\text{H}_2 = 0.00500 \times 24000\text{ cm}^3 = 120\text{ cm}^3\).

Marking scheme

Award 1 mark for calculating the moles of reactants, identifying Mg as the limiting reactant, and converting the resulting moles of hydrogen gas to the correct volume of \(120\text{ cm}^3\).
Question 28 · multiple-choice
1 marks
Two of the Period 3 oxides, \(Y\) and \(Z\), are added separately to water. Oxide \(Y\) reacts vigorously to form a strongly acidic solution of pH 1-2. Oxide \(Z\) does not dissolve in water but reacts with hot aqueous sodium hydroxide to form a soluble salt. Which row correctly identifies the elements forming oxides \(Y\) and \(Z\)?
  1. A.\(Y = \text{Silicon}\), \(Z = \text{Aluminium}\)
  2. B.\(Y = \text{Phosphorus}\), \(Z = \text{Silicon}\)
  3. C.\(Y = \text{Sulfur}\), \(Z = \text{Magnesium}\)
  4. D.\(Y = \text{Sodium}\), \(Z = \text{Silicon}\)
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Worked solution

Phosphorus(V) oxide, \(\text{P}_4\text{O}_{10}\), reacts vigorously with water to form phosphoric(V) acid, which is a strong acid (pH 1-2). Silicon dioxide, \(\text{SiO}_2\), is a giant covalent macromolecular structure that is insoluble in water, but behaves as a weak acidic oxide, reacting with hot concentrated aqueous sodium hydroxide to form soluble sodium silicate, \(\text{Na}_2\text{SiO}_3\). Therefore, \(Y\) is phosphorus and \(Z\) is silicon.

Marking scheme

Award 1 mark for correctly matching the chemical properties of Period 3 oxides to the respective elements phosphorus (Y) and silicon (Z).
Question 29 · multiple-choice
1 marks
Two isomeric halogenoalkanes, 1-chlorobutane and 2-chloro-2-methylpropane, are heated separately with aqueous silver nitrate in ethanol at the same temperature. Which statement correctly compares the rate of hydrolysis and the mechanism involved?
  1. A.2-chloro-2-methylpropane hydrolyzes faster because it reacts via an \(\text{S}_\text{N}1\) mechanism with a stable tertiary carbocation intermediate.
  2. B.1-chlorobutane hydrolyzes faster because it reacts via an \(\text{S}_\text{N}2\) mechanism which has a lower activation energy.
  3. C.2-chloro-2-methylpropane hydrolyzes slower because steric hindrance blocks the nucleophilic attack in the \(\text{S}_\text{N}2\) mechanism.
  4. D.Both hydrolyze at the same rate because the leaving group is a chloride ion in both cases.
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Worked solution

2-chloro-2-methylpropane is a tertiary halogenoalkane and hydrolyzes via the \(\text{S}_\text{N}1\) mechanism. 1-chlorobutane is a primary halogenoalkane and hydrolyzes via the \(\text{S}_\text{N}2\) mechanism. The \(\text{S}_\text{N}1\) mechanism is much faster because the tertiary carbocation intermediate formed in the rate-determining step is highly stabilized by the inductive electron-donating effect of the three methyl groups, leading to a much lower activation energy.

Marking scheme

Award 1 mark for identifying that the tertiary halogenoalkane hydrolyzes faster due to the stable tertiary carbocation intermediate formed in the \(\text{S}_\text{N}1\) mechanism.
Question 30 · multiple-choice
1 marks
An organic compound has the structural formula \(\text{CH}_3\text{CH(OH)CH=CHCH(CH_3)_2}\). How many chiral carbon atoms does this molecule possess, and does it exhibit cis-trans isomerism?
  1. A.1 chiral carbon; exhibits cis-trans isomerism
  2. B.1 chiral carbon; does not exhibit cis-trans isomerism
  3. C.2 chiral carbons; exhibits cis-trans isomerism
  4. D.2 chiral carbons; does not exhibit cis-trans isomerism
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Worked solution

Let us analyze the carbon atoms: Carbon 2, \(\text{-CH(OH)-}\), is bonded to four different groups: \(\text{-H}\), \(\text{-OH}\), \(\text{-CH}_3\), and \(\text{-CH=CHCH(CH_3)_2}\). Thus, it is a chiral center. No other carbon has four different groups (Carbon 5 is bonded to two identical methyl groups, and the double-bonded carbons are \(\text{sp}^2\) hybridized). Therefore, there is exactly 1 chiral carbon. The double bond has different groups attached to each carbon (Carbon 3 has \(\text{-H}\) and \(\text{-CH(OH)CH}_3\); Carbon 4 has \(\text{-H}\) and \(\text{-CH(CH_3)_2}\)), so it exhibits cis-trans isomerism.

Marking scheme

Award 1 mark for correctly determining that there is 1 chiral carbon atom and that the molecule exhibits cis-trans isomerism.
Question 31 · multiple-choice
1 marks
When 2-methylbut-2-ene, \(\text{(CH}_3)_2\text{C=CHCH}_3\), reacts with cold, concentrated hydrobromic acid, \(\text{HBr}\), the major product is 2-bromo-2-methylbutane. Which statement explains why this is the major product?
  1. A.The reaction proceeds via a secondary carbocation intermediate which is more stable than a tertiary carbocation.
  2. B.The reaction proceeds via a tertiary carbocation intermediate which is more stable than a secondary carbocation.
  3. C.Bromide ion is a stronger nucleophile than hydrogen ion, so it attacks the more substituted carbon first.
  4. D.The methyl groups sterically hinder the addition to the C3 carbon, forcing the bromine to add to the C2 carbon.
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Worked solution

The electrophilic addition of \(\text{HBr}\) to an alkene starts with protonation of the double bond. Protonation of 2-methylbut-2-ene can form either a secondary carbocation, \(\text{(CH}_3)_2\text{CH-C}^+\text{HCH}_3\), or a tertiary carbocation, \(\text{(CH}_3)_2\text{C}^+\text{-CH}_2\text{CH}_3\). The tertiary carbocation is more stable than the secondary carbocation due to the greater electron-donating inductive effect of three alkyl groups compared to two. Since the tertiary carbocation is more stable, it forms more rapidly, leading to the major product 2-bromo-2-methylbutane upon attack by the bromide ion.

Marking scheme

Award 1 mark for recognizing that the major product is formed via the more stable tertiary carbocation intermediate.
Question 32 · multiple-choice
1 marks
Acidified potassium manganate(VII) reacts with aqueous sodium sulfite, \(\text{Na}_2\text{SO}_3\), to form manganese(II) ions and sulfate(VI) ions: \[ a\text{MnO}_4^-(\text{aq}) + b\text{SO}_3^{2-}(\text{aq}) + c\text{H}^+(\text{aq}) \rightarrow d\text{Mn}^{2+}(\text{aq}) + e\text{SO}_4^{2-}(\text{aq}) + f\text{H}_2\text{O}(\text{l}) \] What is the ratio of the coefficients \(\frac{b}{a}\) when the equation is balanced using the lowest whole-number coefficients?
  1. A.0.4
  2. B.1.0
  3. C.1.5
  4. D.2.5
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Worked solution

In this redox reaction, manganese is reduced from +7 in \(\text{MnO}_4^-\right.\) to +2 in \(\text{Mn}^{2+}\right.\) (a change of \(-5\) in oxidation number, representing a gain of 5 electrons). Sulfur is oxidized from +4 in \(\text{SO}_3^{2-}\right.\) to +6 in \(\text{SO}_4^{2-}\right.\) (a change of \(+2\) in oxidation number, representing a loss of 2 electrons). To balance the electrons transferred, the oxidation half-reaction must be multiplied by 5 and the reduction half-reaction by 2. This gives \(a = 2\) and \(b = 5\). The ratio \(\frac{b}{a} = \frac{5}{2} = 2.5\).

Marking scheme

Award 1 mark for calculating the changes in oxidation number for Mn and S, using these changes to balance the redox reaction with coefficients \(a=2\) and \(b=5\), and finding the correct ratio of 2.5.
Question 33 · Multiple Choice
1 marks
Solid sodium halide \(NaX\) reacts with concentrated sulfuric acid. A gaseous mixture is produced that contains a choking gas which turns orange acidified dichromate paper green, a purple vapor, a yellow solid, and a gas with a rotten-egg smell.

Which row correctly identifies the halide \(X^-\right.\) and the maximum change in the oxidation number of sulfur in this reaction?
  1. A.Halide \(X^-\right.\) is \(I^-\right.\); Maximum change in oxidation number of sulfur is 8
  2. B.Halide \(X^-\right.\) is \(I^-\right.\); Maximum change in oxidation number of sulfur is 6
  3. C.Halide \(X^-\right.\) is \(Br^-\right.\); Maximum change in oxidation number of sulfur is 2
  4. D.Halide \(X^-\right.\) is \(Br^-\right.\); Maximum change in oxidation number of sulfur is 4
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Worked solution

In the reaction of iodide ions with concentrated sulfuric acid, \(H_2SO_4\) (where sulfur has an oxidation state of +6) is reduced to several products:
- \(SO_2\) (oxidation state +4, a choking gas that turns acidified dichromate paper green)
- \(S\) (oxidation state 0, a yellow solid)
- \(H_2S\) (oxidation state -2, a gas with a rotten-egg smell)

The maximum change in the oxidation state of sulfur is from +6 in \(H_2SO_4\) to -2 in \(H_2S\), which is a total change of \(6 - (-2) = 8\).

For bromide ions, the only sulfur reduction product is \(SO_2\), representing a change of only 2.

Marking scheme

[1 mark] - Correctly identifies the halide as \(I^-\right.\) and the maximum change in oxidation state of sulfur as 8 (Option A).
Question 34 · Multiple Choice
1 marks
The standard enthalpy changes of combustion of methyl ethanoate, ethanoic acid, and methanol are given in the table:

| Compound | \(\Delta H_c^\ominus / \text{kJ mol}^{-1}\) |
|---|---|
| \(CH_3COOCH_3(l)\) | \(-1592\) |
| \(CH_3COOH(l)\) | \(-874\) |
| \(CH_3OH(l)\) | \(-726\) |

What is the standard enthalpy change for the hydrolysis of methyl ethanoate?
\[CH_3COOCH_3(l) + H_2O(l) \rightarrow CH_3COOH(l) + CH_3OH(l)\]
  1. A.\(-8\text{ kJ mol}^{-1}\)
  2. B.\(+8\text{ kJ mol}^{-1}\)
  3. C.\(-3192\text{ kJ mol}^{-1}\)
  4. D.\(+3192\text{ kJ mol}^{-1}\)
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Worked solution

Using Hess's Law and standard enthalpies of combustion:
\(\Delta H^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\)

\(\Delta H^\ominus = [\Delta H_c^\ominus(CH_3COOCH_3(l)) + \Delta H_c^\ominus(H_2O(l))] - [\Delta H_c^\ominus(CH_3COOH(l)) + \Delta H_c^\ominus(CH_3OH(l))]\)

Since water cannot be further combusted, its standard enthalpy of combustion is \(0\text{ kJ mol}^{-1}\):
\(\Delta H^\ominus = -1592 - (-874 - 726)\)
\(\Delta H^\ominus = -1592 - (-1600) = +8\text{ kJ mol}^{-1}\).

Marking scheme

[1 mark] - Correctly applies Hess's law to find the enthalpy of reaction as \(+8\text{ kJ mol}^{-1}\) (Option B).
Question 35 · Multiple Choice
1 marks
A sample of \(10\text{ cm}^3\) of a gaseous hydrocarbon \(C_xH_y\) was exploded with an excess of oxygen (\(80\text{ cm}^3\)). After cooling to room temperature, the total volume of gas remaining was \(65\text{ cm}^3\). On shaking this remaining gas with an excess of aqueous sodium hydroxide, the volume of gas decreased to \(25\text{ cm}^3\).

(All gas volumes are measured at room temperature and pressure).

What is the molecular formula of the hydrocarbon?
  1. A.\(C_3H_6\)
  2. B.\(C_3H_8\)
  3. C.\(C_4H_6\)
  4. D.\(C_4H_8\)
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Worked solution

1. The decrease in volume upon shaking with \(NaOH(aq)\) is due to the absorption of \(CO_2(g)\):
Volume of \(CO_2\) produced = \(65 - 25 = 40\text{ cm}^3\).
Since \(1\text{ mol}\) of \(C_xH_y\) produces \(x\text{ mol}\) of \(CO_2\):
\(10x = 40 \implies x = 4\).

2. The remaining gas (\(25\text{ cm}^3\)) is the unreacted oxygen. Thus, the volume of oxygen used is:
\(80 - 25 = 55\text{ cm}^3\).

3. The combustion equation is:
\(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\)
Using gas volumes at constant temperature and pressure:
\(10(x + \frac{y}{4}) = 55\)
Since \(x = 4\):
\(10(4 + \frac{y}{4}) = 55 \implies 40 + 2.5y = 55 \implies 2.5y = 15 \implies y = 6\).

Therefore, the molecular formula is \(C_4H_6\).

Marking scheme

[1 mark] - Correctly deduces \(x = 4\) and \(y = 6\) from the volume changes to identify the hydrocarbon as \(C_4H_6\) (Option C).
Question 36 · Multiple Choice
1 marks
Two Period 3 elements, \(Y\) and \(Z\), react separately with excess oxygen to form oxides \(Y_2O_3\) and \(ZO_2\) respectively.

- Oxide \(Y_2O_3\) is an amphoteric solid that reacts with both acids and alkalis.
- Oxide \(ZO_2\) is a solid with a giant covalent structure that reacts with hot concentrated sodium hydroxide but is insoluble in water.

What are the identities of elements \(Y\) and \(Z\)?
  1. A.\(Y\) is magnesium; \(Z\) is phosphorus
  2. B.\(Y\) is aluminium; \(Z\) is silicon
  3. C.\(Y\) is aluminium; \(Z\) is sulfur
  4. D.\(Y\) is silicon; \(Z\) is sulfur
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Worked solution

- Aluminium (\(Al\)) forms the amphoteric oxide \(Al_2O_3\) (representing \(Y_2O_3\)).
- Silicon (\(Si\)) forms silicon dioxide, \(SiO_2\) (representing \(ZO_2\)), which has a giant covalent macromolecular structure, is insoluble in water, and reacts with hot concentrated alkalis to form silicates.
- Magnesium forms \(MgO\) (basic, giant ionic), and sulfur forms \(SO_2\) (acidic, simple molecular gas), which do not fit the descriptions.

Marking scheme

[1 mark] - Correctly identifies elements \(Y\) as aluminium and \(Z\) as silicon (Option B).
Question 37 · Multiple Choice
1 marks
Two separate reactions of 2-bromo-2-methylpropane, \((CH_3)_3CBr\), are carried out:

- **Reaction 1:** heating under reflux with aqueous sodium hydroxide.
- **Reaction 2:** heating under reflux with ethanolic sodium hydroxide.

Which row correctly identifies the dominant reaction mechanism and the primary organic product for each reaction?
  1. A.Reaction 1: Nucleophilic substitution; Product: 2-methylpropan-2-ol | Reaction 2: Elimination; Product: methylpropene
  2. B.Reaction 1: Elimination; Product: methylpropene | Reaction 2: Nucleophilic substitution; Product: 2-methylpropan-2-ol
  3. C.Reaction 1: Nucleophilic substitution; Product: 2-methylpropan-2-ol | Reaction 2: Electrophilic addition; Product: 2-methylpropan-2-ol
  4. D.Reaction 1: Electrophilic substitution; Product: methylpropene | Reaction 2: Elimination; Product: methylpropene
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Worked solution

- Reaction 1: In the presence of aqueous sodium hydroxide, the primary pathway is nucleophilic substitution, where the hydroxide ion acts as a nucleophile, replacing the bromide group. Since 2-bromo-2-methylpropane is a tertiary halogenoalkane, this proceeds via the \(S_N1\) mechanism to yield the tertiary alcohol 2-methylpropan-2-ol, \((CH_3)_3COH\).
- Reaction 2: In ethanolic sodium hydroxide under reflux, the hydroxide ion acts as a strong base, favoring elimination of \(HBr\) to form the alkene methylpropene, \((CH_3)_2C=CH_2\).

Marking scheme

[1 mark] - Correctly associates aqueous solvent with nucleophilic substitution to form the alcohol, and ethanolic solvent with elimination to form the alkene (Option A).
Question 38 · Multiple Choice
1 marks
Consider the compound hex-4-en-3-ol, which has the structural formula:
\[CH_3CH=CHCH(OH)CH_2CH_3\]
How many stereoisomers exist for this compound?
  1. A.2
  2. B.4
  3. C.8
  4. D.12
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Worked solution

To find the number of stereoisomers:
1. Check for chiral centers: Carbon-3 is bonded to four different groups (\(-H\), \(-OH\), \(-CH_2CH_3\), and \(-CH=CHCH_3\)). Thus, it is a chiral center, giving \(2^1 = 2\) optical isomers.
2. Check for cis-trans (E/Z) isomerism: The \(C=C\) double bond at Carbon-4 and Carbon-5 has two different groups on each carbon of the double bond (Carbon-4 is bonded to \(-H\) and \(-CH(OH)CH_2CH_3\); Carbon-5 is bonded to \(-H\) and \(-CH_3\)). This allows for cis-trans isomerism, giving \(2^1 = 2\) geometric isomers.

Since the molecule is asymmetric, the total number of stereoisomers is \(2^1 \times 2^1 = 4\) (cis-R, cis-S, trans-R, trans-S).

Marking scheme

[1 mark] - Correctly identifies one chiral center and one double bond capable of geometric isomerism to calculate 4 stereoisomers (Option B).
Question 39 · Multiple Choice
1 marks
An organic compound \(V\) is a hydrocarbon with the molecular formula \(C_6H_{10}\). When \(V\) is heated with hot, concentrated, acidified potassium manganate(VII), the only organic product obtained is a single dicarboxylic acid with the formula \(HOOC-CH_2-CH_2-CH_2-CH_2-COOH\).

What is the IUPAC name of compound \(V\)?
  1. A.cyclohexene
  2. B.methylcyclopentene
  3. C.hex-3-ene
  4. D.hexa-1,5-diene
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Worked solution

1. The molecular formula of \(V\) is \(C_6H_{10}\), representing an alkene with two fewer hydrogens than an acyclic mono-alkene (\(C_6H_{12}\)). This indicates \(V\) is either a cyclic alkene with one double bond or an acyclic diene.
2. Upon vigorous oxidation with hot, concentrated \(KMnO_4\), the \(C=C\) double bond is cleaved. Since a single dicarboxylic acid (hexanedioic acid, a 6-carbon chain) is the only organic product, the double bond must have been part of a six-membered ring.
3. Cleavage of cyclohexene's double bond oxidizes both \(sp^2\) carbons to carboxylic acid groups, forming \(HOOC(CH_2)_4COOH\).
- Methylcyclopentene would form a keto-acid.
- Hex-3-ene (\(C_6H_{12}\)) would form two molecules of propanoic acid.
- Hexa-1,5-diene would yield succinic acid and carbon dioxide.

Marking scheme

[1 mark] - Correctly deduces cyclohexene from the oxidation product and molecular formula (Option A).
Question 40 · Multiple Choice
1 marks
A mixture of \(2.0\text{ mol}\) of ethyl ethanoate, \(2.0\text{ mol}\) of water, \(1.0\text{ mol}\) of ethanoic acid, and \(1.0\text{ mol}\) of ethanol is allowed to reach equilibrium in a sealed vessel of volume \(V\) at a constant temperature.

At equilibrium, it is found that \(1.6\text{ mol}\) of ethanoic acid is present.

The equation for the reaction is:
\[CH_3COOCH_2CH_3(l) + H_2O(l) \rightleftharpoons CH_3COOH(l) + CH_3CH_2OH(l)\]

What is the value of the equilibrium constant, \(K_c\), at this temperature?
  1. A.0.51
  2. B.0.77
  3. C.1.31
  4. D.1.96
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Worked solution

Set up the ICE table (moles):

| | \(CH_3COOCH_2CH_3\) | \(H_2O\) | \(CH_3COOH\) | \(CH_3CH_2OH\) |
|---|---|---|---|---|
| **Initial** | 2.0 | 2.0 | 1.0 | 1.0 |
| **Change** | \(-0.6\) | \(-0.6\) | \(+0.6\) | \(+0.6\) |
| **Equilibrium** | 1.4 | 1.4 | 1.6 | 1.6 |

The change in moles of \(CH_3COOH\) is \(+0.6\text{ mol}\) (since it increases from 1.0 to 1.6). Because the stoichiometry is 1:1:1:1, both reactants decrease by \(0.6\text{ mol}\), and ethanol increases by \(0.6\text{ mol}\).

The expression for \(K_c\) is:
\(K_c = \frac{[CH_3COOH][CH_3CH_2OH]}{[CH_3COOCH_2CH_3][H_2O]}\)

Since the total volume \(V\) cancels out:
\(K_c = \frac{1.6 \times 1.6}{1.4 \times 1.4} = \frac{2.56}{1.96} \approx 1.31\).

Marking scheme

[1 mark] - Calculates equilibrium moles accurately and correctly determines \(K_c \approx 1.31\) (Option C).

Paper 2 (AS Level Structured Questions)

Answer all questions in the spaces provided. Show your working and state units where appropriate.
4 Question · 60 marks
Question 1 · structured
22 marks
Answer all questions in the spaces provided. Show your working and state units where appropriate.

**Part (a)**
Chlorine, \(\text{Cl}_2\), is a highly reactive green gas. In the laboratory, it can be prepared by reacting manganese(IV) oxide, \(\text{MnO}_2\), with concentrated hydrochloric acid, \(\text{HCl}\).

(i) Write a balanced chemical equation for this preparation of chlorine gas. [2]

(ii) State one observation, other than the evolution of a gas, that would be made during this reaction. [1]

(iii) Deduce the oxidation state of manganese in \(\text{MnO}_2\) and in the manganese-containing product. Use these values to explain why this reaction is a redox reaction. [2]

**Part (b)**
A student investigates the relative reactivity of the halogens by carrying out displacement reactions.

(i) Chlorine gas is bubbled through an aqueous solution of sodium bromide.
State the observation made and write an ionic equation, including state symbols, for the reaction that occurs.

Observation:
Ionic equation: [3]

(ii) Explain, in terms of the relative oxidising abilities of the halogens, why no reaction occurs when aqueous bromine is added to aqueous sodium chloride. [2]

**Part (c)**
The hydrogen halides can be prepared by reacting solid sodium halides with concentrated sulfuric acid, \(\text{H}_2\text{SO}_4\).

(i) When solid sodium chloride is reacted with concentrated sulfuric acid, hydrogen chloride gas is produced.
Write an equation for this reaction. State the role of the sulfuric acid in this reaction.

Equation:
Role of sulfuric acid: [2]

(ii) When solid sodium iodide is reacted with concentrated sulfuric acid, a much more complex mixture of products is formed. The observations include:
- a dark purple vapor
- a yellow solid
- a gas with an unpleasant smell of rotten eggs.

Identify the yellow solid and the gas, and write an ionic half-equation to show the formation of the gas with the rotten-egg smell from concentrated sulfuric acid in acidic conditions.

Identity of yellow solid:
Identity of gas:
Ionic half-equation: [4]

**Part (d)**
Chlorine reacts with cold, dilute aqueous sodium hydroxide to form a mixture of two chlorine-containing salts.

(i) Write a balanced equation for this reaction. [1]

(ii) Name the type of redox reaction that occurs, and explain your answer by referring to the oxidation numbers of chlorine in the reactant and products. [2]

**Part (e)**
The concentration of a commercial bleach containing sodium chlorate(I), \(\text{NaClO}\), was determined.
A \(25.0\text{ cm}^3\) sample of the bleach was added to an excess of acidified potassium iodide solution, \(\text{KI}(aq)\).

\(\text{ClO}^-(aq) + 2\text{I}^-(aq) + 2\text{H}^+(aq) \rightarrow \text{Cl}^-(aq) + \text{I}_2(aq) + \text{H}_2\text{O}(l)\)

The liberated iodine, \(\text{I}_2\), was then titrated against \(0.100\text{ mol dm}^{-3}\) aqueous sodium thiosulfate, \(\text{Na}_2\text{S}_2\text{O}_3\).

\(\text{I}_2(aq) + 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow 2\text{I}^-(aq) + \text{S}_4\text{O}_6^{2-}(aq)\)

The mean titre of sodium thiosulfate required was \(20.00\text{ cm}^3\).

Calculate the concentration, in \(\text{mol dm}^{-3}\), of the sodium chlorate(I) in the commercial bleach. Show your working. [3]
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Worked solution

**Part (a)**
(i) The balanced equation for the preparation is:
\(\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}\)

(ii) The black solid (\(\text{MnO}_2\)) dissolves/disappears, or a green/brown solution is formed.

(iii) In \(\text{MnO}_2\), Mn has an oxidation state of +4.
In \(\text{MnCl}_2\), Mn has an oxidation state of +2.
Since the oxidation state of manganese decreases, it is reduced, indicating a redox reaction occurred.

**Part (b)**
(i) Observation: The colourless solution turns orange / yellow / brown.
Ionic equation: \(\text{Cl}_2(g\text{ or }aq) + 2\text{Br}^-(aq) \rightarrow 2\text{Cl}^-(aq) + \text{Br}_2(aq)\)

(ii) Oxidising power decreases down Group 17 (chlorine is a stronger oxidising agent than bromine). Therefore, bromine is not a strong enough oxidising agent to oxidise chloride ions to chlorine.

**Part (c)**
(i) Equation: \(\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}\) (or \(2\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HCl}\))
Role: Acid / proton donor / Br\u00f8nsted-Lowry acid.

(ii) Yellow solid: Sulfur (\(\text{S}\))
Gas: Hydrogen sulfide (\(\text{H}_2\text{S}\))
Half-equation: \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\) or \(\text{SO}_4^{2-} + 10\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\)

**Part (d)**
(i) \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\)

(ii) Disproportionation reaction. Chlorine is simultaneously oxidised and reduced; its oxidation state changes from 0 in \(\text{Cl}_2\) to -1 in \(\text{NaCl}\) (reduction) and to +1 in \(\text{NaClO}\) (oxidation).

**Part (e)**
\(\text{Moles of } \text{S}_2\text{O}_3^{2-} = 0.100\text{ mol dm}^{-3} \times 0.02000\text{ dm}^3 = 2.00 \times 10^{-3}\text{ mol}\)
From the stoichiometry of the titration:
\(1\text{ mol of } \text{I}_2 \equiv 2\text{ mol of } \text{S}_2\text{O}_3^{2-}\)
So, \(\text{moles of } \text{I}_2 = 1.00 \times 10^{-3}\text{ mol}\)
From the reaction of bleach with iodide:
\(1\text{ mol of } \text{ClO}^- \equiv 1\text{ mol of } \text{I}_2\)
So, \(\text{moles of } \text{ClO}^- = 1.00 \times 10^{-3}\text{ mol}\)
Therefore, concentration of \(\text{NaClO} = \frac{1.00 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0400\text{ mol dm}^{-3}\).

Marking scheme

**Part (a)**
- (i) [2 marks] 1 mark for correct species; 1 mark for correct balancing.
- (ii) [1 mark] for stating that the black solid dissolves / disappears.
- (iii) [2 marks] 1 mark for correct oxidation states (+4 and +2); 1 mark for stating that the oxidation state decreases, so manganese undergoes reduction.

**Part (b)**
- (i) [3 marks] 1 mark for correct observation (solution turns orange/yellow-brown); 1 mark for correct ionic species in the equation; 1 mark for correct state symbols.
- (ii) [2 marks] 1 mark for stating that chlorine is a stronger oxidising agent than bromine (or oxidising power decreases down the group); 1 mark for stating that bromine cannot oxidise chloride ions / chlorine has a greater tendency to gain electrons.

**Part (c)**
- (i) [2 marks] 1 mark for correct balanced equation; 1 mark for identifying the role of sulfuric acid as an acid / proton donor / Br\u00f8nsted-Lowry acid (reject: oxidising agent).
- (ii) [4 marks] 1 mark for identifying the yellow solid as sulfur; 1 mark for identifying the gas as hydrogen sulfide; 2 marks for the correct balanced reduction half-equation (1 mark for correct species, 1 mark for correct balancing and charges).

**Part (d)**
- (i) [1 mark] for the correct balanced equation.
- (ii) [2 marks] 1 mark for naming 'disproportionation'; 1 mark for explaining that chlorine's oxidation number changes from 0 to -1 (reduction) and 0 to +1 (oxidation).

**Part (e)**
- [3 marks] 1 mark for calculating moles of thiosulfate (\(2.00 \times 10^{-3}\text{ mol}\)); 1 mark for calculating moles of chlorate(I) (\(1.00 \times 10^{-3}\text{ mol}\)); 1 mark for final concentration of \(0.0400\text{ mol dm}^{-3}\) (accept answer given to 3 significant figures, with units).
Question 2 · structured
11 marks
Nitrosyl chloride, \(\text{NOCl}\), decomposes reversibly at elevated temperatures according to the following equation: \(2\text{NOCl(g)} \rightleftharpoons 2\text{NO(g)} + \text{Cl}_2\text{(g)}\). (a) Write the expression for the equilibrium constant, \(K_c\), for this reaction and state its units. (b) A 2.00 mol sample of \(\text{NOCl}\) is placed in a sealed \(5.00\text{ dm}^3\) container and heated to a constant temperature, \(T\). When equilibrium is established, the mixture is found to contain \(0.45\text{ mol}\) of \(\text{Cl}_2\). (i) Calculate the equilibrium concentrations, in \(\text{mol dm}^{-3}\), of \(\text{NOCl}\), \(\text{NO}\), and \(\text{Cl}_2\). (ii) Calculate the value of \(K_c\) at this temperature, \(T\). (c) The forward reaction is endothermic, with \(\Delta H^\ominus = +76\text{ kJ mol}^{-1}\). (i) State and explain the effect of an increase in temperature on the value of \(K_c\). (ii) State and explain the effect of increasing the pressure (by decreasing the volume of the container) on the position of equilibrium.
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Worked solution

(a) The equilibrium constant expression is Kc = [NO]^2[Cl2] / [NOCl]^2. The units are calculated as: (mol dm-3)^2 * (mol dm-3) / (mol dm-3)^2 = mol dm-3. (b)(i) Let us set up the equilibrium moles: Initial moles: NOCl = 2.00 mol, NO = 0 mol, Cl2 = 0 mol. Since equilibrium Cl2 = 0.45 mol, the change in Cl2 is +0.45 mol. Therefore, the change in NO is +2 * 0.45 = +0.90 mol, and the change in NOCl is -2 * 0.45 = -0.90 mol. Equilibrium moles: NOCl = 2.00 - 0.90 = 1.10 mol; NO = 0.90 mol; Cl2 = 0.45 mol. To find equilibrium concentrations, divide by the volume of 5.00 dm3: [NOCl] = 1.10 / 5.00 = 0.220 mol dm-3; [NO] = 0.90 / 5.00 = 0.180 mol dm-3; [Cl2] = 0.45 / 5.00 = 0.090 mol dm-3. (b)(ii) Substitute the concentrations into the Kc expression: Kc = (0.180)^2 * 0.090 / (0.220)^2 = 0.0324 * 0.090 / 0.0484 = 0.002916 / 0.0484 = 0.0602 mol dm-3 (or 6.02 * 10^-2 mol dm-3). (c)(i) An increase in temperature shifts the equilibrium in the endothermic direction to absorb the added heat. Since the forward reaction is endothermic, the equilibrium shifts to the right, increasing the concentration of products relative to reactants, which increases the value of Kc. (c)(ii) Increasing pressure shifts the position of equilibrium to the side with fewer gas molecules. In this reaction, there are 2 moles of gas on the left (reactant) side and 3 moles of gas on the right (product) side. Thus, the equilibrium shifts to the left (reactant side), decreasing the yield of products.

Marking scheme

(a) [1 mark] for correct Kc expression: Kc = [NO]^2[Cl2] / [NOCl]^2 (must have square brackets). [1 mark] for correct units: mol dm-3. (b)(i) [1 mark] for calculating the correct equilibrium moles of NO (0.90 mol) and NOCl (1.10 mol). [1 mark] for dividing moles by 5.00 dm3 to obtain concentrations. [1 mark] for all three correct concentrations: [NOCl] = 0.220 mol dm-3, [NO] = 0.180 mol dm-3, and [Cl2] = 0.090 mol dm-3. (b)(ii) [1 mark] for correct substitution of their concentrations into their Kc expression. [1 mark] for the correct calculation of Kc = 0.0602 (or 6.02 * 10^-2) (allow 0.06 or 0.060 based on intermediate values, but must be mathematically consistent with their (b)(i)). (c)(i) [1 mark] for stating that Kc increases. [1 mark] for explaining that the forward reaction is endothermic, so equilibrium shifts right to absorb heat. (c)(ii) [1 mark] for stating that the equilibrium shifts to the left / reactant side. [1 mark] for explaining that the left side has fewer moles of gas (2 moles of gas on LHS vs 3 moles on RHS).
Question 3 · structured
11 marks
Answer all parts of the question in the spaces provided.

2-Methylpropan-2-ol, \((CH_3)_3COH\), is an alcohol that reacts rapidly with concentrated hydrochloric acid at room temperature to form 2-chloro-2-methylpropane via a nucleophilic substitution mechanism.

(a) (i) State the class (primary, secondary, or tertiary) of 2-methylpropan-2-ol. [1]
(ii) Suggest why 2-methylpropan-2-ol undergoes substitution primarily via an \(S_N1\) pathway rather than an \(S_N2\) pathway. [1]

(b) The substitution of 2-methylpropan-2-ol by hydrochloric acid occurs in three steps:
1. Protonation of the alcohol hydroxyl group by \(H^+\) to form \((CH_3)_3COH_2^+\).
2. Loss of water to form a carbocation intermediate.
3. Attack of the chloride nucleophile on the carbocation.

Draw the complete mechanism for this reaction. Your mechanism must include:
- curly arrows to show the movement of electron pairs in all three steps,
- any relevant lone pairs and formal charges on intermediates,
- the structure of the carbocation intermediate. [4]

(c) (i) Explain, in terms of the inductive effect, why the tertiary carbocation intermediate formed in (b) is more stable than a primary carbocation. [2]
(ii) Butan-1-ol is a structural isomer of 2-methylpropan-2-ol. When butan-1-ol is treated with concentrated hydrochloric acid, the reaction is extremely slow.
Identify the type of nucleophilic substitution mechanism (\(S_N1\) or \(S_N2\)) that butan-1-ol undergoes, and explain why it does not readily react via the other pathway. [2]

(d) 2-Methylpropan-2-ol can undergo dehydration to form an alkene.
State the reagent(s) and conditions required to carry out this dehydration reaction in a school laboratory. [1]
Show answer & marking scheme

Worked solution

(a) (i) 2-Methylpropan-2-ol is a tertiary alcohol because the carbon atom bonded to the -OH group is itself bonded to three other carbon atoms.
(ii) The three bulky methyl groups create significant steric hindrance around the central carbon atom, preventing the nucleophile from attacking from the back side (precluding \(S_N2\)), and the tertiary carbocation intermediate is highly stable.

(b) Step 1: Draw the structure of 2-methylpropan-2-ol, showing at least one lone pair on the oxygen atom. Draw a curly arrow from the oxygen lone pair to the \(H^+\) ion.
Step 2: In the protonated intermediate \((CH_3)_3C-OH_2^+\), draw a curly arrow starting from the C-O bond and pointing to the oxygen atom, indicating heterolytic cleavage to release \(H_2O\).
Step 3: Draw the tertiary carbocation intermediate, \((CH_3)_3C^+\), clearly showing the positive formal charge on the central carbon atom.
Step 4: Draw a curly arrow from a lone pair on the chloride ion, \(Cl^-\), to the positive central carbon of the carbocation intermediate.

(c) (i) Alkyl (methyl) groups are electron-donating and exert a positive inductive (\(+I\)) effect. In a tertiary carbocation, there are three electron-donating methyl groups, which disperse the positive charge on the central carbon atom more effectively than the single alkyl group in a primary carbocation, stabilizing the intermediate.
(ii) Butan-1-ol is a primary alcohol and reacts via the \(S_N2\) mechanism. It cannot readily react via the \(S_N1\) pathway because the primary carbocation that would be formed is highly unstable due to the lack of stabilizing inductive effects from only one adjacent alkyl group.

(d) Dehydration of 2-methylpropan-2-ol to form methylpropene requires concentrated sulfuric acid (\(H_2SO_4\)) or concentrated phosphoric acid (\(H_3PO_4\)) as a dehydrating agent with heat (typically around 170 °C), or passing the alcohol vapor over a heated aluminium oxide (\(Al_2O_3\)) catalyst at approximately 300 °C.

Marking scheme

(a) (i) Tertiary [1]
(ii) Steric hindrance (due to bulky methyl groups) prevents back-side attack OR the tertiary carbocation intermediate formed is highly stable [1]

(b) 4 marks total:
- Mark 1: Curly arrow from the lone pair on the oxygen atom of 2-methylpropan-2-ol to the \(H^+\) ion (or to the hydrogen of \(H-Cl\) with a second arrow showing the cleavage of the \(H-Cl\) bond) [1]
- Mark 2: Curly arrow from the C-O bond to the oxygen atom in the protonated intermediate, \((CH_3)_3C-OH_2^+\) [1]
- Mark 3: Correct structure of the carbocation intermediate, \((CH_3)_3C^+\), with the positive charge clearly shown on the central carbon [1]
- Mark 4: Curly arrow from a lone pair on the chloride ion, \(Cl^-\), to the positive central carbon of the carbocation intermediate [1]

(c) (i) 2 marks total:
- Mark 1: Methyl/alkyl groups are electron-donating / have a positive inductive (\(+I\)) effect [1]
- Mark 2: This disperses / stabilizes the positive charge on the carbocation carbon atom [1]
(ii) 2 marks total:
- Mark 1: Identifies the mechanism as \(S_N2\) [1]
- Mark 2: States that a primary carbocation (which would form via \(S_N1\)) is highly unstable (or has insufficient inductive stabilization) [1]

(d) Concentrated sulfuric acid (\(H_2SO_4\)) or concentrated phosphoric acid (\(H_3PO_4\)) with heat (accept 170 °C - 180 °C) OR pass vapor over heated aluminium oxide (\(Al_2O_3\)) catalyst at ~300 °C [1]
[Do not accept dilute acid]
Question 4 · structured
16 marks
Propene, \(\text{CH}_3\text{CH}=\text{CH}_2\), is a versatile starting material in organic synthesis.

(a) Propene reacts with gaseous hydrogen bromide, \(\text{HBr}\), at room temperature to form a mixture of two structural isomers: **B** (the major product) and **C** (the minor product).
(i) Draw the structural formulas of both **B** and **C**. [2]
(ii) Name the mechanism of this reaction. Explain why **B** is formed as the major product and **C** is formed as the minor product. [4]

(b) Minor product **C** is isolated and heated under reflux with a reagent in a specific solvent to produce nitrile **D**.
(i) State the reagent and solvent required for this conversion. [2]
(ii) Draw the structural formula of **D** and state its IUPAC name. [2]

(c) Nitrile **D** can be converted into carboxylic acid **E** by heating under reflux with an aqueous acid.
State the essential reagent and conditions for this reaction. [2]

(d) Carboxylic acid **E** reacts with propan-1-ol to form ester **F**.
(i) Draw the structural formula of **F**. [1]
(ii) State the IUPAC name of **F**. [1]
(iii) State two roles of the concentrated sulfuric acid added to the reaction mixture. [2]
Show answer & marking scheme

Worked solution

(a)(i) Propene undergoes hydrohalogenation with \(\text{HBr}\) across the double bond. The major product (Markovnikov's rule) is **B**, 2-bromopropane: \(\text{CH}_3\text{CH(Br)CH}_3\). The minor product is **C**, 1-bromopropane: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\).

(a)(ii) The reaction proceeds via an **electrophilic addition** mechanism.
- The addition of \(\text{H}^+\) to propene can form two carbocation intermediates:
- A secondary (2°) carbocation, \(\text{CH}_3\text{CH}^+\text{CH}_3\), which leads to product **B**.
- A primary (1°) carbocation, \(\text{CH}_3\text{CH}_2\text{CH}_2^+\), which leads to product **C**.
- Secondary carbocations are more stable than primary carbocations.
- This is due to the greater electron-donating (positive inductive, \(+I\)) effect of two alkyl (methyl) groups adjacent to the positive carbon in the secondary carbocation, compared to only one alkyl (ethyl) group in the primary carbocation. This delocalises/disperses the positive charge more effectively.

(b)(i) To convert a halogenoalkane (1-bromopropane) into a nitrile (butanenitrile), nucleophilic substitution is used.
- Reagent: Potassium cyanide (\(\text{KCN}\)) or sodium cyanide (\(\text{NaCN}\))
- Solvent: Ethanol / aqueous ethanol (refluxed)

(b)(ii) Structure of **D**: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}\) (butanenitrile).

(c) Nitriles undergo acid hydrolysis to yield carboxylic acids.
- Reagent: Dilute hydrochloric acid, \(\text{HCl}(aq)\), or dilute sulfuric acid, \(\text{H}_2\text{SO}_4(aq)\).
- Conditions: Heat under reflux.

(d)(i) Structure of **F**: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_2\text{CH}_2\text{CH}_3\) (propyl butanoate).
(d)(ii) Name: propyl butanoate.
(d)(iii) Concentrated sulfuric acid acts as:
1. A catalyst (to speed up the esterification reaction).
2. A dehydrating agent / water-remover (which shifts the position of the dynamic esterification equilibrium to the right, maximising the yield of ester **F**).

Marking scheme

Part (a)(i):
- 1 mark: Structure of **B** correctly drawn as \(\text{CH}_3\text{CH(Br)CH}_3\) (or displayed/skeletal).
- 1 mark: Structure of **C** correctly drawn as \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\) (or displayed/skeletal).

Part (a)(ii):
- 1 mark: Identifies mechanism as 'electrophilic addition'.
- 1 mark: Identifies that **B** is formed via a secondary (2°) carbocation intermediate and **C** via a primary (1°) carbocation.
- 1 mark: Explains that secondary carbocations are more stable than primary carbocations.
- 1 mark: Links stability to the greater positive inductive / electron-donating effect of two alkyl groups (or methyl groups) compared to only one alkyl group.

Part (b)(i):
- 1 mark: Reagent: potassium cyanide / \(\text{KCN}\) or sodium cyanide / \(\text{NaCN}\) (reject hydrogen cyanide).
- 1 mark: Solvent: ethanol / aqueous ethanol (accept 'ethanolic solution').

Part (b)(ii):
- 1 mark: Correct structure of **D** showing the nitrile linkage: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}\).
- 1 mark: Correct IUPAC name: butanenitrile (accept butane-1-nitrile, reject butanylnitrile).

Part (c):
- 1 mark: Reagent: dilute hydrochloric acid / \(\text{HCl}(aq)\) OR dilute sulfuric acid / \(\text{H}_2\text{SO}_4(aq)\). (Do not accept 'concentrated' acids).
- 1 mark: Conditions: heat under reflux / boiling under reflux.

Part (d):
- (i) 1 mark: Correct structural formula of ester **F**: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_2\text{CH}_2\text{CH}_3\) (or displayed/skeletal).
- (ii) 1 mark: Correct IUPAC name: propyl butanoate.
- (iii) 2 marks: Award 1 mark each for any two of:
- Catalyst / increases rate of reaction.
- Dehydrating agent / absorbs/removes water.
- Shifts equilibrium to the right / increases the yield of ester.

Paper 3 (Advanced Practical Skills)

Complete all three practical investigation tasks. Record all data with appropriate precision.
4 Question · 54 marks
Question 1 · practical
11 marks
An experiment is carried out to determine the enthalpy change for the thermal decomposition of calcium carbonate, \(\Delta H_{\text{r}}\):

\(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\)

This enthalpy change cannot be determined directly. Instead, the enthalpy changes for the reactions of calcium carbonate and calcium oxide with excess hydrochloric acid are determined.

**Experiment 1**: A student added \(50.0\text{ cm}^3\) of \(2.0\text{ mol dm}^{-3}\) hydrochloric acid (an excess) to a polystyrene cup. The student weighed a weighing bottle containing anhydrous calcium carbonate. \(3.00\text{ g}\) of \(\text{CaCO}_3\) was added to the acid, and the maximum temperature increase was measured. The empty weighing bottle was reweighed.

**Experiment 2**: The procedure was repeated using \(1.68\text{ g}\) of calcium oxide, \(\text{CaO}\), in place of calcium carbonate.

**Results recorded:**
* **Experiment 1 (\(\text{CaCO}_3\))**: Mass of \(\text{CaCO}_3\) = \(3.00\text{ g}\), Initial temperature of acid = \(21.5\text{ }^\circ\text{C}\), Maximum temperature = \(23.8\text{ }^\circ\text{C}\).
* **Experiment 2 (\(\text{CaO}\))**: Mass of \(\text{CaO}\) = \(1.68\text{ g}\), Initial temperature of acid = \(21.5\text{ }^\circ\text{C}\), Maximum temperature = \(47.8\text{ }^\circ\text{C}\).

*(Assume the density of the acid solution is \(1.00\text{ g cm}^{-3}\), and its specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\). \(A_{\text{r}}\): \(\text{Ca} = 40.1\), \(\text{C} = 12.0\), \(\text{O} = 16.0\))*

**(a)** Calculate the heat energy, \(q\), in joules, released in:
(i) Experiment 1 (\(q_1\))
(ii) Experiment 2 (\(q_2\))
Give both answers to 3 significant figures. [3]

**(b)** Calculate:
(i) the number of moles of \(\text{CaCO}_3\) used in Experiment 1.
(ii) the number of moles of \(\text{CaO}\) used in Experiment 2.
(iii) the enthalpy change, \(\Delta H_1\), in \(\text{kJ mol}^{-1}\), for the reaction of 1 mole of \(\text{CaCO}_3\) with hydrochloric acid.
(iv) the enthalpy change, \(\Delta H_2\), in \(\text{kJ mol}^{-1}\), for the reaction of 1 mole of \(\text{CaO}\) with hydrochloric acid.
Include correct signs and units in your answers. [4]

**(c)** Construct an enthalpy cycle relating the reactions in Experiments 1 and 2 to the thermal decomposition of calcium carbonate, and use it to calculate the enthalpy change of decomposition, \(\Delta H_{\text{r}}\), in \(\text{kJ mol}^{-1}\). [2]

**(d)** Heat loss is a significant source of error in both experiments.
(i) Explain why the percentage error due to heat loss is likely to be greater in Experiment 2 than in Experiment 1.
(ii) Suggest one practical modification to the apparatus that would reduce heat loss. [2]
Show answer & marking scheme

Worked solution

**(a)**
(i) Temperature change \(\Delta T_1 = 23.8 - 21.5 = 2.3\text{ }^\circ\text{C}\).
\(q_1 = m c \Delta T_1 = 50.0 \times 4.18 \times 2.3 = 480.7\text{ J}\). To 3 s.f., \(q_1 = 481\text{ J}\).
(ii) Temperature change \(\Delta T_2 = 47.8 - 21.5 = 26.3\text{ }^\circ\text{C}\).
\(q_2 = m c \Delta T_2 = 50.0 \times 4.18 \times 26.3 = 5496.7\text{ J}\). To 3 s.f., \(q_2 = 5500\text{ J}\).

**(b)**
(i) Molar mass of \(\text{CaCO}_3 = 40.1 + 12.0 + (3 \times 16.0) = 100.1\text{ g mol}^{-1}\).
Moles of \(\text{CaCO}_3 = \frac{3.00}{100.1} = 0.02997\text{ mol}\) (or \(0.0300\text{ mol}\)).
(ii) Molar mass of \(\text{CaO} = 40.1 + 16.0 = 56.1\text{ g mol}^{-1}\).
Moles of \(\text{CaO} = \frac{1.68}{56.1} = 0.02995\text{ mol}\) (or \(0.0300\text{ mol}\)).
(iii) Since the reaction is exothermic, \(\Delta H_1\) is negative.
\(\Delta H_1 = -\frac{480.7 \times 10^{-3}}{0.02997} = -16.0\text{ kJ mol}^{-1}\).
(iv) Since the reaction is exothermic, \(\Delta H_2\) is negative.
\(\Delta H_2 = -\frac{5.4967}{0.02995} = -183.5\text{ kJ mol}^{-1}\) (or \(-183.3\text{ kJ mol}^{-1}\) if using rounded moles of \(0.0300\)).

**(c)**
Applying Hess's Law:
We have:
1) \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})\) with \(\Delta H_1\)
2) \(\text{CaO}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l})\) with \(\Delta H_2\)
By subtracting reaction (2) from reaction (1), we get:
\(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\) with \(\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2\).
\(\Delta H_{\text{r}} = (-16.0) - (-183.5) = +167.5\text{ kJ mol}^{-1}\) (or \(-16.0 - (-183.3) = +167.3\text{ kJ mol}^{-1}\)).

**(d)**
(i) The temperature rise in Experiment 2 is much greater than in Experiment 1 (\(26.3\text{ }^\circ\text{C}\) vs \(2.3\text{ }^\circ\text{C}\)). Therefore, the temperature difference between the cup contents and the room is much larger, leading to a much higher rate of heat loss to the surroundings.
(ii) Put a lid on the polystyrene cup (or wrap the cup in cotton wool / bubble wrap).

Marking scheme

**Part (a)** [3 marks]
* Award 1 mark for correct calculation of \(q_1 = 481\text{ J}\) (accept \(0.481\text{ kJ}\)).
* Award 1 mark for correct calculation of \(q_2 = 5500\text{ J}\) (accept \(5.50\text{ kJ}\)).
* Award 1 mark for expressing both values to 3 significant figures.

**Part (b)** [4 marks]
* Award 1 mark for calculating correct moles: \(\text{CaCO}_3 = 0.0300\text{ mol}\) (or \(0.02997\text{ mol}\)) AND \(\text{CaO} = 0.0300\text{ mol}\) (or \(0.02995\text{ mol}\)).
* Award 1 mark for \(\Delta H_1 = -16.0\text{ kJ mol}^{-1}\) (accept range \(-16.0\) to \(-16.1\); must have negative sign and unit).
* Award 1 mark for \(\Delta H_2 = -183.5\text{ kJ mol}^{-1}\) or \(-183.3\text{ kJ mol}^{-1}\) (accept range \(-183\) to \(-184\); must have negative sign and unit).
* Award 1 mark for showing appropriate working scaling \(q\) by dividing by the number of moles in both calculations.

**Part (c)** [2 marks]
* Award 1 mark for showing the correct relationship: \(\Delta H_{\text{r}} = \Delta H_1 - \Delta H_2\) (either via formula or drawn enthalpy cycle diagram).
* Award 1 mark for correct calculation of \(\Delta H_{\text{r}}\): \(+167.5\text{ kJ mol}^{-1}\) (or \(+167.3\text{ kJ mol}^{-1}\), accept values between \(+167\) and \(+168\); sign must be positive).

**Part (d)** [2 marks]
* Award 1 mark for explaining that the temperature rise (or difference between cup and room temperature) is much greater in Experiment 2, causing a higher rate of heat transfer/loss.
* Award 1 mark for suggesting the use of a lid, wrapping the cup in an insulating material, or using a double-walled cup.
Question 2 · practical
14 marks
### Determination of the Purity of an Impure Sample of Sodium Hydrogencarbonate

In this investigation, you will determine the percentage purity of an impure sample of sodium hydrogencarbonate, \(\text{NaHCO}_3\), by performing a titration with standard hydrochloric acid.

**FA 1** is an impure sample of solid sodium hydrogencarbonate containing an inert soluble impurity.
**FA 2** is \(0.100\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}\).
Methyl orange indicator is provided.

---

### Method

**Preparation of the solution (FA 3):**
1. Weigh a clean, dry weighing bottle. Record its mass in the table below.
2. Add between \(2.00\text{ g}\) and \(2.20\text{ g}\) of **FA 1** into the weighing bottle. Record the total mass.
3. Transfer the solid into a \(100\text{ cm}^3\) beaker. Weigh the empty bottle with any residual solid and record this mass.
4. Add about \(50\text{ cm}^3\) of distilled water to the beaker. Stir with a glass rod until all the solid has dissolved.
5. Transfer the solution quantitatively into a \(250\text{ cm}^3\) volumetric flask. Rinse the beaker and glass rod with distilled water, adding the washings to the flask.
6. Add distilled water to make up the solution exactly to the \(250\text{ cm}^3\) mark. Stopper the flask and invert it several times to ensure thorough mixing. This solution is **FA 3**.

**Titration:**
1. Fill a burette with **FA 2**.
2. Pipette \(25.0\text{ cm}^3\) of **FA 3** into a conical flask.
3. Add 3 to 4 drops of methyl orange indicator.
4. Titrate the solution in the flask with **FA 2** until the end-point is reached. Record your rough titration results.
5. Perform sufficient accurate titrations to obtain concordant results. Record all your titration results in a suitable table.

---

### Tasks

**(a)** Record your weighing data and titration results in the space below. Your tables must show clear headings, correct units, and appropriate precision.

**(b)** Based on your experimental data (assuming a mass of FA 1 transferred of \(2.10\text{ g}\) and a mean titre of \(20.00\text{ cm}^3\) for calculations in this mock environment):
1. Calculate the number of moles of hydrochloric acid, \(\text{HCl}\), in the mean titre.
2. Deduce the number of moles of \(\text{NaHCO}_3\) present in \(25.0\text{ cm}^3\) of **FA 3**.
3. Calculate the number of moles of \(\text{NaHCO}_3\) in the entire \(250\text{ cm}^3\) of **FA 3**.
4. Calculate the mass of pure \(\text{NaHCO}_3\) in the sample of **FA 1** transferred. \([M_r(\text{NaHCO}_3) = 84.0]\)
5. Calculate the percentage purity of the solid **FA 1**.

**(c)** Evaluation:
1. Calculate the percentage uncertainty in your mean titre of \(20.00\text{ cm}^3\), given that the total uncertainty associated with the two burette readings (initial and final) is \(\pm 0.10\text{ cm}^3\).
2. Explain why phenolphthalein would not be a suitable indicator for this titration.
3. A student forgot to shake the volumetric flask containing **FA 3** after making it up to the mark with distilled water. If the first sample was pipetted from the bottom of the flask, explain the effect this would have on the concentration of **FA 3** pipetted and the resulting titre value.
Show answer & marking scheme

Worked solution

### Part (a) Data Recording
- **Weighing table:**
- Mass of weighing bottle + **FA 1** \(= 12.50\text{ g}\)
- Mass of empty weighing bottle + residue \(= 10.40\text{ g}\)
- Mass of **FA 1** transferred \(= 2.10\text{ g}\)
- **Titration table:**
- Must contain: Final burette reading, Initial burette reading, and Titre for rough and at least two accurate titrations.
- Units must be correctly stated as `/ cm3` or `(cm3)`.
- Concordant titres should be within \(\pm 0.10\text{ cm}^3\).
- Theoretical mean titre \(= 20.00\text{ cm}^3\).

### Part (b) Calculations
1. **Moles of \(\text{HCl}\):**
\[ n(\text{HCl}) = 20.00\text{ cm}^3 \times \frac{0.100\text{ mol dm}^{-3}}{1000} = 2.00 \times 10^{-3}\text{ mol} \]
2. **Moles of \(\text{NaHCO}_3\) in \(25.0\text{ cm}^3\):**
The reaction is:
\[ \text{NaHCO}_3(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \]
The molar ratio is 1:1.
\[ n(\text{NaHCO}_3) \text{ in } 25.0\text{ cm}^3 = n(\text{HCl}) = 2.00 \times 10^{-3}\text{ mol} \]
3. **Moles of \(\text{NaHCO}_3\) in \(250\text{ cm}^3\):**
\[ n(\text{NaHCO}_3) \text{ in } 250\text{ cm}^3 = 2.00 \times 10^{-3}\text{ mol} \times \frac{250}{25.0} = 2.00 \times 10^{-2}\text{ mol} \]
4. **Mass of pure \(\text{NaHCO}_3\):**
\[ \text{Mass} = 2.00 \times 10^{-2}\text{ mol} \times 84.0\text{ g mol}^{-1} = 1.68\text{ g} \]
5. **Percentage purity:**
\[ \text{Percentage purity} = \frac{1.68\text{ g}}{2.10\text{ g}} \times 100\% = 80.0\% \]

### Part (c) Evaluation
1. **Percentage uncertainty:**
\[ \text{Percentage uncertainty} = \frac{0.10\text{ cm}^3}{20.00\text{ cm}^3} \times 100\% = 0.50\% \]
2. **Indicator suitability:**
The reaction involves a weak base (\(\text{NaHCO}_3\)) and a strong acid (\(\text{HCl}\)). The equivalence point occurs at an acidic pH (approx. 3.7–4.0). Phenolphthalein has a pH range of 8.2–10.0, which means it changes colour before any hydrochloric acid reacts significantly with the hydrogencarbonate, making it completely unsuitable.
3. **Effect of lack of mixing:**
- Distilled water added last would sit at the top, leaving the more concentrated solute at the bottom.
- Pipetting from the bottom of the flask would draw a solution of higher concentration than average.
- This would require a larger volume of standard acid (**FA 2**) to reach the end-point, resulting in an falsely high titre value.

Marking scheme

### Part (a) [6 marks]
- **M1 (Weighing):** Awarded for a neat weighing table showing all masses recorded to the same decimal precision (2 or 3 d.p.) and a mass of **FA 1** between 2.00 g and 2.20 g. [1]
- **M2 (Titration Table Presentation):** Correct table format with clear column/row headings (Initial, Final, Titre) and appropriate units (/ \(\text{cm}^3\) or in \(\text{cm}^3\)) for rough and accurate trials. [1]
- **M3 (Precision):** All burette readings in accurate titrations are recorded to the nearest \(0.05\text{ cm}^3\). [1]
- **M4 (Concordancy):** At least two accurate titres within \(0.10\text{ cm}^3\) of each other are obtained. [1]
- **M5 & M6 (Accuracy):** Compare the student's average titre to the expected supervisor value (here, \(20.00\text{ cm}^3\)):
- Award 2 marks if difference \(\le 0.10\text{ cm}^3\).
- Award 1 mark if difference is between \(0.11\text{ cm}^3\) and \(0.20\text{ cm}^3\).
- Award 0 marks if difference \(> 0.20\text{ cm}^3\).

### Part (b) [5 marks]
- **M7:** Correctly calculates moles of \(\text{HCl}\) reacting: \(20.00 \times 10^{-3} \times 0.100 = 2.00 \times 10^{-3}\text{ mol}\). (Or matches student's mean titre). [1]
- **M8:** States 1:1 ratio and deduces moles of \(\text{NaHCO}_3\) in \(25.0\text{ cm}^3\) is equal to moles of \(\text{HCl}\) calculated. [1]
- **M9:** Correctly scales up by factor of 10 to obtain moles in \(250\text{ cm}^3\) \((2.00 \times 10^{-2}\text{ mol})\). [1]
- **M10:** Multiplies moles in \(250\text{ cm}^3\) by 84.0 to calculate mass of pure \(\text{NaHCO}_3\) \((1.68\text{ g})\). [1]
- **M11:** Divides calculated mass of pure solid by total mass of **FA 1** transferred (2.10 g) and multiplies by 100 to get percentage purity \((80.0\%)\). Answers must be to 3 significant figures. [1]

### Part (c) [3 marks]
- **M12:** Correctly calculates percentage uncertainty: \(\frac{0.10}{20.00} \times 100\% = 0.50\%\). [1]
- **M13:** Explains that the equivalence point is acidic (or lies below pH 7) because it is a titration of a weak base and a strong acid, whereas phenolphthalein changes colour in an alkaline range (8.2-10.0). [1]
- **M14:** States that the concentration of the pipetted solution is higher because solute is more concentrated at the bottom, leading to a larger titre value than expected. [1]
Question 3 · practical
14 marks
### Determination of the Purity of an Impure Sample of Sodium Hydrogencarbonate

In this investigation, you will determine the percentage purity of an impure sample of sodium hydrogencarbonate, \(\text{NaHCO}_3\), by performing a titration with standard hydrochloric acid.

**FA 1** is an impure sample of solid sodium hydrogencarbonate containing an inert soluble impurity.
**FA 2** is \(0.100\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}\).
Methyl orange indicator is provided.

---

### Method

**Preparation of the solution (FA 3):**
1. Weigh a clean, dry weighing bottle. Record its mass in the table below.
2. Add between \(2.00\text{ g}\) and \(2.20\text{ g}\) of **FA 1** into the weighing bottle. Record the total mass.
3. Transfer the solid into a \(100\text{ cm}^3\) beaker. Weigh the empty bottle with any residual solid and record this mass.
4. Add about \(50\text{ cm}^3\) of distilled water to the beaker. Stir with a glass rod until all the solid has dissolved.
5. Transfer the solution quantitatively into a \(250\text{ cm}^3\) volumetric flask. Rinse the beaker and glass rod with distilled water, adding the washings to the flask.
6. Add distilled water to make up the solution exactly to the \(250\text{ cm}^3\) mark. Stopper the flask and invert it several times to ensure thorough mixing. This solution is **FA 3**.

**Titration:**
1. Fill a burette with **FA 2**.
2. Pipette \(25.0\text{ cm}^3\) of **FA 3** into a conical flask.
3. Add 3 to 4 drops of methyl orange indicator.
4. Titrate the solution in the flask with **FA 2** until the end-point is reached. Record your rough titration results.
5. Perform sufficient accurate titrations to obtain concordant results. Record all your titration results in a suitable table.

---

### Tasks

**(a)** Record your weighing data and titration results in the space below. Your tables must show clear headings, correct units, and appropriate precision.

**(b)** Based on your experimental data (assuming a mass of FA 1 transferred of \(2.10\text{ g}\) and a mean titre of \(20.00\text{ cm}^3\) for calculations in this mock environment):
1. Calculate the number of moles of hydrochloric acid, \(\text{HCl}\), in the mean titre.
2. Deduce the number of moles of \(\text{NaHCO}_3\) present in \(25.0\text{ cm}^3\) of **FA 3**.
3. Calculate the number of moles of \(\text{NaHCO}_3\) in the entire \(250\text{ cm}^3\) of **FA 3**.
4. Calculate the mass of pure \(\text{NaHCO}_3\) in the sample of **FA 1** transferred. \([M_r(\text{NaHCO}_3) = 84.0]\)
5. Calculate the percentage purity of the solid **FA 1**.

**(c)** Evaluation:
1. Calculate the percentage uncertainty in your mean titre of \(20.00\text{ cm}^3\), given that the total uncertainty associated with the two burette readings (initial and final) is \(\pm 0.10\text{ cm}^3\).
2. Explain why phenolphthalein would not be a suitable indicator for this titration.
3. A student forgot to shake the volumetric flask containing **FA 3** after making it up to the mark with distilled water. If the first sample was pipetted from the bottom of the flask, explain the effect this would have on the concentration of **FA 3** pipetted and the resulting titre value.
Show answer & marking scheme

Worked solution

### Part (a) Data Recording
- **Weighing table:**
- Mass of weighing bottle + **FA 1** \(= 12.50\text{ g}\)
- Mass of empty weighing bottle + residue \(= 10.40\text{ g}\)
- Mass of **FA 1** transferred \(= 2.10\text{ g}\)
- **Titration table:**
- Must contain: Final burette reading, Initial burette reading, and Titre for rough and at least two accurate titrations.
- Units must be correctly stated as `/ cm3` or `(cm3)`.
- Concordant titres should be within \(\pm 0.10\text{ cm}^3\).
- Theoretical mean titre \(= 20.00\text{ cm}^3\).

### Part (b) Calculations
1. **Moles of \(\text{HCl}\):**
\[ n(\text{HCl}) = 20.00\text{ cm}^3 \times \frac{0.100\text{ mol dm}^{-3}}{1000} = 2.00 \times 10^{-3}\text{ mol} \]
2. **Moles of \(\text{NaHCO}_3\) in \(25.0\text{ cm}^3\):**
The reaction is:
\[ \text{NaHCO}_3(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \]
The molar ratio is 1:1.
\[ n(\text{NaHCO}_3) \text{ in } 25.0\text{ cm}^3 = n(\text{HCl}) = 2.00 \times 10^{-3}\text{ mol} \]
3. **Moles of \(\text{NaHCO}_3\) in \(250\text{ cm}^3\):**
\[ n(\text{NaHCO}_3) \text{ in } 250\text{ cm}^3 = 2.00 \times 10^{-3}\text{ mol} \times \frac{250}{25.0} = 2.00 \times 10^{-2}\text{ mol} \]
4. **Mass of pure \(\text{NaHCO}_3\):**
\[ \text{Mass} = 2.00 \times 10^{-2}\text{ mol} \times 84.0\text{ g mol}^{-1} = 1.68\text{ g} \]
5. **Percentage purity:**
\[ \text{Percentage purity} = \frac{1.68\text{ g}}{2.10\text{ g}} \times 100\% = 80.0\% \]

### Part (c) Evaluation
1. **Percentage uncertainty:**
\[ \text{Percentage uncertainty} = \frac{0.10\text{ cm}^3}{20.00\text{ cm}^3} \times 100\% = 0.50\% \]
2. **Indicator suitability:**
The reaction involves a weak base (\(\text{NaHCO}_3\)) and a strong acid (\(\text{HCl}\)). The equivalence point occurs at an acidic pH (approx. 3.7–4.0). Phenolphthalein has a pH range of 8.2–10.0, which means it changes colour before any hydrochloric acid reacts significantly with the hydrogencarbonate, making it completely unsuitable.
3. **Effect of lack of mixing:**
- Distilled water added last would sit at the top, leaving the more concentrated solute at the bottom.
- Pipetting from the bottom of the flask would draw a solution of higher concentration than average.
- This would require a larger volume of standard acid (**FA 2**) to reach the end-point, resulting in an falsely high titre value.

Marking scheme

### Part (a) [6 marks]
- **M1 (Weighing):** Awarded for a neat weighing table showing all masses recorded to the same decimal precision (2 or 3 d.p.) and a mass of **FA 1** between 2.00 g and 2.20 g. [1]
- **M2 (Titration Table Presentation):** Correct table format with clear column/row headings (Initial, Final, Titre) and appropriate units (/ \(\text{cm}^3\) or in \(\text{cm}^3\)) for rough and accurate trials. [1]
- **M3 (Precision):** All burette readings in accurate titrations are recorded to the nearest \(0.05\text{ cm}^3\). [1]
- **M4 (Concordancy):** At least two accurate titres within \(0.10\text{ cm}^3\) of each other are obtained. [1]
- **M5 & M6 (Accuracy):** Compare the student's average titre to the expected supervisor value (here, \(20.00\text{ cm}^3\)):
- Award 2 marks if difference \(\le 0.10\text{ cm}^3\).
- Award 1 mark if difference is between \(0.11\text{ cm}^3\) and \(0.20\text{ cm}^3\).
- Award 0 marks if difference \(> 0.20\text{ cm}^3\).

### Part (b) [5 marks]
- **M7:** Correctly calculates moles of \(\text{HCl}\) reacting: \(20.00 \times 10^{-3} \times 0.100 = 2.00 \times 10^{-3}\text{ mol}\). (Or matches student's mean titre). [1]
- **M8:** States 1:1 ratio and deduces moles of \(\text{NaHCO}_3\) in \(25.0\text{ cm}^3\) is equal to moles of \(\text{HCl}\) calculated. [1]
- **M9:** Correctly scales up by factor of 10 to obtain moles in \(250\text{ cm}^3\) \((2.00 \times 10^{-2}\text{ mol})\). [1]
- **M10:** Multiplies moles in \(250\text{ cm}^3\) by 84.0 to calculate mass of pure \(\text{NaHCO}_3\) \((1.68\text{ g})\). [1]
- **M11:** Divides calculated mass of pure solid by total mass of **FA 1** transferred (2.10 g) and multiplies by 100 to get percentage purity \((80.0\%)\). Answers must be to 3 significant figures. [1]

### Part (c) [3 marks]
- **M12:** Correctly calculates percentage uncertainty: \(\frac{0.10}{20.00} \times 100\% = 0.50\%\). [1]
- **M13:** Explains that the equivalence point is acidic (or lies below pH 7) because it is a titration of a weak base and a strong acid, whereas phenolphthalein changes colour in an alkaline range (8.2-10.0). [1]
- **M14:** States that the concentration of the pipetted solution is higher because solute is more concentrated at the bottom, leading to a larger titre value than expected. [1]
Question 4 · practical
15 marks
FA 1 is a hydrated solid salt containing one cation and one anion. FA 2 is an aqueous solution of another ionic salt.

Carry out the following tests and record your observations in the spaces provided. You should test any gas evolved with appropriate reagents.

(a) Heat a small spatula measure of solid FA 1 in a hard-glass test-tube. Record your observations.

(b) Dissolve the remaining FA 1 in about 15 cm³ of distilled water in a beaker. Divide this solution into three portions and perform the following tests:
(i) To the first portion, add aqueous sodium hydroxide dropwise until in excess.
(ii) To the second portion, add aqueous ammonia dropwise until in excess.
(iii) To the third portion, add about 1 cm³ of dilute nitric acid, followed by a few drops of aqueous barium nitrate.

(c) Perform the following tests on FA 2:
(i) To a 2 cm³ portion of FA 2 in a test-tube, add about 2 cm³ of aqueous sodium hydroxide and warm the mixture gently. Test any gas evolved.
(ii) To another 2 cm³ portion of FA 2 in a test-tube, add 1 cm³ of dilute nitric acid, followed by a few drops of aqueous silver nitrate. Then add concentrated aqueous ammonia to the resulting mixture.

(d) Use your observations to identify the ions present in FA 1 and FA 2. Give the formula of each ion.
Show answer & marking scheme

Worked solution

(a) Heating solid FA 1 causes the hydrated salt to lose its water of crystallization. Steam/water vapor is given off, which condenses as liquid droplets near the cooler mouth of the test-tube. The solid changes from a white crystalline appearance to a white anhydrous powder/residue.

(b)(i) Adding \(NaOH(aq)\) to the solution of FA 1 produces a white precipitate of aluminium hydroxide:
\[Al^{3+}(aq) + 3OH^-(aq) \rightarrow Al(OH)_3(s)\]
Because aluminium hydroxide is amphoteric, it dissolves in excess sodium hydroxide to form a colorless solution containing tetrahydroxoaluminate ions:
\[Al(OH)_3(s) + OH^-(aq) \rightarrow [Al(OH)_4]^-(aq)\]

(b)(ii) Adding \(NH_3(aq)\) dropwise also precipitates white \(Al(OH)_3(s)\). However, since aluminium hydroxide does not form a soluble amine complex, the precipitate is insoluble in excess ammonia.

(b)(iii) The addition of nitric acid followed by barium nitrate to the FA 1 solution produces a thick white precipitate of barium sulfate:
\[Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)\]
This precipitate remains insoluble in the acidic solution, confirming the presence of the sulfate ion.

(c)(i) Warming FA 2 with \(NaOH(aq)\) releases ammonia gas from the ammonium ions:
\[NH_4^+(aq) + OH^-(aq) \rightarrow NH_3(g) + H_2O(l)\]
The pungent ammonia gas turns damp red litmus paper blue, confirming the presence of \(NH_4^+\).

(c)(ii) Adding silver nitrate to FA 2 in the presence of nitric acid yields a yellow precipitate of silver iodide:
\[Ag^+(aq) + I^-(aq) \rightarrow AgI(s)\]
Silver iodide is insoluble in both dilute and concentrated aqueous ammonia, confirming the presence of the iodide ion.

(d) Based on these reactions:
- FA 1 contains the aluminium cation, \(Al^{3+}\), and the sulfate anion, \(SO_4^{2-}\).
- FA 2 contains the ammonium cation, \(NH_4^+\), and the iodide anion, \(I^-\).

Marking scheme

Total: 15 Marks

- (a) Heating FA 1 [1 Mark]:
- Award 1 mark for observing liquid droplets/condensation/steam at the top of the tube AND the solid turning into a white powder/residue.

- (b)(i) FA 1 + NaOH [2 Marks]:
- Award 1 mark for observing a white precipitate.
- Award 1 mark for observing that the precipitate dissolves/is soluble in excess to form a colorless solution (Reject: 'clear' instead of 'colorless').

- (b)(ii) FA 1 + NH3 [2 Marks]:
- Award 1 mark for observing a white precipitate.
- Award 1 mark for observing that the precipitate is insoluble in excess.

- (b)(iii) FA 1 + acid + Ba(NO3)2 [2 Marks]:
- Award 1 mark for observing a white precipitate.
- Award 1 mark for stating that the white precipitate is insoluble in nitric acid / remains after acid addition.

- (c)(i) FA 2 + NaOH + heat [2 Marks]:
- Award 1 mark for stating that a pungent gas is evolved OR gas turns damp red litmus paper blue.
- Award 1 mark for identifying the gas as ammonia / \(NH_3\).

- (c)(ii) FA 2 + HNO3 + AgNO3 [2 Marks]:
- Award 1 mark for observing a yellow precipitate.
- Award 1 mark for noting that the yellow precipitate is insoluble in concentrated aqueous ammonia.

- (d) Identifications [4 Marks]:
- Award 1 mark for identifying the cation in FA 1 as \(Al^{3+}\) (accept aluminium).
- Award 1 mark for identifying the anion in FA 1 as \(SO_4^{2-}\) (accept sulfate).
- Award 1 mark for identifying the cation in FA 2 as \(NH_4^+\) (accept ammonium).
- Award 1 mark for identifying the anion in FA 2 as \(I^-\) (accept iodide).
Note: Ionic charges must be correct if formulas are written.

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