Cambridge IAS-Level · Thinka-original Practice Paper

2023 Cambridge IAS-Level Chemistry (9701) Practice Paper with Answers

Thinka Jun 2023 (V3) Cambridge International A Level-Style Mock — Chemistry (9701)

140 marks270 mins2023
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V3) Cambridge International A Level Chemistry (9701) paper. Not affiliated with or reproduced from Cambridge.

Paper 13 (Multiple Choice)

Forty multiple-choice questions with four options. Answer all questions.
40 Question · 40 marks
Question 1 · multipleChoice
1 marks
A \(20\text{ cm}^3\) sample of a gaseous hydrocarbon was completely combusted in \(120\text{ cm}^3\) of oxygen (which is in excess). After cooling to room temperature and pressure, the residual gas had a volume of \(90\text{ cm}^3\). When this residual gas was passed through aqueous sodium hydroxide, the volume of gas remaining was \(50\text{ cm}^3\). What is the molecular formula of the hydrocarbon?
  1. A.\(CH_4\)
  2. B.\(C_2H_4\)
  3. C.\(C_2H_6\)
  4. D.\(C_3H_8\)
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Worked solution

1. The volume decrease when passed through aqueous sodium hydroxide is due to the absorption of carbon dioxide: Volume of \(CO_2\) produced = \(90\text{ cm}^3 - 50\text{ cm}^3 = 40\text{ cm}^3\). 2. Since \(20\text{ cm}^3\) of the hydrocarbon was combusted, the number of carbon atoms per molecule, \(x\), is \(x = 40\text{ cm}^3 / 20\text{ cm}^3 = 2\). 3. The remaining \(50\text{ cm}^3\) of gas is the unreacted (excess) oxygen. 4. Therefore, the volume of oxygen that reacted is: Volume of \(O_2\) reacted = \(120\text{ cm}^3 - 50\text{ cm}^3 = 70\text{ cm}^3\). 5. For the general combustion reaction of a hydrocarbon: \(C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + y/2 H_2O\). The ratio of \(O_2\) reacted to hydrocarbon reacted is: \(x + y/4 = 70\text{ cm}^3 / 20\text{ cm}^3 = 3.5\). 6. Since \(x = 2\): \(2 + y/4 = 3.5\) which gives \(y/4 = 1.5\), so \(y = 6\). Thus, the molecular formula is \(C_2H_6\).

Marking scheme

1 mark for the correct option C. Award 1 mark for calculating the volume of \(CO_2\) as \(40\text{ cm}^3\) and \(O_2\) reacted as \(70\text{ cm}^3\), leading to \(x = 2\) and \(y = 6\). Reject other options as they do not satisfy the gas volume stoichiometry.
Question 2 · multipleChoice
1 marks
Which of the following changes will increase the rate of a gas-phase reaction by increasing the collision frequency of the reactant molecules without changing the activation energy or the Maxwell-Boltzmann energy distribution curve?
  1. A.Increasing the temperature of the reaction mixture
  2. B.Adding a heterogeneous catalyst
  3. C.Decreasing the volume of the reaction vessel
  4. D.Adding an inert gas at constant volume
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Worked solution

Decreasing the volume of the reaction vessel increases the pressure and concentration of the gaseous reactants. This brings the molecules closer together, increasing the frequency of successful collisions and thus increasing the reaction rate. The temperature and the activation energy remain unchanged, so the Maxwell-Boltzmann distribution curve does not change. Increasing the temperature changes the Maxwell-Boltzmann distribution curve by shifting the peak to the right and flattening it. Adding a catalyst lowers the activation energy, but does not increase collision frequency. Adding an inert gas at constant volume does not change the concentrations or collision frequency of the reacting species.

Marking scheme

1 mark for the correct option C. Award 1 mark for identifying that decreasing volume increases concentration and collision frequency without altering temperature (Maxwell-Boltzmann curve) or activation energy. Reject other options.
Question 3 · multipleChoice
1 marks
The table shows the standard enthalpy changes of combustion, \(\Delta H_c^\theta\), for three substances at \(298\text{ K}\). [Substance: \(C(s)\), \(\Delta H_c^\theta = -394\text{ kJ mol}^{-1}\)] [Substance: \(H_2(g)\), \(\Delta H_c^\theta = -286\text{ kJ mol}^{-1}\)] [Substance: \(CH_3OH(l)\), \(\Delta H_c^\theta = -726\text{ kJ mol}^{-1}\)]. What is the standard enthalpy change of formation of liquid methanol, \(CH_3OH(l)\), at \(298\text{ K}\) according to the equation: \(C(s) + 2H_2(g) + \frac{1}{2}O_2(g) \rightarrow CH_3OH(l)\)?
  1. A.\(-240\text{ kJ mol}^{-1}\)
  2. B.\(+240\text{ kJ mol}^{-1}\)
  3. C.\(-966\text{ kJ mol}^{-1}\)
  4. D.\(-1406\text{ kJ mol}^{-1}\)
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Worked solution

Using Hess's Law, the standard enthalpy change of formation can be calculated from enthalpy changes of combustion: \(\Delta H_f^\theta = \sum \Delta H_c^\theta(\text{reactants}) - \sum \Delta H_c^\theta(\text{products})\). Therefore, \(\Delta H_f^\theta(CH_3OH(l)) = [\Delta H_c^\theta(C(s)) + 2 \times \Delta H_c^\theta(H_2(g))] - [\Delta H_c^\theta(CH_3OH(l))] = [-394 + 2(-286)] - [-726] = -966 + 726 = -240\text{ kJ mol}^{-1}\).

Marking scheme

1 mark for the correct option A. Award 1 mark for the correct application of Hess's Law with the correct stoichiometric factor of 2 for hydrogen. Reject B (incorrect sign), C (omitted product combustion), D (added all values directly).
Question 4 · multipleChoice
1 marks
An element \(E\) in Period 3 of the Periodic Table reacts with oxygen to form two different gaseous oxides, \(G\) and \(H\). The oxidation state of \(E\) is \(+4\) in oxide \(G\) and \(+6\) in oxide \(H\). Both oxides react with water to form acidic solutions. What is element \(E\)?
  1. A.Chlorine
  2. B.Phosphorus
  3. C.Silicon
  4. D.Sulfur
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Worked solution

Sulfur is a Period 3 element that forms sulfur dioxide (\(SO_2\), where sulfur is in the \(+4\) oxidation state) and sulfur trioxide (\(SO_3\), where sulfur is in the \(+6\) oxidation state). Both oxides react with water to form acidic solutions (\(H_2SO_3\) and \(H_2SO_4\)). Silicon forms only \(SiO_2\) (oxidation state \(+4\)), which is insoluble. Phosphorus forms oxides in \(+3\) and \(+5\) states. Chlorine forms oxides in \(+1\) and \(+7\) states.

Marking scheme

1 mark for the correct option D. Award 1 mark for identifying sulfur based on its oxides and their respective oxidation states and chemical behavior with water. Reject A, B, and C.
Question 5 · multipleChoice
1 marks
An organic compound \(W\) has the molecular formula \(C_5H_{10}O\). It forms an orange-red precipitate when treated with 2,4-dinitrophenylhydrazine reagent, does not produce a silver mirror when warmed with Tollens' reagent, and does not give a yellow precipitate when treated with alkaline aqueous iodine. What is the identity of compound \(W\)?
  1. A.Pentanal
  2. B.Pentan-2-one
  3. C.Pentan-3-one
  4. D.3-Methylbutanal
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Worked solution

1. The reaction with 2,4-DNPH to form an orange-red precipitate indicates that \(W\) is an aldehyde or a ketone. 2. The negative Tollens' test shows that \(W\) is not an aldehyde, ruling out Pentanal and 3-methylbutanal. 3. The negative triiodomethane (iodoform) test with alkaline aqueous iodine shows that \(W\) does not contain a methyl ketone group (\(CH_3CO-\)). Pentan-2-one is a methyl ketone and would give a positive test, whereas Pentan-3-one does not contain a methyl ketone group and gives a negative test. Therefore, \(W\) is Pentan-3-one.

Marking scheme

1 mark for the correct option C. Award 1 mark for deducing that the compound is a non-methyl ketone based on the three diagnostic tests. Reject A, B, and D.
Question 6 · multipleChoice
1 marks
The following dynamic equilibrium is established in a syringe at room temperature: \(2NO_2(g) \rightleftharpoons N_2O_4(g)\) (where \(NO_2\) is a brown gas and \(N_2O_4\) is a colourless gas). The plunger of the syringe is suddenly pushed in to reduce the volume of the gas mixture to half of its original volume, while the temperature is kept constant. Which statement correctly describes the initial and subsequent colour change of the gas mixture?
  1. A.It immediately becomes darker brown, and then the colour remains constant at this darker shade.
  2. B.It immediately becomes darker brown, and then gradually becomes lighter brown but remains darker than the original equilibrium shade.
  3. C.It immediately becomes lighter brown, and then gradually becomes darker brown.
  4. D.It immediately becomes darker brown, and then gradually becomes lighter than the original equilibrium shade.
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Worked solution

When the volume is halved, the concentration of all gaseous species instantly doubles. Since the concentration of \(NO_2\) doubles, the mixture immediately becomes darker brown. According to Le Chatelier's principle, an increase in pressure shifts the equilibrium to the side with fewer gas molecules (the right-hand side, producing more \(N_2O_4\)). As the system shifts to the right, some \(NO_2\) reacts, so its concentration gradually decreases, making the mixture lighter brown. However, at the new equilibrium, the concentration of \(NO_2\) remains higher than it was in the original equilibrium. Thus, the final colour is still darker than the original equilibrium shade.

Marking scheme

1 mark for the correct option B. Award 1 mark for explaining the initial increase in concentration causing a darker brown colour, followed by a shift in equilibrium to the right reducing the colour intensity, but not below the initial level. Reject other options.
Question 7 · multipleChoice
1 marks
An organic compound \(Y\) is reacted with hot acidified \(\text{KMnO}_4\), yielding a single organic product, \(Z\). When a sample of \(Z\) is added to aqueous sodium carbonate, effervescence is observed and carbon dioxide gas is evolved. Which of the following compounds could be \(Y\)?
  1. A.\(CH_3CH_2CH=CHCH_2CH_3\)
  2. B.\((CH_3)_2C=C(CH_3)_2\)
  3. C.\(CH_3CH_2CH_2CH=CH_2\)
  4. D.\(CH_3CH=C(CH_3)CH_2CH_3\)
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Worked solution

Reaction of an alkene with hot, acidified \(\text{KMnO}_4\) cleaves the \(C=C\) double bond. Since only a single organic product \(Z\) is obtained, the alkene must cleave symmetrically. Since product \(Z\) reacts with \(\text{Na}_2\text{CO}_3\) to evolve \(CO_2\), \(Z\) must be a carboxylic acid. For \(CH_3CH_2CH=CHCH_2CH_3\) (hex-3-ene), symmetrical cleavage yields two molecules of propanoic acid (\(CH_3CH_2COOH\)), which is a carboxylic acid. For \((CH_3)_2C=C(CH_3)_2\), cleavage yields propanone (a ketone), which does not react with carbonates. For \(CH_3CH_2CH_2CH=CH_2\), cleavage yields butanoic acid and \(CO_2\). For \(CH_3CH=C(CH_3)CH_2CH_3\), cleavage yields two different organic products.

Marking scheme

1 mark for the correct option A. Award 1 mark for identifying that hex-3-ene cleaves symmetrically to produce propanoic acid, which is a carboxylic acid. Reject other options.
Question 8 · multipleChoice
1 marks
The successive ionisation energies of a certain Period 3 element, \(T\), are shown in the table. [1st: \(578\text{ kJ mol}^{-1}\)] [2nd: \(1817\text{ kJ mol}^{-1}\)] [3rd: \(2745\text{ kJ mol}^{-1}\)] [4th: \(11577\text{ kJ mol}^{-1}\)] [5th: \(14842\text{ kJ mol}^{-1}\)]. What is the formula of the oxide of \(T\)?
  1. A.\(T_2O\)
  2. B.\(TO\)
  3. C.\(T_2O_3\)
  4. D.\(TO_2\)
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Worked solution

The successive ionisation energies show a massive jump between the 3rd and 4th ionisation energies (from \(2745\) to \(11577\text{ kJ mol}^{-1}\)), indicating that the 4th electron is removed from a closer inner quantum shell. This means element \(T\) has 3 valence electrons and forms a stable \(T^{3+}\) ion. Since oxygen forms a divalent oxide ion (\(O^{2-}\)), the chemical formula of the neutral oxide is \(T_2O_3\).

Marking scheme

1 mark for the correct option C. Award 1 mark for identifying the jump between the 3rd and 4th ionisation energies, deducing a \(+3\) valency, and correctly balancing the charges to write \(T_2O_3\). Reject other options.
Question 9 · multipleChoice
1 marks
When \(10\text{ cm}^3\) of a gaseous hydrocarbon \(\text{C}_x\text{H}_y\) is completely burned in \(70\text{ cm}^3\) of oxygen (an excess), the total volume of gas remaining after combustion and cooling to room temperature is \(50\text{ cm}^3\). Passing this resulting gas mixture through aqueous sodium hydroxide reduces the volume to \(20\text{ cm}^3\). What is the molecular formula of the hydrocarbon?
  1. A.\(\text{CH}_4\)
  2. B.\(\text{C}_2\text{H}_6\)
  3. C.\(\text{C}_3\text{H}_6\)
  4. D.\(\text{C}_3\text{H}_8\)
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Worked solution

1. Under standard room temperature and pressure conditions, water condenses to liquid and has negligible volume.
2. The gas mixture after combustion contains unreacted \(\text{O}_2\) and the product \(\text{CO}_2\). The total volume is \(50\text{ cm}^3\).
3. Aqueous sodium hydroxide absorbs acidic \(\text{CO}_2\) gas. The volume of the remaining unreacted oxygen is \(20\text{ cm}^3\).
4. This means the volume of \(\text{CO}_2\) produced is \(50 - 20 = 30\text{ cm}^3\).
5. Since \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) yielded \(30\text{ cm}^3\) of \(\text{CO}_2\), by mole ratio, \(x = 3\).
6. The volume of \(\text{O}_2\) reacted is \(70 - 20 = 50\text{ cm}^3\). The ratio of hydrocarbon to reacted oxygen is \(10 : 50 = 1 : 5\).
7. Write the balanced equation for combustion: \(\text{C}_3\text{H}_y + (3 + y/4)\text{O}_2 \rightarrow 3\text{CO}_2 + (y/2)\text{H}_2\text{O}\).
8. Set \(3 + y/4 = 5\), which gives \(y/4 = 2\), so \(y = 8\).
9. Therefore, the hydrocarbon is propane, \(\text{C}_3\text{H}_8\).

Marking scheme

Award 1 mark for the correct answer D.
- Reject other choices due to incorrect volume and stoichiometric calculations.
Question 10 · multipleChoice
1 marks
The temperature of a gas mixture is increased from \(T_1\) to \(T_2\). Which statement correctly describes the changes to the Maxwell-Boltzmann distribution curve of molecular kinetic energies and the reaction rate?
  1. A.The peak of the curve shifts to the right and becomes higher; the rate increases because the activation energy decreases.
  2. B.The peak of the curve shifts to the right and becomes lower; the rate increases because a greater fraction of molecules have energy greater than or equal to the activation energy.
  3. C.The area under the curve increases; the rate increases because there are more collisions per unit time.
  4. D.The peak of the curve shifts to the left and becomes lower; the rate increases because the average kinetic energy of the molecules increases.
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Worked solution

When temperature increases:
- The average kinetic energy of the molecules increases, causing the Maxwell-Boltzmann distribution curve to flatten and shift to the right. Consequently, the peak of the curve becomes lower and moves to a higher energy level (to the right).
- The total area under the curve represents the total number of molecules, which remains constant.
- The activation energy (\(E_a\)) is a fixed barrier for a given reaction and does not change with temperature.
- The rate of reaction increases because a significantly larger fraction of molecules now possess kinetic energy equal to or greater than the activation energy (\(E \ge E_a\)), leading to more successful collisions per unit time.

Marking scheme

Award 1 mark for option B.
- Option A is incorrect because the peak becomes lower, not higher, and temperature does not change the activation energy.
- Option C is incorrect because the area under the curve represents the total number of particles, which is constant.
- Option D is incorrect because the peak shifts to the right, not to the left.
Question 11 · multipleChoice
1 marks
Using the standard enthalpy changes of combustion given below, what is the standard enthalpy change of formation of liquid ethanol, \(\text{C}_2\text{H}_5\text{OH}(l)\)?

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

1. Write the target formation equation: \(2\text{C}(s) + 3\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l)\).
2. According to Hess's law, using enthalpy changes of combustion:
\(\Delta H_f^\theta = \sum \Delta H_c^\theta(\text{reactants}) - \sum \Delta H_c^\theta(\text{products})\)
3. Substitute the values:
\(\Delta H_f^\theta = [2 \times (-394) + 3 \times (-286)] - (-1367)\)
\(\Delta H_f^\theta = [-788 - 858] + 1367\)
\(\Delta H_f^\theta = -1646 + 1367 = -279\text{ kJ mol}^{-1}\).

Marking scheme

Award 1 mark for the correct calculation leading to -279 kJ/mol (Option B).
- Option A represents omitting multiplier coefficients: \(-394 - 286 - (-1367) = -687\).
- Option C represents using product minus reactants (wrong sign): \(+279\).
- Option D represents failing to subtract the combustion enthalpy of the product: \(-1646\).
Question 12 · multipleChoice
1 marks
Three Period 3 oxides are added separately to excess water. Which row correctly describes the pH of the resulting mixtures at \(25^\circ\text{C}\)?
  1. A.Oxide 1: \(\text{Na}_2\text{O}\) (pH 13) | Oxide 2: \(\text{Al}_2\text{O}_3\) (pH 5) | Oxide 3: \(\text{P}_4\text{O}_{10}\) (pH 2)
  2. B.Oxide 1: \(\text{MgO}\) (pH 13) | Oxide 2: \(\text{SiO}_2\) (pH 7) | Oxide 3: \(\text{SO}_3\) (pH 1)
  3. C.Oxide 1: \(\text{MgO}\) (pH 9) | Oxide 2: \(\text{Al}_2\text{O}_3\) (pH 9) | Oxide 3: \(\text{P}_4\text{O}_{10}\) (pH 2)
  4. D.Oxide 1: \(\text{MgO}\) (pH 9) | Oxide 2: \(\text{SiO}_2\) (pH 7) | Oxide 3: \(\text{SO}_3\) (pH 1)
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Worked solution

- \(\text{MgO}\) reacts sparingly with water to form a saturated solution of \(\text{Mg(OH)}_2\), which is weakly alkaline with a pH of around 9.
- \(\text{SiO}_2\) is a giant covalent structure that is completely insoluble in water. It does not react, so the pH of the water remains neutral (pH 7).
- \(\text{SO}_3\) reacts violently with water to form the strong acid \(\text{H}_2\text{SO}_4\), resulting in a strongly acidic solution with a pH of around 1.
- Hence, the row in Option D is correct.

Marking scheme

Award 1 mark for Option D.
- Reject option A because \(\text{Al}_2\text{O}_3\) does not dissolve to give an acidic pH of 5.
- Reject option B because \(\text{SiO}_2\) is insoluble and does not give a pH of 5, and \(\text{MgO}\) is not strongly alkaline (pH 13).
- Reject option C because \(\text{Al}_2\text{O}_3\) is insoluble and does not form an alkaline solution (pH 9).
Question 13 · multipleChoice
1 marks
An organic compound **X** has the molecular formula \(\text{C}_4\text{H}_8\text{O}\). **X** reacts with 2,4-dinitrophenylhydrazine reagent to form an orange precipitate, but it does not produce a red precipitate when heated with Fehling's solution. Which structure is a possible isomer of **X**?
  1. A.\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\)
  2. B.\(\text{CH}_3\text{COCH}_2\text{CH}_3\)
  3. C.\(\text{CH}_3\text{CH=CHCH}_2\text{OH}\)
  4. D.\((\text{CH}_3)_2\text{CHCHO}\)
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Worked solution

1. The positive test with 2,4-dinitrophenylhydrazine (2,4-DNPH) indicates that **X** contains a carbonyl group (it is either an aldehyde or a ketone).
2. The negative test with Fehling's solution indicates that **X** is not an aldehyde. Therefore, **X** must be a ketone.
3. The only ketone with the molecular formula \(\text{C}_4\text{H}_8\text{O}\) is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\).
4. Option B shows butanone.

Marking scheme

Award 1 mark for Option B.
- Options A and D are aldehydes, which would react with Fehling's solution to form a red precipitate.
- Option C is an unsaturated alcohol, which does not react with 2,4-DNPH.
Question 14 · multipleChoice
1 marks
For the following dynamic equilibrium:

\(2\text{A}(g) + \text{B}(g) \rightleftharpoons 2\text{C}(g) \quad \Delta H = -92\text{ kJ mol}^{-1}\)

Which change will increase the numerical value of the equilibrium constant, \(K_c\)?
  1. A.decreasing the temperature
  2. B.increasing the temperature
  3. C.increasing the pressure
  4. D.adding a catalyst
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Worked solution

1. The value of the equilibrium constant, \(K_c\), is only affected by a change in temperature.
2. Since the forward reaction is exothermic (\(\Delta H < 0\)), according to Le Chatelier's principle, decreasing the temperature shifts the equilibrium to the right to favor the exothermic direction.
3. Shifting the position of equilibrium to the right increases the concentration of products relative to reactants, which increases the value of \(K_c\).
4. Changing pressure or adding a catalyst has no effect on the value of \(K_c\).

Marking scheme

Award 1 mark for Option A.
- Option B is incorrect because increasing temperature favors the endothermic back-reaction, decreasing \(K_c\).
- Options C and D are incorrect because pressure and catalysts do not change the value of \(K_c\).
Question 15 · multipleChoice
1 marks
An organic compound **Y** with the molecular formula \(\text{C}_4\text{H}_{10}\text{O}_2\) contains two hydroxyl groups. When **Y** is heated under reflux with an excess of acidified potassium dichromate(VI), the organic product formed is a diketone. Which diol is compound **Y**?
  1. A.butane-1,2-diol
  2. B.butane-1,3-diol
  3. C.butane-1,4-diol
  4. D.butane-2,3-diol
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Worked solution

1. Under reflux with excess acidified \(\text{K}_2\text{Cr}_2\text{O}_7\):
- Primary alcohol groups oxidize to carboxylic acids.
- Secondary alcohol groups oxidize to ketones.
- Tertiary alcohol groups do not oxidize.
2. Since the product is a diketone (containing two ketone groups), compound **Y** must contain two secondary alcohol groups.
3. Let us evaluate the options:
- butane-1,2-diol: contains one primary (C1) and one secondary (C2) alcohol group. Its oxidation product would be a keto-acid.
- butane-1,3-diol: contains one primary (C1) and one secondary (C3) alcohol group. Its oxidation product would be a keto-acid.
- butane-1,4-diol: contains two primary alcohol groups. Its oxidation product would be a dicarboxylic acid.
- butane-2,3-diol: \(\text{CH}_3\text{CH(OH)CH(OH)CH}_3\), contains two secondary alcohol groups. Its oxidation product is butane-2,3-dione, which is a diketone.
4. Therefore, compound **Y** is butane-2,3-diol.

Marking scheme

Award 1 mark for Option D.
- Reject options A, B, and C as they contain primary alcohol groups which would oxidize to carboxylic acids rather than ketones.
Question 16 · multipleChoice
1 marks
Equal amounts of four different halogenoalkanes are added to separate test tubes containing aqueous silver nitrate in ethanol at \(50^\circ\text{C}\). Which halogenoalkane will form a precipitate the fastest?
  1. A.1-chlorobutane
  2. B.1-bromobutane
  3. C.1-iodobutane
  4. D.2-iodo-2-methylpropane
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Worked solution

1. The rate of nucleophilic substitution (hydrolysis) of halogenoalkanes depends on two main factors: the strength of the C-X bond and the stability of the intermediate carbocation (tertiary > secondary > primary).
2. Comparing the halogens: The carbon-iodine bond (\(\text{C-I}\)) is weaker and easier to break than the carbon-bromine (\(\text{C-Br}\)) and carbon-chlorine (\(\text{C-Cl}\)) bonds due to the larger size of the iodine atom and less orbital overlap.
3. Comparing the alkyl structures: Tertiary halogenoalkanes (like those containing a tertiary carbon) hydrolyze much faster via the \(\text{S}_\text{N}1\) mechanism because of the stable tertiary carbocation intermediate compared to primary halogenoalkanes.
4. Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) will undergo hydrolysis at the fastest rate, releasing iodide ions quickly to form a yellow precipitate of silver iodide.

Marking scheme

Award 1 mark for Option D.
- Reject options A and B because they contain C-Cl bonds which are much stronger than C-I and C-Br bonds.
- Reject option C because 1-iodobutane is a primary halogenoalkane, which reacts more slowly than the tertiary isomer 2-iodo-2-methylpropane.
Question 17 · multipleChoice
1 marks
An alloy containing only zinc and copper has a mass of 2.00 g. When this sample is treated with an excess of dilute hydrochloric acid, only the zinc reacts, producing \(540\text{ cm}^3\) of hydrogen gas measured at room temperature and pressure (r.t.p.).

What is the percentage by mass of copper in the alloy?
[Molar volume of a gas at r.t.p. = \(24.0\text{ dm}^3\text{ mol}^{-1}\); \(A_r\text{ (Zn) = 65.4}\); \(A_r\text{ (Cu) = 63.5}\)]
  1. A.26.4%
  2. B.73.6%
  3. C.13.2%
  4. D.48.2%
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Worked solution

1. Calculate the number of moles of hydrogen gas produced:
\text{Moles of } \(H_2 = \frac{0.540\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.0225\text{ mol}\)

2. Write the balanced equation for the reaction:
\(Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)\)
Since 1 mole of Zn produces 1 mole of \(H_2\), the moles of Zn in the alloy is also 0.0225 mol.

3. Calculate the mass of zinc:
\text{Mass of Zn} = \(0.0225\text{ mol} \times 65.4\text{ g mol}^{-1} = 1.4715\text{ g}\)

4. Calculate the mass of copper in the alloy:
\text{Mass of Cu} = \(2.00\text{ g} - 1.4715\text{ g} = 0.5285\text{ g}\)

5. Calculate the percentage of copper by mass:
\% \text{ Cu} = \( \frac{0.5285\text{ g}}{2.00\text{ g}} \times 100 = 26.4\\%\)

Marking scheme

1 mark for the correct calculation leading to option A. Option B represents the percentage of zinc. Option C and D represent incorrect stoichiometry calculations.
Question 18 · multipleChoice
1 marks
For many reactions, a modest rise in temperature of 10 K can approximately double the rate of reaction.

Which statement describes the main reason for this observation?
  1. A.The average kinetic energy of the reactant molecules doubles.
  2. B.The activation energy, \(E_a\), of the reaction is significantly lowered.
  3. C.The frequency of collisions between reactant molecules increases by a factor of two.
  4. D.There is a significant increase in the fraction of molecules with energy greater than or equal to the activation energy.
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Worked solution

According to the Maxwell-Boltzmann distribution, a small increase in temperature shifts the curve to the right and flattens it slightly. This leads to a very large increase in the area under the curve to the right of the activation energy (\(E_a\)), meaning a much larger fraction of molecules have energy \( \ge E_a \) and can react upon collision. The average kinetic energy and collision frequency only increase slightly (proportional to \( \sqrt{T} \) and \( T \) respectively), and \(E_a\) remains unchanged.

Marking scheme

1 mark for selecting option D.
Question 19 · multipleChoice
1 marks
Hydrazine, \(N_2H_4(l)\), undergoes combustion according to the equation below:

\(N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(l) \quad \Delta H_c^\ominus = -622\text{ kJ mol}^{-1}\)

The standard enthalpy change of formation of liquid water, \(\Delta H_f^\ominus [H_2O(l)]\), is \(-286\text{ kJ mol}^{-1}\).

What is the standard enthalpy change of formation of liquid hydrazine, \(\Delta H_f^\ominus [N_2H_4(l)]\)?
  1. A.\(-908\text{ kJ mol}^{-1}\)
  2. B.\(-50\text{ kJ mol}^{-1}\)
  3. C.\(+50\text{ kJ mol}^{-1}\)
  4. D.\(+336\text{ kJ mol}^{-1}\)
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Worked solution

Using Hess's Law:
\(\Delta H_c^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})\)

For the combustion of hydrazine:
\(\Delta H_c^\ominus = [2 \times \Delta H_f^\ominus(H_2O(l)) + \Delta H_f^\ominus(N_2(g))] - [\Delta H_f^\ominus(N_2H_4(l)) + \Delta H_f^\ominus(O_2(g))]\)

Given that elements in their standard states have \(\Delta H_f^\ominus = 0\):
\(-622 = [2 \times (-286) + 0] - [\Delta H_f^\ominus(N_2H_4(l)) + 0]\)
\(-622 = -572 - \Delta H_f^\ominus(N_2H_4(l))\)
\(\Delta H_f^\ominus(N_2H_4(l)) = -572 - (-622) = +50\text{ kJ mol}^{-1}\).

Marking scheme

1 mark for the correct enthalpy cycle and calculation leading to +50 kJ/mol (C).
Question 20 · multipleChoice
1 marks
An oxide of a Period 3 element, \(X\), is a solid at room temperature and pressure. When a sample of this oxide is added to water, it reacts vigorously to form a strongly acidic solution.

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

Phosphorus forms the oxide \(P_4O_{10}\), which is a solid at room temperature and pressure. It reacts vigorously with water to form phosphoric(V) acid, \(H_3PO_4\), which is a strongly acidic solution.
Sodium oxide (\(Na_2O\)) reacts to form a strongly alkaline solution. Aluminium oxide (\(Al_2O_3\)) and silicon dioxide (\(SiO_2\)) are insoluble solids that do not react with water.

Marking scheme

1 mark for identifying phosphorus as element X.
Question 21 · multipleChoice
1 marks
An organic compound, \(Y\), reacts with 2,4-dinitrophenylhydrazine reagent to form an orange-red precipitate. However, \(Y\) does not produce a silver mirror when warmed with Tollens' reagent.

Which compound could be \(Y\)?
  1. A.butanal
  2. B.butanone
  3. C.butan-1-ol
  4. D.methylpropan-2-ol
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Worked solution

The reaction with 2,4-dinitrophenylhydrazine (2,4-DNPH) to form an orange-red precipitate indicates that compound \(Y\) contains a carbonyl group (it is either an aldehyde or a ketone). Since it does not react with Tollens' reagent (no silver mirror), it is not an aldehyde. Therefore, \(Y\) must be a ketone. Butanone is a ketone, whereas butanal is an aldehyde, and butan-1-ol and methylpropan-2-ol are alcohols.

Marking scheme

1 mark for selecting the ketone option (B).
Question 22 · multipleChoice
1 marks
Consider the reversible reaction involved in the Contact process:

\(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -197\text{ kJ mol}^{-1}\)

Which change will increase the equilibrium yield of sulfur trioxide, \(SO_3(g)\), but leave the value of the equilibrium constant, \(K_c\), unchanged?
  1. A.decreasing the temperature
  2. B.increasing the pressure of the system
  3. C.adding a vanadium(V) oxide catalyst
  4. D.increasing the volume of the reaction vessel at constant temperature
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Worked solution

The equilibrium constant, \(K_c\), is only affected by temperature. Therefore, changing pressure (option B) or adding a catalyst (option C) will leave \(K_c\) unchanged. Since the forward reaction involves a decrease in the number of moles of gas (3 moles of reactants forming 2 moles of product), increasing the overall pressure will shift the position of equilibrium to the right, thereby increasing the equilibrium yield of \(SO_3(g)\). Decreasing temperature (option A) increases both the yield and \(K_c\) since the forward reaction is exothermic. Adding a catalyst (option C) increases the rate of reaction but does not change the equilibrium yield.

Marking scheme

1 mark for selecting option B.
Question 23 · multipleChoice
1 marks
A gaseous hydrocarbon, \(W\), has a relative molecular mass of 70.0. When a sample of \(W\) is completely burned in excess oxygen, the volume of carbon dioxide gas produced is exactly five times the volume of hydrocarbon \(W\) burned, with both gas volumes measured under the same conditions of temperature and pressure.

What is the molecular formula of hydrocarbon \(W\)?
[\(A_r\text{ (C) = 12.0}\); \(A_r\text{ (H) = 1.0}\)]
  1. A.\(C_5H_8\)
  2. B.\(C_5H_{10}\)
  3. C.\(C_5H_{12}\)
  4. D.\(C_6H_{10}\)
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Worked solution

1. By Avogadro's Law, equal volumes of gases contain equal numbers of moles under the same conditions. Since the volume ratio of \(CO_2\) to \(W\) is 5:1, each molecule of \(W\) must contain exactly 5 carbon atoms (molecular formula \(C_5H_y\)).

2. Calculate the number of hydrogen atoms using the relative molecular mass:
\text{Mass of } \(5 \times C = 5 \times 12.0 = 60.0\)
\text{Mass of H atoms} = \(70.0 - 60.0 = 10.0\)
Since the relative atomic mass of H is 1.0, there must be 10 hydrogen atoms.

3. Therefore, the molecular formula of \(W\) is \(C_5H_{10}\).

Marking scheme

1 mark for selecting option B.
Question 24 · multipleChoice
1 marks
The reaction between peroxodisulfate ions, \(S_2O_8^{2-}\), and iodide ions, \(I^-\), is catalyzed by iron(III) ions, \(Fe^{3+}\):

\(S_2O_8^{2-}(aq) + 2I^-(aq) \xrightarrow{Fe^{3+}} 2SO_4^{2-}(aq) + I_2(aq)\)

Which statement correctly describes a feature of this catalyzed reaction?
  1. A.In the first step of the mechanism, \(Fe^{3+}\) ions are oxidized to \(Fe^{4+}\) ions.
  2. B.The reaction is an example of heterogeneous catalysis.
  3. C.The \(Fe^{3+}\) ions oxidize the \(I^-\) ions to \(I_2\) and are themselves reduced to \(Fe^{2+}\) ions.
  4. D.The activation energy of the catalyzed pathway is higher than that of the uncatalyzed pathway.
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Worked solution

This reaction is an example of homogeneous catalysis because all reactants and the catalyst are in the same phase (aqueous). The uncatalyzed reaction is slow because both reactant ions are negatively charged (\(S_2O_8^{2-}\) and \(I^-\)), resulting in high electrostatic repulsion and a high activation energy. The \(Fe^{3+}\) catalyst provides an alternative pathway with a lower activation energy. In the first step, \(Fe^{3+}\) ions oxidize \(I^-\) ions: \(2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(aq)\), and are reduced to \(Fe^{2+}\). In the second step, \(Fe^{2+}\) is reoxidized back to \(Fe^{3+}\) by peroxodisulfate: \(2Fe^{2+}(aq) + S_2O_8^{2-}(aq) \rightarrow 2Fe^{3+}(aq) + 2SO_4^{2-}(aq)\).

Marking scheme

1 mark for selecting option C.
Question 25 · multipleChoice
1 marks
A \(20.0\text{ cm}^3\) sample of a gaseous hydrocarbon, \(C_xH_y\), was mixed with excess oxygen and exploded. After cooling to room temperature, the decrease in volume of the gas mixture was found to be \(50.0\text{ cm}^3\). When the remaining gas was shaken with excess aqueous sodium hydroxide, there was a further contraction in volume of \(40.0\text{ cm}^3\). What is the molecular formula of the hydrocarbon?
  1. A.\(CH_4\)
  2. B.\(C_2H_4\)
  3. C.\(C_2H_6\)
  4. D.\(C_3H_8\)
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Worked solution

Shaking with excess aqueous NaOH absorbs \(CO_2\). The contraction of \(40.0\text{ cm}^3\) represents the volume of \(CO_2\) produced. Since \(20.0\text{ cm}^3\) of \(C_xH_y\) produced \(40.0\text{ cm}^3\) of \(CO_2\), each molecule of hydrocarbon contains 2 carbon atoms, so \(x = 2\). The combustion equation is \(C_2H_y(g) + (2 + y/4)O_2(g) \rightarrow 2CO_2(g) + y/2 H_2O(l)\). Total initial gaseous volume is \(20.0\) (hydrocarbon) + \(V(O_2)\). Total final gaseous volume (after cooling, before NaOH treatment) is \(40.0\) (\(CO_2\)) + \(V(O_2) - 20(2 + y/4)\). The decrease in volume is \(50.0 = (20.0 + V(O_2)) - (40.0 + V(O_2) - 40.0 - 5y) = 20.0 + 5y\). Solving for \(y\) gives \(5y = 30.0\) which means \(y = 6\). Thus, the formula is \(C_2H_6\).

Marking scheme

Award 1 mark for the correct answer C. Accept only C.
Question 26 · multipleChoice
1 marks
For a gaseous reaction, the temperature is increased from \(T_1\) to \(T_2\). Which statement correctly describes the changes in the Maxwell-Boltzmann distribution curve and the reaction rate?
  1. A.The peak of the curve shifts to a higher energy, the peak height decreases, and the reaction rate increases because the activation energy, \(E_a\), is lowered.
  2. B.The peak of the curve shifts to a higher energy, the peak height increases, and the reaction rate increases because a larger fraction of molecules have energy greater than or equal to \(E_a\).
  3. C.The peak of the curve shifts to a higher energy, the peak height decreases, and the reaction rate increases because a larger fraction of molecules have energy greater than or equal to \(E_a\).
  4. D.The peak of the curve shifts to a lower energy, the peak height decreases, and the reaction rate increases because the frequency of collisions increases.
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Worked solution

When temperature increases, the average kinetic energy of the molecules increases, shifting the peak of the Maxwell-Boltzmann distribution curve to a higher energy (to the right). Since the total number of molecules is constant, the peak height must decrease to maintain the same area under the curve. The activation energy, \(E_a\), is not altered by temperature. The rate of reaction increases because a larger fraction of the molecules possess energy greater than or equal to \(E_a\).

Marking scheme

Award 1 mark for the correct answer C. Accept only C.
Question 27 · multipleChoice
1 marks
The standard enthalpies of combustion, \(\Delta H_c^\theta\), of carbon (graphite), hydrogen gas, and liquid propan-1-ol, \(C_3H_7OH(l)\), are \(-394\text{ kJ mol}^{-1}\), \(-286\text{ kJ mol}^{-1}\), and \(-2021\text{ kJ mol}^{-1}\) respectively. What is the standard enthalpy change of formation, \(\Delta H_f^\theta\), of liquid propan-1-ol?
  1. A.\(-305\text{ kJ mol}^{-1}\)
  2. B.\(-1341\text{ kJ mol}^{-1}\)
  3. C.\(+305\text{ kJ mol}^{-1}\)
  4. D.\(-4347\text{ kJ mol}^{-1}\)
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Worked solution

The chemical equation for the formation of liquid propan-1-ol is \(3C(s) + 4H_2(g) + 0.5O_2(g) \rightarrow C_3H_7OH(l)\). Using Hess's Law with enthalpies of combustion, we have \(\Delta H_f^\theta = 3 \times \Delta H_c^\theta(C) + 4 \times \Delta H_c^\theta(H_2) - \Delta H_c^\theta(C_3H_7OH)\). Substituting the values gives \(\Delta H_f^\theta = [3(-394) + 4(-286)] - [-2021] = [-1182 - 1144] + 2021 = -2326 + 2021 = -305\text{ kJ mol}^{-1}\).

Marking scheme

Award 1 mark for the correct answer A. Accept only A.
Question 28 · multipleChoice
1 marks
An element \(X\) in Period 3 of the Periodic Table forms an oxide \(X_2O_3\). Which statement about this oxide and the element \(X\) is correct?
  1. A.The oxide \(X_2O_3\) reacts with aqueous sodium hydroxide but does not react with dilute hydrochloric acid.
  2. B.The oxide \(X_2O_3\) has a giant covalent macromolecular structure.
  3. C.The element \(X\) has a lower first ionisation energy than the element immediately to its left.
  4. D.The oxide \(X_2O_3\) reacts with water to form an alkaline solution with a pH of approximately 13.
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Worked solution

Element \(X\) is Aluminium (\(Al\)) which forms the amphoteric oxide \(Al_2O_3\). It reacts with both acids and alkalis, making option A incorrect. It has a giant ionic structure with significant covalent character, not giant covalent, making B incorrect. The element immediately to its left is Magnesium (\(Mg\)). Aluminium has a lower first ionisation energy than Magnesium because its outer electron is in a \(3p\) orbital, which is more shielded and higher in energy than the \(3s\) electrons of Magnesium, making C correct. \(Al_2O_3\) is insoluble in water, making D incorrect.

Marking scheme

Award 1 mark for the correct answer C. Accept only C.
Question 29 · multipleChoice
1 marks
An organic compound \(Y\) is treated with HCN in the presence of NaCN. The resulting product is then heated under reflux with dilute hydrochloric acid to yield 2-hydroxy-2-methylbutanoic acid. What is the structure of the starting compound \(Y\)?
  1. A.butanone
  2. B.propanal
  3. C.propanone
  4. D.butanal
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Worked solution

The final product is 2-hydroxy-2-methylbutanoic acid, which is \(CH_3CH_2C(OH)(CH_3)COOH\). This was formed by the acid hydrolysis of the nitrile group in 2-hydroxy-2-methylbutanenitrile, \(CH_3CH_2C(OH)(CH_3)CN\). This hydroxynitrile is formed by the nucleophilic addition of HCN to the carbonyl group of compound \(Y\). Replacing the \(-\text{OH}\) and \(-\text{CN}\) groups on the central carbon atom with a double-bonded oxygen (\(=\text{O}\)) gives butanone, \(CH_3CH_2COCH_3\).

Marking scheme

Award 1 mark for the correct answer A. Accept only A.
Question 30 · multipleChoice
1 marks
In an experiment, \(2.00\text{ mol}\) of nitrogen dioxide, \(NO_2(g)\), is placed in a sealed \(1.00\text{ dm}^3\) flask at a constant temperature. The \(NO_2(g)\) dimerises to form dinitrogen tetroxide, \(N_2O_4(g)\), according to the equation: \(2NO_2(g) \rightleftharpoons N_2O_4(g)\). At equilibrium, the flask contains \(0.60\text{ mol}\) of \(N_2O_4(g)\). What is the value of the equilibrium constant, \(K_c\), at this temperature?
  1. A.\(0.94\text{ dm}^3\text{ mol}^{-1}\)
  2. B.\(0.75\text{ dm}^3\text{ mol}^{-1}\)
  3. C.\(1.07\text{ dm}^3\text{ mol}^{-1}\)
  4. D.\(3.75\text{ dm}^3\text{ mol}^{-1}\)
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Worked solution

Since the volume is \(1.00\text{ dm}^3\), moles and concentrations are numerically equal. At equilibrium, the amount of \(N_2O_4\) formed is \(0.60\text{ mol}\). The amount of \(NO_2\) reacted is \(2 \times 0.60 = 1.20\text{ mol}\). Therefore, the equilibrium amount of \(NO_2\) remaining is \(2.00 - 1.20 = 0.80\text{ mol}\). The equilibrium constant expression is \(K_c = [N_2O_4] / [NO_2]^2\). Substituting the concentrations gives \(K_c = 0.60 / (0.80)^2 = 0.60 / 0.64 = 0.9375 \approx 0.94\text{ dm}^3\text{ mol}^{-1}\).

Marking scheme

Award 1 mark for the correct answer A. Accept only A.
Question 31 · multipleChoice
1 marks
Which molecule or ion has both a bond angle of less than \(109.5^\circ\) and a non-linear shape?
  1. A.\(BF_3\)
  2. B.\(NH_4^+\)
  3. C.\(SF_6\)
  4. D.\(NH_2^-\)
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Worked solution

BF3 is trigonal planar with a bond angle of \(120^\circ\). NH4+ is tetrahedral with a bond angle of \(109.5^\circ\). SF6 is octahedral with bond angles of \(90^\circ\) and \(180^\circ\). NH2- has 2 bonding pairs and 2 lone pairs on the nitrogen atom. Its shape is non-linear (bent), and the repulsion from the 2 lone pairs compresses the H-N-H bond angle to approximately \(104.5^\circ\), which is less than \(109.5^\circ\).

Marking scheme

Award 1 mark for the correct answer D. Accept only D.
Question 32 · multipleChoice
1 marks
An alcohol with the molecular formula \(C_4H_{10}O\) is heated under reflux with acidified potassium dichromate(VI) to form an organic product \(Z\). Product \(Z\) reacts with alkaline aqueous iodine to form a yellow precipitate. Which alcohol is it?
  1. A.butan-1-ol
  2. B.butan-2-ol
  3. C.2-methylpropan-1-ol
  4. D.2-methylpropan-2-ol
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Worked solution

Butan-2-ol is a secondary alcohol that oxidises under reflux with acidified potassium dichromate(VI) to form the ketone butanone, \(CH_3COCH_2CH_3\). Butanone contains a methyl carbonyl group (\(CH_3C=O\)) and therefore undergoes the tri-iodomethane reaction with alkaline aqueous iodine to give a yellow precipitate of \(CHI_3\). Butan-1-ol and 2-methylpropan-1-ol are primary alcohols and oxidise to carboxylic acids which do not react. 2-methylpropan-2-ol is a tertiary alcohol and does not undergo oxidation.

Marking scheme

Award 1 mark for the correct answer B. Accept only B.
Question 33 · multipleChoice
1 marks
A mixture of methane (\(\text{CH}_4\)) and propane (\(\text{C}_3\text{H}_8\)) with a total volume of \(40\text{ cm}^3\) is completely combusted in excess oxygen. The total volume of gaseous products, measured at a high temperature where water is entirely in the gaseous state, is \(180\text{ cm}^3\). All gas volumes are measured at the same temperature and pressure.

What is the volume of propane in the original mixture?
  1. A.\(10\text{ cm}^3\)
  2. B.\(15\text{ cm}^3\)
  3. C.\(25\text{ cm}^3\)
  4. D.\(30\text{ cm}^3\)
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Worked solution

Let \(x\) be the volume of \(\text{CH}_4\) and \(y\) be the volume of \(\text{C}_3\text{H}_8\).
We are given:
\(x + y = 40\) (Equation 1)

The balanced combustion equations are:
1) \(\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)\)
Each mole of \(\text{CH}_4\) produces \(1 + 2 = 3\) moles of gaseous products.
Thus, \(x\text{ cm}^3\) of \(\text{CH}_4\) produces \(3x\text{ cm}^3\) of gaseous products.

2) \(\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)\)
Each mole of \(\text{C}_3\text{H}_8\) produces \(3 + 4 = 7\) moles of gaseous products.
Thus, \(y\text{ cm}^3\) of \(\text{C}_3\text{H}_8\) produces \(7y\text{ cm}^3\) of gaseous products.

The total volume of gaseous products is:
\(3x + 7y = 180\) (Equation 2)

Substitute \(x = 40 - y\) into Equation 2:
\(3(40 - y) + 7y = 180\)
\(120 - 3y + 7y = 180\)
\(4y = 60 \Rightarrow y = 15\text{ cm}^3\).

Therefore, the volume of propane in the original mixture is \(15\text{ cm}^3\).

Marking scheme

Award 1 mark for the correct calculation leading to B (15 cm³).
- Method mark: Set up the simultaneous equations using the stoichiometric ratios of gaseous products (1:3 for methane and 1:7 for propane).
- Accuracy mark: Solve for the volume of propane to get 15 cm³.
Question 34 · multipleChoice
1 marks
A chemical reaction is carried out at temperature \(T_1\). The temperature is then increased to \(T_2\).

Which statement correctly explains the increase in the rate of reaction at the higher temperature, \(T_2\)?
  1. A.The activation energy of the reaction decreases at \(T_2\).
  2. B.The peak of the Maxwell–Boltzmann distribution curve shifts to the right and becomes higher.
  3. C.The total area under the Maxwell–Boltzmann distribution curve increases.
  4. D.The proportion of molecular collisions with energy greater than or equal to the activation energy increases.
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Worked solution

At a higher temperature:
- The activation energy, \(E_a\), of the reaction does not change (ruling out option A).
- The peak of the Maxwell–Boltzmann distribution curve shifts to the right and becomes lower/flatter, not higher (ruling out option B).
- The total area under the curve represents the total number of particles, which remains constant (ruling out option C).
- Because the curve shifts to higher energies, a larger proportion of the molecules possess kinetic energies greater than or equal to the activation energy (\(E \ge E_a\)), leading to a higher frequency of successful collisions and thus an increased rate of reaction (confirming option D).

Marking scheme

Award 1 mark for D.
- Reject A: Activation energy is independent of temperature.
- Reject B: The peak flattens and gets lower at higher temperatures.
- Reject C: The total number of particles is constant, so the area is unchanged.
Question 35 · multipleChoice
1 marks
The standard enthalpy changes of combustion, \(\Delta H_c^\ominus\), for carbon, hydrogen, and propan-1-ol are given in the table.

$$\begin{array}{|c|c|}
\hline
\text{Substance} & \Delta H_c^\ominus / \text{kJ mol}^{-1} \\
\hline
\text{C}(s) & -394 \\
\text{H}_2(g) & -286 \\
\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l) & -2021 \\
\hline
\end{array}$$

What is the standard enthalpy change of formation, \(\Delta H_f^\ominus\), of propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l)\)?
  1. A.\(-4347\text{ kJ mol}^{-1}\)
  2. B.\(-305\text{ kJ mol}^{-1}\)
  3. C.\(-19\text{ kJ mol}^{-1}\)
  4. D.\(+305\text{ kJ mol}^{-1}\)
Show answer & marking scheme

Worked solution

The chemical equation for the standard enthalpy change of formation of propan-1-ol is:
\(3\text{C}(s) + 4\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l)\)

According to Hess's Law, using standard enthalpy changes of combustion:
\(\Delta H_f^\ominus = \sum \Delta H_c^\ominus(\text{reactants}) - \sum \Delta H_c^\ominus(\text{products})\)

\(\Delta H_f^\ominus = [3 \times \Delta H_c^\ominus(\text{C}) + 4 \times \Delta H_c^\ominus(\text{H}_2)] - \Delta H_c^\ominus(\text{propan-1-ol})\)
\(\Delta H_f^\ominus = [3(-394) + 4(-286)] - [-2021]\)
\(\Delta H_f^\ominus = [-1182 - 1144] + 2021\)
\(\Delta H_f^\ominus = -2326 + 2021 = -305\text{ kJ mol}^{-1}\).

Marking scheme

Award 1 mark for B.
- Correct stoichiometry for elements in standard formation: 3 moles of C and 4 moles of H₂.
- Correct Hess's Law equation: Reactants - Products for combustion data.
- Calculation: [3(-394) + 4(-286)] - (-2021) = -305 kJ mol⁻¹.
Question 36 · multipleChoice
1 marks
An element, \(X\), is in Period 3 of the Periodic Table. It forms an oxide, \(X_2\text{O}_3\), which is amphoteric.

Which statement about element \(X\) and its compounds is correct?
  1. A.The chloride of \(X\) has a simple molecular structure and hydrolyses in water to form an acidic solution.
  2. B.The oxide of \(X\) has a giant covalent structure and is soluble in water.
  3. C.Element \(X\) is a non-metal with high electrical conductivity.
  4. D.Element \(X\) reacts vigorously with cold water to form an alkaline solution.
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Worked solution

The element \(X\) in Period 3 that forms an amphoteric oxide of formula \(X_2\text{O}_3\) is aluminium (\(\text{Al}\)).
- Option A: Aluminium chloride (\(\text{Al}_2\text{Cl}_6\) as a dimer in vapor state) has a simple molecular structure and undergoes vigorous hydrolysis in water to produce an acidic solution (pH \(\approx 3\)). This is correct.
- Option B: Aluminium oxide (\(\text{Al}_2\text{O}_3\)) has a giant ionic structure and is insoluble in water. Thus, B is incorrect.
- Option C: Aluminium is a metal, not a non-metal. Silicon is the non-metal/metalloid with a high melting point, and silicon is giant covalent. Thus, C is incorrect.
- Option D: Aluminium does not react with cold water because it is protected by a tough, unreactive layer of aluminium oxide. Thus, D is incorrect.

Marking scheme

Award 1 mark for A.
- Identify element X as aluminium.
- Recall that aluminium chloride dimerizes to form a simple molecular structure and hydrolyses in water to form an acidic solution.
Question 37 · multipleChoice
1 marks
An organic compound \(Y\) has the molecular formula \(\text{C}_4\text{H}_8\text{O}\).

\(Y\) reacts with 2,4-dinitrophenylhydrazine reagent to form an orange precipitate, but does not give a silver mirror when warmed with Tollens' reagent.

Which product is formed when \(Y\) is reduced using \(\text{NaBH}_4\)?
  1. A.butan-1-ol
  2. B.butan-2-ol
  3. C.2-methylpropan-1-ol
  4. D.2-methylpropan-2-ol
Show answer & marking scheme

Worked solution

1. Compound \(Y\) has the formula \(\text{C}_4\text{H}_8\text{O}\).
2. Reaction with 2,4-DNPH indicates that \(Y\) contains a carbonyl group (\(\text{C}=\text{O}\)), meaning it is either an aldehyde or a ketone.
3. The lack of reaction with Tollens' reagent shows that \(Y\) is not an aldehyde. Therefore, \(Y\) must be a ketone.
4. The only ketone with the formula \(\text{C}_4\text{H}_8\text{O}\) is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\).
5. Reducing butanone with \(\text{NaBH}_4\) yields a secondary alcohol, which is butan-2-ol, \(\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3\).

Marking scheme

Award 1 mark for B.
- Identify Y as a ketone (butanone) from the given tests.
- Correctly deduce the reduction product of butanone as butan-2-ol.
Question 38 · multipleChoice
1 marks
A sample of anhydrous magnesium nitrate, \(\text{Mg(NO}_3)_2\), has a mass of \(7.42\text{ g}\).

How many total ions (cations and anions) are present in this sample?

[Assume the Avogadro constant, \(L = 6.02 \times 10^{23}\text{ mol}^{-1}\); \(M_r\) of \(\text{Mg(NO}_3)_2 = 148.3\)]
  1. A.\(3.01 \times 10^{22}\)
  2. B.\(6.02 \times 10^{22}\)
  3. C.\(9.03 \times 10^{22}\)
  4. D.\(2.71 \times 10^{23}\)
Show answer & marking scheme

Worked solution

1. Calculate the number of moles of \(\text{Mg(NO}_3)_2\):
\(\text{moles of Mg(NO}_3)_2 = \frac{7.42\text{ g}}{148.3\text{ g mol}^{-1}} \approx 0.0500\text{ mol}\)

2. Determine the number of ions per formula unit of \(\text{Mg(NO}_3)_2\):
\(\text{Mg(NO}_3)_2(s) \rightarrow \text{Mg}^{2+}(aq) + 2\text{NO}_3^-(aq)\)
Each formula unit dissociates into \(1\text{ Mg}^{2+}\) ion and \(2\text{ NO}_3^-\) ions, giving a total of \(3\) ions.

3. Calculate the total moles of ions:
\(\text{total moles of ions} = 3 \times 0.0500\text{ mol} = 0.150\text{ mol}\)

4. Calculate the total number of ions:
\(\text{number of ions} = 0.150\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 9.03 \times 10^{22}\).

Marking scheme

Award 1 mark for C.
- Calculate moles of Mg(NO₃)₂ = 0.0500 mol.
- Factor in 3 ions per formula unit of Mg(NO₃)₂.
- Calculate total ions = 0.0500 * 3 * 6.02 x 10²³ = 9.03 x 10²².
Question 39 · multipleChoice
1 marks
Which statement correctly describes a heterogeneous catalyst in a reaction?
  1. A.The catalyst is in a different physical state from the reactants and works by adsorbing reactant molecules onto its active sites.
  2. B.The catalyst is in the same physical state as the reactants and works by forming a reactive intermediate.
  3. C.The catalyst decreases the activation energy of the reaction by raising the temperature of the local environment.
  4. D.The catalyst is consumed in the first step of the reaction mechanism and is regenerated in a subsequent step.
Show answer & marking scheme

Worked solution

- Option A: A heterogeneous catalyst exists in a different physical state from the reactants (e.g., solid catalyst with gaseous reactants) and operates by adsorbing reactant molecules onto active sites on its solid surface. This is correct.
- Option B: This describes a homogeneous catalyst, which is in the same physical state and forms a reactive intermediate.
- Option C: Catalysts lower the activation energy by providing an alternative reaction pathway, but they do not alter the temperature of the reaction mixture.
- Option D: Heterogeneous catalysts act through physical adsorption on surface sites, and are not chemically consumed and regenerated in sequence like homogeneous transition-metal ions in solution.

Marking scheme

Award 1 mark for A.
- Recall definition of a heterogeneous catalyst (different phase/state) and its mode of action (adsorption of reactants).
Question 40 · multipleChoice
1 marks
Some average bond energies are given in the table.

$$\begin{array}{|c|c|}
\hline
\text{Bond} & \text{Average bond energy} / \text{kJ mol}^{-1} \\
\hline
\text{C}-\text{H} & 413 \\
\text{C}-\text{Cl} & 346 \\
\text{H}-\text{Cl} & 432 \\
\text{Cl}-\text{Cl} & 242 \\
\hline
\end{array}$$

The reaction between methane and chlorine to form chloromethane is shown.

$$\text{CH}_4(g) + \text{Cl}_2(g) \rightarrow \text{CH}_3\text{Cl}(g) + \text{HCl}(g)$$

What is the enthalpy change, \(\Delta H\), for this reaction?
  1. A.\-123\text{ kJ mol}^{-1}\
  2. B.\-85\text{ kJ mol}^{-1}\
  3. C.\+85\text{ kJ mol}^{-1}\
  4. D.\+123\text{ kJ mol}^{-1}\
Show answer & marking scheme

Worked solution

To calculate the enthalpy change of the reaction:
\(\Delta H = \sum \text{Bond energy of reactants (broken)} - \sum \text{Bond energy of products (formed)}\)

Bonds broken:
- 1 mol of \(\text{C}-\text{H}\) bonds: \(1 \times 413 = +413\text{ kJ}\)
- 1 mol of \(\text{Cl}-\text{Cl}\) bonds: \(1 \times 242 = +242\text{ kJ}\)
Total bonds broken = \(+655\text{ kJ}\)

Bonds formed:
- 1 mol of \(\text{C}-\text{Cl}\) bonds: \(1 \times 346 = 346\text{ kJ}\)
- 1 mol of \(\text{H}-\text{Cl}\) bonds: \(1 \times 432 = 432\text{ kJ}\)
Total bonds formed = \(346 + 432 = 778\text{ kJ}\)

\(\Delta H = 655 - 778 = -123\text{ kJ mol}^{-1}\).

Marking scheme

Award 1 mark for A.
- Method: Enthalpy change = bonds broken - bonds formed.
- Bonds broken = 413 + 242 = 655 kJ mol⁻¹.
- Bonds formed = 346 + 432 = 778 kJ mol⁻¹.
- Accuracy: 655 - 778 = -123 kJ mol⁻¹.

Paper 23 (Structured Questions)

Structured AS Level theory questions. Answer all questions. Show your working and use appropriate units.
6 Question · 60 marks
Question 1 · structuredShort
10 marks
An alloy of zinc and copper has a mass of 2.45 g. When this alloy sample is reacted with an excess of dilute hydrochloric acid, only the zinc reacts, producing zinc chloride and hydrogen gas:

$$\text{Zn(s)} + 2\text{HCl(aq)} \rightarrow \text{ZnCl}_2\text{(aq)} + \text{H}_2\text{(g)}$$

The hydrogen gas produced is collected and its volume is measured as $685\text{ cm}^3$ at a temperature of $20\text{ }^\circ\text{C}$ and a pressure of $101\text{ kPa}$.

(a) Define the term relative atomic mass.
(b) Use the ideal gas equation, $pV = nRT$, to calculate the number of moles of hydrogen gas collected. Give your answer to 3 significant figures. [$R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}$]
(c) Deduce the number of moles of zinc that reacted, and calculate the mass of zinc in the sample. [$A_r$: $\text{Zn} = 65.4$]
(d) Calculate the percentage by mass of copper in the alloy sample.
(e) Write the ionic equation, including state symbols, for the reaction of zinc with dilute hydrochloric acid.
Show answer & marking scheme

Worked solution

**(a)** Relative atomic mass ($A_r$) is the weighted average mass of an atom of an element compared to $1/12\text{th}$ of the mass of an atom of carbon-12.

**(b)** Using $pV = nRT$:
- $p = 101 \times 10^3\text{ Pa}$
- $V = 685 \times 10^{-6}\text{ m}^3 = 6.85 \times 10^{-4}\text{ m}^3$
- $T = 20 + 273.15 = 293.15\text{ K}$ (or $293\text{ K}$)

$$n = \frac{pV}{RT} = \frac{101 \times 10^3 \times 6.85 \times 10^{-4}}{8.31 \times 293.15} = 0.0284\text{ mol}$$

**(c)** From the chemical equation, 1 mole of $\text{Zn}$ reacts to produce 1 mole of $\text{H}_2$.
Therefore, moles of $\text{Zn} = 0.0284\text{ mol}$.
$$\text{Mass of Zn} = 0.028415 \times 65.4 = 1.86\text{ g}$$

**(d)** Mass of copper in the alloy = $2.45 - 1.858 = 0.592\text{ g}$ (or $2.45 - 1.86 = 0.59\text{ g}$).
$$\text{Percentage of copper} = \frac{0.592}{2.45} \times 100\% = 24.2\%$$

**(e)** $\text{Zn(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{H}_2\text{(g)}$

Marking scheme

(a) 2 marks:
- 1 mark for: weighted average mass of an atom of an element compared to 1/12th of the mass of an atom of carbon-12.
- 1 mark for: explicit mention of carbon-12 / 12C isotope.

(b) 3 marks:
- 1 mark for converting all units correctly: $p = 101000\text{ Pa}$, $V = 6.85 \times 10^{-4}\text{ m}^3$, $T = 293\text{ K}$.
- 1 mark for rearrangement of formula: $n = pV / RT$.
- 1 mark for correct calculation to 3 significant figures: $0.0284\text{ mol}$ (accept $0.0284$ - $0.0285$).

(c) 2 marks:
- 1 mark for identifying the 1:1 mole ratio, thus moles of $\text{Zn} = \text{moles of } \text{H}_2 = 0.0284\text{ mol}$.
- 1 mark for calculating mass of $\text{Zn} = 1.86\text{ g}$ (accept $1.85\text{ g}$ to $1.86\text{ g}$ depending on rounding).

(d) 2 marks:
- 1 mark for calculating the mass of copper: $2.45 - 1.86 = 0.59\text{ g}$ (or $0.592\text{ g}$).
- 1 mark for correct percentage by mass: $24.2\%$ (accept $24.1\%$ to $24.3\%$ depending on intermediate rounding).

(e) 1 mark:
- $\text{Zn(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{H}_2\text{(g)}$ (equation must be fully balanced with correct state symbols).
Question 2 · structuredShort
10 marks
(a) The reaction between peroxodisulfate ions, $\text{S}_2\text{O}_8^{2-}$, and iodide ions, $\text{I}^-$, is catalysed by $\text{Fe}^{3+}$(aq) ions.

(i) State why this reaction is very slow in the absence of a catalyst.
(ii) Explain why $\text{Fe}^{3+}$(aq) acts as a homogeneous catalyst in this reaction.
(iii) Write two ionic equations to show how $\text{Fe}^{3+}$(aq) catalyses this reaction.

(b) The decomposition of ammonia on a hot tungsten surface is an example of heterogeneous catalysis.

(i) Describe the general mechanism of heterogeneous catalysis on a solid surface.
(ii) Describe the Maxwell–Boltzmann distribution of molecular energies and explain how a catalyst increases the rate of reaction. In your answer, refer to the relative positions of the uncatalysed activation energy ($E_{\text{a, uncatalysed}}$) and the catalysed activation energy ($E_{\text{a, catalysed}}$).
Show answer & marking scheme

Worked solution

**(a)(i)** Both reactant ions ($\text{S}_2\text{O}_8^{2-}$ and $\text{I}^-$) are negatively charged anions. Therefore, they repel each other, resulting in a very high activation energy for the uncatalysed reaction.

**(a)(ii)** The catalyst $\text{Fe}^{3+}$(aq) is in the same physical state / phase (aqueous) as the reactants.

**(a)(iii)**
1. $2\text{Fe}^{3+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + \text{I}_2\text{(aq)}$
2. $2\text{Fe}^{2+}\text{(aq)} + \text{S}_2\text{O}_8^{2-}\text{(aq)} \rightarrow 2\text{Fe}^{3+}\text{(aq)} + 2\text{SO}_4^{2-}\text{(aq)}$

**(b)(i)** Heterogeneous catalysis proceeds via three main steps:
1. **Adsorption**: Reactant molecules diffuse to and form weak bonds with active sites on the solid catalyst surface.
2. **Reaction**: The bonds within the reactant molecules are weakened and broken on the surface, and new bonds are formed to construct the product molecules.
3. **Desorption**: The product molecules break their bonds with the catalyst surface and diffuse away.

**(b)(ii)**
- In a Maxwell–Boltzmann distribution, the molecular energies of a gas are spread over a range.
- A catalyst provides an alternative pathway with a lower activation energy, meaning $E_{\text{a, catalysed}}$ is located to the left of $E_{\text{a, uncatalysed}}$ on the energy axis.
- Consequently, a much larger fraction of molecules possess energy greater than or equal to $E_{\text{a, catalysed}}$ compared to $E_{\text{a, uncatalysed}}$. This significantly increases the frequency of successful collisions, thereby increasing the rate of reaction.

Marking scheme

(a)(i) 1 mark:
- Repulsion between two negatively charged ions / anions.

(a)(ii) 1 mark:
- Catalyst and reactants are in the same phase / physical state.

(a)(iii) 2 marks:
- 1 mark for: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$
- 1 mark for: $2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}$

(b)(i) 3 marks:
- 1 mark for describing adsorption (bonding of reactants to active sites of catalyst surface).
- 1 mark for describing the reaction step (weakening/breaking of bonds in reactants, forming new bonds on the surface).
- 1 mark for describing desorption (products leaving the surface).

(b)(ii) 3 marks:
- 1 mark for stating that a catalyst provides an alternative reaction pathway with lower activation energy ($E_{\text{a, catalysed}} < E_{\text{a, uncatalysed}}$).
- 1 mark for explaining that a greater proportion/fraction of molecules have energy $\ge E_{\text{a, catalysed}}$.
- 1 mark for stating that this leads to more successful collisions per unit time / higher frequency of successful collisions.
Question 3 · structuredShort
10 marks
(a) Define standard enthalpy change of formation, $\Delta H_{\text{f}}^\ominus$.
(b) Write the chemical equation for the reaction representing the standard enthalpy change of formation of liquid propan-1-ol, $\text{C}_3\text{H}_7\text{OH(l)}$. Include state symbols.
(c) The standard enthalpy changes of combustion, $\Delta H_{\text{c}}^\ominus$, for carbon, hydrogen, and propan-1-ol are given below:
- $\Delta H_{\text{c}}^\ominus[\text{C(s, graphite)}] = -394\text{ kJ mol}^{-1}$
- $\Delta H_{\text{c}}^\ominus[\text{H}_2\text{(g)}] = -286\text{ kJ mol}^{-1}$
- $\Delta H_{\text{c}}^\ominus[\text{C}_3\text{H}_7\text{OH(l)}] = -2021\text{ kJ mol}^{-1}$

(i) State Hess's Law and construct an enthalpy cycle to relate these values to the enthalpy of formation of propan-1-ol.
(ii) Calculate the standard enthalpy change of formation of liquid propan-1-ol. Show your working.
(d) Explain why the experimental value of the enthalpy change of combustion of propan-1-ol determined using a simple laboratory calorimeter is usually less exothermic than the theoretical value.
Show answer & marking scheme

Worked solution

**(a)** The standard enthalpy change of formation, $\Delta H_{\text{f}}^\ominus$, is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (100 kPa and 298 K).

**(b)** The chemical equation is:
$$3\text{C(s, graphite)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_7\text{OH(l)}$$

**(c)(i)** Hess's Law states that the enthalpy change of a reaction is independent of the route taken, provided the initial and final states are the same.

Enthalpy Cycle:
$$3\text{C(s)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \xrightarrow{\Delta H_{\text{f}}^\ominus} \text{C}_3\text{H}_7\text{OH(l)}$$
Both sides point downwards towards the combustion products:
$$3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}$$
- Left-to-bottom arrow: $3 \times \Delta H_{\text{c}}^\ominus[\text{C}] + 4 \times \Delta H_{\text{c}}^\ominus[\text{H}_2] = 3(-394) + 4(-286) = -2326\text{ kJ mol}^{-1}$
- Right-to-bottom arrow: $\Delta H_{\text{c}}^\ominus[\text{C}_3\text{H}_7\text{OH(l)}] = -2021\text{ kJ mol}^{-1}$

**(c)(ii)** According to Hess's Law:
$$\Delta H_{\text{f}}^\ominus = \sum \Delta H_{\text{c}}^\ominus(\text{reactants}) - \sum \Delta H_{\text{c}}^\ominus(\text{products})$$
$$\Delta H_{\text{f}}^\ominus = [3(-394) + 4(-286)] - [-2021]$$
$$\Delta H_{\text{f}}^\ominus = -2326 + 2021 = -305\text{ kJ mol}^{-1}$$

**(d)** Two main experimental sources of error:
1. Heat energy is lost to the surroundings/beaker/calorimeter rather than being fully absorbed by the water.
2. Incomplete combustion of the alcohol occurs, which produces carbon (soot) or carbon monoxide rather than carbon dioxide, releasing less heat.

Marking scheme

(a) 2 marks:
- 1 mark for: enthalpy change when 1 mole of a compound is formed from its constituent elements.
- 1 mark for: elements/compounds in their standard states under standard conditions.

(b) 2 marks:
- 1 mark for correct species and balancing: $3\text{C} + 4\text{H}_2 + 0.5\text{O}_2 \rightarrow \text{C}_3\text{H}_7\text{OH}$.
- 1 mark for all correct state symbols: $\text{C(s)}$, $\text{H}_2\text{(g)}$, $\text{O}_2\text{(g)}$, and $\text{C}_3\text{H}_7\text{OH(l)}$.

(c)(i) 2 marks:
- 1 mark for correct statement of Hess's Law.
- 1 mark for drawing/describing a correct cycle linking reactants, products and combustion products ($3\text{CO}_2 + 4\text{H}_2\text{O}$).

(c)(ii) 2 marks:
- 1 mark for correct mathematical equation / working: $\Delta H_{\text{f}}^\ominus = [3(-394) + 4(-286)] - (-2021)$.
- 1 mark for correct evaluation with correct sign and unit: $-305\text{ kJ mol}^{-1}$.

(d) 2 marks:
- 1 mark for heat loss to the surroundings / calorimeter.
- 1 mark for incomplete combustion of the alcohol (forming soot/CO).
Question 4 · structuredShort
10 marks
(a) Phosphorus(V) chloride, $\text{PCl}_5$, and silicon(IV) chloride, $\text{SiCl}_4$, both react vigorously with water.

(i) Write a balanced chemical equation for the reaction of $\text{SiCl}_4$ with water.
(ii) State the approximate pH of the resulting solution when $\text{PCl}_5$ reacts with excess water. Explain this observation by identifying the species responsible for the acidity.

(b) Period 3 oxides show a trend from basic to acidic behavior across the period.

(i) Contrast the structures and bonding of sodium oxide, $\text{Na}_2\text{O}$, and sulfur trioxide, $\text{SO}_3$.
(ii) Describe what is observed when a sample of solid $\text{Na}_2\text{O}$ is added to water, and write an equation for the reaction.
(iii) Write a balanced chemical equation for the reaction of aluminium oxide, $\text{Al}_2\text{O}_3$, with hot aqueous sodium hydroxide, and state the term used to describe this type of oxide behavior.
Show answer & marking scheme

Worked solution

**(a)(i)**
$$\text{SiCl}_4\text{(l)} + 2\text{H}_2\text{O(l)} \rightarrow \text{SiO}_2\text{(s)} + 4\text{HCl(aq)}$$
(Note: $\text{Si(OH)}_4\text{(aq)}$ or $\text{H}_4\text{SiO}_4\text{(aq)}$ are also accepted as products instead of $\text{SiO}_2\text{(s)}$).

**(a)(ii)** The pH is approximately 1 to 2 (accept 0 - 2).
Explanation: $\text{PCl}_5$ reacts with water to form phosphoric(V) acid ($\text{H}_3\text{PO}_4$) and hydrochloric acid ($\text{HCl}$), both of which are strong/moderate acids that dissociate to release hydrogen ions, $\text{H}^+$(aq).

**(b)(i)**
- **Sodium oxide, $\text{Na}_2\text{O}$**: Giant ionic structure consisting of a regular 3D lattice of sodium ions ($\text{Na}^+$) and oxide ions ($\text{O}^{2-}$), held together by strong electrostatic attractions (ionic bonding).
- **Sulfur trioxide, $\text{SO}_3$**: Simple molecular structure with covalent bonding (shared pairs of electrons) holding the sulfur and oxygen atoms together. Weak intermolecular forces (van der Waals forces) exist between individual molecules.

**(b)(ii)**
- Observation: The white solid dissolves to form a colourless solution.
- Equation: $\text{Na}_2\text{O(s)} + \text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)}$

**(b)(iii)**
- Equation:
$$\text{Al}_2\text{O}_3\text{(s)} + 2\text{NaOH(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2\text{NaAl(OH)}_4\text{(aq)}$$
(or ionic equivalent: $\text{Al}_2\text{O}_3\text{(s)} + 2\text{OH}^-\text{(aq)} + 3\text{H}_2\text{O(l)} \rightarrow 2[\text{Al(OH)}_4]^-\text{(aq)}$)
- Behaviour: Amphoteric.

Marking scheme

(a)(i) 1 mark:
- $\text{SiCl}_4 + 2\text{H}_2\text{O} \rightarrow \text{SiO}_2 + 4\text{HCl}$ (allow balanced equation forming $\text{Si(OH)}_4$).

(a)(ii) 2 marks:
- 1 mark for pH 0, 1, or 2.
- 1 mark for: $\text{HCl}$ / $\text{H}_3\text{PO}_4$ formed which dissociate to release $\text{H}^+$ / $\text{H}_3\text{O}^+$ ions.

(b)(i) 3 marks:
- 1 mark for: $\text{Na}_2\text{O}$ is giant ionic lattice / has ionic bonding.
- 1 mark for: $\text{SO}_3$ is simple molecular / has covalent bonding.
- 1 mark for correct contrast (e.g. ions vs. molecules / description of electrostatic attraction vs. shared pairs of electrons).

(b)(ii) 2 marks:
- 1 mark for observation: white solid dissolves / disappears to form a colourless solution.
- 1 mark for equation: $\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$ (state symbols not required unless specified, but must be balanced).

(b)(iii) 2 marks:
- 1 mark for equation: $\text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{NaAl(OH)}_4$ (accept ionic equations forming aluminate).
- 1 mark for: amphoteric.
Question 5 · structuredShort
10 marks
(a) Propanal and propanone are structural isomers with the molecular formula $\text{C}_3\text{H}_6\text{O}$.

(i) State the type of structural isomerism shown by these two compounds.
(ii) Describe a chemical test, including reagents and observations, that can be used to distinguish propanal from propanone.

(b) Propanal reacts with hydrogen cyanide, $\text{HCN}$, in the presence of a sodium cyanide, $\text{NaCN}$, catalyst to form 2-hydroxybutanenitrile.

(i) Write the mechanism for this nucleophilic addition reaction. Show all relevant lone pairs, dipoles, and curly arrows.
(ii) Explain why the product 2-hydroxybutanenitrile is formed as a racemic mixture of optical isomers.

(c) Identify the reagent and conditions required to reduce propanone to propan-2-ol.
Show answer & marking scheme

Worked solution

**(a)(i)** Functional group isomerism.

**(a)(ii)**
- **Test**: Add Tollens' reagent (ammoniacal silver nitrate) and warm.
- **Observation with propanal**: A silver mirror is formed on the inside of the test tube.
- **Observation with propanone**: No change is observed (solution remains colourless).
- *(Alternative test: Fehling's solution. Propanal: blue solution forms a red precipitate; Propanone: remains a blue solution).*

**(b)(i)** Mechanism:
1. The nucleophile $:\text{CN}^-$ (cyanide ion) has a lone pair on the carbon atom. Draw a curly arrow from this lone pair to the carbonyl carbon atom in propanal.
2. The carbonyl carbon atom has a $\delta+$ charge and the oxygen has a $\delta-$ charge. Draw a curly arrow from the $\text{C}=\text{O}$ double bond to the oxygen atom.
3. This leads to the intermediate species, $\text{CH}_3\text{CH}_2\text{CH(O}^-\text{)CN}$.
4. Finally, draw a curly arrow from the lone pair on the negative oxygen atom of the intermediate to the $\text{H}^+$ (or to the hydrogen of an $\text{HCN}$ molecule) to form the final product 2-hydroxybutanenitrile.

**(b)(ii)**
- The carbonyl carbon atom in propanal has a planar geometry (trigonal planar).
- The nucleophilic cyanide ion ($:\text{CN}^-$) can attack this planar carbon atom with equal probability from either side (above or below the plane).
- This results in the formation of equal amounts of the two enantiomers (optical isomers), producing an optically inactive racemic mixture.

**(c)**
- **Reagent**: Sodium tetrahydridoborate(III) / $\text{NaBH}_4$.
- **Conditions**: In aqueous or alcoholic solution.

Marking scheme

(a)(i) 1 mark:
- Functional group isomerism.

(a)(ii) 2 marks:
- 1 mark for correct reagent: Tollens' reagent (accept Fehling's solution / acidified potassium dichromate(VI)).
- 1 mark for correct contrasting observations (e.g. Tollens': silver mirror with propanal, no change/no mirror with propanone; Fehling's: red ppt with propanal, remains blue with propanone).

(b)(i) 4 marks:
- 1 mark for correct dipole $\text{C}^{\delta+}=\text{O}^{\delta-}$ and curly arrow from the double bond to the oxygen.
- 1 mark for curly arrow from the lone pair on the carbon of $:\text{CN}^-$ to the carbonyl carbon.
- 1 mark for correct structure of the intermediate: $\text{CH}_3\text{CH}_2\text{CH(O}^-\text{)CN}$.
- 1 mark for curly arrow from the lone pair on $\text{O}^-$ of the intermediate to $\text{H}^+$ (or $\text{H-CN}$ with arrow breaking the H-C bond).

(b)(ii) 2 marks:
- 1 mark for stating that the carbonyl carbon/group is planar.
- 1 mark for stating that the nucleophile / cyanide ion can attack from either side with equal probability / frequency.

(c) 1 mark:
- $\text{NaBH}_4$ in aqueous/alcoholic solution (or $\text{LiAlH}_4$ in dry ether, or $\text{H}_2$ with $\text{Ni}$ catalyst and heat).
Question 6 · structuredShort
10 marks
Methanol is synthesised industrially by the reversible reaction of carbon monoxide with hydrogen in the presence of a catalyst:

$$\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)} \quad \Delta H = -91\text{ kJ mol}^{-1}$$

(a) State two features of a system in dynamic equilibrium.
(b) Write an expression for the equilibrium constant, $K_c$, for this reaction, and state its units.
(c) In an experiment, $1.20\text{ mol}$ of $\text{CO}$ and $2.20\text{ mol}$ of $\text{H}_2$ are placed in a sealed container of volume $5.00\text{ dm}^3$ at a constant temperature. When equilibrium is reached, the mixture contains $0.40\text{ mol}$ of $\text{CH}_3\text{OH}$.

(i) Calculate the equilibrium amounts, in moles, of $\text{CO}$ and $\text{H}_2$.
(ii) Calculate the value of $K_c$ at this temperature. Show your working and include units.

(d) State and explain the effect of increasing the temperature on the yield of methanol, $\text{CH}_3\text{OH}$, at equilibrium.
Show answer & marking scheme

Worked solution

**(a)** Two features of dynamic equilibrium are:
1. The rate of the forward reaction equals the rate of the reverse/backward reaction.
2. The concentrations of reactants and products remain constant in a closed system.

**(b)**
$$K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}$$
Units:
$$\text{Units} = \frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2} = \text{mol}^{-2}\text{ dm}^6$$

**(c)(i)** Using the stoichiometry of the equation:
- Initial moles: $\text{CO} = 1.20\text{ mol}$, $\text{H}_2 = 2.20\text{ mol}$, $\text{CH}_3\text{OH} = 0\text{ mol}$.
- At equilibrium, $\text{CH}_3\text{OH} = 0.40\text{ mol}$. This means $0.40\text{ mol}$ of $\text{CO}$ and $2 \times 0.40 = 0.80\text{ mol}$ of $\text{H}_2$ have reacted.
- Equilibrium moles:
- $\text{CO} = 1.20 - 0.40 = 0.80\text{ mol}$
- $\text{H}_2 = 2.20 - 0.80 = 1.40\text{ mol}$

**(c)(ii)** Equilibrium concentrations (Volume $= 5.00\text{ dm}^3$):
- $[\text{CO}] = \frac{0.80}{5.00} = 0.160\text{ mol dm}^{-3}$
- $[\text{H}_2] = \frac{1.40}{5.00} = 0.280\text{ mol dm}^{-3}$
- $[\text{CH}_3\text{OH}] = \frac{0.40}{5.00} = 0.080\text{ mol dm}^{-3}$

Substitute concentrations into the $K_c$ expression:
$$K_c = \frac{0.080}{0.160 \times (0.280)^2} = \frac{0.080}{0.160 \times 0.0784} = 6.38\text{ mol}^{-2}\text{ dm}^6$$

**(d)** The yield of methanol decreases. Since the forward reaction is exothermic ($\Delta H$ is negative), increasing the temperature causes the equilibrium position to shift to the left (in the endothermic direction) to absorb the added heat, according to Le Chatelier's principle.

Marking scheme

(a) 2 marks:
- 1 mark for: rate of forward reaction equals rate of reverse/backward reaction.
- 1 mark for: concentration of reactants and products remains constant (accept system is closed).

(b) 2 marks:
- 1 mark for correct $K_c$ expression.
- 1 mark for correct units: $\text{mol}^{-2}\text{ dm}^6$.

(c)(i) 2 marks:
- 1 mark for equilibrium moles of $\text{CO} = 0.80\text{ mol}$.
- 1 mark for equilibrium moles of $\text{H}_2 = 1.40\text{ mol}$.

(c)(ii) 3 marks:
- 1 mark for dividing moles by volume ($5.00\text{ dm}^3$) to obtain concentrations: $[\text{CO}] = 0.16$, $[\text{H}_2] = 0.28$, $[\text{CH}_3\text{OH}] = 0.08\text{ mol dm}^{-3}$.
- 1 mark for correct substitution into expression: $K_c = 0.08 / (0.16 \times 0.28^2)$.
- 1 mark for correct value: $6.38$ (accept $6.4$ or $6.378$).

(d) 1 mark:
- Yield decreases because the forward reaction is exothermic, so equilibrium shifts to the left / in the endothermic direction.

Paper 33 (Practical Skills)

Perform quantitative analysis (titrations and kinetics) and qualitative analysis. Record results to the correct decimal precision.
3 Question · 40 marks
Question 1 · titrationStoichiometry
13 marks
In this experiment, you will determine the value of \(x\) in hydrated sodium carbonate, \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\). You are provided with: **FB 1**: Hydrated sodium carbonate, \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\) (solid); **FB 2**: \(0.100\text{ mol dm}^{-3}\) hydrochloric acid, \(\text{HCl}\); Methyl orange indicator. **Procedure:** 1. Weigh the container with **FB 1** and record its mass. 2. Dissolve all the **FB 1** in about \(100\text{ cm}^3\) of distilled water in a beaker. 3. Transfer the solution quantitatively to a \(250.0\text{ cm}^3\) volumetric flask, make up to the mark with distilled water, and mix thoroughly. Label this solution **FB 3**. 4. Fill a burette with **FB 2**. 5. Pipette a \(25.0\text{ cm}^3\) portion of **FB 3** into a conical flask. 6. Add a few drops of methyl orange indicator. 7. Titrate the solution in the flask with **FB 2** until the end-point is reached. 8. Repeat the titration as many times as necessary to obtain concordant results. **Sample Results:** For the purpose of the calculations in parts (b) to (c), use the following sample data: Mass of container + **FB 1** = \(14.82\text{ g}\); Mass of empty container = \(11.24\text{ g}\); Mean volume of **FB 2** used = \(25.00\text{ cm}^3\). **Questions:** (a) Prepare a results table showing all titration details. (b) (i) Calculate the amount, in moles, of \(\text{HCl}\) present in the mean volume of **FB 2** used. (ii) Using the equation: \(\text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\), calculate the amount, in moles, of sodium carbonate present in the \(25.0\text{ cm}^3\) portion of **FB 3** pipetted. (iii) Calculate the amount, in moles, of sodium carbonate present in the \(250.0\text{ cm}^3\) volumetric flask. (iv) Calculate the relative formula mass, \(M_{\text{r}}\), of the hydrated sodium carbonate, \(\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}\). (v) Determine the value of \(x\) to the nearest integer. (c) (i) Each weighing on the balance used has an uncertainty of \(\pm 0.01\text{ g}\). Calculate the maximum percentage uncertainty in the mass of **FB 1** used. (ii) Suggest one modification to the titration method, other than repeating the titration, that would reduce the percentage uncertainty in the volume of the titre.
Show answer & marking scheme

Worked solution

(a) Titration table should include final and initial burette readings and the volume of titre for each titration, with all readings recorded to 2 decimal places. (b) (i) Moles of \(\text{HCl} = 25.00 \times 10^{-3} \times 0.100 = 2.50 \times 10^{-3}\text{ mol}\). (ii) Moles of \(\text{Na}_2\text{CO}_3\) in \(25.0\text{ cm}^3 = \frac{2.50 \times 10^{-3}}{2} = 1.25 \times 10^{-3}\text{ mol}\). (iii) Moles of \(\text{Na}_2\text{CO}_3\) in \(250.0\text{ cm}^3 = 1.25 \times 10^{-3} \times 10 = 1.25 \times 10^{-2}\text{ mol}\). (iv) Mass of **FB 1** = \(14.82 - 11.24 = 3.58\text{ g}\). \(M_{\text{r}} = \frac{3.58}{1.25 \times 10^{-2}} = 286.4\text{ g mol}^{-1}\) (or 286 to 3 SF). (v) \(M_{\text{r}}(\text{Na}_2\text{CO}_3) = 2(23.0) + 12.0 + 3(16.0) = 106.0\). Mass of water per mole = \(286.4 - 106.0 = 180.4\text{ g}\). \(x = \frac{180.4}{18.0} = 10.02\), which to the nearest integer is 10. (c) (i) Two weighings are required, so total uncertainty = \(2 \times 0.01 = 0.02\text{ g}\). Percentage uncertainty = \(\frac{0.02}{3.58} \times 100 = 0.56\%\). (ii) Use a more dilute solution of standard acid in the burette, or use a larger pipette volume (such as \(50.0\text{ cm}^3\)), which increases the required titre volume and thus reduces the percentage uncertainty.

Marking scheme

Part (a) [3 Marks]: 1 mark for structured table with appropriate headings and units (e.g., /cm3). 1 mark for all burette readings recorded to 2 decimal places. 1 mark for concordant titres within 0.10 cm3. Part (b) [7 Marks total]: (i) 1 mark for correct calculation of moles of HCl: 2.50 x 10^-3 mol. (ii) 1 mark for correct calculation of moles of carbonate in 25 cm3: 1.25 x 10^-3 mol. (iii) 1 mark for correct calculation of moles of carbonate in 250 cm3: 1.25 x 10^-2 mol. (iv) 1 mark for showing division of mass by moles in (iii), and 1 mark for correct Mr = 286.4 (or 286). (v) 1 mark for subtracting 106.0 from calculated Mr, and 1 mark for final integer x = 10. Part (c) [3 Marks total]: (i) 1 mark for identifying total uncertainty of 0.02 g, and 1 mark for calculating 0.56% (allow 0.559%). (ii) 1 mark for suggesting a more dilute standard acid in the burette or using a larger aliquot of carbonate solution to increase titre volume.
Question 2 · kineticsInvestigation
12 marks
In this experiment, you will investigate how the concentration of sodium thiosulfate affects the rate of its reaction with hydrochloric acid.

The equation for the reaction is:
\[\text{Na}_2\text{S}_2\text{O}_3\text{(aq)} + 2\text{HCl(aq)} \rightarrow 2\text{NaCl(aq)} + \text{S(s)} + \text{SO}_2\text{(g)} + \text{H}_2\text{O(l)}\]

You are provided with:
* \text{FA 1}: \(0.10\text{ mol dm}^{-3}\) sodium thiosulfate solution
* \text{FA 2}: \(2.0\text{ mol dm}^{-3}\) hydrochloric acid
* Distilled water

**Method:**
1. Draw a dark black cross on a piece of white paper and place a \(250\text{ cm}^3\) conical flask directly over the cross.
2. Add \(50.0\text{ cm}^3\) of \text{FA 1} to the flask.
3. Measure \(10.0\text{ cm}^3\) of \text{FA 2} using a measuring cylinder.
4. Add the \text{FA 2} to the flask, immediately start a stopclock, and swirl the mixture once to mix.
5. Look down vertically through the mouth of the flask at the cross. Record the time, \(t\), in seconds, taken for the cross to become completely obscured by the sulfur precipitate.
6. Empty, thoroughly rinse, and dry the flask.
7. Repeat the experiment using different mixtures of \text{FA 1} and distilled water to make a total volume of \(50.0\text{ cm}^3\) of thiosulfate solution. In each run, keep the volume of \text{FA 2} constant at \(10.0\text{ cm}^3\). Carry out four further experiments using different concentrations of \text{FA 1}.

**(a)** Record your results in a single, clearly structured table. Include columns for the volume of \text{FA 1}, volume of water, time \(t\), and the rate represented by \(1/t\) (calculated to 3 significant figures with appropriate units).

**(b)** Plot a graph of \(1/t\) (y-axis) against the volume of \text{FA 1} (x-axis) on a grid. Draw a line of best fit.

**(c)** (i) Using your graph, describe the relationship between the concentration of sodium thiosulfate and the rate of reaction.
(ii) Deduce the order of reaction with respect to sodium thiosulfate, justifying your answer.
(iii) State how the rate of reaction would change if the concentration of \text{FA 2} (hydrochloric acid) was doubled, assuming the reaction is zero-order with respect to \text{HCl}.

**(d)** (i) State one significant source of experimental error in this method that arises from the observation of the cross.
(ii) Suggest a specific improvement to the apparatus or method to minimize this source of error.
Show answer & marking scheme

Worked solution

**Part (a): Sample Results Table**

\[\begin{array}{|c|c|c|c|c|} \hline \text{Experiment} & \text{Volume of FA 1 / cm}^3 & \text{Volume of H}_2\text{O / cm}^3 & \text{Time, } t \text{ / s} & \text{Rate, } 1/t \text{ / s}^{-1} \\hline 1 & 50.0 & 0.0 & 25 & 0.0400 \\hline 2 & 40.0 & 10.0 & 31 & 0.0323 \\hline 3 & 30.0 & 20.0 & 42 & 0.0238 \\hline 4 & 20.0 & 30.0 & 63 & 0.0159 \\hline 5 & 10.0 & 40.0 & 125 & 0.00800 \\hline \end{array}\]

**Part (b): Graph Plotting**
* The y-axis is labeled as \(1/t\text{ / s}^{-1}\) and the x-axis is labeled as \(\text{Volume of FA 1 / cm}^3\).
* Scale is chosen so that points span at least half of the grid in both dimensions (e.g., x-axis from 0 to 50, y-axis from 0.000 to 0.050).
* Plotted points fall on a straight line passing through the origin \((0,0)\).

**Part (c): Interpretation**
* **(i)** Since the line of best fit is a straight line passing through the origin, the rate of reaction is directly proportional to the volume of \text{FA 1} (and hence the concentration of sodium thiosulfate).
* **(ii)** The reaction is **first-order** with respect to sodium thiosulfate because doubling the concentration (volume) of \text{FA 1} doubles the rate of reaction (direct proportionality).
* **(iii)** There would be **no change** to the rate of reaction because the reaction is zero-order with respect to \text{HCl}.

**Part (d): Evaluation**
* **(i)** Subjective judgment of when the cross is completely obscured, which varies between trials or different observers.
* **(ii)** Use a light sensor and data logger placed beneath/opposite a light source to measure the absorbance or transmittance over time, stopping at a precise percentage transmission threshold.

Marking scheme

**Part (a) [4 Marks]**
* **M1 (Table Layout):** Single table showing results for 5 experiments with appropriate headers and units: Volume of \text{FA 1} / \text{cm}^3\, Volume of water / \text{cm}^3\, Time, \(t\) / \text{s}\, Rate, \(1/t\) / \text{s}^{-1}\. (1 mark)
* **M2 (Precision):** Volumes of \text{FA 1} and water recorded to \(0.1\text{ cm}^3\) (if burette is used) or consistently to nearest \(1\text{ cm}^3\). Time recorded to the nearest second. (1 mark)
* **M3 (Calculations):** All values of \(1/t\) calculated correctly to 3 significant figures. (1 mark)
* **M4 (Trend):** Recorded times must increase as the volume of \text{FA 1} decreases. (1 mark)

**Part (b) [3 Marks]**
* **M1 (Axes & Scales):** Axes correctly labelled with units: \(1/t\text{ / s}^{-1}\) on the y-axis and volume of \text{FA 1 / cm}^3\) on the x-axis. Scales must allow plotted points to occupy more than half of the grid. (1 mark)
* **M2 (Accuracy):** All points plotted accurately to within half a small square. (1 mark)
* **M3 (Best-fit line):** Draws a single straight line of best fit passing through (or pointing directly to) the origin. (1 mark)

**Part (c) [3 Marks]**
* **(i)** States that the rate of reaction is directly proportional to the volume of \text{FA 1} / concentration of sodium thiosulfate. (1 mark)
* **(ii)** Deduces **first-order** AND justifies by referencing the straight-line graph through the origin / direct proportionality. (1 mark)
* **(iii)** States that there will be **no change** in rate. (1 mark)

**Part (d) [2 Marks]**
* **(i)** Identifies the key source of error as the subjective nature of deciding when the cross disappears / human reaction time. (1 mark) *[Do not accept 'timing error' or 'parallax error' without qualification.]*
* **(ii)** Suggests using a colorimeter / light sensor / light meter (connected to a data logger) to measure light transmission objectively. (1 mark) *[Do not accept 'use a digital stopwatch' or 'do more repeats'.]*
Question 3 · qualitativeAnalysis
15 marks
You are provided with two aqueous solutions, FA 1 and FA 2. Each solution contains one cation and one anion from the list in the Qualitative Analysis Notes.

(a) Carry out the following tests on FA 1 and record your observations in a single, clearly structured table. Include tests with aqueous sodium hydroxide, aqueous ammonia, and aqueous silver nitrate followed by dilute aqueous ammonia.

Identify the cation and anion present in FA 1. Support your identifications with reasoning from your observations.

(b) Carry out the following tests on FA 2 and record your observations:
- Test 1: To a 1 cm depth of FA 2 in a boiling tube, add 1 cm depth of aqueous sodium hydroxide, \(\text{NaOH}\). Warm the mixture gently and test any gas evolved with damp red litmus paper.
- Test 2: To a 1 cm depth of FA 2 in a test-tube, add 1 cm depth of aqueous barium nitrate, followed by 1 cm depth of dilute nitric acid.

Identify the cation and anion present in FA 2.

(c) Describe a chemical test that would allow you to distinguish between aqueous solutions containing sulfite, \(\text{SO}_3^{2-}\), and sulfate, \(\text{SO}_4^{2-}\) ions. State the reagents you would use and the expected observations for each ion.
Show answer & marking scheme

Worked solution

For part (a):
- Adding dropwise aqueous \(\text{NaOH}\) to FA 1 produces a white precipitate of magnesium hydroxide, \(\text{Mg(OH)}_2\), which is insoluble in excess.
- Adding dropwise aqueous \(\text{NH}_3\) to FA 1 produces a white precipitate, which is also insoluble in excess. This confirms the cation is \(\text{Mg}^{2+}\).
- Adding aqueous \(\text{AgNO}_3\) produces a cream precipitate of silver bromide, \(\text{AgBr}\).
- Adding dilute aqueous ammonia to this mixture does not dissolve the cream precipitate (it remains insoluble). This confirms the anion is \(\text{Br}^-\).

For part (b):
- Heating FA 2 with aqueous \(\text{NaOH}\) produces ammonia gas, \(\text{NH}_3\), which turns damp red litmus paper blue. This confirms the cation is \(\text{NH}_4^+\).
- Adding aqueous barium nitrate to FA 2 produces a white precipitate of barium sulfate, \(\text{BaSO}_4\). This precipitate remains completely insoluble upon addition of dilute nitric acid, confirming the anion is \(\text{SO}_4^{2-}\).

For part (c):
- Add dilute hydrochloric acid or nitric acid to the precipitate formed after adding barium ions. The barium sulfite precipitate will dissolve and effervesce, producing sulfur dioxide gas (which turns acidified potassium manganate(VII) from purple to colourless), while the barium sulfate precipitate is insoluble and remains unchanged.
- Alternatively, add acidified potassium manganate(VII) directly to the aqueous solutions. The sulfite solution will decolourise the purple solution, whereas the sulfate solution will show no visible change.

Marking scheme

Part (a) [6 marks]:
- 1 mark: Records a white precipitate, insoluble in excess, when aqueous \(\text{NaOH}\) is added to FA 1.
- 1 mark: Records a white precipitate, insoluble in excess, when aqueous \(\text{NH}_3\) is added to FA 1.
- 1 mark: Records a cream precipitate when aqueous \(\text{AgNO}_3\) is added to FA 1.
- 1 mark: Records that the cream precipitate does not dissolve / remains insoluble in dilute aqueous ammonia.
- 1 mark: Correctly identifies the cation as magnesium / \(\text{Mg}^{2+}\) and states the reason (white precipitate with both \(\text{NaOH}\) and \(\text{NH}_3\), both insoluble in excess).
- 1 mark: Correctly identifies the anion as bromide / \(\text{Br}^-\) and states the reason (cream precipitate with silver ions, insoluble in dilute ammonia).

Part (b) [6 marks]:
- 1 mark: Records that a gas is evolved that turns damp red litmus paper blue.
- 1 mark: Identifies the gas evolved in Test 1 as ammonia / \(\text{NH}_3\).
- 1 mark: Records a white precipitate on addition of aqueous barium nitrate / barium chloride.
- 1 mark: Records that the white precipitate remains insoluble in dilute nitric acid / hydrochloric acid.
- 1 mark: Correctly identifies the cation in FA 2 as ammonium / \(\text{NH}_4^+\).
- 1 mark: Correctly identifies the anion in FA 2 as sulfate / \(\text{SO}_4^{2-}\).

Part (c) [3 marks]:
- 1 mark: Identifies a suitable distinguishing reagent (e.g., adding dilute acid to the barium precipitate, OR direct addition of acidified potassium manganate(VII) / potassium dichromate(VI) to the solutions).
- 1 mark: Correct observation for sulfite (precipitate dissolves with acid / decolourises acidified manganate(VII) / turns dichromate(VI) green).
- 1 mark: Correct observation for sulfate (precipitate does not dissolve with acid / no reaction or no color change with acidified manganate(VII) or dichromate(VI)).

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