2023 AQA AS-Level Chemistry 7404 模擬試題連答案詳解

(a) An empirical formula is the simplest whole number ratio of atoms of each element present in a compound.

(b) Step 1: Calculate the masses.
- Mass of hydrated salt = \(24.51\text{ g} - 20.34\text{ g} = 4.17\text{ g}\)
- Mass of anhydrous salt (\(\text{FeSO}_4\)) = \(22.62\text{ g} - 20.34\text{ g} = 2.28\text{ g}\)
- Mass of water lost = \(4.17\text{ g} - 2.28\text{ g} = 1.89\text{ g}\)

Step 2: Calculate moles.
- \(M_r(\text{FeSO}_4) = 55.8 + 32.1 + (16.0 \times 4) = 151.9\text{ g mol}^{-1}\)
- \(n(\text{FeSO}_4) = \frac{2.28\text{ g}}{151.9\text{ g mol}^{-1}} = 0.01501\text{ mol}\)
- \(n(\text{H}_2\text{O}) = \frac{1.89\text{ g}}{18.0\text{ g mol}^{-1}} = 0.105\text{ mol}\)

Step 3: Find the ratio of moles.
- Ratio \(x = \frac{0.105\text{ mol}}{0.01501\text{ mol}} = 6.995 \approx 7\)

(c) Equation: \(2\text{FeSO}_4(\text{s}) \rightarrow \text{Fe}_2\text{O}_3(\text{s}) + \text{SO}_2(\text{g}) + \text{SO}_3(\text{g})\)

(d) If some of the anhydrous salt decomposes to form gases (\(\text{SO}_2\) and \(\text{SO}_3\)), these gases escape into the atmosphere. This causes the measured mass of the anhydrous residue to be lower than it should be. Consequently, the calculated mass of water lost (which is hydrated mass minus residue mass) appears larger than it actually is, resulting in a higher calculated value of \(x\).

**Part (a) [1 mark]**
- M1: Simplest whole number ratio of atoms of each element in a compound. (1) (Reject: 'simplest ratio of elements' without mentioning atoms/moles).

**Part (b) [4 marks]**
- M1: Calculation of mass of \(\text{FeSO}_4\) (\(2.28\text{ g}\)) and mass of water (\(1.89\text{ g}\)). (1)
- M2: Calculation of moles of \(\text{FeSO}_4\) (\(0.0150\text{ mol}\)) (using \(M_r = 151.9\)). (1)
- M3: Calculation of moles of water (\(0.105\text{ mol}\)). (1)
- M4: Ratio \(x = 7\) (from correct working, allow range 6.9 to 7.1). (1)

**Part (c) [1.125 marks]**
- M1: Balanced chemical equation with correct formulas: \(2\text{FeSO}_4 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3\). (1.125) (Allow state symbols but not required. Ignore incorrect state symbols).

**Part (d) [2 marks]**
- M1: Mass of residue is lower / loss of sulfur oxide gases. (1)
- M2: Calculated mass of water lost is too high, so the calculated value of \(x\) increases. (1)

(a)(i) Shape: Tetrahedral. Bond angle: \(109.5^\circ\).
Explanation: The central phosphorus atom has 4 bonding pairs and 0 lone pairs of electrons in its outer shell (as the positive charge means it has lost 1 valence electron, leaving 4 valence electrons to bond with 4 chlorine atoms). Electron pairs repel each other to achieve maximum separation / minimum repulsion, resulting in a symmetrical tetrahedral shape.

(ii) Shape: Trigonal bipyramidal. Bond angles: \(120^\circ\) (equatorial-equatorial) and \(90^\circ\) (axial-equatorial).

(b) Chlorine is more electronegative than phosphorus, so the \(\text{P-Cl}\) bonds are polar.
- In \(\text{PCl}_3\), the shape is trigonal pyramidal due to the presence of one lone pair on the phosphorus atom. The molecule is asymmetrical, meaning the individual bond dipoles do not cancel, resulting in a net dipole (polar molecule).
- In \(\text{PCl}_5\), the shape is trigonal bipyramidal, which is highly symmetrical. The bond dipoles cancel each other out completely, resulting in no net molecular dipole (non-polar molecule).

**Part (a)(i) [4 marks]**
- M1: Shape is tetrahedral AND bond angle is \(109.5^\circ\). (1)
- M2: Phosphorus has 4 bonding pairs and 0 lone pairs. (1)
- M3: Electron pairs repel each other to get as far apart as possible / to a position of minimum repulsion. (1)
- M4: Symmetrical arrangement / equal repulsion between all 4 bonding pairs. (1)

**Part (a)(ii) [2 marks]**
- M1: Shape is trigonal bipyramidal. (1)
- M2: Bond angles are \(120^\circ\) and \(90^\circ\) (both required). (1)

**Part (b) [2.125 marks]**
- M1: Chlorine is more electronegative than phosphorus, leading to polar bonds. (1)
- M2: \(\text{PCl}_3\) is asymmetrical (trigonal pyramidal) so bond dipoles do not cancel, whereas \(\text{PCl}_5\) is highly symmetrical (trigonal bipyramidal) so bond dipoles cancel out. (1.125)

(a) Expression:
\[K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}\]

Units:
\[\text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2 (\text{mol dm}^{-3})} = \text{mol}^{-1}\text{ dm}^3\]

(b) Step 1: Set up the equilibrium moles table (I.C.E.).
- Initial moles: \(n(\text{SO}_2) = 1.50\), \(n(\text{O}_2) = 1.00\), \(n(\text{SO}_3) = 0.00\).
- Change in moles: \(n(\text{SO}_3)\) increased by \(+0.60\text{ mol}\).
Therefore, \(n(\text{SO}_2)\) decreased by \(-0.60\text{ mol}\), and \(n(\text{O}_2)\) decreased by \(-0.30\text{ mol}\).
- Equilibrium moles:
- \(n(\text{SO}_3) = 0.60\text{ mol}\)
- \(n(\text{SO}_2) = 1.50 - 0.60 = 0.90\text{ mol}\)
- \(n(\text{O}_2) = 1.00 - 0.30 = 0.70\text{ mol}\)

Step 2: Convert to concentrations using volume \(V = 5.00\text{ dm}^3\).
- \([\text{SO}_3]_{eq} = \frac{0.60\text{ mol}}{5.00\text{ dm}^3} = 0.12\text{ mol dm}^{-3}\)
- \([\text{SO}_2]_{eq} = \frac{0.90\text{ mol}}{5.00\text{ dm}^3} = 0.18\text{ mol dm}^{-3}\)
- \([\text{O}_2]_{eq} = \frac{0.70\text{ mol}}{5.00\text{ dm}^3} = 0.14\text{ mol dm}^{-3}\)

Step 3: Calculate \(K_c\).
- \(K_c = \frac{(0.12)^2}{(0.18)^2 \times 0.14} = \frac{0.0144}{0.0324 \times 0.14} = \frac{0.0144}{0.004536} = 3.1737 \approx 3.17\text{ mol}^{-1}\text{ dm}^3\)

(c) The forward reaction is exothermic (\(\Delta H < 0\)). According to Le Chatelier's principle, an increase in temperature will cause the equilibrium to shift in the endothermic direction (to the left) to absorb the heat. Because the equilibrium position shifts to the left, the concentration of the product (\(\text{SO}_3\)) decreases and the concentrations of the reactants increase, which decreases the value of \(K_c\).

**Part (a) [2 marks]**
- M1: Correct expression for \(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]}\) (square brackets are essential). (1)
- M2: Correct units: \(\text{mol}^{-1}\text{ dm}^3\) (or \(\text{dm}^3\text{ mol}^{-1}\)). (1)

**Part (b) [4.125 marks]**
- M1: Correct equilibrium moles of \(\text{SO}_2\) (\(0.90\text{ mol}\)) and \(\text{O}_2\) (\(0.70\text{ mol}\)). (1)
- M2: Correct equilibrium concentrations (divided by 5): \([\text{SO}_3] = 0.12\), \([\text{SO}_2] = 0.18\), \([\text{O}_2] = 0.14\). (1)
- M3: Correct substitution of values into the \(K_c\) expression: \(K_c = \frac{0.12^2}{0.18^2 \times 0.14}\). (1)
- M4: Correct calculation of \(K_c = 3.17\) (accept range 3.17 - 3.20, must be 3 significant figures). (1.125)
*(Note: If volume is omitted in calculation, max 2 marks for Part b)*

**Part (c) [2 marks]**
- M1: \(K_c\) decreases. (1)
- M2: Forward reaction is exothermic, so equilibrium shifts to the left to oppose temperature increase. (1)

(a) Use \(q = m c \Delta T\):
- \(m = 150.0\text{ g}\) (mass of water)
- \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\)
- \(\Delta T = 56.2 - 20.2 = 36.0\ ^{\circ}\text{C} = 36.0\text{ K}\)
- \(q = 150.0 \times 4.18 \times 36.0 = 22572\text{ J} = 22.572\text{ kJ}\) (or \(22.6\text{ kJ}\))

(b) Step 1: Calculate the moles of propan-1-ol burned.
- \(n = \frac{\text{mass}}{M_r} = \frac{0.900\text{ g}}{60.0\text{ g mol}^{-1}} = 0.0150\text{ mol}\)

Step 2: Calculate \(\Delta H_c\).
- \(\Delta H_c = -\frac{q}{n} = -\frac{22.572\text{ kJ}}{0.0150\text{ mol}} = -1504.8\text{ kJ mol}^{-1}\)

To 3 significant figures and with a sign:
- \(\Delta H_c = -1500\text{ kJ mol}^{-1}\) (or \(-1.50 \times 10^3\text{ kJ mol}^{-1}\))

(c) Two practical reasons:
1. Heat loss to the surroundings / copper beaker / air.
2. Loss of propan-1-ol by evaporation from the wick of the spirit burner after weighing.
(Also accept: Heat capacity of the copper calorimeter is ignored).

(d) The standard enthalpy of combustion is the enthalpy change when one mole of a substance is burned completely in oxygen under standard conditions (\(298\text{ K}\), \(100\text{ kPa}\)), with all reactants and products in their standard states.

**Part (a) [2 marks]**
- M1: Calculation of \(\Delta T = 36.0\text{ K}\). (1)
- M2: Calculation of \(q = 22.6\text{ kJ}\) (or \(22.572\text{ kJ}\)). (1)

**Part (b) [3 marks]**
- M1: Moles of propan-1-ol = \(0.0150\text{ mol}\). (1)
- M2: Calculation of \(\Delta H = 1504.8\text{ kJ mol}^{-1}\). (1)
- M3: Correct final answer to 3 significant figures with negative sign: \(-1500\text{ kJ mol}^{-1}\) (or \(-1.50 \times 10^3\text{ kJ mol}^{-1}\)). (1)
*(Note: positive sign or missing sign loses M3)*

**Part (c) [2 marks]**
- M1: Heat loss to surroundings / copper container. (1)
- M2: Loss of fuel by evaporation from wick. (1) (Accept: 'heat capacity of the container was ignored').
*(Do not accept: incomplete combustion, human error, or systematic error)*

**Part (d) [1.125 marks]**
- M1: Enthalpy change when 1 mole of a substance is burned completely in oxygen, under standard conditions of \(298\text{ K}\) and \(100\text{ kPa}\) (or \(1\text{ bar}\)), with all reactants and products in their standard states. (1.125)

(a)(i) Product: Sulfur dioxide (\(\text{SO}_2\)).
Observation: Choking / pungent gas (or turns acidified potassium dichromate paper green).

(ii) Equation:
\[2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\]
(or ionic equivalent: \(2\text{Br}^- + 2\text{H}_2\text{SO}_4 + 2\text{H}^+ \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\))

(b) Test procedure and observations:
1. Dissolve each solid separately in deionised water to form solutions.
2. Add dilute nitric acid (\(\text{HNO}_3\)) followed by aqueous silver nitrate (\(\text{AgNO}_3\)).
3. Sodium chloride forms a white precipitate of silver chloride (\(\text{AgCl}\)).
4. Sodium iodide forms a yellow precipitate of silver iodide (\(\text{AgI}\)).
5. Confirmatory test: Add dilute ammonia (\(\text{NH}_3\)) solution.
- The white precipitate (\(\text{AgCl}\)) dissolves to form a colorless solution.
- The yellow precipitate (\(\text{AgI}\)) is insoluble in dilute ammonia (and remains insoluble even in concentrated ammonia).

**Part (a)(i) [2 marks]**
- M1: Sulfur dioxide (or \(\text{SO}_2\)). (1)
- M2: Choking gas / pungent smell / acidic gas (or turns acidified potassium dichromate paper green). (1) (Reject: 'bad egg smell' / 'yellow solid' - these are for hydrogen sulfide and sulfur respectively).

**Part (a)(ii) [2 marks]**
- M1: Correct reactants and products: \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Br}_2 + \text{SO}_2 + \text{H}_2\text{O}\). (1)
- M2: Correctly balanced: \(2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\). (1)
*(Note: Allow ionic equation: \(2\text{Br}^- + 2\text{H}_2\text{SO}_4 + 2\text{H}^+ \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\). Allow hydrogen bromide reactant: \(2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\))*

**Part (b) [4.125 marks]**
- M1: Add acidified silver nitrate solution (acidified with nitric acid). (1) (Reject hydrochloric acid or sulfuric acid as the acidifier).
- M2: Sodium chloride gives a white precipitate AND sodium iodide gives a yellow precipitate. (1)
- M3: Add dilute ammonia (or concentrated ammonia) solution. (1)
- M4: The white precipitate (silver chloride) dissolves, whereas the yellow precipitate (silver iodide) does not dissolve. (1.125)

(a) Iron has atomic number 26. The iron atom configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\).
When forming the \(\text{Fe}^{2+}\) ion, the two \(4s\) electrons are lost first.
Therefore, the configuration of \(\text{Fe}^{2+}\) is:
\(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6\) (or \([\text{Ar}] 3d^6\)).

(b)
- Phosphorus has its outer electrons in the \(3p\) subshell, with three singly-occupied orbitals (\(3p^3\)).
- Sulfur has its outer electrons in the \(3p\) subshell, with one doubly-occupied orbital and two singly-occupied orbitals (\(3p^4\)).
- In sulfur, the paired electrons in the same \(3p\) orbital repel each other. This repulsion makes it easier to remove one of these paired electrons, requiring less energy than removing an electron from a singly occupied orbital in phosphorus.

(c)(i) Equation: \(\text{Ga}(\text{g}) + \text{e}^- \rightarrow \text{Ga}^+(\text{g}) + 2\text{e}^-\) (or \(\text{Ga}(\text{g}) \rightarrow \text{Ga}^+(\text{g}) + \text{e}^-\)).

(c)(ii) Step 1: Calculate the mass of a single \({}^{69}\text{Ga}^+\) ion in kg.
- \(m = \frac{\text{mass of 1 mole of Ga ion in kg}}{L} = \frac{69 \times 10^{-3}\text{ kg mol}^{-1}}{6.022 \times 10^{23}\text{ mol}^{-1}} = 1.1458 \times 10^{-25}\text{ kg}\)

Step 2: Calculate the velocity, \(v\).
- \(KE = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2KE}{m}}\)
- \(v = \sqrt{\frac{2 \times 1.15 \times 10^{-16}\text{ J}}{1.1458 \times 10^{-25}\text{ kg}}} = \sqrt{2.00733 \times 10^9} = 4.480 \times 10^4\text{ m s}^{-1}\)

Step 3: Calculate the time of flight, \(t\).
- \(v = \frac{d}{t} \Rightarrow t = \frac{d}{v}\)
- \(t = \frac{1.20\text{ m}}{4.480 \times 10^4\text{ m s}^{-1}} = 2.678 \times 10^{-5}\text{ s}\)

To 3 significant figures, \(t = 2.68 \times 10^{-5}\text{ s}\).

**Part (a) [1 mark]**
- M1: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6\) (allow \([\text{Ar}] 3d^6\)). (1) (Reject if \(4s^2\) is present).

**Part (b) [3 marks]**
- M1: Sulfur has paired electrons in a \(3p\) orbital (or sulfur is \(3p^4\) and phosphorus is \(3p^3\)). (1)
- M2: These paired electrons repel each other. (1)
- M3: Repulsion makes it easier to remove the electron (less energy required). (1)

**Part (c)(i) [1.125 marks]**
- M1: Equation: \(\text{Ga}(\text{g}) \rightarrow \text{Ga}^+(\text{g}) + \text{e}^-\) with correct state symbols. (1.125) (Allow \(\text{Ga}(\text{g}) + \text{e}^- \rightarrow \text{Ga}^+(\text{g}) + 2\text{e}^-\)).

**Part (c)(ii) [3 marks]**
- M1: Calculate mass of one ion in kg: \(1.15 \times 10^{-25}\text{ kg}\) (allow \(1.146 \times 10^{-25}\)). (1)
- M2: Calculate velocity: \(4.48 \times 10^4\text{ m s}^{-1}\) (consequential on mass). (1)
- M3: Calculate time: \(2.68 \times 10^{-5}\text{ s}\) (consequential on velocity, must be 3 sig figs). (1)

(a) Solubility of Group 2 hydroxides increases down the group.
(b) Solubility of Group 2 sulfates decreases down the group.
(c) Equation: \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\)
(d)
- Hydrochloric acid is added to react with and remove any carbonate (\(\text{CO}_3^{2-}\)) or sulfite (\(\text{SO}_3^{2-}\)) impurities that might be present in the solution. These impurities would otherwise react with barium ions to form white precipitates (\(\text{BaCO}_3\) or \(\text{BaSO}_3\)), giving a false positive.
- Sulfuric acid cannot be used because it contains sulfate ions (\(\text{SO}_4^{2-}\)), which would react directly with the barium chloride to produce a white precipitate of barium sulfate, rendering the test useless.
(e)
- Magnesium hydroxide: Used as an antacid to relieve indigestion/neutralise excess stomach acid. It is safe because it is only sparingly soluble, so it does not produce a highly alkaline solution that would damage tissues.
- Barium sulfate: Used as a 'barium meal' (contrast agent) for X-ray imaging of the digestive system. It is safe because it is highly insoluble, so toxic barium ions (\(\text{Ba}^{2+}\)) are not absorbed into the bloodstream.

**Part (a) [1 mark]**
- M1: Solubility increases down the group. (1)

**Part (b) [1 mark]**
- M1: Solubility decreases down the group. (1)

**Part (c) [2 marks]**
- M1: Correct species: \(\text{Ba}^{2+}\) + \(\text{SO}_4^{2-}\) \(\rightarrow\) \(\text{BaSO}_4\). (1)
- M2: Correct state symbols: \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\). (1)

**Part (d) [2.125 marks]**
- M1: Hydrochloric acid reacts with / removes carbonate/sulfite impurities (which would otherwise form false positive precipitates). (1)
- M2: Sulfuric acid contains sulfate ions, which would react with barium ions to form a white precipitate (false positive). (1.125)

**Part (e) [2 marks]**
- M1: Magnesium hydroxide is an antacid, safe because it is sparingly soluble / weakly alkaline (does not damage stomach). (1)
- M2: Barium sulfate is a barium meal / contrast agent, safe because it is insoluble (so toxic barium ions cannot be absorbed). (1)

(a)(i) \(\text{MnO}_4^-\): \(x + 4(-2) = -1 \Rightarrow x = +7\).
(ii) Half-equation: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
(iii) Half-equation: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\)
(iv) Multiply the iron half-equation by 5 and add:
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\)

(b)(i) Equation: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\)

(ii) Disproportionation is a reaction in which an element is simultaneously oxidised and reduced.
- In \(\text{Cl}_2\), chlorine has an oxidation state of \(0\).
- In \(\text{NaCl}\), chlorine has an oxidation state of \(-1\) (reduced).
- In \(\text{NaClO}\), chlorine has an oxidation state of \(+1\) (oxidised).

**Part (a) [5.125 marks]**
- (i) M1: \(+7\) (or 7). (1)
- (ii) M1: Correct reactants and products: \(\text{MnO}_4^- + \text{H}^+ \rightarrow \text{Mn}^{2+} + \text{H}_2\text{O}\). (1)
- (ii) M2: Correct balancing and electrons: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\). (1)
- (iii) M1: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\). (1)
- (iv) M1: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\). (1.125)

**Part (b) [3.125 marks]**
- (i) M1: Correct species: \(\text{Cl}_2 + \text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). (1)
- (i) M2: Correctly balanced: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\). (1)
- (ii) M1: Disproportionation definition: same element (chlorine) is simultaneously oxidised and reduced, with oxidation state changing from 0 in \(\text{Cl}_2\) to -1 in \(\text{NaCl}\) and +1 in \(\text{NaClO}\). (1.125)

The amide ion, \( \text{NH}_2^- \), has 7 (nitrogen) + 2 (hydrogen) + 1 (negative charge) = 10 electrons. Its electron configuration is \( 1s^2 2s^2 2p^6 \). The hydronium ion, \( \text{H}_3\text{O}^+ \), has 8 (oxygen) + 3 (hydrogen) - 1 (positive charge) = 10 electrons, giving it the identical electron configuration \( 1s^2 2s^2 2p^6 \). Other options have different electron numbers: \( \text{CH}_3^+ \) has 8 electrons, \( \text{Li}^+ \) has 2 electrons, and \( \text{He} \) has 2 electrons.

1 mark for identifying that both \( \text{NH}_2^- \) and \( \text{H}_3\text{O}^+ \) have 10 electrons and therefore the same electron configuration.

Using the ideal gas equation, \( PV = nRT \), we must convert all units to SI: pressure \( P = 100 \times 10^3\text{ Pa} \), volume \( V = 240 \times 10^{-6}\text{ m}^3 \), and temperature \( T = 37.0 + 273.15 = 310.15\text{ K} \). Rearranging to solve for \( n \) gives: \( n = \frac{PV}{RT} = \frac{100 \times 10^3 \times 240 \times 10^{-6}}{8.31 \times 310.15} = \frac{24.0}{2577.3} \approx 9.3 \times 10^{-3}\text{ mol} \).

1 mark for the correct calculation showing unit conversions for pressure, volume, and temperature to yield \( 9.3 \times 10^{-3}\text{ mol} \).

Boron trichloride (\( \text{BCl}_3 \)) has 3 bonding pairs and 0 lone pairs around the central boron atom, giving it a trigonal planar geometry with bond angles of exactly \( 120^\circ \). Nitrogen trichloride (\( \text{NCl}_3 \)) and phosphorus trichloride (\( \text{PCl}_3 \)) both have 3 bonding pairs and 1 lone pair, resulting in a trigonal pyramidal shape with angles of approximately \( 107^\circ \). The tetrachloroaluminate ion (\( \text{AlCl}_4^- \)) has 4 bonding pairs and 0 lone pairs, giving a tetrahedral shape with bond angles of \( 109.5^\circ \).

1 mark for correctly identifying the trigonal planar species with bond angles of exactly \( 120^\circ \).

Silicon (\( \text{Si} \)) has a giant covalent structure with very strong covalent bonds, giving it the highest melting point of the period. Sulfur exists as large \( \text{S}_8 \) molecules with relatively strong van der Waals forces. Phosphorus exists as smaller \( \text{P}_4 \) molecules with weaker van der Waals forces. Chlorine exists as diatomic \( \text{Cl}_2 \) molecules, which are smaller still and have the weakest intermolecular attractions of the group. Therefore, the correct decreasing order is \( \text{Si} > \text{S}_8 > \text{P}_4 > \text{Cl}_2 \).

1 mark for selecting the correct sequence of decreasing melting points based on structure and bonding trends across Period 3.

As you descend Group 2 from magnesium to barium: (1) The solubility of the hydroxides increases (magnesium hydroxide is sparingly soluble, whereas barium hydroxide is much more soluble). (2) The solubility of the sulfates decreases (magnesium sulfate is highly soluble, whereas barium sulfate is virtually insoluble).

1 mark for identifying the correct solubility trends for both Group 2 hydroxides and sulfates down the group.

Iodide ions are strong reducing agents and reduce the sulfur in concentrated sulfuric acid (oxidation state +6) to several states, including sulfur dioxide (oxidation state +4, a choking gas), sulfur (oxidation state 0, which is observed as a yellow solid precipitate), and hydrogen sulfide (oxidation state -2, a toxic gas with a bad egg smell). Iodine is also formed, but it is the oxidation product of iodide, not a reduction product of sulfuric acid, and is a grey-black solid.

1 mark for identifying sulfur as the yellow solid reduction product of concentrated sulfuric acid.

The equilibrium constant expression is \( K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} \). Substituting the standard concentration unit \( \text{mol dm}^{-3} \) gives: \( \text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})} = \frac{1}{\text{mol dm}^{-3}} = \text{mol}^{-1}\text{dm}^3 \).

1 mark for correctly determining the units of \( K_c \) by working through the equilibrium concentration expression.

A catalyst increases the reaction rate by providing an alternative reaction pathway with a lower activation energy. This allows a greater proportion of colliding reactant molecules to have energy equal to or greater than the activation energy. Catalysts do not increase collision frequency, change the average kinetic energy of molecules (which is determined solely by temperature), or shift the position of an equilibrium.

1 mark for selecting the correct thermodynamic/kinetic pathway explanation for the action of a catalyst.

Let b be the percentage abundance of \(^{25}\text{X}\) and c be the percentage abundance of \(^{26}\text{X}\). Since the total percentage must equal 100%: \(78.6 + b + c = 100 \implies c = 21.4 - b\). Using the formula for relative atomic mass (\(A_r\)): \(A_r = \frac{24 \times 78.6 + 25b + 26c}{100} = 24.32\). Multiplying by 100: \(1886.4 + 25b + 26(21.4 - b) = 2432\). Expanding the terms: \(1886.4 + 25b + 556.4 - 26b = 2432\), which simplifies to \(2442.8 - b = 2432\). This gives \(b = 10.8\%\). Therefore, the abundance of \(^{25}\text{X}\) is 10.8%.

1 mark: Correct option C chosen.

First, use the ideal gas equation to find the number of moles (n): \(pV = nRT \implies n = \frac{pV}{RT}\). Converting pressure into Pascals: \(p = 120\text{ kPa} = 120,000\text{ Pa}\). Substituting the values: \(n = \frac{120,000 \times (4.50 \times 10^{-4})}{8.31 \times 310} = \frac{54}{2576.1} = 0.02096\text{ mol}\). Now, calculate the mass (m): \(m = n \times M_r = 0.02096 \times 44.0 = 0.922\text{ g}\). This rounds to 0.92 g.

1 mark: Correct option B chosen.

We determine the shapes of the species using VSEPR theory. For \(\text{BF}_3\), Boron has 3 valence electrons + 3 from F = 3 bonding pairs and no lone pairs, so the shape is trigonal planar. For \(\text{ClF}_3\), Chlorine has 7 valence electrons + 3 from F = 5 electron pairs with 3 bonding pairs and 2 lone pairs, so the shape is T-shaped (planar). For \(\text{PCl}_3\), Phosphorus has 5 valence electrons + 3 from Cl = 4 electron pairs with 3 bonding pairs and 1 lone pair, so the shape is trigonal pyramidal (non-planar). For \(\text{XeF}_4\), Xenon has 8 valence electrons + 4 from F = 6 electron pairs with 4 bonding pairs and 2 lone pairs, so the shape is square planar.

1 mark: Correct option C chosen.

Option A is incorrect because the solubility of Group 2 sulfates decreases down the group. Option B is correct because adding acidified \(\text{BaCl}_2\) to a solution containing \(\text{SO}_4^{2-}\) ions forms an insoluble white precipitate of \(\text{BaSO}_4\), which is the standard test for sulfate ions. Option C is incorrect because reactivity with water increases down Group 2, so barium reacts much more rapidly with cold water than magnesium does. Option D is incorrect because since sulfate solubility decreases down the group, strontium sulfate is less soluble than calcium sulfate.

1 mark: Correct option B chosen.

When solid sodium chloride reacts with concentrated sulfuric acid, only an acid-base reaction occurs because chloride ions are weak reducing agents; steamy fumes of \(\text{HCl}\) gas are produced, but no reduction occurs. When solid sodium bromide reacts with concentrated sulfuric acid, bromide ions are strong enough to reduce the sulfuric acid. In addition to the acid-base reaction that produces \(\text{HBr}\) gas, a redox reaction occurs where bromide ions reduce \(\text{H}_2\text{SO}_4\) to sulfur dioxide (\(\text{SO}_2\)) gas while being oxidized to orange bromine fumes (\(\text{Br}_2\)). Therefore, \(\text{SO}_2\) gas is produced in the reaction with \(\text{NaBr}\) but not with \(\text{NaCl}\).

1 mark: Correct option B chosen.

Let us set up the changes in amounts to reach equilibrium. The initial moles are A = 2.00 mol, B = 1.50 mol, and C = 0.00 mol. At equilibrium, C = 0.80 mol, which represents a change in C of +0.80 mol. According to the balanced equation, 2 moles of B react to form 2 moles of C (a 1:1 ratio of change). Therefore, the change in B is -0.80 mol. The equilibrium moles of B is equal to the initial moles minus 0.80, which is \(1.50 - 0.80 = 0.70\text{ mol}\).

1 mark: Correct option B chosen.

First ionization energy generally increases across a period due to increasing nuclear charge and constant shielding. However, there are two exceptions in Period 3. There is a dip from magnesium to aluminium because the outer electron in aluminium is in a 3p sub-shell, which is higher in energy and partially shielded by 3s electrons, making it easier to remove than an electron from magnesium's outer 3s sub-shell. Silicon has a higher nuclear charge than magnesium and aluminium, so it has a higher first ionization energy than both. Therefore, the correct increasing order is: \(\text{Na} < \text{Al} < \text{Mg} < \text{Si}\).

1 mark: Correct option A chosen.

a) The mechanism is electrophilic addition. Curly arrow starts from the double bond of but-1-ene pointing to the delta-positive hydrogen of H-Br. Simultaneously, a curly arrow starts from the H-Br bond pointing to the delta-negative bromine. This generates a secondary carbocation (CH3-CH2-CH+-CH3) and a bromide ion (Br-). A curly arrow then starts from a lone pair on the bromide ion pointing to the positively charged carbon. b) The major product is 2-bromobutane. This is because the reaction proceeds via the secondary carbocation intermediate, which is more stable than the primary carbocation intermediate due to the electron-releasing inductive effect of two alkyl groups compared to only one in the primary carbocation. c) The repeating unit of poly(but-1-ene) is drawn by opening the double bond: -[CH(C2H5)-CH2]- with single bonds extending through the brackets.

a) M1: Curly arrow from C=C double bond to H of H-Br (with correct partial charges shown on H-Br). M2: Curly arrow from H-Br bond to Br. M3: Structure of secondary carbocation intermediate shown with positive charge on carbon-2. M4: Curly arrow from lone pair on Br- to carbocation carbon. b) M5: IUPAC name: 2-bromobutane. M6: Explanation of carbocation stability: major product is formed via a secondary carbocation. M7: Explanation of why: secondary carbocations are more stable than primary carbocations due to the inductive effect of two alkyl groups (releasing electron density). c) M8: Correct repeating unit of poly(but-1-ene) showing polymer brackets and open bonds: -[CH(CH2CH3)-CH2]-.

a) The Maxwell-Boltzmann distribution at temperature T1 starts at the origin, rises to a peak, and then decreases asymmetrically towards the x-axis, which it never touches. When temperature increases to T2: the peak shifts to the right (higher energy) and is lower (fewer molecules at peak energy). The total area under both curves remains identical. b) At T2, the distribution curve shifts to the right, meaning the fraction of molecules with energy greater than or equal to the activation energy (Ea) increases significantly. This leads to a greater frequency of successful collisions per unit time, which increases the reaction rate. c) A catalyst works by providing an alternative reaction pathway with a lower activation energy (Ea). Consequently, a larger proportion of molecules have energy greater than the new activation energy, increasing the rate of successful collisions.

a) M1: Curve starts at origin and asymmetric shape. M2: At T2, the peak shifts to the right. M3: At T2, the peak is lower than at T1. M4: Curve at T2 must cross T1 curve only once and remain higher than T1 at high energy. b) M5: More molecules have energy greater than or equal to the activation energy (Ea) at higher temperature. M6: This leads to a higher frequency of successful collisions. c) M7: Catalyst provides an alternative pathway. M8: This alternative pathway has a lower activation energy.

a) To obtain propanal, propan-1-ol is heated with acidified potassium dichromate(VI) using a distillation apparatus, which allows the volatile propanal to distil off as it forms, preventing further oxidation. Equation: CH3CH2CH2OH + [O] -> CH3CH2CHO + H2O. b) To obtain propanoic acid, excess acidified potassium dichromate(VI) is used under reflux, ensuring any vaporised intermediate propanal condenses and falls back into the reaction flask to complete the oxidation. Equation: CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O. c) To distinguish, add Fehling's solution (or Tollens' reagent) to both. Propanal reacts to form a brick-red precipitate (or silver mirror for Tollens'), while propanoic acid shows no change. Alternatively, add sodium hydrogencarbonate; propanoic acid will effervesce (releasing CO2) while propanal will not.

a) M1: Acidified potassium dichromate(VI) (or H+/Cr2O7^2-) and distillation setup. M2: Balanced equation: CH3CH2CH2OH + [O] -> CH3CH2CHO + H2O. b) M3: Excess acidified potassium dichromate(VI) and reflux setup. M4: Balanced equation: CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O. c) M5: Suitable test reagent (e.g., Tollens' reagent, Fehling's solution, or NaHCO3/Na2CO3). M6: Correct matching observation for both compounds (e.g., Fehling's: propanal gives red precipitate, propanoic acid remains blue; NaHCO3: propanoic acid gives bubbles, propanal does not).

a) The hydroxide ion acts as a nucleophile. A curly arrow starts from the lone pair on the oxygen of the OH- ion and points to the electron-deficient carbon atom (C delta-positive) bonded to the bromine atom. At the same time, a curly arrow starts from the C-Br bond and points to the bromine atom (Br delta-negative). This results in the formation of butan-1-ol and a bromide ion. b) 1-Iodobutane hydrolyses faster than 1-bromobutane. This is because the C-I bond is longer and weaker (has a lower bond enthalpy) than the C-Br bond, meaning less energy is required to break the C-I bond. Bond enthalpy is the dominant factor determining rate of hydrolysis, overriding bond polarity. c) The hydroxide ion acts as a nucleophile (electron pair donor).

a) M1: Correct dipoles shown on C-Br bond (C delta-positive, Br delta-negative). M2: Curly arrow from the lone pair on the hydroxide oxygen to the carbon atom of the C-Br bond. M3: Curly arrow from the C-Br bond to the bromine atom. M4: Correct structures of products (butan-1-ol and Br-). b) M5: 1-Iodobutane is faster / hydrolyses at a higher rate. M6: The C-I bond is weaker than the C-Br bond / has a lower bond enthalpy. M7: The C-I bond breaks more easily than the C-Br bond. c) M8: Nucleophile / electron pair donor.

a) The broad absorption peak at 2500-3000 cm^-1 is characteristic of the O-H stretching vibration in a carboxylic acid. The sharp peak at 1715 cm^-1 represents the C=O stretch of a carbonyl group. Together, these indicate that the functional group present is a carboxylic acid. b) With three carbons, the carboxylic acid is propanoic acid. Its displayed formula is CH3-CH2-COOH showing all single and double bonds explicitly (C-H, C-C, C=O, C-O, O-H). c) The fragment ion with m/z = 45 is the carboxyl cation, [COOH]+. The equation for its formation from the molecular ion [C3H6O2].+ is: [C3H6O2].+ -> [COOH]+ + .CH2CH3 (or .C2H5 radical).

a) M1: Identifies carboxylic acid group. M2: Attributes 2500-3000 cm^-1 peak to O-H stretch (in carboxylic acid). M3: Attributes 1715 cm^-1 peak to C=O stretch. b) M4: Correct displayed structure showing all bonds (including the O-H and C=O bonds). M5: Correct IUPAC name: propanoic acid. c) M6: Identifies fragment ion formula as [COOH]+ (must include positive charge). M7: Correct reactant in equation: molecular ion [C3H6O2].+ (must have charge and radical dot, or accept [C3H6O2]+). M8: Equation balances in terms of atoms and charge (shows radical .C2H5 as the neutral product).

a) Ultraviolet light provides the activation energy required to break the covalent chlorine-chlorine bond homolytically to form chlorine free radicals. b) Initiation: Cl2 -> 2Cl. Propagation 1: CH4 + Cl. -> .CH3 + HCl. Propagation 2: .CH3 + Cl2 -> CH3Cl + Cl. c) Termination forming a larger hydrocarbon: 2.CH3 -> C2H6 (ethane). d) Further propagation reactions can occur because the product, chloromethane, still has hydrogen atoms that can be abstracted by chlorine radicals, leading to dichloromethane, trichloromethane, etc. To maximise the yield of chloromethane, a large excess of methane relative to chlorine should be used, making collisions between chlorine radicals and methane much more likely than collisions with chloromethane.

a) M1: Provides energy to break the Cl-Cl bond (or for homolytic fission of chlorine molecule). b) M2: Cl2 -> 2Cl. (must show radical dots). M3: CH4 + Cl. -> .CH3 + HCl. M4: .CH3 + Cl2 -> CH3Cl + Cl. c) M5: 2.CH3 -> C2H6 (or CH3CH3). d) M6: Chloromethane reacts further with chlorine radicals (or undergoes further substitution). M7: Use a large excess of methane. M8: This makes collisions between Cl. and CH4 more likely than with CH3Cl.

a) Mass of solution, m = 50.0 cm3 + 50.0 cm3 = 100.0 g (since density = 1.00 g/cm3). Temperature change, dT = 25.3 - 18.5 = 6.8 K. q = mc(dT) = 100.0 g * 4.18 J g^-1 K^-1 * 6.8 K = 2842.4 J. b) Moles of HCl = conc * vol = 1.00 * (50.0/1000) = 0.0500 mol. Moles of NaOH = conc * vol = 1.00 * (50.0/1000) = 0.0500 mol. Since they react in a 1:1 ratio, moles of H2O formed = 0.0500 mol. c) Enthalpy change of neutralisation is the energy released per mole of water formed. dH = -q / (moles of H2O * 1000) = -2842.4 J / 0.0500 mol = -56848 J/mol = -56.8 kJ/mol (to 3 sig figs, negative sign required as reaction is exothermic). d) Major source of error is heat loss to the surroundings. This can be minimised by using a polystyrene cup as a calorimeter, adding a lid, or plotting a cooling curve to extrapolate the true maximum temperature.

a) M1: Identifies correct mass m = 100 g and temperature change dT = 6.8 K. M2: q = 2842.4 J (or 2.84 kJ). b) M3: Correct calculation of moles of HCl and NaOH (0.0500 mol). M4: Moles of water formed = 0.0500 mol. c) M5: Division of energy (in kJ) by moles of water (2.8424 / 0.0500). M6: Evaluates to 56.8. M7: Correct negative sign and units: -56.8 kJ/mol (allow -57). d) M8: Heat loss to surroundings AND use a polystyrene cup/lid/windshield (or incomplete mixing/slow reaction AND plot a cooling curve to extrapolate maximum temperature).

a) Initial moles: SO2 = 2.00, O2 = 1.00, SO3 = 0.00. Equilibrium moles: SO2 = 0.40. Change in SO2 moles = 0.40 - 2.00 = -1.60 mol. From the stoichiometry (2:1:2 ratio): Change in O2 moles = -1.60 / 2 = -0.80 mol. Change in SO3 moles = +1.60 mol. Equilibrium moles: O2 = 1.00 - 0.80 = 0.20 mol. SO3 = 0.00 + 1.60 = 1.60 mol. b) Kc = [SO3]^2 / ([SO2]^2 * [O2]). c) Divide equilibrium moles by volume (2.00 dm3) to find concentrations: [SO2] = 0.40 / 2.00 = 0.20 mol dm^-3; [O2] = 0.20 / 2.00 = 0.10 mol dm^-3; [SO3] = 1.60 / 2.00 = 0.80 mol dm^-3. Kc = (0.80)^2 / ((0.20)^2 * 0.10) = 0.64 / (0.04 * 0.10) = 0.64 / 0.004 = 160. Units of Kc = (mol dm^-3)^2 / ((mol dm^-3)^2 * mol dm^-3) = (mol dm^-3)^-1 = mol^-1 dm^3.

a) M1: Shows change in SO2 is -1.60 mol. M2: Equilibrium moles of O2 = 0.20 mol. M3: Equilibrium moles of SO3 = 1.60 mol. b) M4: Kc = [SO3]^2 / ([SO2]^2 * [O2]) (must use square brackets). c) M5: Calculates correct equilibrium concentrations by dividing by volume (V = 2.00 dm3). M6: Substitutes concentrations correctly into the Kc expression. M7: Evaluation of Kc = 160. M8: Units = mol^-1 dm^3.

First, calculate the molar mass of \(C_3H_6O\):
\(M_r = (3 \times 12.0) + (6 \times 1.0) + 16.0 = 58.0 \text{ g mol}^{-1}\).

Next, calculate the number of moles of \(X\):
\(n = \frac{\text{mass}}{M_r} = \frac{0.116}{58.0} = 0.00200 \text{ mol}\).

Use the ideal gas equation \(PV = nRT\) to find the volume in \(\text{m}^3\):
\(V = \frac{nRT}{P} = \frac{0.00200 \times 8.31 \times 373}{100 \times 10^3} = 6.20 \times 10^{-5} \text{ m}^3\).

Convert \(\text{m}^3\) to \(\text{cm}^3\) by multiplying by \(10^6\):
\(V = 6.20 \times 10^{-5} \times 10^6 = 62.0 \text{ cm}^3\).

1 mark for the correct calculation leading to 62.0 (Option B). Award 0 marks for common conversion errors such as 6.20 (Option A) or 620 (Option C).

In a Maxwell-Boltzmann distribution, the curve is asymmetrical and skewed to the right. Because of this right-skewed tail, the mean (average) energy of the molecules is located to the right of the peak (which represents the most probable energy). Thus, the average energy is higher than the most probable energy.

Option A is incorrect because a catalyst decreases the activation energy but has no effect on the distribution of molecular energies. Option B is incorrect because the area under the curve to the right of the activation energy represents only the number of molecules with energy greater than or equal to the activation energy. Option C is incorrect because at higher temperatures the peak shifts to the right and is lower.

1 mark for identifying D as the correct statement. Deduct 0 marks for incorrect options.

The rate of nucleophilic substitution of halogenoalkanes depends on the strength of the carbon-halogen bond (bond enthalpy). The C–I bond is the weakest of the carbon-halogen bonds listed, meaning it requires the least energy to break. Therefore, iodoalkanes react much faster than bromoalkanes and chloroalkanes. Among the options, 2-iodobutane contains the weakest C–Halogen bond and will hydrolyse the fastest.

1 mark for selecting D as the correct option, recognizing that C-I bond fission is the fastest due to lowest bond enthalpy.

The electrophilic addition of HBr to 2-methylbut-2-ene \(((CH_3)_2C=CHCH_3)\) proceeds via the most stable carbocation intermediate.

Adding \(H^+\) to C3 produces a tertiary carbocation at C2: \((CH_3)_2C^+-CH_2CH_3\).
Adding \(H^+\) to C2 produces a secondary carbocation at C3: \((CH_3)_2CH-CH^+-CH_3\).

Since a tertiary carbocation is more stable than a secondary carbocation (due to the electron-releasing inductive effect of three alkyl groups), the reaction proceeds via the tertiary carbocation. Subsequent attack by the bromide ion (\(Br^-\)) yields 2-bromo-2-methylbutane as the major product.

1 mark for choosing option A. Correctly identifying the major product based on carbocation stability.

Compound \(Y\) is an alcohol.
- Primary alcohols (butan-1-ol and 2-methylpropan-1-ol) oxidize under reflux to carboxylic acids (butanoic acid and 2-methylpropanoic acid, respectively), which would react with sodium carbonate to produce carbon dioxide gas.
- Secondary alcohols (butan-2-ol) oxidize to ketones (butanone), which do not react with sodium carbonate.
- Tertiary alcohols (2-methylpropan-2-ol) resist oxidation under these conditions, so they would not yield an oxidized organic product.

Since the organic product does not react with sodium carbonate, \(Y\) must be the secondary alcohol, butan-2-ol.

1 mark for identifying B as the correct answer. Rejecting A and C as they form carboxylic acids, and D because it does not oxidize.

The infrared absorption spectrum exhibits two key features:
1. A very broad, strong band in the 2500–3000 \(\text{cm}^{-1}\) range, which is characteristic of the O–H bond stretch in a carboxylic acid.
2. A strong band at 1715 \(\text{cm}^{-1}\), which is characteristic of the C=O stretch.

These combined features identify the compound as a carboxylic acid. Propanoic acid is the only carboxylic acid listed.

1 mark for identifying propanoic acid (A) as the correct option.

The value of the equilibrium constant \(K_c\) is only affected by changes in temperature. Changes in pressure or concentration, or the addition of a catalyst, do not affect \(K_c\).

Since the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature shifts the equilibrium position to the right to oppose the change, producing more product and reducing reactant concentrations. This increases the value of \(K_c\).

1 mark for identifying option A. Award 0 marks if a change in pressure or catalyst is chosen, as they do not affect Kc.

Using a Hess's law cycle based on enthalpy of combustion data:
\(\Delta_r H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)

\(\Delta_r H^\theta = [\Delta_c H^\theta(\text{propene}) + \Delta_c H^\theta(\text{hydrogen})] - \Delta_c H^\theta(\text{propane})\)

\(\Delta_r H^\theta = [(-2058) + (-286)] - [-2220]\)
\(\Delta_r H^\theta = -2344 + 2220 = -124 \text{ kJ mol}^{-1}\).

1 mark for calculating the correct enthalpy change of -124 (Option A). Award 0 marks for common arithmetic or sign errors leading to Option B (+124).

Reaction of 2-methylbut-1-ene (\(\text{CH}_2=\text{C(CH}_3)\text{CH}_2\text{CH}_3\)) with \(\text{HBr}\) proceeds via electrophilic addition. According to Markovnikov's rule, the electrophilic hydrogen ion adds to the double-bonded carbon with more hydrogen atoms (C1), forming the more stable tertiary carbocation intermediate: \(\text{(CH}_3)_2\text{C}^+-\text{CH}_2\text{CH}_3\). The bromide ion then attacks this carbocation, forming 2-bromo-2-methylbutane as the major product.

[1 mark] B - 2-bromo-2-methylbutane is the major product because it is formed via the more stable tertiary carbocation intermediate.

A primary alcohol is oxidized under reflux with excess acidified potassium dichromate(VI) to a carboxylic acid. A carboxylic acid is acidic enough to turn blue litmus paper red, but it does not react with Tollens' reagent (which only reacts with aldehydes). Among the options, 2,2-dimethylpropan-1-ol is a primary alcohol, 2-methylbutan-2-ol is a tertiary alcohol (resistant to oxidation), and both 3-methylbutan-2-ol and pentan-3-ol are secondary alcohols (oxidized to ketones, which are not acidic and do not turn litmus paper red).

[1 mark] C - 2,2-dimethylpropan-1-ol is a primary alcohol and is oxidized to a carboxylic acid, which turns litmus red.

Heating a halogenoalkane with ethanolic potassium hydroxide (KOH) favors elimination, yielding an alkene (propene). Aqueous KOH, aqueous ammonia, and aqueous potassium cyanide all favor nucleophilic substitution, producing an alcohol, an amine, and a nitrile, respectively.

[1 mark] B - Ethanolic KOH under heat promotes elimination.

When temperature increases, the Maxwell-Boltzmann distribution curve flattens and shifts to the right, meaning the peak (which represents the most probable energy, \(E_{\text{mp}}\)) shifts to a higher energy. The activation energy (\(E_{\text{a}}\)) depends only on the chemical reaction pathway and is not affected by temperature.

[1 mark] B - \(E_{\text{mp}}\) increases because the peak of the Maxwell-Boltzmann curve shifts to the right, and \(E_{\text{a}}\) is independent of temperature.

A propagation step must have a radical on both the reactant side and the product side. In free-radical chlorination of alkanes, hydrogen radicals (\(\text{H}^\bullet\)) are not formed, making option A incorrect. Option C is a termination step because two radicals react to form a non-radical. Option D is an initiation step because a non-radical molecule undergoes homolytic fission to form two radicals. Option B is a correct propagation step where the secondary butyl radical reacts with a chlorine molecule to produce 2-chlorobutane and regenerate the chlorine radical.

[1 mark] B - Correctly identifies a propagation step with a radical on both sides, yielding the chlorinated product and regenerating the halogen radical.

The infrared absorption band in the range \(2500\text{--}3000\text{ cm}^{-1}\) is characteristic of the O-H stretch in a carboxylic acid (which overlaps the C-H stretching region). The strong absorption band at approximately \(1715\text{ cm}^{-1}\) corresponds to the C=O stretch. A compound with both of these features and the formula \(\text{C}_3\text{H}_6\text{O}_2\) is a carboxylic acid, namely propanoic acid.

[1 mark] D - Propanoic acid is the only carboxylic acid, matching the characteristic O-H (acid) and C=O stretches.

For stereoisomerism (\(E\)-\(Z\) isomerism) to occur, each of the two carbon atoms in the carbon-carbon double bond must be bonded to two different groups. Let us analyze each option:
- 1,1-dichlorobut-1-ene: C1 is bonded to two identical chlorine atoms. No \(E\)-\(Z\) isomerism.
- 2-methylbut-2-ene: C2 is bonded to two identical methyl groups. No \(E\)-\(Z\) isomerism.
- 1,2-dichlorobut-2-ene (\(\text{ClCH}_2\text{C(Cl)}=\text{CHCH}_3\)): C2 is bonded to \(-\text{CH}_2\text{Cl}\) and \(-\text{Cl}\) (different groups); C3 is bonded to \(-\text{H}\) and \(-\text{CH}_3\) (different groups). This compound can exist as \(E\) and \(Z\) isomers.
- 2,3-dimethylbut-2-ene: both double-bonded carbons are bonded to identical methyl groups. No \(E\)-\(Z\) isomerism.

[1 mark] C - 1,2-dichlorobut-2-ene has two different groups attached to each carbon of the double bond, satisfying the criteria for \(E\)-\(Z\) isomerism.