2024 OCR AS Level Chemistry A - H032 模擬試題連答案詳解

Let \(x\) be the abundance of \(^{25}X\) and \(y\) be the abundance of \(^{26}X\).
Since the total abundance must equal \(100\%\):
\(78.99 + x + y = 100 \implies x = 21.01 - y\)

Now set up the relative atomic mass equation:
\(A_r = \frac{(24 \times 78.99) + 25x + 26y}{100} = 24.31\)
\(1895.76 + 25(21.01 - y) + 26y = 2431\)
\(1895.76 + 525.25 - 25y + 26y = 2431\)
\(2421.01 + y = 2431\)
\(y = 9.99\%\)

Therefore, the percentage abundance of the \(^{26}X\) isotope is \(9.99\%\).

1 mark for the correct calculation leading to 9.99%. Allow 1 mark for correct setting up of simultaneous equations followed by arithmetic error.

First, calculate the number of moles of \(\text{CO}_2\) gas collected:
\(n(\text{CO}_2) = \frac{300\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0125\text{ mol}\)

From the 1:1 stoichiometry of the reaction, the moles of \(\text{MCO}_3\) is also \(0.0125\text{ mol}\).

Now calculate the molar mass of \(\text{MCO}_3\):
\(M_r(\text{MCO}_3) = \frac{1.25\text{ g}}{0.0125\text{ mol}} = 100\text{ g mol}^{-1}\)

Subtract the relative formula mass of the carbonate group (\(\text{CO}_3^{2-}\)):
\(M_r(\text{CO}_3) = 12.0 + (3 \times 16.0) = 60.0\text{ g mol}^{-1}\)
\(A_r(\text{M}) = 100 - 60 = 40\text{ g mol}^{-1}\)

This corresponds to calcium (\(\text{Ca}\)).

1 mark for identifying Calcium (B) by calculating the correct molar mass of 100 g mol^-1.

Let's determine the electron pair geometry and bond angles for each species:
1. \(\text{NH}_2^-\) has 2 bonding pairs and 2 lone pairs around the central nitrogen atom. Due to lone pair-lone pair repulsion, the bond angle is reduced from the tetrahedral angle to approximately \(104.5^\circ\).
2. \(\text{NH}_3\) has 3 bonding pairs and 1 lone pair around the central nitrogen atom, resulting in a pyramidal shape with a bond angle of approximately \(107^\circ\).
3. \(\text{NH}_4^+\) has 4 bonding pairs and 0 lone pairs around the central nitrogen atom, resulting in a perfect tetrahedral shape with a bond angle of \(109.5^\circ\).

Therefore, the correct order is \(\text{NH}_2^- < \text{NH}_3 < \text{NH}_4^+\).

1 mark for identifying the correct order based on the number of lone pairs repelling bonding pairs.

The equation for the formation of propanone is:
\(3\text{C}(\text{s}) + 3\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{COCH}_3(\text{l})\)

Using Hess's Law with enthalpy changes of combustion:
\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)
\(\Delta_f H^\theta = [3 \times \Delta_c H^\theta(\text{C}) + 3 \times \Delta_c H^\theta(\text{H}_2)] - \Delta_c H^\theta(\text{propanone})\)
\(\Delta_f H^\theta = [3(-394) + 3(-286)] - [-1786]\)
\(\Delta_f H^\theta = [-1182 - 858] + 1786\)
\(\Delta_f H^\theta = -2040 + 1786 = -254\text{ kJ mol}^{-1}\).

1 mark for correct calculation leading to -254 kJ mol^-1.

When temperature is increased, the peak of the Boltzmann distribution shifts to a higher energy (to the right) and a lower height (flattens), because the average kinetic energy of the molecules increases and the spread of energies is wider.
Adding a catalyst provides an alternative pathway with a lower activation energy, effectively moving the activation energy value, \(E_a\), to the left on the distribution axis, while leaving the distribution curve itself unchanged.

1 mark for correctly identifying both the shift in the peak of the distribution curve and the change in activation energy.

For \(E/Z\) isomerism to occur, both carbon atoms of the double bond must be attached to two different groups.
- In 1-bromo-2-methylbut-2-ene (\(\text{BrCH}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3\)), C2 is bonded to \(-\text{CH}_2\text{Br}\) and \(-\text{CH}_3\) (two different groups), and C3 is bonded to \(-\text{H}\) and \(-\text{CH}_3\) (two different groups). Hence, it exhibits stereoisomerism.
- In 2-methylbut-2-ene (\(\text{(CH}_3)_2\text{C}=\text{CHCH}_3\)), one of the carbons has two identical methyl groups attached, so no E/Z isomerism.
- In 1,1-dichloroprop-1-ene (\(\text{Cl}_2\text{C}=\text{CHCH}_3\)), C1 is bonded to two identical chlorine atoms.
- In 2-ethylbut-1-ene (\(\text{CH}_2=\text{C}(\text{C}_2\text{H}_5)_2\)), C1 has two hydrogens and C2 has two identical ethyl groups.

1 mark for identifying that 1-bromo-2-methylbut-2-ene has two different groups attached to each carbon of the double bond.

The rate of hydrolysis of haloalkanes is determined by the strength of the carbon–halogen bond (bond enthalpy), rather than the bond polarity.
Because bond enthalpy decreases down the group (\(\text{C–Cl} > \text{C–Br} > \text{C–I}\)), the \(\text{C–I}\) bond is the weakest and requires the least energy to break. Consequently, 1-iodobutane undergoes nucleophilic substitution fastest, yielding a precipitate of silver iodide almost immediately.

1 mark for identifying 1-iodobutane as the fastest because of its low C-I bond enthalpy.

Let's analyze the peak assignments:
- The broad absorption at \(3200-3600\text{ cm}^{-1}\) corresponds to an alcohol \(\text{O–H}\) bond.
- The strong, sharp absorption at \(1715\text{ cm}^{-1}\) corresponds to a carbonyl \(\text{C=O}\) bond.

Therefore, the compound must contain both a carbonyl group and an alcohol group.

Looking at the options:
- 3-hydroxybutan-2-one has the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\) and contains both a ketone group (\(\text{C=O}\)) and an alcohol group (\(\text{O–H}\)).
- Ethyl ethanoate is an ester and has no \(\text{O–H}\) group.
- Butane-1,3-diol has the formula \(\text{C}_4\text{H}_{10}\text{O}_2\) and has no \(\text{C=O}\) group.
- Butanoic acid is a carboxylic acid, where the characteristic carboxylic acid \(\text{O–H}\) peak would be extremely broad and centered at \(2500-3300\text{ cm}^{-1}\), not \(3200-3600\text{ cm}^{-1}\).

1 mark for identifying 3-hydroxybutan-2-one by matching the molecular formula and both functional groups correctly.

Let the fractional abundance of the heavier isotope, \(^{65}\text{X}\), be \(x\). The abundance of the lighter isotope, \(^{63}\text{X}\), is \(1 - x\). We set up the relative atomic mass equation: \(63(1 - x) + 65x = 63.6\). Simplifying this yields: \(63 - 63x + 65x = 63.6 \Rightarrow 2x = 0.6 \Rightarrow x = 0.3\). This corresponds to an abundance of 30%.

Award 1 mark for the correct calculation of 30% abundance for the heavier isotope. Correct answer only.

\(\text{BF}_3\), \(\text{SO}_3\), and \(\text{CO}_3^{2-}\) all have three bonding regions around the central atom with no lone pairs, resulting in a trigonal planar shape with bond angles of \(120^{\circ}\). \(\text{NH}_3\) has three bonding pairs and one lone pair, which gives it a trigonal pyramidal shape with a bond angle of approximately \(107^{\circ}\).

Award 1 mark for identifying that ammonia has a different shape. Correct answer only.

First, calculate the mass of water lost: \(4.93\text{ g} - 2.41\text{ g} = 2.52\text{ g}\). Next, calculate the amount in moles of anhydrous magnesium sulfate: \(n(\text{MgSO}_4) = \frac{2.41}{120.4} = 0.0200\text{ mol}\). Then, calculate the amount in moles of water lost: \(n(\text{H}_2\text{O}) = \frac{2.52}{18.0} = 0.140\text{ mol}\). Determine the ratio of moles of water to moles of salt: \(\frac{0.140}{0.0200} = 7\). Therefore, \(x = 7\).

Award 1 mark for the correct identification of the integer value of x as 7.

In \(\text{HNO}_2\), hydrogen has an oxidation state of +1 and each oxygen has an oxidation state of -2. For a neutral molecule: \(+1 + \text{N} + 2(-2) = 0 \Rightarrow \text{N} - 3 = 0 \Rightarrow \text{N} = +3\). In \(\text{N}_2\text{O}\), nitrogen is +1. In \(\text{NH}_4\text{NO}_3\), nitrogen has oxidation states of -3 and +5. In \(\text{HNO}_3\), nitrogen is +5.

Award 1 mark for identifying the correct substance with nitrogen in the +3 oxidation state.

Dehydration of pentan-2-ol involves the elimination of water by removing the -OH group from carbon-2 and a hydrogen atom from either carbon-1 or carbon-3. If hydrogen is removed from carbon-1, pent-1-ene is formed. If hydrogen is removed from carbon-3, pent-2-ene is formed. Pent-1-ene and pent-2-ene are structural isomers (position isomers). Pent-2-ene exists as stereoisomers (E/Z), but these are not structural isomers. Therefore, there are exactly 2 structural isomers.

Award 1 mark for identifying that exactly 2 structurally isomeric alkenes are formed.

The equation for the formation of propan-1-ol is: \(3\text{C}(\text{s}) + 4\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{C}_3\text{H}_7\text{OH}(\text{l})\). Using Hess's Law with standard enthalpy changes of combustion: \(\Delta_{\text{f}}H^{\ominus} = \sum \Delta_{\text{c}}H^{\ominus}(\text{reactants}) - \sum \Delta_{\text{c}}H^{\ominus}(\text{products})\). Therefore: \(\Delta_{\text{f}}H^{\ominus} = 3(-394) + 4(-286) - (-2021)\). \(\Delta_{\text{f}}H^{\ominus} = -1182 - 1144 + 2021 = -305\text{ kJ mol}^{-1}\).

Award 1 mark for the correct calculation of standard enthalpy of formation with negative sign.

A catalyst provides an alternative reaction pathway with a lower activation energy, which shifts the activation energy barrier to a lower value (to the left on a Boltzmann distribution). A catalyst does not change the temperature of the system, so the shape and position of the Boltzmann distribution curve of molecular energies remains completely unchanged.

Award 1 mark for identifying that the curve is unchanged and activation energy is lowered.

The production of ammonia gas (which turns damp red litmus blue) upon heating with NaOH confirms the presence of the ammonium cation, \(\text{NH}_4^+\). The formation of a cream precipitate that is insoluble in dilute aqueous ammonia but soluble in concentrated aqueous ammonia confirms the presence of the bromide anion, \(\text{Br}^-\). Thus, the salt is ammonium bromide, \(\text{NH}_4\text{Br}\).

Award 1 mark for identifying the correct compound from qualitative analysis.

According to Avogadro's law, equal volumes of gases under the same conditions contain equal numbers of molecules (and thus moles). Therefore, the ratio of the volume of reacting gas to the volumes of product gases is equal to their mole ratio in the balanced equation. The volume ratio is: \(20\text{ cm}^3\) of \(N_xO_y\) to \(20\text{ cm}^3\) of \(N_2\) to \(10\text{ cm}^3\) of \(O_2\). This simplifies to a mole ratio of \(2 : 2 : 1\). Thus, the balanced equation for the decomposition is \(2N_xO_y \rightarrow 2N_2 + O_2\). Balancing the atoms on both sides: there are \(4\) nitrogen atoms on the right, so \(2x = 4\) which gives \(x = 2\). There are \(2\) oxygen atoms on the right, so \(2y = 2\) which gives \(y = 1\). The molecular formula of the nitrogen oxide is therefore \(N_2O\).

1 mark for the correct option B. Reject all other options.

The central nitrogen atom in the amide ion, \(NH_2^-\), has \(5\) valence electrons plus \(1\) extra electron from the negative charge, making a total of \(6\) outer-shell electrons. It forms \(2\) covalent bonds with hydrogen atoms, which leaves \(4\) non-bonding electrons as \(2\) lone pairs. The total of \(4\) electron pairs (\(2\) bonding pairs and \(2\) lone pairs) arrange themselves in a tetrahedral tetrahedral parent geometry to minimize repulsion. Since lone pairs repel more than bonding pairs, the bond angle is reduced from the tetrahedral angle of \(109.5^\circ\) by approximately \(2.5^\circ\) for each lone pair, resulting in a non-linear (bent) shape with a bond angle of approximately \(104.5^\circ\).

1 mark for the correct option B. Reject all other options.

The equation for the standard enthalpy change of formation of propane is: \(3C(s) + 4H_2(g) \rightarrow C_3H_8(g)\). According to Hess's Law, the enthalpy change of formation can be calculated from enthalpy changes of combustion using: \(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\). Therefore: \(\Delta_f H^\theta = [3 \times \Delta_c H^\theta(C(s)) + 4 \times \Delta_c H^\theta(H_2(g))] - [\Delta_c H^\theta(C_3H_8(g))]\). Substituting the given values: \(\Delta_f H^\theta = [3(-394) + 4(-286)] - [-2220] = [-1182 - 1144] + 2220 = -2326 + 2220 = -106\text{ kJ mol}^{-1}\).

1 mark for the correct option A. Reject all other options.

The rate of hydrolysis of haloalkanes depends on the identity of the halogen atom and the structure of the carbon skeleton. Bromoalkanes react faster than chloroalkanes because the \(C-Br\) bond is weaker (has a lower bond enthalpy) than the \(C-Cl\) bond, making it easier to break. Additionally, tertiary haloalkanes react much faster than secondary and primary haloalkanes because they hydrolyze via an \(S_N1\) mechanism, which involves the formation of a highly stable tertiary carbocation intermediate. Comparing the options, 2-bromo-2-methylpropane is a tertiary bromoalkane, 2-bromobutane is a secondary bromoalkane, 1-bromobutane is a primary bromoalkane, and 1-chlorobutane is a primary chloroalkane. Therefore, 2-bromo-2-methylpropane reacts the fastest.

1 mark for the correct option D. Reject all other options.

1. Calculate the mass of water lost:
\(\text{Mass of water} = 5.56\text{ g} - 3.04\text{ g} = 2.52\text{ g}\).

2. Calculate the amount, in moles, of anhydrous \(\text{FeSO}_4\):
\(M_r(\text{FeSO}_4) = 55.8 + 32.1 + (4 \times 16.0) = 151.9\text{ g mol}^{-1}\).
\(n(\text{FeSO}_4) = \frac{3.04\text{ g}}{151.9\text{ g mol}^{-1}} = 0.0200\text{ mol}\).

3. Calculate the amount, in moles, of water lost:
\(M_r(\text{H}_2\text{O}) = 18.0\text{ g mol}^{-1}\).
\(n(\text{H}_2\text{O}) = \frac{2.52\text{ g}}{18.0\text{ g mol}^{-1}} = 0.140\text{ mol}\).

4. Find the simplest whole-number ratio:
\(x = \frac{0.140}{0.0200} = 7\).

- Mark 1: For calculating the moles of anhydrous \(\text{FeSO}_4\) as 0.0200 mol (or 2.52 g of water lost).
- Mark 2: For calculating the moles of water as 0.140 mol.
- Mark 3: For the final integer value of \(x = 7\) from correct mathematical working.

1. Central nitrogen atom has 5 valence electrons and forms 3 single covalent bonds with fluorine atoms, leaving 1 lone pair. This gives a total of 4 electron pairs (3 bonding pairs, 1 lone pair).
2. Electron pairs repel each other to get as far apart as possible to minimise repulsion.
3. Lone pairs repel more strongly than bonding pairs, which reduces the tetrahedral bond angle of \(109.5^\circ\) by approximately \(2.5^\circ\) to \(107^\circ\).
4. This results in a trigonal pyramidal shape.

- Mark 1: Identifies 3 bonding pairs and 1 lone pair around the central nitrogen atom.
- Mark 2: States that electron pairs repel to minimise repulsion, with lone pairs repelling more than bonding pairs.
- Mark 3: Identifies the shape as trigonal pyramidal and states the bond angle as \(107^\circ\) (accept range \(106^\circ - 108^\circ\)).

1. Calculate heat energy released, \(q\):
\(q = m c \Delta T = 150.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 22.5\text{ K} = 14107.5\text{ J} = 14.108\text{ kJ}\).

2. Calculate the amount, in moles, of propan-1-ol burned:
\(n(\text{propan-1-ol}) = \frac{0.550\text{ g}}{60.0\text{ g mol}^{-1}} = 9.167 \times 10^{-3}\text{ mol}\).

3. Calculate the enthalpy change of combustion, \(\Delta H_c\):
\(\Delta H_c = -\frac{q}{n} = -\frac{14.108\text{ kJ}}{9.167 \times 10^{-3}\text{ mol}} = -1539\text{ kJ mol}^{-1}\).
Rounding to 3 significant figures gives \(-1540\text{ kJ mol}^{-1}\).

- Mark 1: Correct calculation of heat energy transferred, \(q = 14.1\text{ kJ}\) (or 14107.5 J).
- Mark 2: Correct calculation of moles of propan-1-ol, \(n = 9.17 \times 10^{-3}\text{ mol}\).
- Mark 3: Correct value of \(-1540\text{ kJ mol}^{-1}\) (must include the negative sign and be rounded to 3 significant figures. Accept -1539).

The major product is 2-bromopropane. This reaction proceeds via an electrophilic addition mechanism involving carbocation intermediates. The addition of \(\text{H}^+\) to the double bond can form either a secondary carbocation (\(\text{CH}_3\text{C}^+\text{HCH}_3\)) or a primary carbocation (\(\text{CH}_3\text{CH}_2\text{C}^+\text{H}_2\)). The secondary carbocation is more stable than the primary carbocation because it has two electron-releasing alkyl groups attached to the positively charged carbon, which disperse the positive charge more effectively. Therefore, the secondary carbocation is formed preferentially, leading to 2-bromopropane as the major product.

- Mark 1: Identifies 2-bromopropane as the major product.
- Mark 2: Explains that the reaction goes via the more stable secondary carbocation intermediate (as opposed to the less stable primary carbocation).
- Mark 3: Links the greater stability of the secondary carbocation to the inductive/electron-releasing effect of the two alkyl groups.

To compare the rates of hydrolysis, add aqueous silver nitrate and ethanol (as a mutual solvent) to separate test tubes containing each haloalkane. Place them in a water bath at a constant temperature (e.g., 50–60 \(^\circ\text{C}\)) and measure the time taken for a halide precipitate to form. The rate of hydrolysis increases in the order: 1-chlorobutane < 1-bromobutane < 1-iodobutane. This is because the C-I bond is the weakest (has the lowest bond enthalpy) and requires the least energy to break, making it the most reactive, while the C-Cl bond is the strongest.

- Mark 1: Identifies reagents as aqueous silver nitrate and ethanol (solvent), with heating in a water bath.
- Mark 2: Correctly identifies the trend: rate of hydrolysis increases in the order 1-chlorobutane < 1-bromobutane < 1-iodobutane (or 1-iodobutane reacts the fastest).
- Mark 3: Explains the trend in terms of decreasing carbon-halogen bond strength/enthalpy down the group (C-I is the weakest bond, so breaks easiest).

1. Determine equilibrium moles using a ICE table:
- Initial: \(n(\text{SO}_2) = 0.800\text{ mol}\), \(n(\text{O}_2) = 0.500\text{ mol}\), \(n(\text{SO}_3) = 0\text{ mol}\).
- Change: \(n(\text{O}_2)\) reacted = \(0.500 - 0.200 = 0.300\text{ mol}\).
- Since the mole ratio is 2:1:2, \(n(\text{SO}_2)\) reacted = \(2 \times 0.300 = 0.600\text{ mol}\). \(n(\text{SO}_3)\) produced = \(2 \times 0.300 = 0.600\text{ mol}\).
- Equilibrium moles: \(n(\text{SO}_2) = 0.800 - 0.600 = 0.200\text{ mol}\); \(n(\text{O}_2) = 0.200\text{ mol}\); \(n(\text{SO}_3) = 0.600\text{ mol}\).

2. Convert to equilibrium concentrations (Volume = 2.00 \(\text{dm}^3\)):
- \([\text{SO}_2] = \frac{0.200\text{ mol}}{2.00\text{ dm}^3} = 0.100\text{ mol dm}^{-3}\).
- \([\text{O}_2] = \frac{0.200\text{ mol}}{2.00\text{ dm}^3} = 0.100\text{ mol dm}^{-3}\).
- \([\text{SO}_3] = \frac{0.600\text{ mol}}{2.00\text{ dm}^3} = 0.300\text{ mol dm}^{-3}\).

3. Calculate \(K_c\):
\(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} = \frac{(0.300)^2}{(0.100)^2 \times 0.100} = \frac{0.0900}{0.00100} = 90.0\).

4. Determine units:
\(\text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2 \times \text{mol dm}^{-3}} = \text{mol}^{-1}\text{ dm}^3\) (or \(\text{dm}^3\text{ mol}^{-1}\)).

- Mark 1: For finding correct equilibrium moles of \(\text{SO}_2\) (0.200 mol) and \(\text{SO}_3\) (0.600 mol).
- Mark 2: For converting moles to concentrations (dividing by 2) and substituting into the correct \(K_c\) expression to get 90.0.
- Mark 3: For the correct units: \(\text{dm}^3\text{ mol}^{-1}\) (or \(\text{mol}^{-1}\text{ dm}^3\)).

1. The molecular formula \(\text{C}_3\text{H}_6\text{O}\) indicates one degree of unsaturation.
2. The strong infrared absorption peak at 1715 \(\text{cm}^{-1}\) is characteristic of a carbonyl group (\(\text{C}=\text{O}\)).
3. The absence of a broad peak at 3200–3600 \(\text{cm}^{-1}\) shows there is no \(\text{O}-\text{H}\) group, ruling out unsaturated alcohols.
4. Compound **X** must be a carbonyl compound, either propanal or propanone.
5. In the mass spectrum, propanone (\(\text{CH}_3\text{COCH}_3\)) undergoes fragmentation by cleavage of a C-C bond adjacent to the carbonyl group, forming the acylium ion \(\text{CH}_3\text{CO}^+\) with an \(m/z\) value of \(15 + 12 + 16 = 43\). Propanal would fragment to give peaks at \(m/z = 29\) (\(\text{CHO}^+\) or \(\text{C}_2\text{H}_5^+\)) or \(57\) (\(\text{C}_2\text{H}_5\text{CO}^+\)). Therefore, compound **X** is propanone, and the fragment ion is \(\text{CH}_3\text{CO}^+\).

- Mark 1: Deduces that the peak at 1715 \(\text{cm}^{-1}\) indicates a \(\text{C}=\text{O}\) bond and that no \(\text{O}-\text{H}\) group is present.
- Mark 2: Correctly identifies compound **X** as propanone.
- Mark 3: Identifies the fragment ion at \(m/z = 43\) as \(\text{CH}_3\text{CO}^+\) (must include positive charge).

1. The chemical equation for the reaction is:
\(\text{Cl}_2\text{(g)} + 2\text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{NaClO(aq)} + \text{H}_2\text{O(l)}\)

2. Disproportionation is a reaction in which the same element is simultaneously oxidised and reduced.
- In the reactant, molecular chlorine (\(\text{Cl}_2\)) has an oxidation number of 0.
- In sodium chloride (\(\text{NaCl}\)), chlorine has an oxidation number of -1 (reduction from 0 to -1).
- In sodium chlorate(I) (\(\text{NaClO}\)), chlorine has an oxidation number of +1 (oxidation from 0 to +1).

- Mark 1: For the correct balanced equation: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (or ionic equivalent).
- Mark 2: Defines disproportionation as the simultaneous oxidation and reduction of the same element in a reaction.
- Mark 3: States the oxidation states of Cl clearly: 0 in \(\text{Cl}_2\), -1 in \(\text{NaCl}\) (reduced), and +1 in \(\text{NaClO}\) (oxidised).

Let the percentage abundance of \(^{69}\text{Ga}\) be \(x\%\). Then the abundance of \(^{71}\text{Ga}\) is \((100 - x)\%\).
\(69.723 = \frac{69x + 71(100 - x)}{100}\)
\(6972.3 = 69x + 7100 - 71x\)
\(2x = 127.7\)
\(x = 63.85\).

Therefore, the percentage abundance of the \(^{69}\text{Ga}\) isotope is 63.85%.

1 mark for setting up the algebraic equation: \(69.723 = \frac{69x + 71(100 - x)}{100}\) (or equivalent).
1 mark for solving to find \(2x = 127.7\) (or equivalent).
0.63 marks for the final correct answer: 63.85% (must be given to 2 decimal places).

Mass of water lost = \(4.758\text{ g} - 2.598\text{ g} = 2.160\text{ g}\).

Molar mass of \(\text{CoCl}_2 = 58.9 + (2 \times 35.5) = 129.9\text{ g mol}^{-1}\).

Amount of anhydrous \(\text{CoCl}_2 = \frac{2.598}{129.9} = 0.0200\text{ mol}\).

Molar mass of \(\text{H}_2\text{O} = 18.0\text{ g mol}^{-1}\).

Amount of \(\text{H}_2\text{O} = \frac{2.160}{18.0} = 0.120\text{ mol}\).

Mole ratio of \(\text{H}_2\text{O} : \text{CoCl}_2 = \frac{0.120}{0.0200} = 6\).

Therefore, \(x = 6\).

1 mark for calculating the mass of water lost (2.160 g) and the amount of anhydrous cobalt chloride (0.0200 mol) and water (0.120 mol).
1 mark for determining the mole ratio of 6:1.
0.63 marks for the final integer value of \(x = 6\).

Sulfur has 6 outer shell electrons and forms 4 covalent bonds with fluorine, leaving 1 lone pair. This gives a total of 5 electron pairs (4 bonding pairs, 1 lone pair).

According to VSEPR theory, these 5 electron pairs arrange themselves in a trigonal bipyramidal base to minimize repulsion. The lone pair occupies an equatorial position, resulting in a seesaw shape.

Due to the greater repulsion of the lone pair compared to bonding pairs, the bond angles are reduced from the ideal trigonal bipyramidal angles: the equatorial-equatorial angle is approximately \(102^\circ\) (less than \(120^\circ\)) and the axial-axial angle is approximately \(173^\circ\) (less than \(180^\circ\)).

1 mark for stating that sulfur has 4 bonding pairs and 1 lone pair, leading to a seesaw shape.
1 mark for stating that the bond angles are less than \(120^\circ\) and less than \(90^\circ\) / \(180^\circ\) (accept specific values around \(102^\circ\) and \(173^\circ\)).
0.63 marks for explaining that lone pairs repel more than bonding pairs, thereby reducing the bond angles.

1. Heat energy absorbed by the water: \(q = m c \Delta T\) where \(\Delta T = 46.3 - 18.5 = 27.8\text{ K}\).
\(q = 150.0 \times 4.18 \times 27.8 = 17430.6\text{ J} = 17.43\text{ kJ}\).

2. Amount of methanol burned: \(n = \frac{\text{mass}}{M_r}\) where \(M_r(\text{CH}_3\text{OH}) = 12.0 + (4 \times 1.0) + 16.0 = 32.0\text{ g mol}^{-1}\).
\(n = \frac{0.832}{32.0} = 0.0260\text{ mol}\).

3. Enthalpy change of combustion: \(\Delta_c H = -\frac{q}{n} = -\frac{17.4306}{0.0260} = -670.4\text{ kJ mol}^{-1}\).

To 3 significant figures, \(\Delta_c H = -670\text{ kJ mol}^{-1}\).

1 mark for calculating the heat change of water: \(17.43\text{ kJ}\).
1 mark for calculating the moles of methanol burned: \(0.0260\text{ mol}\).
0.63 marks for dividing the energy by moles with a negative sign and rounding to 3 significant figures: \(-670\text{ kJ mol}^{-1}\) (allow \(-671\text{ kJ mol}^{-1}\) if intermediate rounding was done).

The catalyst works by providing an alternative reaction pathway with a lower activation energy.

Step 1: \(\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + 2\text{Fe}^{3+}(\text{aq})\)
Step 2: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)

The iron(II) catalyst is regenerated. Without a catalyst, two negatively charged anions (\(\text{S}_2\text{O}_8^{2-}\) and \(\text{I}^-\)) must collide, which is energetically unfavorable due to electrostatic repulsion. The positive iron ions attract the negative reactant ions, significantly lowering the activation energy.

1 mark for two correct equations showing the oxidation of \(\text{Fe}^{2+}\) and the subsequent reduction of \(\text{Fe}^{3+}\).
1 mark for stating that \(\text{Fe}^{2+}\) is regenerated.
0.63 marks for explaining that the uncatalyzed reaction involves electrostatic repulsion between two anions, which is avoided in the catalyzed steps.

Compound **A** has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\). Since it is a straight-chain compound and can be oxidized under reflux to a carboxylic acid **B** (\(\text{C}_4\text{H}_8\text{O}_2\)) and under distillation to an aldehyde **C** (\(\text{C}_4\text{H}_8\text{O}\)), **A** must be a primary alcohol.

Therefore, **A** is butan-1-ol with the structural formula \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\).

Compound **B** is the corresponding carboxylic acid, butanoic acid, with the structural formula \(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\).

1 mark for identifying **A** as butan-1-ol and drawing or writing its structure (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\)).
1 mark for identifying **B** as butanoic acid and drawing or writing its structure (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}\)).
0.63 marks for correct IUPAC names: 'butan-1-ol' and 'butanoic acid'.

(i) The mechanism is nucleophilic substitution.

(ii) In 1-chlorobutane, the \(\text{C}-\text{Cl}\) bond is polar, with the carbon being \(\delta+\) and chlorine being \(\delta-\).

The hydroxide ion (\(\text{OH}^-\)) acts as a nucleophile. A curly arrow starts from the lone pair on the oxygen of the \(\text{OH}^-\) ion and points to the \(\delta+\) carbon atom of the \(\text{C}-\text{Cl}\) bond.

Concurrently, a curly arrow starts from the \(\text{C}-\text{Cl}\) bond and points to the chlorine atom, breaking the bond heterolytically to release a chloride ion (\(\text{Cl}^-\)).

1 mark for identifying the mechanism as nucleophilic substitution.
1 mark for describing the curly arrow from the polar \(\text{C}^{\delta+}-\text{Cl}^{\delta-}\) bond to the chlorine atom.
0.63 marks for describing the curly arrow from the lone pair on the hydroxide ion to the carbon atom.

1. From the infrared spectrum:
- The broad absorption peak at \(3350\text{ cm}^{-1}\) indicates the presence of an \(\text{O}-\text{H}\) group (alcohol).
- The absorption at \(1645\text{ cm}^{-1}\) indicates the presence of a \(\text{C}=\text{C}\) bond (alkene).
- The absence of any peak in the range \(1680\text{--}1750\text{ cm}^{-1}\) confirms there is no \(\text{C}=\text{O}\) bond (carbonyl group).

2. From the mass spectrum:
- The molecular ion peak at \(m/z = 58\) indicates that the relative molecular mass of the compound is 58.
- For a compound containing an \(\text{O}-\text{H}\) and a \(\text{C}=\text{C}\) bond, let's determine the molecular formula. If we have one oxygen (16), we are left with \(58 - 16 = 42\) for carbon and hydrogen.
- \(3 \times 12 + 6 = 42\), which gives a molecular formula of \(\text{C}_3\text{H}_6\text{O}\).

3. Combining these features, the compound is an unsaturated alcohol with 3 carbon atoms: prop-2-en-1-ol, which has the structural formula \(\text{CH}_2=\text{CHCH}_2\text{OH}\).

1 mark for identifying the alcohol group (\(\text{O}-\text{H}\) at \(3350\text{ cm}^{-1}\)) and alkene group (\(\text{C}=\text{C}\) at \(1645\text{ cm}^{-1}\)) and confirming the absence of a carbonyl group (\(\text{C}=\text{O}\)).
1 mark for deducing the molecular formula \(\text{C}_3\text{H}_6\text{O}\) using \(m/z = 58\).
0.63 marks for drawing or writing the correct structural formula of prop-2-en-1-ol (\(\text{CH}_2=\text{CHCH}_2\text{OH}\)).

1. Identify the electron pairs around the central nitrogen atom: Nitrogen is in Group 15 and has 5 valence electrons. It forms 3 single covalent bonds with fluorine atoms, leaving 1 lone pair. Total of 4 electron pairs (3 bonding pairs, 1 lone pair). 2. Explain repulsion theory: Electron pairs repel each other to minimise repulsion (or get as far apart as possible). Lone pairs repel more than bonding pairs. 3. State shape and bond angle: The arrangement of 4 electron pairs is based on a tetrahedron, but the presence of 1 lone pair makes the molecular geometry trigonal pyramidal. The greater repulsion from the lone pair reduces the bond angle from \(109.5^\circ\) to \(107^\circ\) (accept range \(106^\circ - 108^\circ\)).

Mark 1: States shape as trigonal pyramidal AND bond angle as \(107^\circ\) (accept \(106^\circ\) to \(108^\circ\)). Mark 2: Identifies 3 bonding pairs and 1 lone pair of electrons around the central N atom. Mark 3: States that electron pairs repel to get as far apart as possible AND lone pairs repel more than bonding pairs. (Marks scaled to 2.63)

1. Calculate the mass of water lost: \(\text{Mass of water} = 5.56\text{ g} - 3.04\text{ g} = 2.52\text{ g}\). 2. Calculate the amount, in moles, of anhydrous \(\text{FeSO}_4\): \(\text{M}_r(\text{FeSO}_4) = 55.8 + 32.1 + (16.0 \times 4) = 151.9\text{ g mol}^{-1}\). \(\text{Moles of FeSO}_4 = 3.04 / 151.9 = 0.0200\text{ mol}\). 3. Calculate the amount, in moles, of water: \(\text{M}_r(\text{H}_2\text{O}) = (1.0 \times 2) + 16.0 = 18.0\text{ g mol}^{-1}\). \(\text{Moles of H}_2\text{O} = 2.52 / 18.0 = 0.140\text{ mol}\). 4. Determine the molar ratio of water to anhydrous salt: \(x = 0.140 / 0.0200 = 7\).

Mark 1: Calculates moles of anhydrous \(\text{FeSO}_4\) correctly as \(0.0200\text{ mol}\) (from \(3.04 / 151.9\)). Mark 2: Calculates mass of water lost as \(2.52\text{ g}\) and moles of water as \(0.140\text{ mol}\) (from \(2.52 / 18.0\)). Mark 3: Calculates the ratio to find \(x = 7\). (Marks scaled to 2.63)

1. Write the equation for the formation of pentane: \(5\text{C(s)} + 6\text{H}_2\text{(g)} \rightarrow \text{C}_5\text{H}_{12}\text{(l)}\). 2. Set up the Hess's Law equation using combustion data: \(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\). 3. Substitute the values into the equation: \(\Delta_f H^\theta = [5 \times \Delta_c H^\theta(\text{C}) + 6 \times \Delta_c H^\theta(\text{H}_2)] - \Delta_c H^\theta(\text{C}_5\text{H}_{12})\). \(\Delta_f H^\theta = [5 \times (-394) + 6 \times (-286)] - (-3509)\). \(\Delta_f H^\theta = [-1970 - 1716] + 3509\). \(\Delta_f H^\theta = -3686 + 3509 = -177\text{ kJ mol}^{-1}\).

Mark 1: Correctly applies Hess's law multipliers to reactant combustions: \(5 \times (-394)\) and \(6 \times (-286)\). Mark 2: Calculates sum of reactants' combustions as \(-3686\text{ kJ mol}^{-1}\). Mark 3: Obtains final value of \(-177\text{ kJ mol}^{-1}\) (must have the correct negative sign and units/value). Reject \(+177\). (Marks scaled to 2.63)

Silicon dioxide (\(\text{SiO}_2\)) forms a giant covalent lattice (macromolecular) structure. Each silicon atom is tetrahedrally bonded to four oxygen atoms, and each oxygen atom is bonded to two silicon atoms by strong covalent bonds.

1. Melting Point: To melt silicon dioxide, a vast number of these strong covalent bonds must be broken. This process requires a very large amount of thermal energy, resulting in a very high melting point.
2. Electrical Conductivity: All of the valence electrons of silicon and oxygen are localized and involved in covalent bonding. Because there are no free-moving delocalized electrons or mobile ions available to carry an electrical charge, silicon dioxide cannot conduct electricity in any state.

Award up to 4 marks (scaled to 3.87) as follows:
* **Mark 1**: Identifies the structure as a giant covalent lattice / macromolecular structure.
* **Mark 2**: States that each silicon is bonded to 4 oxygen atoms and/or each oxygen is bonded to 2 silicon atoms (or describes many strong covalent bonds throughout the structure).
* **Mark 3**: Explains high melting point in terms of requiring a large amount of energy to break these strong covalent bonds.
* **Mark 4**: Explains non-conductivity due to the absence of mobile charge carriers (no delocalized electrons or free-moving ions).

Step 1: Calculate the heat energy absorbed by the water using \(q = m c \Delta T\):
\(q = 250\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 36.5\text{ K} = 38142.5\text{ J} = 38.1425\text{ kJ}\)

Step 2: Calculate the number of moles of pentane (\(\text{C}_5\text{H}_{12}\)) burned:
\(M_r(\text{C}_5\text{H}_{12}) = (5 \times 12.0) + (12 \times 1.0) = 72.0\text{ g mol}^{-1}\)
\(n(\text{C}_5\text{H}_{12}) = \frac{1.08\text{ g}}{72.0\text{ g mol}^{-1}} = 0.0150\text{ mol}\)

Step 3: Calculate the enthalpy change of combustion per mole:
\(\Delta_c H = -\frac{q}{n} = -\frac{38.1425\text{ kJ}}{0.0150\text{ mol}} = -2542.83\text{ kJ mol}^{-1}\)

Step 4: Round to 3 significant figures:
\(\Delta_c H = -2540\text{ kJ mol}^{-1}\)

Award up to 4 marks (scaled to 3.87) as follows:
* **Mark 1**: Correct calculation of heat energy, \(q = 38.1\text{ kJ}\) (or \(38142.5\text{ J}\)).
* **Mark 2**: Correct calculation of moles of pentane, \(n = 0.0150\text{ mol}\).
* **Mark 3**: Division of energy by moles to obtain \(2543\text{ kJ mol}^{-1}\) (or ecf from previous values).
* **Mark 4**: Correct negative sign and rounding to 3 significant figures: \(-2540\text{ kJ mol}^{-1}\).

During the electrophilic addition of \(\text{HBr}\) to but-1-ene, the hydrogen ion (\(\text{H}^+\)) adds to the double bond to form a carbocation intermediate.

There are two possible intermediates:
1. A secondary carbocation (\(\text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_3\)), which leads to the formation of 2-bromobutane.
2. A primary carbocation (\(\text{C}^+\text{H}_2\text{CH}_2\text{CH}_2\text{CH}_3\)), which leads to the formation of 1-bromobutane.

Secondary carbocations are more stable than primary carbocations. This is because alkyl groups have an electron-donating inductive effect, which spreads out the positive charge on the carbon atom and stabilizes the ion. Since the secondary carbocation has two alkyl groups attached to the positively charged carbon compared to only one in the primary carbocation, it is more stable and forms preferentially as the major intermediate.

Award up to 4 marks (scaled to 3.87) as follows:
* **Mark 1**: Mentions that the reaction proceeds via a carbocation intermediate.
* **Mark 2**: Identifies that 2-bromobutane is formed from a secondary carbocation and 1-bromobutane is formed from a primary carbocation.
* **Mark 3**: States that secondary carbocations are more stable than primary carbocations.
* **Mark 4**: Explains stability in terms of the electron-donating inductive effect of the alkyl groups (two vs one).

Step 1: Calculate the mass of anhydrous \(\text{FeSO}_4\) and water lost:
- Mass of anhydrous \(\text{FeSO}_4 = 25.38\text{ g} - 22.34\text{ g} = 3.04\text{ g}\)
- Mass of water lost = \(27.90\text{ g} - 25.38\text{ g} = 2.52\text{ g}\)

Step 2: Calculate the relative formula mass (\(M_r\)) of \(\text{FeSO}_4\) and \(\text{H}_2\text{O}\):
- \(M_r(\text{FeSO}_4) = 55.8 + 32.1 + (4 \times 16.0) = 151.9\text{ g mol}^{-1}\)
- \(M_r(\text{H}_2\text{O}) = (2 \times 1.0) + 16.0 = 18.0\text{ g mol}^{-1}\)

Step 3: Calculate the moles of anhydrous \(\text{FeSO}_4\) and water:
- \(n(\text{FeSO}_4) = \frac{3.04\text{ g}}{151.9\text{ g mol}^{-1}} = 0.0200\text{ mol}\)
- \(n(\text{H}_2\text{O}) = \frac{2.52\text{ g}}{18.0\text{ g mol}^{-1}} = 0.140\text{ mol}\)

Step 4: Find the simplest whole-number ratio of \(\text{H}_2\text{O} : \text{FeSO}_4\):
- Ratio \(x = \frac{0.140}{0.0200} = 7\)

Therefore, the value of \(x = 7\).

Award up to 4 marks (scaled to 3.87) as follows:
* **Mark 1**: Calculates the mass of anhydrous salt (\(3.04\text{ g}\)) and mass of water (\(2.52\text{ g}\)) correctly.
* **Mark 2**: Calculates the moles of anhydrous \(\text{FeSO}_4\) correctly (\(0.0200\text{ mol}\)).
* **Mark 3**: Calculates the moles of \(\text{H}_2\text{O}\) correctly (\(0.140\text{ mol}\)).
* **Mark 4**: Correctly deduces the ratio to find \(x = 7\).

1. **Functional Group**: The broad absorption band at \(3350\text{ cm}^{-1}\) is characteristic of an \(\text{O-H}\) alcohol bond. The absorption at \(1050\text{ cm}^{-1}\) belongs to the \(\text{C-O}\) bond. The absence of any band in the range \(1630\text{--}1820\text{ cm}^{-1}\) confirms there is no carbonyl (\(\text{C=O}\)) group. Hence, the functional group is an alcohol (hydroxyl group).

2. **Molecular Formula**: For a mono-alcohol, the general molecular formula is \(\text{C}_n\text{H}_{2n+2}\text{O}\).
\(12.0n + 1.0(2n+2) + 16.0 = 74\)
\(14n + 18 = 74 \implies 14n = 56 \implies n = 4\).
Thus, the molecular formula is \(\text{C}_4\text{H}_{10}\text{O}\).

3. **Structural Formula**: Since **X** is a branched tertiary alcohol with 4 carbons, it must have the tertiary structure 2-methylpropan-2-ol. The structural formula is \(\text{(CH}_3\text{)}_3\text{COH}\).

Award up to 4 marks (scaled to 3.87) as follows:
* **Mark 1**: Identifies the functional group as an alcohol / hydroxyl / \(\text{-OH}\) group and link this to the absorption at \(3350\text{ cm}^{-1}\).
* **Mark 2**: Correctly calculates/deduces the molecular formula as \(\text{C}_4\text{H}_{10}\text{O}\) from the molecular ion peak \(m/z = 74\).
* **Mark 3**: Identifies the structure of **X** as 2-methylpropan-2-ol (or tert-butanol) because it is a tertiary branched alcohol.
* **Mark 4**: Writes the correct structural formula \(\text{(CH}_3\text{)}_3\text{COH}\) (or draw an appropriate skeletal/displayed structure).

Step 1: Set up the equilibrium table to find the equilibrium moles of reactants:
- Moles of \(\text{PCl}_5\) at equilibrium = \(0.60\text{ mol}\)
- Moles of \(\text{PCl}_3\) reacted = \(0.60\text{ mol}\) \(\implies\) equilibrium moles of \(\text{PCl}_3 = 1.00 - 0.60 = 0.40\text{ mol}\)
- Moles of \(\text{Cl}_2\) reacted = \(0.60\text{ mol}\) \(\implies\) equilibrium moles of \(\text{Cl}_2 = 1.50 - 0.60 = 0.90\text{ mol}\)

Step 2: Calculate the equilibrium concentrations using \(V = 2.00\text{ dm}^3\):
- \([\text{PCl}_3] = \frac{0.40\text{ mol}}{2.00\text{ dm}^3} = 0.20\text{ mol dm}^{-3}\)
- \([\text{Cl}_2] = \frac{0.90\text{ mol}}{2.00\text{ dm}^3} = 0.45\text{ mol dm}^{-3}\)
- \([\text{PCl}_5] = \frac{0.60\text{ mol}}{2.00\text{ dm}^3} = 0.30\text{ mol dm}^{-3}\)

Step 3: Write the expression for \(K_c\) and calculate its value:
- \(K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]} = \frac{0.30}{0.20 \times 0.45} = \frac{0.30}{0.090} = 3.33\) (or \(3.3\))

Step 4: Determine the units:
- \(\text{Units} = \frac{\text{mol dm}^{-3}}{\text{mol dm}^{-3} \times \text{mol dm}^{-3}} = \text{dm}^3\text{ mol}^{-1}\)

Award up to 4 marks (scaled to 3.87) as follows:
* **Mark 1**: Calculates the correct equilibrium moles of reactants (\(\text{PCl}_3 = 0.40\text{ mol}\), \(\text{Cl}_2 = 0.90\text{ mol}\)).
* **Mark 2**: Divides the equilibrium moles by the volume (\(2.00\text{ dm}^3\)) to find concentrations.
* **Mark 3**: Substitutes the concentrations correctly into the \(K_c\) expression and calculates \(3.33\) (allow \(3.3\)).
* **Mark 4**: States correct units: \(\text{dm}^3\text{ mol}^{-1}\).

The order of the rate of hydrolysis from fastest to slowest is:
1-iodobutane > 1-bromobutane > 1-chlorobutane.

Explanation:
Hydrolysis involves the nucleophilic attack of the water molecule on the carbon atom attached to the halogen, which results in the cleavage of the carbon-halogen (\(\text{C-X}\)) bond.

Down Group 17 (7), as the halogen atoms increase in size, the orbital overlap becomes less effective, resulting in longer and weaker carbon-halogen bonds. Therefore, the bond enthalpy decreases in the order: \(\text{C-Cl} > \text{C-Br} > \text{C-I}\).

Because the \(\text{C-I}\) bond has the lowest bond enthalpy, it is the easiest/weakest to break, requiring the least activation energy. Thus, 1-iodobutane is hydrolyzed the fastest. The \(\text{C-Cl}\) bond has the highest bond enthalpy, requiring the most energy to break, so 1-chlorobutane is hydrolyzed the slowest.

Award up to 4 marks (scaled to 3.87) as follows:
* **Mark 1**: Correct order: 1-iodobutane (fastest) followed by 1-bromobutane and then 1-chlorobutane (slowest).
* **Mark 2**: Explains that the reaction rate depends on the ease of breaking the carbon-halogen (\(\text{C-X}\)) bond (not bond polarity).
* **Mark 3**: States that bond enthalpy decreases down the group (\(\text{C-Cl} > \text{C-Br} > \text{C-I}\)).
* **Mark 4**: Explains that because the \(\text{C-I}\) bond is the weakest / has the lowest bond enthalpy, it breaks most easily, leading to the fastest rate of reaction.

1. **Role of a Catalyst**: A catalyst increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy (\(E_c\) compared to the original \(E_a\)).

2. **Effect on the Boltzmann Distribution**:
- The Boltzmann distribution curve itself remains unchanged because the temperature is constant.
- On the energy axis (x-axis), the activation energy value with a catalyst (\(E_c\)) lies to the left of the uncatalyzed activation energy (\(E_a\)).
- Because \(E_c\) is shifted to the left, the shaded area under the curve to the right of the activation energy line increases.
- This indicates that a greater proportion (or fraction) of molecules now possess kinetic energy greater than or equal to the activation energy (\(E \ge E_c\)).
- Consequently, there is a higher frequency of successful collisions (more successful collisions per unit time), which increases the overall rate of reaction.

Award up to 4 marks (scaled to 3.87) as follows:
* **Mark 1**: States that a catalyst provides an alternative pathway with a lower activation energy.
* **Mark 2**: Describes that on the Boltzmann distribution, the catalyzed activation energy (\(E_c\)) is located to the left of the uncatalyzed activation energy (\(E_a\)).
* **Mark 3**: Explains that a larger area under the curve is to the right of \(E_c\) / a greater proportion of molecules have energy greater than or equal to the activation energy.
* **Mark 4**: Connects this to a higher frequency of successful collisions (or more successful collisions per second).

First, calculate the heat energy released using q = m * c * delta T. Here, m = 200.0 g, c = 4.18 J g-1 K-1, and delta T = 28.5 K. This gives q = 200.0 * 4.18 * 28.5 = 23826 J = 23.826 kJ. Next, calculate the number of moles of ethanol burned: n = mass / molar mass = 1.15 g / 46.0 g mol-1 = 0.0250 mol. The enthalpy change of combustion is given by delta H_c = -q / n = -23.826 kJ / 0.0250 mol = -953.04 kJ mol-1. Rounding to 3 significant figures gives -953 kJ mol-1.

M1: Calculation of heat energy q = 23.826 kJ (or 23826 J). M2: Calculation of moles of ethanol = 0.0250 mol. M3: Evaluation of enthalpy change with negative sign (-953.04 kJ mol-1). M4: Final answer of -953 (kJ mol-1) to 3 significant figures with correct units.

First, calculate the mass of water lost during heating: 4.76 g - 2.60 g = 2.16 g. Second, calculate the moles of anhydrous CoCl2: the molar mass of CoCl2 is 58.9 + (2 * 35.5) = 129.9 g mol-1. Moles of CoCl2 = 2.60 g / 129.9 g mol-1 = 0.0200 mol. Third, calculate the moles of water lost: the molar mass of H2O is 18.0 g mol-1. Moles of H2O = 2.16 g / 18.0 g mol-1 = 0.120 mol. Finally, find the ratio of moles of H2O to moles of CoCl2: 0.120 mol / 0.0200 mol = 6. Thus, x = 6.

M1: Calculates mass of water lost as 2.16 g. M2: Calculates moles of CoCl2 as 0.0200 mol and moles of H2O as 0.120 mol. M3: Determines the ratio of moles of water to moles of CoCl2 as 6 : 1. M4: State the final value of x = 6.

(a) The reaction of 2-methylbut-2-ene with HBr proceeds via electrophilic addition. The major product is 2-bromo-2-methylbutane. (b) The addition of H+ to the double bond can form two carbocation intermediates. A tertiary carbocation is formed when H+ adds to carbon-3, and a secondary carbocation is formed when H+ adds to carbon-2. The tertiary carbocation is more stable than the secondary carbocation due to the electron-donating inductive effect of the three alkyl groups. The major product is formed via this more stable tertiary carbocation intermediate.

M1: State the name of the major product as 2-bromo-2-methylbutane. M2: Identify that the major pathway involves a tertiary carbocation intermediate, while the minor pathway involves a secondary carbocation. M3: State that tertiary carbocations are more stable than secondary carbocations. M4: Explain the higher stability in terms of greater inductive effect / electron donation from more alkyl groups.

Phosphorus is in Group 15 and has 5 valence electrons. In PCl3, it forms 3 single covalent bonds with chlorine atoms, leaving one lone pair. This gives 4 regions of electron density (3 bonding pairs and 1 lone pair). According to electron pair repulsion theory, electron pairs repel each other to get as far apart as possible. Lone pairs repel more than bonding pairs, which pushes the bonding pairs closer together, resulting in a trigonal pyramidal shape and a bond angle of 107 degrees (reduced from the tetrahedral 109.5 degrees).

M1: Identifies the shape as trigonal pyramidal. M2: States the bond angle is 107 degrees (accept 106 to 108 degrees). M3: Explains that there are 3 bonding pairs and 1 lone pair around the central phosphorus atom. M4: States that electron pairs repel to get as far apart as possible, and that lone pairs repel more than bonding pairs.

(a) The broad absorption at 3200-3600 cm-1 indicates an O-H group (alcohol), and the lack of a peak at 1630-1820 cm-1 confirms the absence of a C=O (carbonyl) group. Thus, X is an alcohol. (b) Given the molecular ion peak is at m/z = 74, the molar mass is 74 g mol-1. For a mono-alcohol with the general formula CnH(2n+2)O, we have 12n + (2n + 2) + 16 = 74, which simplifies to 14n = 56, so n = 4. The molecular formula is therefore C4H10O. (c) Possible structures for X include butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, or 2-methylpropan-2-ol.

M1: Identifies the functional group as an alcohol / O-H group based on the IR peak at 3200-3600 cm-1 and absence of C=O. M2: Deduces the molecular formula as C4H10O. M3: Shows working for molecular formula (e.g., matching mass of 74). M4: Suggests a correct name or structure (e.g., butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, or 2-methylpropan-2-ol).

First, write the expression: Kc = [C] / ([A][B]^2). At equilibrium, [C] = 0.200 mol dm-3, so equilibrium moles of C = 0.200 mol dm-3 * 2.00 dm3 = 0.400 mol. According to the stoichiometry: Equilibrium moles of A = 1.00 - 0.400 = 0.600 mol, so [A] = 0.600 mol / 2.00 dm3 = 0.300 mol dm-3. Equilibrium moles of B = 1.20 - (2 * 0.400) = 0.400 mol, so [B] = 0.400 mol / 2.00 dm3 = 0.200 mol dm-3. Substitute concentrations into the Kc expression: Kc = 0.200 / (0.300 * 0.200^2) = 0.200 / 0.012 = 16.67. Units of Kc = (mol dm-3) / (mol dm-3 * (mol dm-3)^2) = dm6 mol-2. To 3 significant figures, Kc = 16.7 dm6 mol-2.

M1: Correct expression for Kc = [C] / ([A][B]^2). M2: Correct calculation of equilibrium concentrations: [A] = 0.300 mol dm-3 and [B] = 0.200 mol dm-3. M3: Correct calculation of the numerical value of Kc = 16.7 (accept 16.67). M4: Correct units: dm6 mol-2.

A catalyst increases the rate of a chemical reaction by providing an alternative reaction pathway that has a lower activation energy. On a Boltzmann distribution of molecular energies, the activation energy of the catalysed reaction, Ec, is positioned to the left (at a lower energy value) compared to the uncatalysed activation energy, Ea. Because the activation energy is lower, a much larger proportion of the reactant molecules have energy greater than or equal to this new activation energy (represented by a larger shaded area under the curve to the right of Ec). This leads to a higher frequency of successful collisions, increasing the reaction rate.

M1: Catalyst provides an alternative pathway with a lower activation energy. M2: On the Boltzmann distribution, the catalysed activation energy (Ec) is located to the left of the uncatalysed activation energy (Ea). M3: Explain that a larger proportion of molecules have energy greater than or equal to the catalysed activation energy. M4: Consequently, there is a higher frequency of successful collisions (or more successful collisions per unit time).

**Indicative scientific content:**

**Chemical Tests:**
1. **To identify pent-1-ene (alkene):**
- Reagent: Bromine water, \( \text{Br}_2(aq) \).
- Observation: The orange/yellow solution is decolourised (turns colourless).
- Equation: \( \text{CH}_2=\text{CHCH}_2\text{CH}_2\text{CH}_3 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCHBrCH}_2\text{CH}_2\text{CH}_3 \) (or molecular formula: \( \text{C}_5\text{H}_{10} + \text{Br}_2 \rightarrow \text{C}_5\text{H}_{10}\text{Br}_2 \)).

2. **To identify 1-bromopentane (haloalkane):**
- Reagents/Conditions: Add aqueous silver nitrate, \( \text{AgNO}_3(aq) \), in the presence of ethanol (solvent) and warm in a water bath.
- Observation: A cream precipitate forms (silver bromide, \( \text{AgBr} \)).
- Equations:
Hydrolysis: \( \text{C}_5\text{H}_{11}\text{Br} + \text{H}_2\text{O} \rightarrow \text{C}_5\text{H}_{11}\text{OH} + \text{H}^+ + \text{Br}^- \)
Precipitation: \( \text{Ag}^+(aq) + \text{Br}^-(aq) \rightarrow \text{AgBr}(s) \)

**Infrared (IR) Spectroscopy:**
1. **To identify pentan-1-ol (alcohol):**
- The infrared spectrum of pentan-1-ol will show a characteristic broad absorption peak due to the alcohol \( \text{O-H} \) bond in the range \( 3200-3600\text{ cm}^{-1} \).
- This broad peak will be absent in both pent-1-ene and 1-bromopentane.
- Additionally, pent-1-ene has a characteristic alkene \( \text{C}=\text{C} \) stretch in the range \( 1620-1680\text{ cm}^{-1} \), which is absent in pentan-1-ol and 1-bromopentane.

**Level 3 (5-6 marks):**
- Detailed and accurate chemical tests described for both pent-1-ene and 1-bromopentane, including correct reagents, observations, and balanced equations.
- Correct use of IR spectroscopy to identify pentan-1-ol, including the correct O-H absorption range and at least one other distinguishing IR feature (e.g., C=C stretch in pent-1-ene or lack of O-H in the other two).
- Logic is well-structured and uses clear chemical terminology.

**Level 2 (3-4 marks):**
- Identifies correct chemical tests for both compounds with correct reagents and observations, but may miss conditions or have minor errors in equations.
- Uses IR spectroscopy to identify pentan-1-ol with correct absorption range.
- Description is clear but may lack completeness in some areas.

**Level 1 (1-2 marks):**
- Describes at least one correct chemical test with reagents and observations OR provides the correct IR range for the O-H group in pentan-1-ol.
- Logic may be fragmented.

**Level 0 (0 marks):**
- No response or no creditworthy chemistry.

*Key marking points/notes:*
- Alkene test: Bromine water turns from orange/yellow to colourless. Equation: \( \text{C}_5\text{H}_{10} + \text{Br}_2 \rightarrow \text{C}_5\text{H}_{10}\text{Br}_2 \) (Accept correct structural/skeletal formulae).
- Haloalkane test: Aqueous silver nitrate, ethanol, heat/warm. Cream ppt. Equation: \( \text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr} \).
- IR: Alcohol \( \text{O-H} \) at \( 3200-3600\text{ cm}^{-1} \). Alkene \( \text{C}=\text{C} \) at \( 1620-1680\text{ cm}^{-1} \).

**Indicative scientific content:**

**Experimental Procedure:**
1. Measure a known volume (e.g., \( 50.0\text{ cm}^3 \)) of copper(II) sulfate solution of a known concentration (e.g., \( 0.500\text{ mol dm}^{-3} \)) using a volumetric pipette or measuring cylinder.
2. Transfer the solution into a polystyrene cup, which serves as a calorimeter to minimise heat loss to the surroundings. Place the cup inside a glass beaker for stability, and use a plastic lid with a hole for the thermometer to reduce heat loss by convection.
3. Weigh an excess of zinc powder (e.g., \( \sim 2\text{ g} \)) on a balance to ensure all the copper(II) ions in solution react completely.
4. Start a timer. Record the temperature of the copper(II) sulfate solution every minute for 3 minutes.
5. At the 4th minute, add the zinc powder to the cup and stir the mixture continuously, but do NOT record the temperature at this minute.
6. Record the temperature of the mixture every minute from the 5th minute up to the 10th minute.

**Plotting and Extrapolating the Graph:**
1. Plot a graph of temperature (y-axis) against time (x-axis).
2. Draw two best-fit lines: one through the temperature points before mixing (0-3 minutes) and another through the cooling points after mixing (5-10 minutes).
3. Extrapolate both lines to the 4th minute (the time of mixing).
4. The vertical distance between these two extrapolated lines at the 4th minute gives the accurate maximum temperature rise, \( \Delta T \), compensating for heat loss during the reaction.

**Processing of Results:**
1. Calculate the heat energy released, \( q \), in joules using:
\( q = m \times c \times \Delta T \)
where:
- \( m \) is the mass of the solution (assuming a density of \( 1.0\text{ g cm}^{-3} \), so a volume of \( 50.0\text{ cm}^3 \) has a mass of \( 50.0\text{ g} \)),
- \( c \) is the specific heat capacity of water (\( 4.18\text{ J g}^{-1}\text{ K}^{-1} \)),
- \( \Delta T \) is the temperature rise obtained from the graph.
2. Calculate the amount in moles, \( n \), of \( \text{CuSO}_4 \) used:
\( n = \text{concentration} \times \text{volume (in dm}^3\text{)} \) (as Zn is in excess, \( \text{CuSO}_4 \) is the limiting reactant).
3. Calculate the enthalpy change of reaction, \( \Delta H_r \), in \( \text{kJ mol}^{-1} \):
\( \Delta H_r = -\frac{q}{1000 \times n} \)
(The negative sign must be included as the reaction is exothermic).

**Level 3 (5-6 marks):**
- Describes a complete, practical, and logical laboratory procedure using appropriate apparatus (polystyrene cup, lid, thermometer, pipette/measuring cylinder) and systematic temperature measurements over time.
- Explains clearly how to plot a temperature-time graph and extrapolate the cooling curve to the time of mixing to find the theoretical maximum temperature change, \( \Delta T \).
- Provides correct mathematical steps to calculate \( q \), moles of limiting reactant, and \( \Delta H_r \) (including conversion to kJ and the negative sign for an exothermic reaction).
- The response is well-structured and uses highly accurate scientific terminology.

**Level 2 (3-4 marks):**
- Describes a workable practical method, but may omit details like the lid, regular stirring, or measuring temperature at systematic time intervals before addition.
- Mentions plotting a graph and finding \( \Delta T \), but may be vague on how extrapolation to the time of mixing corrects for heat loss.
- Provides most of the calculation steps for \( q \) and \( \Delta H_r \), but may omit the negative sign or the conversion from J to kJ.

**Level 1 (1-2 marks):**
- Identifies some key practical steps (e.g., mixing Zn and \( \text{CuSO}_4 \), measuring temperature rise) or writes down the formula \( q = mc\Delta T \).
- Explanation of the graph or the moles calculation is incomplete, incorrect, or missing.

**Level 0 (0 marks):**
- No response or no creditworthy chemistry.

*Key marking points/notes:*
- Calorimeter: Must specify a polystyrene cup (with lid/beaker) to minimise heat loss.
- Graph: Must mention plotting temperature against time and extrapolating back to the time of mixing (e.g. 4th minute).
- Equation for heat: \( q = mc\Delta T \) and conversion of volume to mass using density of \( 1.0\text{ g cm}^{-3} \).
- Moles calculation: \( n = c \times V \) for \( \text{CuSO}_4 \).
- Enthalpy: \( \Delta H_r = -\frac{q}{n} \) (divided by 1000) with a negative sign.