2024 Edexcel IAS-Level Chemistry (XCH11) 模擬試題連答案詳解

First, find the molar mass (\(M\)) of the hydrocarbon:
\(M = \frac{\text{mass}}{\text{moles}} = \frac{5.6\text{ g}}{0.10\text{ mol}} = 56\text{ g mol}^{-1}\).

Next, calculate the mass of carbon in one mole of the hydrocarbon using the percentage abundance:
\(56\text{ g mol}^{-1} \times 0.857 = 48.0\text{ g mol}^{-1}\).

Calculate the number of carbon atoms per molecule:
\(\frac{48.0}{12.0} = 4\).

The remaining mass of the molecule is due to hydrogen:
\(56.0 - 48.0 = 8.0\text{ g mol}^{-1}\).

Calculate the number of hydrogen atoms per molecule:
\(\frac{8.0}{1.0} = 8\).

Therefore, the molecular formula is \(C_4H_8\).

1 mark: B - Correct answer based on molar mass calculation and percentage composition.

The successive ionization energies show a very large increase (a 'jump') between the third and fourth ionization energies (from \(2745\text{ kJ mol}^{-1}\) to \(11578\text{ kJ mol}^{-1}\)). This indicates that the fourth electron is removed from a shell closer to the nucleus, which experiences a much stronger electrostatic attraction. Therefore, there are three electrons in the outer shell of element \(X\), placing it in Group 3.

1 mark: C - Correctly identifies the group based on the position of the largest jump in ionization energy.

Let's analyze the shapes and bond angles using electron pair repulsion theory:

- \(NH_4^+\): Nitrogen has 5 valence electrons minus 1 electron for the positive charge, giving 4 valence electrons. It forms 4 bonding pairs with hydrogen, resulting in a tetrahedral shape with a bond angle of \(109.5^\circ\).
- \(NH_3\): Nitrogen has 5 valence electrons. It forms 3 bonding pairs and has 1 lone pair, resulting in a trigonal pyramidal shape with a bond angle of \(107^\circ\).
- \(NH_2^-\): Nitrogen has 5 valence electrons plus 1 electron for the negative charge, giving 6 valence electrons. It forms 2 bonding pairs and has 2 lone pairs, resulting in a non-linear (bent) shape. The two lone pairs repel more strongly than the bonding pairs, reducing the bond angle to approximately \(104.5^\circ\).
- \(CO_2\): Carbon forms 2 double bonds with 0 lone pairs on the carbon atom, resulting in a linear shape with a bond angle of \(180^\circ\).

1 mark: C - Correctly identifies the species with 2 bonding pairs and 2 lone pairs resulting in a \(104.5^\circ\) bond angle.

Let's evaluate each option:
- Option A is incorrect because hydrogen radicals (\(H \cdot\)) do not form in free-radical substitution. Instead, a hydrogen atom is abstracted to form \(HCl\).
- Option B represents a termination step where two free radicals combine to form a stable molecule.
- Option C represents a valid propagation step where an ethyl radical reacts with a chlorine molecule to produce chloroethane and regenerate a chlorine radical.
- Option D represents the initiation step, which involves homolytic fission of the chlorine-chlorine bond under UV light.

1 mark: C - Correctly identifies the propagation step in the mechanism.

For an alkene to exhibit E/Z isomerism, each carbon atom of the C=C double bond must be attached to two different groups.
- A: 2-methylbut-2-ene is \((CH_3)_2C=CH(CH_3)\). The left carbon is bonded to two identical methyl groups, so E/Z isomerism is not possible.
- B: 1,1-dichloroprop-1-ene is \(Cl_2C=CH(CH_3)\). The left carbon is bonded to two identical chlorine atoms, so E/Z isomerism is not possible.
- C: 3-methylpent-2-ene is \(CH_3CH=C(CH_3)CH_2CH_3\). The left carbon is bonded to \(-H\) and \(-CH_3\) (different groups). The right carbon is bonded to \(-CH_3\) and \(-CH_2CH_3\) (different groups). Therefore, it can exist as E/Z isomers.
- D: 2-ethylbut-1-ene is \(CH_2=C(CH_2CH_3)_2\). The left carbon has two identical hydrogen atoms, and the right carbon has two identical ethyl groups, so E/Z isomerism is not possible.

1 mark: C - Correctly identifies the alkene structure that meets the criteria for E/Z stereoisomerism.

The formula for percentage atom economy is:
\(\text{Atom Economy} = \frac{\text{Total molar mass of desired product(s)}}{\text{Total molar mass of all reactants}} \times 100\%\)

Here, the desired product is ethanol (\(C_2H_5OH\)).
From the equation, 1 mole of glucose produces 2 moles of ethanol.

Total mass of desired product = \(2 \times 46.0\text{ g mol}^{-1} = 92.0\text{ g mol}^{-1}\).
Total mass of reactants = \(180.0\text{ g mol}^{-1}\).

\(\text{Atom Economy} = \frac{92.0}{180.0} \times 100\% \approx 51.1\%\).

1 mark: B - Correct calculation of percentage atom economy.

Iron (\(Fe\)) has an atomic number of 26. Its ground state electronic configuration is:
\(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2\).

When a transition metal atom forms a positive ion, the electrons from the outermost subshell (\(4s\)) are lost first. Therefore, when forming \(Fe^{2+}\), the two \(4s\) electrons are removed first, leaving:
\(1s^2 2s^2 2p^6 3s^2 3p^6 3d^6\).

1 mark: D - Correctly identifies the electronic configuration of \(Fe^{2+}\), showing loss of \(4s\) electrons before \(3d\).

Let's analyze each substance:
- **Graphite** has a giant covalent structure and a very high melting point, but it conducts electricity due to delocalized electrons.
- **Magnesium oxide** has a giant ionic lattice structure and a high melting point, but it conducts electricity when liquid (molten).
- **Silicon dioxide** has a giant covalent structure, a very high melting point, and has no free-moving ions or delocalized electrons, so it does not conduct electricity in either solid or liquid states.
- **Iodine** has a simple molecular structure with a low melting point.

1 mark: C - Correctly identifies the giant covalent structure and physical properties of silicon dioxide.

1. Calculate the number of moles of \(\text{CO}_2\) produced:
\(\text{Moles of CO}_2 = \frac{600\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.025\text{ mol}\).

2. Determine the moles of \(\text{MCO}_3\) using the decomposition equation:
\(\text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})\).
Since the mole ratio is 1:1, there are \(0.025\text{ mol}\) of \(\text{MCO}_3\).

3. Calculate the molar mass of \(\text{MCO}_3\):
\(M_r = \frac{2.11\text{ g}}{0.025\text{ mol}} = 84.4\text{ g mol}^{-1}\).

4. Calculate the relative atomic mass of \(\text{M}\):
\(A_r(\text{M}) = 84.4 - [12.0 + (3 \times 16.0)] = 24.4\text{ g mol}^{-1}\).
This value is closest to the relative atomic mass of magnesium (\(24.3\text{ g mol}^{-1}\)).

A - correct answer (1 mark).

The third ionisation energy is the energy required for the process: \(\text{X}^{2+}(\text{g}) \rightarrow \text{X}^{3+}(\text{g}) + \text{e}^-\).
- For sodium and magnesium, the third electron is removed from the stable inner \(2\text{p}\) subshell (shell change).
- For aluminium and silicon, the third electron is removed from the outer \(3\text{s}\) subshell.
- Comparing sodium and magnesium, magnesium has a higher nuclear charge (\(Z=12\)) than sodium (\(Z=11\)), meaning its outer electrons experience a stronger nuclear pull. Thus, magnesium has the highest third ionisation energy.

B - correct answer (1 mark).

In the carbonate ion, \(\text{CO}_3^{2-}\), there are three bonding regions of electron density around the central carbon atom and no lone pairs. These repel each other equally to adopt a trigonal planar geometry with bond angles of exactly \(120^\circ\).
- \(\text{H}_3\text{O}^+\) has 3 bonding pairs and 1 lone pair (trigonal pyramidal, \(\approx 107^\circ\)).
- \(\text{NH}_4^+\) has 4 bonding pairs and no lone pairs (tetrahedral, \(109.5^\circ\)).
- \(\text{SF}_6\) has 6 bonding pairs and no lone pairs (octahedral, \(90^\circ\) and \(180^\circ\)).

A - correct answer (1 mark).

1. Moles of ethanol reacted = \(\frac{4.60\text{ g}}{46.0\text{ g mol}^{-1}} = 0.100\text{ mol}\).
2. Based on the 1:1 stoichiometry, the theoretical yield of ethyl ethanoate = \(0.100\text{ mol}\).
3. Theoretical mass of ethyl ethanoate = \(0.100\text{ mol} \times 88.0\text{ g mol}^{-1} = 8.80\text{ g}\).
4. Percentage yield = \(\frac{5.28\text{ g}}{8.80\text{ g}} \times 100\% = 60.0\%\).

B - correct answer (1 mark).

- Option A represents a termination step (two radicals combining).
- Option B shows the formation of a hydrogen radical (\(\text{H}^\bullet\)), which does not occur as hydrogen radicals are too unstable.
- Option C represents a termination step (butyl radical and chlorine radical combining).
- Option D represents a correct propagation step where a chlorine radical abstracts a hydrogen atom from butane to form a butyl radical and hydrogen chloride.

D - correct answer (1 mark).

For an alkene to exhibit E/Z isomerism, both carbon atoms of the double bond must be bonded to two different groups.
- In 2-methylbut-2-ene, one double-bonded carbon has two identical methyl groups.
- In 2-methylbut-1-ene and pent-1-ene, one double-bonded carbon has two identical hydrogen atoms.
- In pent-2-ene (\(\text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3\)), Carbon 2 is bonded to \(-\text{H}\) and \(-\text{CH}_3\), and Carbon 3 is bonded to \(-\text{H}\) and \(-\text{CH}_2\text{CH}_3\). Therefore, it can exist as E and Z isomers.

D - correct answer (1 mark).

A neutral chromium atom (\(\text{Cr}\)) has the exceptional configuration: \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^5 4\text{s}^1\).
When transition metal atoms form ions, the \(4\text{s}\) electrons are lost first, followed by the \(3\text{d}\) electrons.
To form \(\text{Cr}^{3+}\), three electrons are removed: one from the \(4\text{s}\) subshell and two from the \(3\text{d}\) subshell, resulting in \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^3\).

A - correct answer (1 mark).

According to Fajans' rules, covalent character is maximised when there is high polarisation of the anion by the cation. This occurs with:
1. A highly polarizing cation, which has a small ionic radius and high charge density. Comparing \(\text{Mg}^{2+}\) and \(\text{Ba}^{2+}\), the magnesium ion is much smaller and therefore more polarizing.
2. A highly polarizable anion, which is large. Comparing \(\text{I}^-\) and \(\text{Cl}^-\), the iodide ion is larger and its outer electron cloud is more easily deformed.
Therefore, \(\text{MgI}_2\) has the greatest degree of covalent character.

A - correct answer (1 mark).

1. Find the amount of \(\text{CO}_2\) gas produced: \(\text{Moles of CO}_2 = \frac{1.08\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.0450\text{ mol}\). 2. Determine the moles of pure \(\text{CaCO}_3\) decomposed: since the thermal decomposition equation is \(\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)\), the mole ratio of \(\text{CaCO}_3 : \text{CO}_2\) is \(1 : 1\). Thus, the amount of pure \(\text{CaCO}_3\) in the sample is \(0.0450\text{ mol}\). 3. Calculate the mass of pure \(\text{CaCO}_3\): \(\text{Mass} = 0.0450\text{ mol} \times 100.1\text{ g mol}^{-1} = 4.5045\text{ g}\). 4. Calculate the percentage purity: \(\text{Percentage purity} = \frac{4.5045\text{ g}}{5.00\text{ g}} \times 100\% = 90.09\%\), which rounds to \(90.1\%\).

1 Mark: Correctly calculates the moles of \(\text{CO}_2\) (\(0.0450\text{ mol}\)) and the corresponding mass of \(\text{CaCO}_3\) (\(4.505\text{ g}\)), leading to the correct option C (\(90.1\%\)). Other options are the result of stoichiometry errors or inverted ratios.

1. Analyze the differences between successive ionisation energies: from 1st to 2nd is \(1239\text{ kJ mol}^{-1}\), from 2nd to 3rd is \(928\text{ kJ mol}^{-1}\), and from 3rd to 4th is \(8833\text{ kJ mol}^{-1}\). The extremely large increase between the 3rd and 4th ionisation energies indicates that the 4th electron is removed from a shell closer to the nucleus (an inner shell). 2. This shows that element \(X\) has 3 valence electrons and belongs to Group 3 (Group 13) of the periodic table, forming a stable \(X^{3+}\) ion. 3. Oxygen forms an oxide ion, \(\text{O}^{2-}\). 4. Combining these to form an electrically neutral compound gives \(X_2\text{O}_3\).

1 Mark: Identifies the large jump between the 3rd and 4th ionisation energies, deduces that the stable ion of the element is \(X^{3+}\), and determines that the oxide formula is \(X_2\text{O}_3\) (Option C). Other options represent elements belonging to different groups.

1. \(\text{BF}_3\) has 3 bonding pairs and 0 lone pairs around the central boron atom. This gives a trigonal planar shape with a bond angle of \(120^\circ\). The symmetrical distribution of the polar B-F bonds results in the dipoles cancelling out, making the molecule non-polar. 2. \(\text{PF}_3\) has a trigonal pyramidal geometry due to the presence of a lone pair, making it polar with bond angles of approximately \(107^\circ\). 3. \(\text{SO}_2\) is a bent molecule with a bond angle of about \(119^\circ\) and is polar. 4. \(\text{CF}_4\) is tetrahedral and non-polar, but has bond angles of \(109.5^\circ\).

1 Mark: Selects Option A, identifying that \(\text{BF}_3\) has trigonal planar geometry (\(120^\circ\) bond angles) and is non-polar. Other options are incorrect because they are either polar molecules or do not exhibit \(120^\circ\) bond angles.

1. 2-methylbut-2-ene has the structure \(\text{(CH}_3)_2\text{C}=\text{CHCH}_3\). 2. During electrophilic addition of \(\text{HBr}\), the proton (\(\text{H}^+\)) adds to the double bond to form the more stable carbocation intermediate. 3. Addition of \(\text{H}^+\) to the C3 carbon yields a tertiary carbocation at C2, \(\text{(CH}_3)_2\text{C}^+-\text{CH}_2\text{CH}_3\), which is stabilized by the electron-donating inductive effect of three alkyl groups. 4. Addition of \(\text{H}^+\) to the C2 carbon yields a less stable secondary carbocation at C3, \(\text{(CH}_3)_2\text{CH}-\text{CH}^+-\text{CH}_3\). 5. The bromide ion (\(\text{Br}^-\)) attacks the more stable tertiary carbocation, forming the major product, 2-bromo-2-methylbutane.

1 Mark: Correctly names 2-bromo-2-methylbutane (Option B) as the major product based on Markovnikov's rule and carbocation stability. Option A is the minor product, Option C represents incorrect addition regiochemistry, and Option D is the dibromo product from reaction with bromine gas.

(a) 'Heating to constant mass' means heating the crucible and its contents, allowing them to cool, and weighing them. This process is repeated until two consecutive mass readings are identical (or within experimental error), ensuring all water of crystallisation has been driven off.

(b) (i)
- Mass of anhydrous \(\text{CoSO}_4\) = \(20.41\text{ g} - 18.24\text{ g} = 2.17\text{ g}\)
- Mass of water lost = \(22.17\text{ g} - 20.41\text{ g} = 1.76\text{ g}\)
- Moles of anhydrous \(\text{CoSO}_4\) (Molar mass = \(58.9 + 32.1 + (16.0 \times 4) = 155.0\text{ g mol}^{-1}\)):
\(\text{Moles} = \frac{2.17\text{ g}}{155.0\text{ g mol}^{-1}} = 0.0140\text{ mol}\)
- Moles of \(\text{H}_2\text{O}\) (Molar mass = \(18.0\text{ g mol}^{-1}\)):
\(\text{Moles} = \frac{1.76\text{ g}}{18.0\text{ g mol}^{-1}} = 0.0978\text{ mol}\)
- Mole ratio of \(\text{H}_2\text{O} : \text{CoSO}_4\) = \(\frac{0.0978}{0.0140} = 6.99 \approx 7\)
- Therefore, \(x = 7\).

(b) (ii) \(\text{CoSO}_4 \cdot 7\text{H}_2\text{O} \rightarrow \text{CoSO}_4 + 7\text{H}_2\text{O}\)

(c) (i)
- Moles of \(\text{Co}^{2+}\) ions = Moles of \(\text{CoSO}_4\) dissolved = \(0.0140\text{ mol}\n- Volume of solution = \)250\text{ cm}^3 = 0.250\text{ dm}^3\)
- Concentration of \(\text{Co}^{2+}\) = \(\frac{0.0140\text{ mol}}{0.250\text{ dm}^3} = 0.0560\text{ mol dm}^{-3}\)

(c) (ii)
1. Dissolve the anhydrous solid from the crucible in a beaker using a volume of distilled (or deionised) water that is significantly less than \(250\text{ cm}^3\) (e.g. \(100\text{ cm}^3\)), stirring with a glass rod.
2. Transfer the solution quantitatively to a \(250\text{ cm}^3\) volumetric flask. Rinse the beaker, glass rod, and funnel with distilled water and add the washings to the flask.
3. Add distilled water to the flask until the bottom of the meniscus is exactly on the \(250\text{ cm}^3\) graduation mark. Stopper the flask and invert it several times to ensure a homogeneous solution.

**Part (a):**
- **M1:** Heat, cool, and weigh the crucible and contents (1)
- **M2:** Repeat this process until two consecutive mass readings are identical / constant (1)

**Part (b)(i):**
- **M1:** Calculation of mass of anhydrous cobalt sulfate (\(2.17\text{ g}\)) AND mass of water (\(1.76\text{ g}\)) (1)
- **M2:** Calculation of moles of anhydrous \(\text{CoSO}_4\) = \(0.0140\text{ mol}\) (1)
- **M3:** Calculation of moles of \(\text{H}_2\text{O}\) = \(0.0978\text{ mol}\) (1)
- **M4:** Calculation of simplest ratio \(x = 7\) (must be a whole number) (1)
*Allow ECF at each step.*

**Part (b)(ii):**
- **M1:** Correct balanced equation: \(\text{CoSO}_4 \cdot 7\text{H}_2\text{O} \rightarrow \text{CoSO}_4 + 7\text{H}_2\text{O}\) (1)
*Allow ECF from (b)(i) for their value of x.*

**Part (c)(i):**
- **M1:** States/uses that moles of \(\text{Co}^{2+}\) = \(0.0140\text{ mol}\) (1)
- **M2:** Correct calculation of concentration = \(0.0560\text{ mol dm}^{-3}\) (1)
*Allow ECF from (b)(i).*

**Part (c)(ii):**
- **M1:** Dissolve the solid in distilled water in a beaker first (using less than 250 cm3) (1)
- **M2:** Transfer to a 250 cm3 volumetric flask and add washings from the beaker/stirring rod (1)
- **M3:** Fill to the graduation mark with distilled water (meniscus level) and invert to mix (1)

(a) (i)
Observation: The solution changes from colourless to orange/yellow/brown.
Ionic equation: \(\text{Cl}_2\text{(aq)} + 2\text{Br}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{Br}_2\text{(aq)}\)

(ii)
- Chlorine has fewer electron shells and a smaller atomic radius than bromine (outer electrons are closer to the nucleus).
- Chlorine experiences less shielding of its outer shell electrons from the nucleus than bromine.
- Consequently, there is a stronger electrostatic attraction between the chlorine nucleus and an incoming electron, making it easier for chlorine to gain an electron and act as a stronger oxidising agent than bromine.

(b) (i)
- Yellow solid: Sulfur (\(\text{S}\))
- Purple vapour: Iodine (\(\text{I}_2\))

(ii) The reduction half-equation is:
\(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\) (or \(\text{SO}_4^{2-} + 10\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\))

(iii)
- Oxidation state of sulfur in \(\text{H}_2\text{SO}_4\): \(+6\)
- Oxidation state of sulfur in \(\text{H}_2\text{S}\): \(-2\)

**Part (a)(i):**
- **M1:** Observation: Colourless solution turns orange / yellow / brown (1) *[Reject: red-brown precipitate]*
- **M2:** Correct formulae in equation: \(\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2\) (1)
- **M3:** Correct state symbols: all \(\text{(aq)}\) (1) *[Dependent on correct chemical species]*

**Part (a)(ii):**
- **M1:** Chlorine has fewer electron shells / smaller atomic radius (1)
- **M2:** Chlorine has less shielding (1)
- **M3:** Stronger attraction between nucleus and incoming electron / easier to gain an electron (1)

**Part (b)(i):**
- **M1:** Yellow solid is sulfur / \(\text{S}\) (1)
- **M2:** Purple vapour is iodine / \(\text{I}_2\) (1)

**Part (b)(ii):**
- **M1:** Correct species on left (\(\text{H}_2\text{SO}_4\) or \(\text{SO}_4^{2-}\)) and right (\(\text{H}_2\text{S}\)) (1)
- **M2:** Correctly balanced equation with electrons (\(8\text{e}^-\)) (1)
*e.g. \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\)*

**Part (b)(iii):**
- **M1:** In \(\text{H}_2\text{SO}_4\): \(+6\) (1)
- **M2:** In \(\text{H}_2\text{S}\): \(-2\) (1)

(a) (i)
- \(q = m \times c \times \Delta T\)
- \(q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\,^\circ\text{C}^{-1} \times 5.6\,^\circ\text{C} = 1170.4\text{ J}\) (or \(1.17\text{ kJ}\))

(a) (ii)
- Moles of \(\text{K}_2\text{CO}_3 = \frac{3.45\text{ g}}{138.2\text{ g mol}^{-1}} = 0.02496\text{ mol}\)
- \(\Delta H_1 = -\frac{q}{\text{moles}} = -\frac{1.1704\text{ kJ}}{0.02496\text{ mol}} = -46.88\text{ kJ mol}^{-1}\)
- Rounded to 3 significant figures = \(-46.9\text{ kJ mol}^{-1}\) (allow \(-46.9\) to \(-47.0\) depending on intermediate rounding)

(b) (i)
A correct Hess's Law cycle shows:
- Top reactants: \(2\text{KHCO}_3\text{(s)}\)
- Top products: \(\text{K}_2\text{CO}_3\text{(s)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)
- Bottom products/intermediates: \(2\text{KCl(aq)} + 2\text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\)
- Arrow from \(2\text{KHCO}_3\text{(s)}\) to bottom intermediate labeled \(2 \Delta H_2\) (or \(2 \times (+28.5)\))
- Arrow from \(\text{K}_2\text{CO}_3\text{(s)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\) to bottom intermediate labeled \(\Delta H_1\) (or \(-46.9\))
- Arrow from reactants to products labeled \(\Delta H_{\text{reaction}}\).

(b) (ii)
- From cycle: \(\Delta H_{\text{reaction}} + \Delta H_1 = 2 \Delta H_2\)
- \(\Delta H_{\text{reaction}} = 2 \Delta H_2 - \Delta H_1\)
- \(\Delta H_{\text{reaction}} = 2 \times (+28.5) - (-46.9) = 57.0 + 46.9 = +103.9\text{ kJ mol}^{-1}\)

(c)
- Heat loss to the surroundings / air.
- Heat capacity of the polystyrene cup / thermometer is neglected.

**Part (a)(i):**
- **M1:** Correct formula used with mass of solution (50.0 g) (1)
- **M2:** Correct calculation: \(1170.4\text{ J}\) (or \(1.17\text{ kJ}\)) (1)

**Part (a)(ii):**
- **M1:** Calculation of moles of \(\text{K}_2\text{CO}_3\) = \(0.02496\text{ mol}\) (1)
- **M2:** Division of energy by moles (1)
- **M3:** Correct value to 3 s.f. and negative sign: \(-46.9\text{ kJ mol}^{-1}\) (1)
*(Allow ECF from (a)(i). Accept range -46.9 to -47.0)*

**Part (b)(i):**
- **M1:** Correct triangle cycle with correct chemical species in three positions (1)
- **M2:** Arrows pointing downwards from both reactants and products to the bottom intermediate (1)
- **M3:** Correctly labeled arrows with stoichiometry: \(2 \Delta H_2\) on left and \(\Delta H_1\) on right (1)

**Part (b)(ii):**
- **M1:** Correct application of Hess's Law: \(\Delta H_{\text{reaction}} = 2 \Delta H_2 - \Delta H_1\) (1)
- **M2:** Correct evaluation: \(+103.9\text{ kJ mol}^{-1}\) (1)
*(Accept range +103.9 to +104.0 depending on ECF from (a)(ii). Correct sign required)*

**Part (c):**
- **M1:** Heat loss to the surroundings (1)
- **M2:** The heat capacity of the polystyrene cup/thermometer is ignored (1)

(a) (i)
- Skeletal formula: A chain of 4 carbons with a double bond between C2 and C3, and a methyl group on C2.
- Explanation: One of the carbon atoms in the double bond (C2) is attached to two identical groups (two methyl groups), which means E/Z isomerism is not possible.

(b) (i)
- Isomers: 2-bromo-2-methylbutane and 2-bromo-3-methylbutane.
- Major product: 2-bromo-2-methylbutane.

(b) (ii)
- Step 1: Curly arrow starts from the double bond of 2-methylbut-2-ene to the hydrogen atom of the \(\text{H-Br}\) molecule.
- A curly arrow from the \(\text{H-Br}\) bond to the bromine atom.
- Correct dipoles on the hydrogen bromide molecule (\(\text{H}^{\delta+}-\text{Br}^{\delta-}\)).
- Step 2: Draw the tertiary carbocation intermediate: \(\text{(CH}_3\text{)}_2\text{C}^+-\text{CH}_2\text{CH}_3\), and the bromide ion (\(\text{Br}^-\)) with a lone pair.
- Step 3: Curly arrow from the lone pair on the bromide ion (\(\text{Br}^-\)) to the positively charged carbon of the carbocation to form 2-bromo-2-methylbutane.

(b) (iii)
- The major product is formed via a tertiary carbocation intermediate, while the minor product is formed via a secondary carbocation intermediate.
- Tertiary carbocations are more stable than secondary carbocations.
- This is because tertiary carbocations have three electron-donating alkyl groups that release electron density towards the positively charged carbon (positive inductive effect), dispersing the positive charge more effectively.

**Part (a)(i):**
- **M1:** Correct skeletal formula of 2-methylbut-2-ene (1)
- **M2:** Explanation: Carbon-2 is bonded to two identical groups / two methyl groups (so cannot show stereoisomerism) (1)

**Part (b)(i):**
- **M1:** 2-bromo-2-methylbutane (1)
- **M2:** 2-bromo-3-methylbutane (accept 3-bromo-2-methylbutane) (1)
- **M3:** Identifies 2-bromo-2-methylbutane as the major product (1)

**Part (b)(ii):**
- **M1:** Curly arrow from C=C double bond to H of H-Br AND correct dipoles on \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) (1)
- **M2:** Curly arrow from H-Br bond to Br (1)
- **M3:** Correct tertiary carbocation intermediate \(\text{(CH}_3\text{)}_2\text{C}^+-\text{CH}_2\text{CH}_3\) (1)
- **M4:** Curly arrow from lone pair on \(\text{Br}^-\) to \(\text{C}^+\) of carbocation (1)

**Part (b)(iii):**
- **M1:** Major product is formed via a tertiary carbocation intermediate and minor via a secondary carbocation (1)
- **M2:** Tertiary carbocation is more stable than secondary carbocation (1)
- **M3:** Due to the electron-donating inductive effect of three alkyl groups compared to two (which stabilizes/disperses the positive charge) (1)

(a) (i) Relative atomic mass is the weighted average mass of an atom of an element compared to 1/12th of the mass of an atom of carbon-12.

(a) (ii)
- \(\text{RAM} = \frac{(20 \times 90.48) + (21 \times 0.27) + (22 \times 9.25)}{100}\)
- \(\text{RAM} = \frac{1809.6 + 5.67 + 203.5}{100} = \frac{2018.77}{100} = 20.19\)

(b) (i)
\(\text{Na(g)} \rightarrow \text{Na}^+\text{(g)} + \text{e}^-\)

(b) (ii)
The general trend is that the first ionization energy increases across Period 3.

(b) (iii)
- Phosphorus has the outer electronic configuration \(3\text{s}^2 3\text{p}^3\) with three singly occupied p-orbitals.
- Sulfur has the outer electronic configuration \(3\text{s}^2 3\text{p}^4\) with one doubly occupied p-orbital.
- The repulsion between the two paired electrons in the same p-orbital of sulfur makes it easier to remove one of these electrons compared to removing an electron from a singly occupied orbital in phosphorus.

(b) (iv)
- Magnesium has a higher nuclear charge / more protons than sodium.
- The outer electrons of both elements are in the same electron shell (3s) and experience similar shielding, so the outer electron in magnesium is more strongly attracted to the nucleus.

**Part (a)(i):**
- **M1:** Weighted average mass of an atom of an element (1)
- **M2:** Compared to 1/12th of the mass of an atom of carbon-12 (1)

**Part (a)(ii):**
- **M1:** Correct expression for weighted average (1)
- **M2:** \(20.19\) (must be to 2 decimal places) (1)

**Part (b)(i):**
- **M1:** Correct species: \(\text{Na}\) and \(\text{Na}^+ + \text{e}^-\)\ (1)
- **M2:** Correct gas state symbols on both sodium species: \(\text{(g)}\) (1)

**Part (b)(ii):**
- **M1:** Increases (1)

**Part (b)(iii):**
- **M1:** Phosphorus has outer p-orbitals that are singly occupied / \(3\text{p}^3\) (1)
- **M2:** Sulfur has one pair of electrons in a p-orbital / \(3\text{p}^4\) (1)
- **M3:** Mutual repulsion between the paired electrons in the sulfur orbital makes that electron easier to remove (1)

**Part (b)(iv):**
- **M1:** Magnesium has more protons / greater nuclear charge (1)
- **M2:** Shielding is similar / outer electron is in the same shell (so stronger electrostatic attraction) (1)

1-iodobutane has the highest boiling temperature because iodine is the largest halogen atom with the greatest number of electrons. This results in the strongest London forces (instantaneous dipole-induced dipole interactions) between the molecules, which require the most thermal energy to overcome, despite 1-fluorobutane having the most polar carbon-halogen bond.

D is the correct answer. 1 mark is awarded for the correct option.

During the reaction, iodide ions are oxidized to iodine, where the oxidation state of iodine increases from -1 to 0. Sulfuric acid acts as an oxidizing agent and is reduced to form sulfur dioxide, sulfur, and hydrogen sulfide, which are all reduction products.

A is the correct answer. 1 mark is awarded for identifying the correct oxidation product.

First calculate heat transferred using q = m * c * delta T, where m = 50.0 g + 50.0 g = 100.0 g. This gives q = 100.0 * 4.18 * 6.7 = 2800.6 J = 2.8006 kJ. The number of moles of water formed is n = c * V = 1.00 * 0.0500 = 0.0500 mol. The enthalpy change of neutralisation is delta H = -q / n = -2.8006 / 0.0500 = -56.0 kJ mol-1.

A is the correct answer. 1 mark is awarded for the correct calculation of neutralisation enthalpy.

The rate of hydrolysis of halogenoalkanes is determined by the strength of the carbon-halogen bond. The C-Cl bond is the strongest (highest bond enthalpy) and requires the most energy to break, making 1-chlorobutane hydrolyse the slowest. The C-I bond is the weakest (lowest bond enthalpy) and breaks most easily, making 1-iodobutane hydrolyse the fastest. Therefore, the rate of precipitate formation increases in the order: 1-chlorobutane < 1-bromobutane < 1-iodobutane.

B is the correct answer. 1 mark is awarded for identifying the correct relative rates based on carbon-halogen bond strength.

The forward reaction is exothermic, so decreasing the temperature shifts the equilibrium to the right to produce heat. There are fewer moles of gas on the product side (2 moles) than on the reactant side (3 moles), so increasing the pressure shifts the equilibrium to the right to decrease pressure. Therefore, decreasing the temperature and increasing the pressure will both shift the position of equilibrium to the right.

C is the correct answer. 1 mark is awarded for identifying the correct conditions.

In hot, concentrated sodium hydroxide, chlorine undergoes disproportionation: 3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O. In sodium chloride (NaCl), the oxidation state of chlorine is -1. In sodium chlorate(V) (NaClO3), the oxidation state of chlorine is +5.

C is the correct answer. 1 mark is awarded for the correct oxidation states.

Compound Y reacts with sodium carbonate to produce carbon dioxide, indicating Y is a carboxylic acid. Since Y is formed by oxidation of X under reflux, X must be a primary alcohol. A carboxylic acid contains a carbonyl group (C=O), which gives a characteristic strong, sharp peak at 1680-1750 cm-1. This peak is absent in the alcohol X.

B is the correct answer. 1 mark is awarded for identifying the correct IR absorption band for the carbonyl group.

Enthalpy change can be calculated as: delta H = sum of bond enthalpies of bonds broken - sum of bond enthalpies of bonds formed. Bonds broken: 1 mol of C=C (614 kJ) and 1 mol of H-Cl (431 kJ), giving a total energy input of 1045 kJ. Bonds formed: 1 mol of C-C (347 kJ), 1 mol of C-H (413 kJ), and 1 mol of C-Cl (346 kJ), giving a total energy output of 1106 kJ. Therefore, delta H = 1045 - 1106 = -61 kJ mol-1.

B is the correct answer. 1 mark is awarded for the correct calculation of enthalpy change.

Using Hess's Law: \(\Delta_f H^\ominus(\text{C}_3\text{H}_6\text{, g}) = 3 \times \Delta_c H^\ominus(\text{C, s}) + 3 \times \Delta_c H^\ominus(\text{H}_2\text{, g}) - \Delta_c H^\ominus(\text{C}_3\text{H}_6\text{, g})\). Substituting the values: \(\Delta_f H^\ominus = 3(-394) + 3(-286) - (-2058) = -1182 - 858 + 2058 = +18\text{ kJ mol}^{-1}\).

1 mark: Correctly calculates the value with the correct positive sign (+18).

Propan-1-ol has the highest boiling temperature because it forms extensive intermolecular hydrogen bonds. Compared to propan-2-ol, it has a straight chain allowing for stronger London forces due to greater surface area contact. Methoxyethane has only dipole-dipole and London forces, while butane has only weak London forces.

1 mark: Correct choice of Propan-1-ol.

Thermal stability of Group 2 carbonates increases down the group. As the cation size increases from \(\text{Mg}^{2+}\) to \(\text{Ba}^{2+}\), its charge density decreases. Thus, it polarizes the carbonate anion less, making the C-O bond harder to break and requiring more thermal energy to decompose.

1 mark: Correctly identifies barium carbonate (BaCO3) as the most thermally stable.

The strong, sharp absorption band at \(1715\text{ cm}^{-1}\) indicates a carbonyl (C=O) group. The absence of a broad band above \(3200\text{ cm}^{-1}\) rules out the presence of an alcohol (O-H) group. Propanone, a ketone, fits this description and molecular formula.

1 mark: Correctly identifies propanone based on the spectral analysis.

A catalyst provides an alternative pathway with a lower activation energy, meaning the activation energy boundary (E_a) shifts to the left on the energy axis. It does not alter the actual distribution of molecular energies, so the shape and position of the curve itself remain completely unchanged.

1 mark: Correctly identifies that the activation energy position shifts to the left relative to the distribution.

In the reaction between KI and concentrated H2SO4, sulfuric acid is reduced to sulfur dioxide (SO2, where S is +4), sulfur (S, where S is 0), and hydrogen sulfide (H2S, where S is -2). Iodine (I2) is formed by the oxidation of iodide ions, and hydrogen iodide (HI) is formed by an acid-base reaction, not redox.

1 mark: Correctly identifies the list containing only reduction products of sulfuric acid.

The rate of hydrolysis of halogenoalkanes depends on bond strength (C-I is weaker than C-Cl, hence reacts faster) and class. Tertiary halogenoalkanes (like 2-iodo-2-methylpropane) react faster than primary ones because they undergo the SN1 mechanism, which proceeds via a stable tertiary carbocation.

1 mark: Correctly identifies the tertiary iodoalkane as the fastest-reacting species.

Total mass of solution \(m = 50.0 + 50.0 = 100.0\text{ g}\). Heat energy released \(q = m c \Delta T = 100.0 \times 4.18 \times 6.50 = 2717\text{ J} = 2.717\text{ kJ}\). Moles of acid reacting = \(1.00 \times 0.0500 = 0.0500\text{ mol}\). Since neutralization is exothermic, \(\Delta_{\text{neut}}H = -\frac{2.717\text{ kJ}}{0.0500\text{ mol}} = -54.34\text{ kJ mol}^{-1}\), which rounds to \(-54.3\text{ kJ mol}^{-1}\).

1 mark: Correctly calculates \(-54.3\text{ kJ mol}^{-1}\) including the negative sign.

The equation for the standard enthalpy change of formation of propane is: \(3\text{C(graphite)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\) Using a Hess cycle with combustion products: \(\Delta_f H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})\) \(\Delta_f H^\ominus = [3 \times (-394) + 4 \times (-286)] - [-2219]\) \(\Delta_f H^\ominus = [-1182 - 1144] + 2219 = -2326 + 2219 = -107\text{ kJ mol}^{-1}\). Therefore, the correct option is A.

1 mark: Correct calculation of the standard enthalpy change of formation with the correct sign (-107 kJ mol^-1).

Down Group 2, the cationic radius of the metal ion increases while its charge remains constant (+2). This leads to a decrease in charge density and consequently a decrease in the polarizing power of the metal cation. As a result, the nitrate ion is polarized (distorted) to a lesser extent, making it more thermally stable and requiring a higher temperature for decomposition. Thus, barium nitrate decomposes at a higher temperature than calcium nitrate.

1 mark: Correct statement identified (D).

The rate of hydrolysis of halogenoalkanes depends on both the strength of the C-X bond (C-I is the weakest, so iodoalkanes hydrolyse much faster than bromoalkanes and chloroalkanes) and the structure of the halogenoalkane (tertiary halogenoalkanes react via the faster \text{S}_\text{N}1 mechanism compared to the primary halogenoalkanes which react via the slower \text{S}_\text{N}2 pathway). Since 2-iodo-2-methylpropane is a tertiary iodoalkane, it will undergo hydrolysis and form the silver iodide precipitate significantly faster than any of the other options.

1 mark: Correct identification of the fastest-reacting halogenoalkane (C).

Propan-1-ol has a polar -OH group which enables it to form intermolecular hydrogen bonds. Hydrogen bonds are the strongest type of intermolecular force. Propanal and propanone only experience permanent dipole-dipole forces and London forces, while propane is non-polar and only experiences weak London forces. Therefore, propan-1-ol requires the most energy to overcome its intermolecular forces and has the highest boiling temperature.

1 mark: Correct organic compound identified (A).

### Solution

**(a)(i)**
To compare the rates of hydrolysis of halogenoalkanes, we use aqueous silver nitrate solution in the presence of an ethanol solvent. The mixture is heated in a water bath. Ethanol acts as a mutual solvent, allowing both the halogenoalkanes and aqueous silver ions to mix.

**(a)(ii)**
The rate of hydrolysis increases in the order: 1-chlorobutane < 1-bromobutane < 1-iodobutane.
This is because the carbon-halogen bond strength (bond enthalpy) decreases down the group: \(\text{C-Cl} > \text{C-Br} > \text{C-I}\). A weaker bond requires less energy to break, so the C-I bond breaks fastest in the rate-determining step.

**(b)(i)**
- Mechanism name: \(\text{S}_\text{N}1\) (Nucleophilic Substitution Unimolecular).
- Mechanism steps:
1. The polar C-Br bond in 2-bromo-2-methylpropane breaks heterolytically. Draw a curly arrow from the C-Br bond to the Br atom. Label the C atom with \(\delta+\) and the Br atom with \(\delta-\).
2. This forms a tertiary carbocation intermediate, \(\text{(CH}_3)_3\text{C}^+\), and a bromide ion, \(\text{Br}^-\).
3. The hydroxide ion, \(\text{OH}^-\), acts as a nucleophile. Draw a curly arrow from a lone pair on the oxygen of the \(\text{OH}^-\) ion to the positively charged carbon atom of the carbocation.
4. The final product, 2-methylpropan-2-ol, is formed.

**(b)(ii)**
- Tertiary carbocations are highly stable due to the positive inductive effect of the three electron-donating alkyl (methyl) groups, which stabilizes the positive charge.
- The bulky methyl groups surrounding the central carbon atom cause significant steric hindrance, preventing the nucleophile from attacking from the rear as required in the \(\text{S}_\text{N}2\) mechanism.

**(c)(i)**
The organic product of the elimination reaction is methylpropene. In skeletal formula, this is drawn as a central carbon double-bonded to a \(\text{CH}_2\) group and single-bonded to two methyl groups (a "Y" shape with one double bond: \(\text{(CH}_3)_2\text{C=CH}_2\)).

**(c)(ii)**
In this elimination reaction, the hydroxide ion acts as a **base** (by accepting a proton from one of the methyl groups).

**(a)(i)**
- **Mark 1:** Silver nitrate (aqueous) [1]
- **Mark 2:** Ethanol (as solvent) AND heating / warm water bath [1]

**(a)(ii)**
- **Mark 1:** Rate increases from chlorine to iodine (or 1-chlorobutane is slowest, 1-iodobutane is fastest) [1]
- **Mark 2:** Down the group, C-Halogen bond enthalpy decreases / bond strength decreases [1]
- **Mark 3:** Bond enthalpy is the dominant factor (not polarity), so the weaker C-I bond breaks easiest/fastest [1]

**(b)(i)**
- **Mark 1:** Correct name: \(\text{S}_\text{N}1\) / nucleophilic substitution [1]
- **Mark 2:** First step: Curly arrow from C-Br bond to Br, with correct dipoles \(\text{C}^{\delta+}\) and \(\text{Br}^{\delta-}\) shown [1]
- **Mark 3:** Intermediates: Correctly drawn tertiary carbocation structure and intermediate \(\text{Br}^-\), and curly arrow from lone pair on \(\text{OH}^-\) to the carbocation carbon [1]
- **Mark 4:** Correct structure of 2-methylpropan-2-ol [1]

**(b)(ii)**
- **Mark 1:** The tertiary carbocation intermediate is highly stable due to the inductive effect of three alkyl groups [1]
- **Mark 2:** Steric hindrance from the bulky methyl groups prevents back-attack (ruling out \(\text{S}_\text{N}2\)) [1]

**(c)(i)**
- **Mark 1:** Correct skeletal structure of methylpropene [1]

**(c)(ii)**
- **Mark 1:** Base / proton acceptor (Do NOT accept 'nucleophile') [1]

### Solution

**(a)(i)**
The standard enthalpy change of combustion, \(\Delta_c H^\ominus\), is the enthalpy change when **one mole** of a substance is **completely burned in oxygen** under standard conditions (298 K, 100 kPa / 1 bar), with all reactants and products in their standard states.

**(a)(ii)**
- **Hess's Law Cycle:**
Top-left: \(\text{CO(g)} + 2\text{H}_2\text{(g)}\)
Top-right: \(\text{CH}_3\text{OH(l)}\)
Bottom: Combustion products: \(\text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\)

Arrows point downwards from both top boxes to the bottom box.
- Left arrow (reactants to combustion products): \(\Delta H_1 = \Delta_c H^\ominus[\text{CO(g)}] + 2\Delta_c H^\ominus[\text{H}_2\text{(g)}]\)
- Right arrow (product to combustion products): \(\Delta H_2 = \Delta_c H^\ominus[\text{CH}_3\text{OH(l)}]\)

- **Calculation:**
According to Hess's Law:
$$\Delta_r H^\ominus = \Delta H_1 - \Delta H_2$$
$$\Delta H_1 = -283.0 + 2(-285.8) = -283.0 - 571.6 = -854.6\text{ kJ mol}^{-1}$$
$$\Delta H_2 = -726.0\text{ kJ mol}^{-1}$$
$$\Delta_r H^\ominus = -854.6 - (-726.0) = -128.6\text{ kJ mol}^{-1}$$

**(b)(i)**
- **Axes:** The y-axis must be labeled 'Number of molecules' (or 'Fraction of molecules') and the x-axis must be labeled 'Energy' (or 'Kinetic energy').
- **Curves:** Both curves must start at the origin \((0,0)\). The curve for 500 K has a higher peak and is narrower. The curve for 600 K has a lower peak, shifted to the right, and is flatter/wider, crossing the 500 K curve only once. Neither curve should touch the x-axis at high energy.
- **Activation Energy (\(E_a\)):** Marked as a vertical line on the x-axis to the right of both peaks.

**(b)(ii)**
At the higher temperature of 600 K, the distribution curve shifts to the right, meaning the molecules have a higher average kinetic energy. Consequently, a significantly larger fraction of molecules have energy greater than or equal to the activation energy (\(\ge E_a\)), as shown by the larger area under the 600 K curve to the right of the \(E_a\) line. This leads to a greater frequency of successful collisions per unit time.

**(b)(iii)**
Adding a catalyst has **no effect** on the value of the equilibrium constant, \(K_c\). This is because the catalyst decreases the activation energy equally for both the forward and reverse reactions, increasing both reaction rates by the exact same factor without changing the position of equilibrium.

**(a)(i)**
- **Mark 1:** Enthalpy change when one mole of a substance is burned completely in oxygen [1]
- **Mark 2:** Under standard conditions (298 K, 100 kPa / 1 bar) with all species in their standard states [1]

**(a)(ii)**
- **Mark 1:** Correct Hess cycle structure showing reactants, products, and correct combustion products \(\text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\) at the bottom [1]
- **Mark 2:** Both arrows pointing downwards from reactants and product to the combustion products, labeled correctly with coefficients [1]
- **Mark 3:** Correct method/calculation step: \([-283.0 + 2(-285.8)] - [-726.0]\) [1]
- **Mark 4:** Final answer of \(-128.6\text{ kJ mol}^{-1}\) (must include correct sign, value, and units) [1]

**(b)(i)**
- **Mark 1:** Correct labels on both axes: y-axis = 'Number of molecules' (or fraction), x-axis = 'Energy' [1]
- **Mark 2:** 500 K curve has a higher peak and is shifted left compared to the 600 K curve; both start at origin and do not touch x-axis at high energy [1]
- **Mark 3:** 600 K curve peak is lower and shifted right, crossing the 500 K curve only once [1]
- **Mark 4:** \(E_a\) marked on the x-axis to the right of the peaks [1]

**(b)(ii)**
- **Mark 1:** At 600 K, the average energy of molecules increases / curve shifts to the right [1]
- **Mark 2:** A greater fraction/number of molecules have energy \(\ge E_a\) (larger area to the right of \(E_a\)), resulting in more successful collisions per unit time / more frequent successful collisions [1]

**(b)(iii)**
- **Mark 1:** No effect on \(K_c\) [0.33]
- **Mark 2:** Explanation: It increases the rates of both the forward and reverse reactions by the same factor / lowers the activation energy for both forward and reverse reactions by the same amount [1]

### Solution

**(a)(i)**
The balanced equation for the thermal decomposition of barium nitrate is:
$$2\text{Ba(NO}_3)_2\text{(s)} \rightarrow 2\text{BaO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}$$
(Alternative balanced equation using fractional coefficients is also correct: \(\text{Ba(NO}_3)_2\text{(s)} \rightarrow \text{BaO(s)} + 2\text{NO}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)}\)).

**(a)(ii)**
- **Trend:** Thermal stability increases down the group.
- **Explanation:**
1. Down Group 2, the cationic radius of the metal ion (\(\text{M}^{2+}\)) increases.
2. Consequently, the charge density of the metal cation decreases.
3. This leads to a weaker polarizing effect on the nitrate anion (specifically, the C-O/N-O bonds are less distorted).
4. Therefore, more thermal energy (higher temperature) is required to break the covalent bonds within the nitrate ion to cause decomposition.

**(b)(i)**
- **Gas:** Hydrogen chloride, \(\text{HCl(g)}\).
- **Equation:**
$$\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)}$$
(or \(2\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{Na}_2\text{SO}_4\text{(s)} + 2\text{HCl(g)}\)).
- **Why not redox:** No elements change their oxidation states. The oxidation state of chlorine remains at -1, sodium at +1, hydrogen at +1, oxygen at -2, and sulfur at +6. It is an acid-base (proton transfer) reaction.

**(b)(ii)**
- **Purple vapor:** Iodine, \(\text{I}_2\text{(g)}\)
- **Yellow solid:** Sulfur, \(\text{S(s)}\)
- **Gas with the smell of bad eggs:** Hydrogen sulfide, \(\text{H}_2\text{S(g)}\)
- **Oxidation state of sulfur in \(\text{H}_2\text{S}\):** -2.

**(b)(iii)**
The half-equation for the reduction of sulfuric acid to hydrogen sulfide is:
$$\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}$$
(or \(\text{SO}_4^{2-} + 10\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\)).

**(a)(i)**
- **Mark 1:** Correct formulas and balancing of all reactants and products: \(2\text{Ba(NO}_3)_2 \rightarrow 2\text{BaO} + 4\text{NO}_2 + \text{O}_2\) [1]
- **Mark 2:** Correct state symbols: \(\text{(s)}\), \(\text{(s)}\), \(\text{(g)}\), \(\text{(g)}\) [1]

**(a)(ii)**
- **Mark 1:** Thermal stability increases down the group [1]
- **Mark 2:** Cationic radius of the metal ion increases / charge density of the metal ion decreases [1]
- **Mark 3:** Polarizing power of the cation decreases [1]
- **Mark 4:** Polarisation/distortion of the nitrate ion decreases, requiring more energy to break bonds [1]

**(b)(i)**
- **Mark 1:** Gas is hydrogen chloride / \(\text{HCl}\) [1]
- **Mark 2:** Balanced equation: \(\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}\) (allow state symbols omitted if not specified) [1]
- **Mark 3:** No change in oxidation states of any elements / it is an acid-base reaction [1]

**(b)(ii)**
- **Mark 1:** Purple vapor is iodine / \(\text{I}_2\) [1]
- **Mark 2:** Yellow solid is sulfur / \(\text{S}\) [1]
- **Mark 3:** Gas with bad egg smell is hydrogen sulfide / \(\text{H}_2\text{S}\) [1]
- **Mark 4:** Oxidation state of sulfur in \(\text{H}_2\text{S}\) is -2 [0.33]

**(b)(iii)**
- **Mark 1:** Correctly balanced ionic half-equation: \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\) [1]

### Part (a)

**(i)**
1. Calculate mass of 2-methylpropan-1-ol burned:
$$\text{mass} = 186.45 - 185.81 = 0.64 \text{ g}$$

2. Calculate moles of 2-methylpropan-1-ol:
$$\text{moles} = \frac{0.64 \text{ g}}{74.1 \text{ g mol}^{-1}} = 8.6369 \times 10^{-3} \text{ mol}$$

3. Calculate heat energy transferred to water (\(q\)):
$$\Delta T = 51.6 - 20.2 = 31.4 \text{ }^\circ\text{C}$$
$$q = m c \Delta T = 150.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ }^\circ\text{C}^{-1} \times 31.4 \text{ }^\circ\text{C} = 19687.8 \text{ J} = 19.6878 \text{ kJ}$$

4. Calculate enthalpy of combustion (\(\Delta H_c\)):
$$\Delta H_c = -\frac{q}{\text{moles}} = -\frac{19.6878 \text{ kJ}}{8.6369 \times 10^{-3} \text{ mol}} = -2279.48 \text{ kJ mol}^{-1}$$

Rounded to 3 significant figures:
$$\Delta H_c = -2280 \text{ kJ mol}^{-1}$$

**(ii)**
Any two of:
- Heat loss to the surrounding air / calorimeter.
- Heat absorbed by the copper calorimeter / thermometer is not accounted for.
- Evaporation of the alcohol from the wick of the burner (while weighing or cooling).
- Reaction not carried out under standard conditions.

---

### Part (b)

**(i)**
$$(CH_3)_3COH + HCl \rightarrow (CH_3)_3CCl + H_2O$$
*(Accept molecular formula: \(C_4H_{10}O + HCl \rightarrow C_4H_9Cl + H_2O\))*

**(ii)**
**Mechanism:**
1. **Protonation of alcohol:**
Curly arrow from the lone pair on the oxygen atom of \((CH_3)_3C-OH\) to the hydrogen ion \(H^+\) (or to the \(H\) of \(H-Cl\) with a simultaneous curly arrow breaking the \(H-Cl\) bond to the chlorine atom).

2. **Formation of Carbocation:**
Structure of protonated intermediate shown: \((CH_3)_3C-OH_2^+\).
Curly arrow from the \(C-O\) bond to the positive oxygen atom to show loss of \(H_2O\).

3. **Nucleophilic Attack:**
Structure of tertiary carbocation shown: \((CH_3)_3C^+\).
Curly arrow from the lone pair on a chloride ion (\(Cl^-\)) to the positively charged carbon atom of the carbocation.

**(iii)**
- 2-methylpropan-2-ol is a tertiary alcohol and forms a highly stable tertiary carbocation, \((CH_3)_3C^+\), because of the electron-donating inductive effect of the three methyl groups. This stabilizes the carbocation intermediate, making the \(\text{S}_\text{N}1\) route energetically favorable.
- 2-methylpropan-1-ol is a primary alcohol, which would form an extremely unstable primary carbocation. It has minimal steric hindrance around the reaction center, allowing direct back-side attack via the \(\text{S}_\text{N}2\) transition state.

---

### Part (c)

**(i)**
- The broad absorption peak due to the \(O-H\) stretch in alcohols at \(3230-3550 \text{ cm}^{-1}\) (or the peak due to the \(C-O\) stretch at \(1000-1300 \text{ cm}^{-1}\)) decreases in intensity and eventually disappears.
- A new sharp absorption peak due to the \(C-Cl\) stretch appears at \(600-800 \text{ cm}^{-1}\).

**(ii)**
- Species: The carbocation \([C_4H_9]^+\) or \([(CH_3)_3C]^+\).
- Stability: It is a tertiary carbocation, stabilized by the positive inductive effect (electron release) of the three methyl groups attached to the positively charged carbon atom.

---

### Part (d)

**(i)**
- Ethanol acts as a mutual solvent (cosolvent) to dissolve both the halogenoalkane (which is insoluble in water) and the aqueous silver nitrate solution so they can react in a single phase.

**(ii)**
- Observation: A white precipitate is formed.
- Ionic equation: \(Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)\) *(state symbols must be correct)*

### Part (a) [Total: 6 Marks]

**(i) [4 Marks]**
- **M1:** Correct calculation of mass of 2-methylpropan-1-ol burned: \(0.64 \text{ g}\) **AND** the temperature change \(\Delta T = 31.4\text{ }^\circ\text{C}\).
- **M2:** Correct calculation of heat energy transferred:
\(q = 150.0 \times 4.18 \times 31.4 = 19687.8 \text{ J} = 19.69 \text{ kJ}\).
- **M3:** Correct calculation of moles of alcohol:
\(\text{moles} = 0.64 / 74.1 = 8.64 \times 10^{-3} \text{ mol}\).
- **M4:** Correct final value of enthalpy of combustion with minus sign, rounded to 3 significant figures:
\(\Delta H_c = -2280 \text{ kJ mol}^{-1}\) (accept \(-2279\) to \(-2282 \text{ kJ mol}^{-1}\) depending on intermediate rounding).
*Allow ECF at each step.*
*Deduct 1 mark if unit is missing or incorrect, or if negative sign is omitted.*

**(ii) [2 Marks]**
- Any two correct reasons (1 mark each):
- Heat loss to the surrounding air/calorimeter.
- Heat absorbed by the copper calorimeter / beaker / thermometer is neglected.
- Evaporation of alcohol from the wick after extinguishing / while cooling before weighing.
- Non-standard conditions.
*Do not accept 'incomplete combustion' as it is excluded by the question.*

---

### Part (b) [Total: 7 Marks]

**(i) [1 Mark]**
- **M1:** Correct balanced equation:
\((CH_3)_3COH + HCl \rightarrow (CH_3)_3CCl + H_2O\) (or molecular equivalents).

**(ii) [4 Marks]**
- **M1:** Curly arrow from the oxygen lone pair of \((CH_3)_3COH\) to \(H^+\) (or showing full proton transfer from \(HCl\) with arrow to \(H\) and from \(H-Cl\) bond to \(Cl\)).
- **M2:** Correct structure of the protonated intermediate \((CH_3)_3C-OH_2^+\) and a curly arrow from the \(C-O\) bond to the oxygen.
- **M3:** Correct structure of the tertiary carbocation intermediate: \((CH_3)_3C^+\).
- **M4:** Curly arrow from the lone pair of \(Cl^-\) to the carbon with positive charge.

**(iii) [2 Marks]**
- **M1:** 2-methylpropan-2-ol forms a tertiary carbocation, which is stable due to the positive inductive effect (electron-donating effect) of the three methyl groups.
- **M2:** 2-methylpropan-1-ol is a primary alcohol, which would form a very unstable primary carbocation, and has minimal steric hindrance, facilitating direct back-side nucleophilic attack (\(\text{S}_\text{N}2\)).

---

### Part (c) [Total: 4 Marks]

**(i) [2 Marks]**
- **M1:** The broad absorption peak/band due to \(O-H\) stretch of alcohol at \(3230-3550 \text{ cm}^{-1}\) (or \(C-O\) stretch at \(1000-1300 \text{ cm}^{-1}\)) disappears/decreases.
- **M2:** A new peak due to the \(C-Cl\) stretch appears at \(600-800 \text{ cm}^{-1}\).
*Wavenumber ranges must be mentioned and correct as per the data booklet.*

**(ii) [2 Marks]**
- **M1:** Identify the species as the butyl carbocation, \([C_4H_9]^+\) or \([(CH_3)_3C]^+\) (positive charge required).
- **M2:** Explain that it is highly stable because it is a tertiary carbocation stabilized by the positive inductive effect of three methyl/alkyl groups.

---

### Part (d) [Total: 3 Marks]

**(i) [1 Mark]**
- **M1:** Ethanol acts as a mutual solvent to allow the halogenoalkane and the aqueous silver nitrate to mix and react in a single phase.

**(ii) [2 Marks]**
- **M1:** Observation: White precipitate.
- **M2:** Ionic equation: \(Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)\) (state symbols required).

(a) Zinc is in excess to ensure that all the copper(II) ions in solution react completely.

(b) Recording the temperature before addition establishes an accurate initial temperature baseline and confirms that the solution is in thermal equilibrium with the surroundings.

(c) Plot temperature on the y-axis against time on the x-axis. Draw a line of best fit through the cooling points from $4$ to $8$ minutes and extrapolate this line back to $3$ minutes (the time of mixing).
Extrapolation:
The rate of cooling is $(31.8 - 27.4) / (8 - 4) = 1.1\ ^\circ\text{C min}^{-1}$.
Extrapolated temperature at $3$ minutes $= 31.8 + 1.1 = 32.9\ ^\circ\text{C}$.
$\Delta T = 32.9\ ^\circ\text{C} - 21.0\ ^\circ\text{C} = 11.9\ ^\circ\text{C}$.

(d) Mass of solution, $m = 50.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g}$.
$q = m \times c \times \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times 11.9\ ^\circ\text{C} = 2487.1\text{ J} \approx 2490\text{ J}$ (or $2.49\text{ kJ}$).

(e) Moles of $\text{CuSO}_4$ reacted $= 0.200\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0100\text{ mol}$.
$\Delta H = -\frac{q}{n} = -\frac{2.4871\text{ kJ}}{0.0100\text{ mol}} = -248.71\text{ kJ mol}^{-1} \approx -249\text{ kJ mol}^{-1}$ (to 3 significant figures).

(f) Source 1: Heat loss to the surroundings. This can be minimized by using a lid on the polystyrene cup and placing the cup inside a glass beaker for extra insulation.
Source 2: The heat capacity of the polystyrene cup and thermometer are ignored.

(a) [1 mark] For stating that zinc is in excess to ensure all $\text{Cu}^{2+}$ reacts.

(b) [1 mark] For stating it establishes a reliable/stable initial temperature baseline.

(c) [3 marks total]
* [1 mark] For description of plotting a graph and drawing a line of best fit for the cooling curve.
* [1 mark] For stating the line is extrapolated back to the time of mixing ($3$ minutes).
* [1 mark] For calculating $\Delta T = 11.9\ ^\circ\text{C}$ (Allow $11.8 - 12.0$).

(d) [2 marks total]
* [1 mark] For correct use of $q = mc\Delta T$ (e.g. $50.0 \times 4.18 \times 11.9$).
* [1 mark] For correct value of $2490\text{ J}$ or $2487\text{ J}$ (Accept $2.49\text{ kJ}$).

(e) [3 marks total]
* [1 mark] For calculation of moles of $\text{CuSO}_4 = 0.0100\text{ mol}$.
* [1 mark] For $\Delta H$ magnitude of $249\text{ kJ mol}^{-1}$ (Allow ECF from part d).
* [1 mark] For the negative sign (essential for exothermic reactions).

(f) [3 marks total]
* [1 mark] For identifying heat loss to the surroundings.
* [1 mark] For suggesting a lid / wrapping the cup in cotton wool / putting cup in a beaker.
* [1 mark] For identifying a second error: heat capacity of cup/thermometer is neglected or zinc absorbs heat.

(a) Concentrated phosphoric(V) acid is a weaker oxidizing agent than concentrated sulfuric acid, which minimizes side reactions, charring, and the production of toxic sulfur dioxide gas.

(b) Carbon dioxide gas is produced during the neutralization reaction of acid impurities with sodium hydrogencarbonate. Venting releases this gas to prevent pressure build-up inside the separating funnel.

(c) Cyclohexene is the upper layer because its density ($0.81\text{ g cm}^{-3}$) is lower than that of the aqueous layer ($1.00\text{ g cm}^{-3}$), and it is immiscible with water.

(d) The drying agent stops clumping together and disperses as a fine, free-flowing powder when shaken, and the liquid changes from cloudy to clear.

(e) Moles of cyclohexanol $= \frac{10.0\text{ g}}{100.0\text{ g mol}^{-1}} = 0.100\text{ mol}$.
Theoretical moles of cyclohexene $= 0.100\text{ mol}$.
Theoretical mass of cyclohexene $= 0.100\text{ mol} \times 82.0\text{ g mol}^{-1} = 8.20\text{ g}$.
Percentage yield $= \left(\frac{4.51\text{ g}}{8.20\text{ g}}\right) \times 100 = 55.0\%$.

(f) Add a few drops of bromine water to the sample. The orange/brown solution is decolourised (turns colourless).

(a) [1 mark] For stating that phosphoric(V) acid is a less powerful oxidizing agent / does not cause charring / does not form sulfur dioxide.

(b) [1 mark] For stating it releases pressure built up by carbon dioxide gas.

(c) [2 marks total]
* [1 mark] For identifying it is the upper layer.
* [1 mark] For stating its density is lower than water / it is immiscible.

(d) [2 marks total]
* [1 mark] For stating that the solid does not clump / remains free-flowing.
* [1 mark] For stating that the liquid becomes clear/transparent.

(e) [3 marks total]
* [1 mark] For calculating moles of cyclohexanol $= 0.100\text{ mol}$.
* [1 mark] For theoretical mass of cyclohexene $= 8.20\text{ g}$.
* [1 mark] For percentage yield $= 55.0\%$.

(f) [3 marks total]
* [1 mark] For identifying bromine water (or acidified potassium manganate(VII)).
* [1 mark] For starting colour: orange/yellow/brown (or purple).
* [1 mark] For ending colour: colourless (or decolourised) (Reject "clear").

(a) Dilute sulfuric acid provides the $\text{H}^+$ ions required for the reduction of $\text{MnO}_4^-$. Hydrochloric acid cannot be used because the chloride ions, $\text{Cl}^-$, would be oxidized to toxic chlorine gas, $\text{Cl}_2$, by the strongly oxidizing manganate(VII) ions.

(b) The reaction mixture goes from colourless to a permanent pale pink.

(c) Moles of $\text{MnO}_4^-$ used $= 0.00500\text{ mol dm}^{-3} \times \left(\frac{24.00}{1000}\right)\text{ dm}^3 = 1.20 \times 10^{-4}\text{ mol}$.

(d) From the equation, $1\text{ mol}$ of $\text{MnO}_4^-$ reacts with $5\text{ mol}$ of $\text{Fe}^{2+}$.
Moles of $\text{Fe}^{2+}$ in $25.0\text{ cm}^3 = 5 \times 1.20 \times 10^{-4}\text{ mol} = 6.00 \times 10^{-4}\text{ mol}$.
Moles of $\text{Fe}^{2+}$ in the $250.0\text{ cm}^3$ volumetric flask $= 10 \times 6.00 \times 10^{-4}\text{ mol} = 6.00 \times 10^{-3}\text{ mol}$.
Therefore, total mass of $\text{FeSO}_4$ in $5$ tablets $= 6.00 \times 10^{-3}\text{ mol} \times 151.9\text{ g mol}^{-1} = 0.9114\text{ g}$.
Mass of $\text{FeSO}_4$ in $1$ tablet $= \frac{0.9114\text{ g}}{5} = 0.18228\text{ g} \approx 0.182\text{ g}$ (to 3 significant figures).

(e) Since a titre value is obtained from two burette readings (initial and final):
Total uncertainty $= 2 \times 0.05\text{ cm}^3 = 0.10\text{ cm}^3$.
Percentage uncertainty $= \left(\frac{0.10\text{ cm}^3}{24.00\text{ cm}^3}\right) \times 100\% = 0.417\%$.

(f) The calculated mass of iron(II) sulfate would increase. Distilled water remaining in the burette dilutes the potassium manganate(VII) solution. This means a larger volume (titre) of the solution is required to reach the end-point, which leads to a larger calculated amount of iron(II) ions.

(a) [2 marks total]
* [1 mark] For stating hydrochloric acid would be oxidized / $\text{Cl}^-$ oxidized to $\text{Cl}_2$.
* [1 mark] For stating sulfuric acid does not get oxidized by $\text{MnO}_4^-$ / provides $\text{H}^+$.

(b) [1 mark] For stating colourless to permanent pale pink.

(c) [1 mark] For $1.20 \times 10^{-4}\text{ mol}$ (or $0.00012\text{ mol}$).

(d) [5 marks total]
* [1 mark] For ratio of $1\text{ MnO}_4^- : 5\text{ Fe}^{2+}$ applied to get $6.00 \times 10^{-4}\text{ mol}$ in $25\text{ cm}^3$.
* [1 mark] For scaling up by $10$ to find $6.00 \times 10^{-3}\text{ mol}$ in $250\text{ cm}^3$.
* [1 mark] For calculating total mass of $\text{FeSO}_4 = 0.9114\text{ g}$ using $M_r = 151.9$.
* [1 mark] For dividing by $5$ to find mass per tablet $= 0.182\text{ g}$.
* [1 mark] For correct final answer to $3$ significant figures with units.

(e) [2 marks total]
* [1 mark] For recognizing two readings are taken, giving a total uncertainty of $\pm 0.10\text{ cm}^3$.
* [1 mark] For percentage uncertainty $= 0.417\%$.

(f) [2 marks total]
* [1 mark] For stating calculated mass of $\text{FeSO}_4$ increases.
* [1 mark] For explaining dilution of $\text{KMnO}_4$ leads to a larger titre.

(a) Lilac flame indicates the presence of potassium ions. Formula: $\text{K}^+$.

(b) The anion is bromide, $\text{Br}^-$.
* Silver nitrate reacts with bromide ions to form a cream precipitate of silver bromide, $\text{AgBr}$ (which distinguishes it from white $\text{AgCl}$ and yellow $\text{AgI}$).
* Silver bromide dissolves in concentrated ammonia but is insoluble in dilute ammonia.
* Nitric acid is added to remove any carbonate, $\text{CO}_3^{2-}$, or hydrogencarbonate, $\text{HCO}_3^-$, impurities, which would otherwise react with silver ions to form a white precipitate of silver carbonate, giving a false positive.

(c) $\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr(s)}$

(d) Chlorine is more reactive than bromine and displaces bromide ions from solution. This produces molecular bromine, $\text{Br}_2$. Bromine is non-polar and preferentially dissolves in the top organic hexane layer, turning it orange.
Equation: $\text{Cl}_2(\text{aq}) + 2\text{Br}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{Br}_2(\text{aq})$

(a) [2 marks total]
* [1 mark] For potassium.
* [1 mark] For $\text{K}^+$.

(b) [4 marks total]
* [1 mark] For identifying bromide / $\text{Br}^-$.
* [1 mark] For explaining that cream precipitate with $\text{AgNO}_3$ suggests bromide (or iodide).
* [1 mark] For explaining that solubility in concentrated ammonia confirms bromide.
* [1 mark] For stating nitric acid removes carbonate/hydrogencarbonate ions.

(c) [2 marks total]
* [1 mark] For correct species in the equation: $\text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr}$.
* [1 mark] For correct state symbols: $(\text{aq})$, $(\text{aq})$, $(\text{s})$.

(d) [4 marks total]
* [1 mark] For stating chlorine displaces bromide ions (or chlorine oxidizes bromide to bromine).
* [1 mark] For explaining that the orange layer is due to bromine dissolving in hexane.
* [2 marks] For correct balanced ionic equation: $\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2$ (1 mark for correct species, 1 mark for balancing; ignore state symbols).