2022 Edexcel AS Level Chemistry (8CH0) 模擬試題連答案詳解

The successive ionisation energies show a very large increase between the fifth and sixth ionisation energies (from \(6274\text{ kJ mol}^{-1}\) to \(21269\text{ kJ mol}^{-1}\)). This indicates that the sixth electron is removed from a shell closer to the nucleus, meaning the element has five electrons in its outer shell and belongs to Group 15. Phosphorus is in Group 15 and has five valence electrons.

1 mark: Correct option B.

The amide ion, \(\text{NH}_2^-\), has a nitrogen atom with 5 valence electrons plus 1 from the negative charge, giving 6 outer electrons. It forms 2 single covalent bonds to hydrogen atoms, leaving 2 lone pairs on the nitrogen atom. This results in a tetrahedral arrangement of electron pairs (4 electron domains). The repulsion between the two lone pairs reduces the bond angle from the tetrahedral angle of \(109.5^\circ\) by approximately \(2 \times 2.5^\circ = 5^\circ\), resulting in a non-linear shape with a bond angle of approximately \(104.5^\circ\).

1 mark: Correct option C.

First, convert all quantities to SI units: \(p = 102 \times 10^3\text{ Pa}\), \(V = 80.0 \times 10^{-6}\text{ m}^3\), and \(T = 97 + 273 = 370\text{ K}\).

Using the ideal gas equation: \(pV = nRT \implies n = \frac{pV}{RT}\).

\(n = \frac{102 \times 10^3 \times 80.0 \times 10^{-6}}{8.31 \times 370} = \frac{8.16}{3074.7} \approx 2.654 \times 10^{-3}\text{ mol}\).

Now, calculate the relative molecular mass (\(M_r\)): \(M_r = \frac{mass}{n} = \frac{0.232}{2.654 \times 10^{-3}} \approx 87.4\text{ g mol}^{-1}\).

1 mark: Correct option C.

Down Group 2, the cationic radius increases, so the charge density of the \(M^{2+}\) ion decreases. This means the cation has less polarising power and distorts the electron cloud of the carbonate ion less easily. Consequently, the \(C-O\) bond within the carbonate ion is weakened less, making the carbonate compound more thermally stable down the group.

1 mark: Correct option A.

When chlorine reacts with hot, concentrated sodium hydroxide, it disproportionates to form chloride (\(\text{Cl}^-\), oxidation state \(-1\)) and chlorate(V) (\(\text{ClO}_3^-\), oxidation state \(+5\)). The balanced ionic equation is: 3\(\text{Cl}_2 + 6\text{OH}^- \rightarrow 5\text{Cl}^- + \text{ClO}_3^- + 3\text{H}_2\text{O}\).

1 mark: Correct option B.

The atomic number of Cobalt (Co) is 27, so a neutral cobalt atom has 27 electrons. Its electronic configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2\). When transition metal atoms form ions, they lose electrons from the \(4s\) orbital before the \(3d\) orbitals. Therefore, a \(\text{Co}^{2+}\) ion loses the two \(4s\) electrons, giving the electronic configuration \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^7\).

1 mark: Correct option B.

Hydrogen fluoride (\(\text{HF}\)) has the highest boiling temperature among the hydrogen halides. This is due to the presence of intermolecular hydrogen bonds, which are significantly stronger than the permanent dipole-dipole forces and London forces present in \(\text{HCl}\), \(\text{HBr}\), and \(\text{HI}\). Although \(\text{HI}\) has stronger London forces than \(\text{HF}\) due to having more electrons, the hydrogen bonding in \(\text{HF}\) requires far more energy to overcome.

1 mark: Correct option A.

1. Calculate moles of ethanol used: \(n(\text{C}_2\text{H}_5\text{OH}) = \frac{9.2}{46.0} = 0.20\text{ mol}\).

2. Calculate the theoretical yield of bromoethane: Since the molar ratio of reactant to product is \(1:1\), the theoretical yield of \(\text{C}_2\text{H}_5\text{Br}\) is \(0.20\text{ mol}\).

3. Calculate the theoretical mass of bromoethane: \(\text{Theoretical mass} = 0.20\text{ mol} \times 109.0\text{ g mol}^{-1} = 21.8\text{ g}\).

4. Calculate percentage yield: \(\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{13.1}{21.8} \times 100 \approx 60.1\%\).

1 mark: Correct option B.

The largest increase (jump) in successive ionisation energies occurs between the third and fourth ionisation energies (from \(2745\text{ kJ mol}^{-1}\) to \(11577\text{ kJ mol}^{-1}\)). This indicates that the fourth electron is being removed from an inner quantum shell, which is closer to the nucleus and less shielded. Therefore, the element has three electrons in its outer shell (valence shell). Since the element is in Period 3, it must be aluminium (group 13/3, electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^1\)).

Award [1] for the correct identification of aluminium based on the jump between the third and fourth ionisation energies.

Going down Group 2, the ionic radius of the metal cation increases while the 2+ charge remains the same. This leads to a decrease in charge density down the group, meaning the \(\text{Ba}^{2+}\) ion has a much lower charge density than the \(\text{Mg}^{2+}\) ion. A lower charge density results in a lower polarising power, so \(\text{Ba}^{2+}\) polarises (distorts) the electron cloud of the carbonate ion less. Consequently, the \(\text{C-O}\) bond in the carbonate ion is weakened to a lesser extent, making barium carbonate more stable to thermal decomposition.

Award [1] for the correct option selection explaining the trend in thermal stability using charge density and polarisation.

Phosphorus has the outer electronic configuration \(3s^2 3p^3\) and sulfur has the outer electronic configuration \(3s^2 3p^4\). In sulfur, there is a pair of electrons in one of the \(3p\) orbitals, whereas in phosphorus, all three \(3p\) orbitals are singly occupied. The mutual repulsion between these paired electrons in the same \(3p\) orbital of sulfur (spin-pair repulsion) makes it easier to remove one electron from sulfur than from the singly occupied orbitals of phosphorus, resulting in a lower first ionisation energy.

MP1: State or write the outer electronic configurations: \(P\) is \(3p^3\) and \(S\) is \(3p^4\) (or equivalent full configurations). (1 mark)
MP2: State that sulfur has paired electrons in a \(3p\) orbital, while phosphorus has only singly occupied \(3p\) orbitals. (1 mark)
MP3: Explain that repulsion between these paired electrons (spin-pair repulsion) in sulfur makes the outer electron easier to remove. (1 mark)

In a \(\text{ClF}_3\) molecule, there are 5 electron pairs around the central chlorine atom: 3 bonding pairs and 2 lone pairs. According to electron pair repulsion theory, electron pairs arrange themselves to minimise repulsion, with lone pairs repelling more than bonding pairs. This pushes the three \(\text{Cl-F}\) bonds closer together, resulting in a T-shaped geometry with an approximate bond angle of \(87.5^\circ\) (accept values in the range \(86^\circ\) to \(88^\circ\)).

MP1: Identify that the central chlorine atom has 3 bonding pairs and 2 lone pairs of electrons. (1 mark)
MP2: State that lone pairs repel more than bonding pairs. (1 mark)
MP3: Correctly identify the shape as T-shaped and state a bond angle of \(87.5^\circ\) (allow any angle from \(86^\circ\) to \(88^\circ\) or 'slightly less than \(90^\circ\)'). (1 mark)

Step 1: Calculate the amount, in moles, of \(\text{Na}_2\text{CO}_3\):
\(n(\text{Na}_2\text{CO}_3) = \frac{4.24\text{ g}}{106.0\text{ g mol}^{-1}} = 0.0400\text{ mol}\)

Step 2: Determine the moles of \(\text{CO}_2\) produced:
According to the stoichiometry of the equation, \(1\text{ mol}\) of \(\text{Na}_2\text{CO}_3\) produces \(1\text{ mol}\) of \(\text{CO}_2\).
Therefore, \(n(\text{CO}_2) = 0.0400\text{ mol}\).

Step 3: Calculate the volume of \(\text{CO}_2\):
\(V = 0.0400\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.960\text{ dm}^3\).

MP1: Calculate the number of moles of \(\text{Na}_2\text{CO}_3\) as \(0.0400\text{ mol}\). (1 mark)
MP2: State or use the 1:1 molar ratio to show that \(0.0400\text{ mol}\) of \(\text{CO}_2\) is produced. (1 mark)
MP3: Calculate the volume of \(\text{CO}_2\) as \(0.960\text{ dm}^3\) (or \(0.96\text{ dm}^3\)), including correct units. (1 mark)
[Allow full transferred error (TE) from incorrect moles in MP1]

The magnesium ion (\(\text{Mg}^{2+}\)) is smaller than the barium ion (\(\text{Ba}^{2+}\)), meaning the magnesium ion has a higher charge density. Consequently, \(\text{Mg}^{2+}\) polarises (distorts the electron cloud of) the carbonate ion to a greater extent. This polarisation weakens the carbon-oxygen covalent bond within the carbonate ion, lowering the activation energy for thermal decomposition and requiring less thermal energy to break the bond.

MP1: Identify that the magnesium ion is smaller than the barium ion, or has a higher charge density. (1 mark)
MP2: State that the magnesium ion polarises / distorts the electron cloud of the carbonate ion to a greater extent. (1 mark)
MP3: Explain that this polarising action weakens the carbon-oxygen bond (making it break more easily on heating). (1 mark)

In the reactant molecule \(\text{Cl}_2\), the oxidation number of chlorine is \(0\). In the product \(\text{NaCl}\), the oxidation number of chlorine is \(-1\) (reduction). In the product \(\text{NaClO}\), the oxidation number of chlorine is \(+1\) (oxidation). Because the same element is simultaneously oxidised and reduced, this is a disproportionation reaction.

MP1: State that the oxidation number of chlorine changes from \(0\) to \(-1\) in \(\text{NaCl}\). (1 mark)
MP2: State that the oxidation number of chlorine changes from \(0\) to \(+1\) in \(\text{NaClO}\). (1 mark)
MP3: Identify the reaction type as a disproportionation reaction. (1 mark)

Step 1: Calculate the mass of water lost during heating:
\(m(\text{H}_2\text{O}) = 5.00\text{ g} - 4.26\text{ g} = 0.74\text{ g}\)

Step 2: Calculate the number of moles of anhydrous \(\text{BaCl}_2\) and of water:
\(n(\text{BaCl}_2) = \frac{4.26\text{ g}}{208.2\text{ g mol}^{-1}} = 0.02046\text{ mol}\)
\(n(\text{H}_2\text{O}) = \frac{0.74\text{ g}}{18.0\text{ g mol}^{-1}} = 0.04111\text{ mol}\)

Step 3: Determine the molar ratio of water to anhydrous salt:
\(\text{Ratio} = \frac{0.04111}{0.02046} = 2.01\)

To the nearest whole number, \(x = 2\).

MP1: Calculate the mass of water lost (\(0.74\text{ g}\)). (1 mark)
MP2: Calculate the correct moles of both \(\text{BaCl}_2\) (\(0.02046\text{ mol}\)) and \(\text{H}_2\text{O}\) (\(0.04111\text{ mol}\)). (1 mark)
MP3: Calculate the ratio of water to salt to find \(x = 2\). (1 mark)
[Accept alternative standard empirical formula calculations. Allow TE at each stage.]

Fluorine is highly electronegative, creating a strong permanent dipole in the \(\text{H-F}\) bond. This allows hydrogen fluoride molecules to form intermolecular hydrogen bonds. Hydrogen chloride molecules do not form hydrogen bonds; instead, they are held together by weaker London forces and permanent dipole-dipole interactions. Because hydrogen bonds are significantly stronger than these other intermolecular forces, more thermal energy is needed to separate \(\text{HF}\) molecules, resulting in a higher boiling temperature.

MP1: State that there are hydrogen bonds between \(\text{HF}\) molecules. (1 mark)
MP2: State that \(\text{HCl}\) molecules are held together by London forces / permanent dipole-dipole forces (not hydrogen bonds). (1 mark)
MP3: Explain that hydrogen bonds are much stronger than London forces / permanent dipole-dipole forces, requiring significantly more energy to overcome. (1 mark)

The reaction of barium ions with sulfate ions produces insoluble barium sulfate, which is seen as a white precipitate. The ionic equation is \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\). Hydrochloric acid is added before the test to react with and remove any carbonate (\(\text{CO}_3^{2-}\)) or sulfite (\(\text{SO}_3^{2-}\)) impurities. If not removed, these ions would also react with barium ions to form white precipitates (such as \(\text{BaCO}_3\)), which would give a false positive result.

MP1: Write the correct ionic equation with state symbols: \(\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})\). (1 mark)
MP2: Identify the observation as a white precipitate. (1 mark)
MP3: Explain that hydrochloric acid is added to remove carbonate/sulfite ions (to prevent the formation of other insoluble barium salts/false positives). (1 mark)

Let \(x\) be the percentage abundance of \(^{71}\text{Ga}\). The percentage abundance of \(^{69}\text{Ga}\) is therefore \(100 - x\).

Set up the equation for the relative atomic mass:
\(\frac{69(100 - x) + 71x}{100} = 69.80\)

Multiply by 100:
\(6900 - 69x + 71x = 6980\)

Simplify:
\(2x = 80\)
\(x = 40\)

So, the percentage abundance of the \(^{71}\text{Ga}\) isotope is 40%.

M1: Setting up a correct algebraic equation for the relative atomic mass (e.g., \(69(100 - x) + 71x = 6980\) or equivalent) (1 mark)
M2: Correct rearrangement to isolate the variable (e.g., \(2x = 80\)) (1 mark)
M3: Correct final answer: 40% (or 40) (1 mark)

1. Phosphorus is in Group 5 and has 5 valence electrons. The \(+\) charge indicates the loss of 1 electron, leaving 4 valence electrons. These 4 electrons form single covalent bonds with 4 fluorine atoms. Thus, there are 4 bonding pairs and 0 lone pairs around the central phosphorus atom.
2. According to electron pair repulsion theory, these 4 bonding pairs repel each other equally to be as far apart as possible to minimize repulsion.
3. This equal repulsion results in a tetrahedral shape with a bond angle of \(109.5^\circ\).

M1: State that there are 4 bonding pairs and 0 lone pairs around the central phosphorus atom (1 mark)
M2: State that the electron pairs repel each other to minimize repulsion / maximize separation (1 mark)
M3: Identify the shape as tetrahedral and state the bond angle as \(109.5^\circ\) (accept \(109^\circ\) to \(110^\circ\)) (1 mark)

1. Calculate the amount in moles of \(\text{Cu}_2\text{CO}_3(\text{OH})_2\):
\(n = \frac{\text{mass}}{\text{Molar Mass}} = \frac{2.21\text{ g}}{221.0\text{ g mol}^{-1}} = 0.0100\text{ mol}\)

2. Determine the total moles of gas produced:
From the balanced equation, \(1\text{ mol}\) of \(\text{Cu}_2\text{CO}_3(\text{OH})_2\) produces \(1\text{ mol}\) of \(\text{CO}_2(\text{g})\) and \(1\text{ mol}\) of \(\text{H}_2\text{O}(\text{g})\), giving a total of \(2\text{ moles}\) of gas.
\(n(\text{gas}) = 2 \times 0.0100\text{ mol} = 0.0200\text{ mol}\)

3. Calculate the total volume of gas at r.t.p.:
\(V = 0.0200\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.480\text{ dm}^3\)

M1: Correct calculation of moles of copper carbonate = \(0.0100\text{ mol}\) (1 mark)
M2: Correct determination of total moles of gas produced = \(0.0200\text{ mol}\) (consequential on M1) (1 mark)
M3: Correct calculation of volume = \(0.480\text{ dm}^3\) (allow \(0.48\text{ dm}^3\) or \(480\text{ cm}^3\) if unit is clearly stated) (1 mark)

1. The magnesium ion (\(\text{Mg}^{2+}\)) has a smaller ionic radius than the barium ion (\(\text{Ba}^{2+}\)) while having the same \(+2\) charge, meaning \(\text{Mg}^{2+}\) has a higher charge density.
2. As a result, the \(\text{Mg}^{2+}\) ion polarises (distorts the electron cloud of) the carbonate ion (\(\text{CO}_3^{2-}\)) more strongly than the \(\text{Ba}^{2+}\) ion.
3. This polarising effect weakens the carbon-oxygen (\(\text{C-O}\)) bond within the carbonate ion, requiring less thermal energy to break and therefore a lower temperature for decomposition.

M1: State that the magnesium ion is smaller / has a higher charge density than the barium ion (1 mark)
M2: State that the magnesium ion polarises / distorts the electron cloud of the carbonate ion more effectively (1 mark)
M3: State that this weakens the carbon-oxygen bond (within the carbonate group), requiring less energy to break (1 mark)

1. Identify the oxidation numbers and their changes:
- Iodine: \(\text{I}_2\) (0) to \(\text{HIO}_3\) (+5). The change per iodine atom is \(+5\). Since there are 2 iodine atoms in \(\text{I}_2\), the total increase in oxidation number is \(2 \times 5 = +10\).
- Nitrogen: \(\text{HNO}_3\) (+5) to \(\text{NO}_2\) (+4). The change per nitrogen atom is \(-1\).

2. Balance the changes in oxidation number:
To balance the \(+10\) increase from the iodine, we need 10 nitrogen atoms to undergo the \(-1\) decrease. This establishes a stoichiometry of 1 mole of \(\text{I}_2\) to 10 moles of \(\text{HNO}_3\):
\(\text{I}_2 + 10\text{HNO}_3 \rightarrow 2\text{HIO}_3 + 10\text{NO}_2 + x\text{H}_2\text{O}\)

3. Balance the remaining hydrogen and oxygen atoms:
- There are 10 hydrogen atoms on the left-hand side.
- There are 2 hydrogen atoms in the \(2\text{HIO}_3\) on the right-hand side, meaning 8 hydrogen atoms are needed for water. This requires \(4\text{H}_2\text{O}\).
- Check oxygen balance: Left side has 30 O. Right side has \((2 \times 3) + (10 \times 2) + 4 = 30\) O. The fully balanced equation is:
\(\text{I}_2 + 10\text{HNO}_3 \rightarrow 2\text{HIO}_3 + 10\text{NO}_2 + 4\text{H}_2\text{O}\)

M1: Identify the correct changes in oxidation number: iodine increases by 5 (or 10 for \(\text{I}_2\)) and nitrogen decreases by 1 (1 mark)
M2: Deduce the 1:10 mole ratio between \(\text{I}_2\) and \(\text{HNO}_3\) to balance the electron transfer (1 mark)
M3: Provide the fully balanced equation: \(\text{I}_2 + 10\text{HNO}_3 \rightarrow 2\text{HIO}_3 + 10\text{NO}_2 + 4\text{H}_2\text{O}\) (1 mark)

1. Calculate the amount in moles of \(\text{H}_2\text{SO}_4\) used in the titration:
\(n(\text{H}_2\text{SO}_4) = \text{concentration} \times \text{volume} = 0.125\text{ mol dm}^{-3} \times \frac{18.40}{1000}\text{ dm}^3 = 2.30 \times 10^{-3}\text{ mol}\)

2. Use the stoichiometric ratio from the balanced equation (\(2\text{NaOH} : 1\text{H}_2\text{SO}_4\)) to find the moles of \(\text{NaOH}\):
\(n(\text{NaOH}) = 2 \times 2.30 \times 10^{-3}\text{ mol} = 4.60 \times 10^{-3}\text{ mol}\)

3. Calculate the concentration of the sodium hydroxide solution:
\(c(\text{NaOH}) = \frac{\text{moles}}{\text{volume}} = \frac{4.60 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.184\text{ mol dm}^{-3}\)

M1: Calculate the moles of \(\text{H}_2\text{SO}_4 = 2.30 \times 10^{-3}\text{ mol}\) (1 mark)
M2: Calculate the moles of \(\text{NaOH} = 4.60 \times 10^{-3}\text{ mol}\) (consequential on M1) (1 mark)
M3: Calculate the concentration of \(\text{NaOH} = 0.184\text{ mol dm}^{-3}\) (consequential on M2, must be 3 significant figures) (1 mark)

1. The third ionisation energy is represented by the equation showing the removal of one mole of electrons from one mole of gaseous \(\text{Al}^{2+}\) ions:
\(\text{Al}^{2+}(\text{g}) \rightarrow \text{Al}^{3+}(\text{g}) + \text{e}^-\)

2. The electron configuration of aluminium is \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^1\).
- The third electron is removed from the outer \(3\text{s}\) subshell.
- The fourth electron is removed from the inner \(2\text{p}\) subshell.

3. The \(2\text{p}\) subshell is in a lower principal quantum shell closer to the nucleus, experiencing significantly less shielding and a much stronger electrostatic attraction to the positive nucleus. This requires a much larger amount of energy to overcome.

M1: Write the correct equation with state symbols: \(\text{Al}^{2+}(\text{g}) \rightarrow \text{Al}^{3+}(\text{g}) + \text{e}^-\) (1 mark)
M2: Explain that the fourth electron is removed from an inner shell / \(2\text{p}\) subshell, whereas the third is removed from the outer shell / \(3\text{s}\) subshell (1 mark)
M3: Explain that this inner shell experiences significantly less shielding / is closer to the nucleus, resulting in a much stronger electrostatic attraction (1 mark)

1. Chlorine is a stronger oxidising agent (more reactive) than iodine, so it oxidises iodide ions in solution to elemental iodine.
2. The ionic equation for this displacement reaction is:
\(\text{Cl}_2(\text{g}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{I}_2(\text{aq})\)
3. Elemental iodine dissolving in aqueous potassium iodide solution forms a brown (or dark yellow/orange) solution.

M1: State that the solution turns brown / dark yellow / orange-brown (1 mark)
M2: Write the correct species for the ionic equation: \(\text{Cl}_2 + 2\text{I}^- \rightarrow 2\text{Cl}^- + \text{I}_2\) (1 mark)
M3: Provide correct state symbols for all species: \(\text{Cl}_2(\text{g}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Cl}^-(\text{aq}) + \text{I}_2(\text{aq})\) (allow \(\text{Cl}_2(\text{aq})\)) (1 mark)

Step 1: Calculate the moles of hydrogen gas produced. \( n(\text{H}_2) = \frac{\text{Volume in cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = \frac{90.0}{24000} = 0.00375\text{ mol} \). Step 2: Determine the moles of the Group 2 metal, \( \text{M} \). From the equation: \( \text{M(s)} + 2\text{HCl(aq)} \rightarrow \text{MCl}_2\text{(aq)} + \text{H}_2\text{(g)} \), the molar ratio of \( \text{M} : \text{H}_2 \) is \( 1 : 1 \). Therefore, \( n(\text{M}) = 0.00375\text{ mol} \). Step 3: Calculate the relative atomic mass of \( \text{M} \) and identify the metal. \( A_r = \frac{\text{mass}}{n} = \frac{0.150\text{ g}}{0.00375\text{ mol}} = 40.0 \). Looking at the Periodic Table, the Group 2 metal with a relative atomic mass closest to \( 40.0 \) is calcium (\( \text{Ca} \)).

M1: Calculates moles of hydrogen gas as \( 0.00375\text{ mol} \) (1)
M2: Calculates \( A_r \) of \( \text{M} \) as \( 40.0 \) (1)
M3: Identifies the metal as calcium / \( \text{Ca} \) (1)

Step 1: Write down the electronic configurations of the outer shell. Phosphorus (P) has the configuration \( [Ne] 3s^2 3p^3 \). Sulfur (S) has the configuration \( [Ne] 3s^2 3p^4 \). Step 2: Describe the distribution of electrons in the \( 3p \) subshell. In phosphorus, the three \( 3p \) electrons occupy three separate \( 3p \) orbitals singly (half-filled subshell). In sulfur, there are four \( 3p \) electrons, meaning two electrons must pair up in one of the \( 3p \) orbitals. Step 3: Explain the effect of electron pairing on ionization energy. The pairing of electrons in the same orbital results in mutual repulsion (spin-pair repulsion). This repulsion makes it easier to remove one of these paired electrons from sulfur, requiring less energy than removing an unpaired electron from phosphorus.

M1: Identifies the outer subshell configurations: P is \( 3p^3 \) and S is \( 3p^4 \) (or full configurations / orbital diagrams showing electron arrangement) (1)
M2: States that sulfur has paired electrons in a \( 3p \) orbital, whereas phosphorus has only singly occupied \( 3p \) orbitals (1)
M3: Explains that mutual repulsion / spin-pair repulsion between the paired electrons in sulfur's orbital makes the electron easier to remove (1)

**(a) Ionic equation:**
\(\text{CaCO}_3\text{(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)

**(b) Practical steps to ensure homogeneity:**
1. Insert the stopper into the neck of the volumetric flask securely.
2. Invert the flask several times and shake/mix thoroughly.

**(c) Calculation:**
1. Calculate the moles of \(\text{NaOH}\) used in the titration:
\(n(\text{NaOH}) = \frac{18.20}{1000} \times 0.100 = 0.00182\text{ mol}\)

2. Moles of \(\text{HCl}\) in the 25.0 cm^{3} portion = 0.00182 mol (since the molar ratio of reacted HCl to NaOH is 1:1).

3. Calculate the moles of excess \(\text{HCl}\) in the 250.0 cm^{3} volumetric flask:
\(n(\text{HCl})_{\text{excess}} = 0.00182 \times \frac{250.0}{25.0} = 0.0182\text{ mol}\)

4. Calculate the initial moles of \(\text{HCl}\) added to the sample:
\(n(\text{HCl})_{\text{initial}} = \frac{50.0}{1000} \times 1.00 = 0.0500\text{ mol}\)

5. Calculate the moles of \(\text{HCl}\) that reacted with \(\text{CaCO}_3\):
\(n(\text{HCl})_{\text{reacted}} = 0.0500 - 0.0182 = 0.0318\text{ mol}\)

6. Calculate the moles of \(\text{CaCO}_3\) in the sample:
Since \(\text{CaCO}_3\text{(s)} + 2\text{HCl}\text{(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\), the ratio is 1:2.
\(n(\text{CaCO}_3) = \frac{0.0318}{2} = 0.0159\text{ mol}\)

7. Calculate the mass of \(\text{CaCO}_3\):
\(\text{mass} = 0.0159 \times 100.1 = 1.59159\text{ g}\)

8. Calculate the percentage by mass:
\(\text{Percentage by mass} = \frac{1.59159}{2.15} \times 100 = 74.027... \approx 74.0\%\)

**Part (a) [2 marks]**
- **1 mark** for correct chemical formulas of reactants and products: \(\text{CaCO}_3\text{(s)} + 2\text{H}^+\text{(aq)} \rightarrow \text{Ca}^{2+}\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)
- **1 mark** for correct state symbols throughout (all must be correct).
*Accept* the full molecular equation: \(\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\) for a maximum of 1 mark if state symbols are correct.

**Part (b) [2 marks]**
- **1 mark** for stating to stopper / seal the volumetric flask.
- **1 mark** for stating to invert the flask multiple times / shake thoroughly.

**Part (c) [4 marks]**
- **1 mark** for calculating moles of \(\text{NaOH}\): \(0.100 \times 0.01820 = 0.00182\text{ mol}\).
- **1 mark** for scaling up to find excess moles of \(\text{HCl}\) in the volumetric flask: \(0.00182 \times 10 = 0.0182\text{ mol}\).
- **1 mark** for finding moles of \(\text{HCl}\) reacted: \(0.0500 - 0.0182 = 0.0318\text{ mol}\), and dividing by 2 to find moles of \(\text{CaCO}_3\): \(0.0159\text{ mol}\).
- **1 mark** for calculating the mass of \(\text{CaCO}_3\) and final percentage purity to 3 significant figures: \(1.59\text{ g}\) (or \(1.59159\text{ g}\)) and \(74.0\%\) (Allow TE for error in moles).

**(a) Halide ion and ionic equation:**
- The halide ion is bromide, \(\text{Br}^-\), because a cream precipitate is formed with acidified silver nitrate.
- Ionic equation: \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\)

**(b) Role of sulfuric acid and relative reducing powers:**
- Concentrated sulfuric acid acts as an **oxidising agent** (it oxidises \(\text{Br}^-\) to \(\text{Br}_2\)).
- Bromide ions are larger than chloride ions because they have more electron shells, resulting in a larger ionic radius.
- This means there is more shielding of the outer electrons from the nucleus, resulting in a weaker electrostatic attraction, making it easier for bromide ions to lose an electron (making bromide a stronger reducing agent than chloride).
- Therefore, bromide ions are strong enough to reduce the sulfur in concentrated sulfuric acid from +6 to +4 (in \(\text{SO}_2\)), whereas chloride ions are too weak as reducing agents to do this.

**(c) Metal cation and flame test:**
- (i) A brick-red flame indicates the presence of **calcium ions**, \(\text{Ca}^{2+}\). The formula of the anhydrous halide is \(\text{CaBr}_2\).
- (ii) Electrons absorb thermal energy from the flame and are excited/promoted to higher energy levels. As they relax/fall back to lower energy levels, they emit light/photons of specific energy/wavelength within the visible region (corresponding to brick-red).

**Part (a) [2 marks]**
- **1 mark** for identifying the halide ion as bromide / \(\text{Br}^-\).
- **1 mark** for the correct ionic equation with state symbols: \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\).

**Part (b) [3 marks]**
- **1 mark** for identifying the role of concentrated sulfuric acid as an oxidising agent.
- **1 mark** for stating that bromide ions are larger / have more shells / have greater shielding than chloride ions.
- **1 mark** for explaining that the attraction to outer electrons is weaker, making bromide a stronger reducing agent than chloride (capable of reducing sulfur from +6 to +4).

**Part (c)(i) [1 mark]**
- **1 mark** for identifying the metal cation as calcium / \(\text{Ca}^{2+}\) AND giving the formula \(\text{CaBr}_2\).

**Part (c)(ii) [2 marks]**
- **1 mark** for stating that electrons absorb thermal energy and are promoted / excited to higher energy levels.
- **1 mark** for explaining that as electrons fall back to lower energy levels, they emit light / photons of energy corresponding to the visible (brick-red) wavelength.

First, calculate the heat energy transferred to the water, \(q\):
\(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1\ \circ}\text{C}^{-1} \times 40.0^\circ\text{C} = 16720\text{ J} = 16.72\text{ kJ}\).
Next, calculate the amount in moles of propan-1-ol burned:
\(n = \frac{\text{mass}}{M_r} = \frac{0.600\text{ g}}{60.0\text{ g mol}^{-1}} = 0.0100\text{ mol}\).
Finally, calculate the enthalpy of combustion, \(\Delta H_c\):
\(\Delta H_c = -\frac{q}{n} = -\frac{16.72\text{ kJ}}{0.0100\text{ mol}} = -1672\text{ kJ mol}^{-1}\).
Rounding to three significant figures gives \(-1670\text{ kJ mol}^{-1}\).

1 mark: Correct calculation of heat and moles to arrive at option A.

A catalyst provides an alternative reaction pathway with a lower activation energy, so the activation energy \(E_a\) decreases. The Maxwell-Boltzmann distribution curve represents the distribution of molecular kinetic energies, which depends solely on temperature. Therefore, adding a catalyst does not change the shape or position of the distribution curve itself.

1 mark: Correctly identifies that the curve shape remains unchanged and activation energy decreases (Option B).

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

1 mark: Correct derivation of units to arrive at option B.

The reaction of a tertiary halogenoalkane (2-bromo-2-methylpropane) with aqueous sodium hydroxide to form an alcohol (2-methylpropan-2-ol) is a nucleophilic substitution reaction, where the hydroxide ion (\(\text{OH}^-\)) acts as a nucleophile to substitute the bromine atom.

1 mark: Correctly identifies the reaction mechanism as nucleophilic substitution.

The structural isomers for \(\text{C}_4\text{H}_8\) are:
1. But-1-ene (alkene)
2. But-2-ene (alkene)
3. 2-methylpropene (alkene)
4. Cyclobutane (cyclic alkane)
5. Methylcyclopropane (cyclic alkane)
Note: cis-but-2-ene and trans-but-2-ene are stereoisomers, not structural isomers. Therefore, there are exactly 5 structural isomers.

1 mark: Correct count of 5 structural isomers.

The absorption at \(3350\text{ cm}^{-1}\) indicates the presence of an \(\text{O-H}\) bond (alcohol). The absorption at \(1645\text{ cm}^{-1}\) indicates a \(\text{C=C}\) double bond. The lack of peak at \(1700-1750\text{ cm}^{-1}\) confirms there is no \(\text{C=O}\) carbonyl group. Prop-2-en-1-ol containing both a \(\text{C=C}\) double bond and an \(\text{O-H}\) group satisfies all spectral requirements and has the molecular formula \(\text{C}_3\text{H}_6\text{O}\).

1 mark: Correctly matches the infrared bands to Prop-2-en-1-ol.

First, calculate the molar mass (\(M_r\)) of the hydrocarbon:
\(M_r = \frac{\text{mass}}{\text{moles}} = \frac{5.61\text{ g}}{0.100\text{ mol}} = 56.1\text{ g mol}^{-1}\).
Next, find the mass of carbon in one mole of the compound:
\(56.1\text{ g mol}^{-1} \times 0.857 = 48.1\text{ g}\).
Number of carbon atoms per molecule = \(\frac{48.1}{12.0} \approx 4\).
Remaining mass is due to hydrogen: \(56.1 - 48.1 = 8.0\text{ g}\).
Number of hydrogen atoms per molecule = \(\frac{8.0}{1.0} = 8\).
Therefore, the molecular formula is \(\text{C}_4\text{H}_8\).

1 mark: Correct calculation of molar mass and molecular formula.

Propan-1-ol contains an \(\text{-OH}\) group which allows it to form intermolecular hydrogen bonds. Hydrogen bonds are the strongest intermolecular forces. Propanal and propanone form permanent dipole-dipole forces, and propane only forms London forces, which are weaker. Thus, propan-1-ol requires the most energy to overcome intermolecular forces and has the highest boiling temperature.

1 mark: Correct selection of propan-1-ol.

In a Maxwell-Boltzmann distribution, temperature represents the average kinetic energy of the particles. An increase in temperature shifts the peak of the curve to the right (higher energy) and flattens it (lower peak), whilst keeping the area under the curve constant. Adding a catalyst provides an alternative pathway with a lower activation energy (\(E_\text{a}\)), but does not change the energies of the reacting molecules, so the shape of the Maxwell-Boltzmann curve remains unchanged.

[1 mark] Correctly identifies B as the only option that correctly describes both effects. Reject A and C because they state the catalyst shifts the curve. Reject D because it states temperature does not change the curve shape and that temperature lowers the activation energy.

The rate of hydrolysis of halogenoalkanes is determined by bond enthalpy rather than bond polarity. The \(\text{C}-\text{I}\) bond is the weakest (lowest bond enthalpy) compared to \(\text{C}-\text{Br}\) and \(\text{C}-\text{Cl}\) because iodine has a larger atomic radius, leading to a longer and weaker covalent bond. Therefore, 1-iodobutane reacts the fastest, forming a yellow precipitate of silver iodide (\(\text{AgI}\)).

[1 mark] Correctly identifies D as the option showing silver iodide is formed fastest due to the lower bond enthalpy of the \(\text{C}-\text{I}\) bond. Reject A and C because silver chloride is formed the slowest. Reject B because although silver iodide is formed fastest, the explanation involving bond polarity is incorrect.

In this nucleophilic substitution mechanism, the nucleophile is the hydroxide ion, OH^-. Curly arrows in mechanisms are used to represent the movement of a pair of electrons. The attack of the nucleophile occurs at the carbon atom bonded to bromine because the C-Br bond is highly polar. Bromine is more electronegative than carbon, which creates a partial positive charge (delta +) on the carbon atom and a partial negative charge (delta -) on the bromine atom. The negatively charged hydroxide ion (nucleophile) is attracted to this electron-deficient carbon atom.

M1: Identifies hydroxide ion (OH^-) as the nucleophile (1 mark). M2: States that curly arrows represent the movement of a pair of electrons (1 mark). M3: Explains that the C-Br bond is polar due to electronegativity difference, creating a delta + on carbon and delta - on bromine (1 mark). M4: States that the nucleophile is attracted to the delta + / electron-deficient carbon (0.5 marks).

First, calculate the heat energy change (q) using q = m * c * delta T: q = 50.0 g * 4.18 J g^-1 K^-1 * 6.5 K = 1358.5 J = 1.3585 kJ. Next, calculate the amount in moles of CuSO4 used: n = mass / Mr = 2.00 g / 159.6 g mol^-1 = 0.01253 mol. Finally, calculate the molar enthalpy change of solution: delta H = -q / n = -1.3585 kJ / 0.01253 mol = -108.42 kJ mol^-1. Expressed to 3 significant figures, this is -108 kJ mol^-1.

M1: Calculates heat energy transferred correctly, q = 50.0 * 4.18 * 6.5 = 1358.5 J or 1.36 kJ (1 mark). M2: Calculates moles of CuSO4 correctly, n = 2.00 / 159.6 = 0.01253 mol (1 mark). M3: Divides energy by moles to find enthalpy change magnitude, 1.3585 / 0.01253 = 108.4 kJ mol^-1 (1 mark). M4: Applies correct negative sign (for exothermic reaction) and rounds to 3 significant figures to give -108 kJ mol^-1 (0.5 marks).

A catalyst provides an alternative reaction pathway with a lower activation energy (Ec) compared to the original activation energy (Ea). On a Maxwell-Boltzmann distribution curve, the activation energy line shifts to the left. This means that a larger fraction of the molecules now possess kinetic energy greater than or equal to this lower activation energy. Consequently, there is a greater frequency of successful collisions per unit time, which increases the rate of the reaction.

M1: States that a catalyst provides an alternative reaction pathway with a lower activation energy (1 mark). M2: Explains that on a Maxwell-Boltzmann distribution curve, the activation energy shifts to a lower value / to the left (1 mark). M3: Explains that a larger fraction or proportion of molecules have energy greater than or equal to this new activation energy (1 mark). M4: Concludes that this results in an increased frequency of successful collisions (0.5 marks).

To maximize the yield of propanal, propan-1-ol is heated gently with acidified potassium dichromate(VI) and the product is distilled off immediately. Propanal has a lower boiling temperature than propan-1-ol and propanoic acid because propanal molecules only experience permanent dipole-dipole forces and London forces, whereas propan-1-ol and propanoic acid can form stronger hydrogen bonds. Distilling the propanal as soon as it forms physically removes it from the reaction mixture, preventing it from remaining in contact with the oxidizing agent and being further oxidized to propanoic acid.

M1: States use of acidified potassium dichromate(VI) with distillation (or immediate distillation) (1 mark). M2: Identifies that propanal has a lower boiling temperature than propan-1-ol / propanoic acid (1 mark). M3: Explains that propanal has weaker intermolecular forces (dipole-dipole/London) than the hydrogen bonding in propan-1-ol/propanoic acid (1 mark). M4: States that distillation removes propanal from the reaction mixture, preventing further oxidation to propanoic acid (0.5 marks).

The strong, sharp absorption peak at 1715 cm^-1 corresponds to the C=O stretch of a carbonyl group. The absence of a broad absorption peak in the region 3200-3600 cm^-1 indicates that there is no O-H bond, which rules out an alcohol. This means compound X is either an aldehyde (propanal) or a ketone (propanone). In the mass spectrum, the major fragment peak at m/z = 43 corresponds to the acetyl ion, [CH3CO]^+, which is formed by the loss of a methyl group (mass of 15) from the molecular ion of propanone (m/z = 58). Therefore, compound X is propanone.

M1: Identifies the peak at 1715 cm^-1 as representing a C=O carbonyl group (1 mark). M2: Explains that the absence of a peak at 3200-3600 cm^-1 rules out an O-H group / alcohol (1 mark). M3: Identifies the m/z = 43 peak as the [CH3CO]^+ fragment (1 mark). M4: Correctly names compound X as propanone (0.5 marks).

According to Le Chatelier's principle, increasing the temperature of a system in equilibrium will cause the position of equilibrium to shift in the direction that absorbs heat (the endothermic direction) to oppose the temperature increase. Since the forward reaction is endothermic (dH = +57 kJ mol^-1), the equilibrium shifts to the right (producing more NO2). Because the concentration of the product (NO2) increases and the reactant (N2O4) decreases, the value of the equilibrium constant, Kc = [NO2]^2 / [N2O4], increases.

M1: States that the position of equilibrium shifts to the right (towards products) (1 mark). M2: Explains that the forward reaction is endothermic, so the system shifts to absorb the added heat (1 mark). M3: States that the value of Kc increases (1 mark). M4: Explains that Kc increases because the concentration of products increases relative to reactants (0.5 marks).

Although butane and propan-1-ol have very similar molar masses (and thus similar strength of London forces), they have different intermolecular forces. Butane is a non-polar hydrocarbon and only forms weak London forces (instantaneous dipole-induced dipole interactions) between its molecules. Propan-1-ol contains a polar -O-H group, which allows it to form strong hydrogen bonds between its molecules (in addition to London forces and permanent dipole-dipole forces). Hydrogen bonds are significantly stronger than London forces and require much more thermal energy to overcome, resulting in a much higher boiling temperature for propan-1-ol.

M1: Identifies that butane only has London forces between its molecules (1 mark). M2: Identifies that propan-1-ol has hydrogen bonds between its molecules due to the O-H group (1 mark). M3: States that hydrogen bonds are significantly stronger than London forces (1 mark). M4: Explains that more energy is required to overcome the stronger intermolecular forces in propan-1-ol, leading to a higher boiling temperature (0.5 marks).

First, find the moles of carbon: moles of CO2 = mass / Mr = 4.40 g / 44.0 g mol^-1 = 0.100 mol. Since each CO2 molecule contains one carbon atom, moles of C = 0.100 mol. Next, find the moles of hydrogen: moles of H2O = mass / Mr = 2.16 g / 18.0 g mol^-1 = 0.120 mol. Since each H2O molecule contains two hydrogen atoms, moles of H = 2 * 0.120 mol = 0.240 mol. Determine the simplest whole-number ratio of C to H by dividing by the smallest value: C : H = 0.100 / 0.100 : 0.240 / 0.100 = 1 : 2.4. Multiply both by 5 to obtain whole numbers: C : H = 5 : 12. Therefore, the empirical formula of the alkane is C5H12.

M1: Calculates the moles of carbon correctly, n(C) = 4.40 / 44.0 = 0.100 mol (1 mark). M2: Calculates the moles of hydrogen correctly, n(H) = 2 * (2.16 / 18.0) = 0.240 mol (1 mark). M3: Determines the ratio of C : H is 1 : 2.4 and scales to whole numbers (5 : 12) (1 mark). M4: States correct empirical formula as C5H12 (0.5 marks).

Calculate the energy transferred: \(q = m c \Delta T = 50.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ °C}^{-1} \times (45.6 - 20.2) \text{ °C} = 5308.6 \text{ J} = 5.3086 \text{ kJ}\). Calculate the moles of propan-1-ol burned: \(n = \frac{0.35 \text{ g}}{60.0 \text{ g mol}^{-1}} = 0.005833 \text{ mol}\). Calculate the enthalpy change: \(\Delta H_c = -\frac{5.3086 \text{ kJ}}{0.005833 \text{ mol}} = -910.05 \text{ kJ mol}^{-1}\). Rounded to 3 significant figures, the final value is \(-910 \text{ kJ mol}^{-1}\).

M1: Calculation of energy transferred, \(q = 5.31 \text{ kJ}\) [1 mark]. M2: Calculation of moles of propan-1-ol, \(n = 0.00583 \text{ mol}\) [1 mark]. M3: Division of energy by moles with a negative sign [1 mark]. M4: Final answer rounded to 3 significant figures with correct units: \(-910 \text{ kJ mol}^{-1}\) [0.5 marks].

In the nucleophilic substitution of 1-bromobutane: 1. The polar \(C-Br\) bond contains a partial positive carbon (\(C^{\delta+}\)) and a partial negative bromine (\(Br^{\delta-}\)). 2. A curly arrow is drawn from a lone pair of electrons on the hydroxide ion (\(OH^-\)) to the carbon atom bonded to the bromine. 3. Another curly arrow is drawn from the \(C-Br\) bond to the bromine atom, representing heterolytic fission. 4. The organic product formed is butan-1-ol, alongside a bromide ion.

M1: Correct identification or representation of dipoles (\(C^{\delta+}\) and \(Br^{\delta-}\)) on the carbon-halogen bond [1 mark]. M2: Curly arrow drawn from the lone pair of the oxygen on \(OH^-\). to the carbon atom [1 mark]. M3: Curly arrow drawn from the carbon-bromine bond to the bromine atom [1 mark]. M4: Identification of the organic product as butan-1-ol [0.5 marks].

The equilibrium constant expression is \(K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}\). According to Le Chatelier's principle, because the forward reaction is exothermic, increasing the temperature favors the endothermic reverse reaction to absorb the added thermal energy. Therefore, the position of equilibrium shifts to the left (reactants side), reducing the yields of products relative to reactants. As a result, the value of the equilibrium constant \(K_c\) decreases.

M1: Correct mathematical expression for \(K_c\) [1 mark]. M2: Clear statement that the equilibrium position shifts to the left [1 mark]. M3: Clear statement that the value of \(K_c\) decreases [1 mark]. M4: Explaining that the forward reaction is exothermic, so increased temperature favors the reverse endothermic reaction [0.5 marks].

When temperature increases: 1. The peak of the Maxwell-Boltzmann distribution shifts to the right (higher energy) and becomes lower. 2. The curve flattens out, but the area underneath remains the same as the total number of particles is constant. A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy (\(E_{cat}\)). On the distribution graph, this shifts the activation energy boundary to the left, which significantly increases the proportion of molecules that possess sufficient energy to react upon collision.

M1: State that the peak of the curve shifts to the right and becomes lower [1 mark]. M2: State that the area under the curve remains constant [1 mark]. M3: Explain that a catalyst provides an alternative pathway with a lower activation energy [1 mark]. M4: Explain that this results in a larger fraction of molecules having energy greater than or equal to this lower activation energy [0.5 marks].

The broad absorption peak at 2500-3300 cm\(^{-1}\) indicates an \(O-H\) stretch of a carboxylic acid, while the sharp peak at 1715 cm\(^{-1}\) indicates a carbonyl \(C=O\) stretch. Together, these confirm a carboxylic acid functional group. Since the molecular formula is \(C_3H_6O_2\), compound Y must be propanoic acid, \(CH_3CH_2COOH\). The fragment peak at m/z = 45 corresponds to the carboxyl cation, \([COOH]^+\) (mass: 12 + 16 + 16 + 1 = 45).

M1: Deduce the presence of a carboxylic acid group from the IR stretching frequencies (O-H and C=O) [1 mark]. M2: Draw or write the structural formula of propanoic acid (\(CH_3CH_2COOH\)) [1 mark]. M3: Identify the m/z = 45 fragment as \(COOH^+\) [1 mark]. M4: Include the positive charge on the fragment ion [0.5 marks].

Propan-1-ol (\(M_r = 60.0\)) contains an \(O-H\) bond, enabling it to form intermolecular hydrogen bonds. Hydrogen bonds are the strongest type of intermolecular force. Propanone (\(M_r = 58.0\)) contains a polar carbonyl group (\(C=O\)) but no hydrogen atom bonded directly to a highly electronegative atom, so its strongest intermolecular forces are permanent dipole-dipole forces. Because hydrogen bonds are significantly stronger than permanent dipole-dipole forces, much more thermal energy is needed to overcome these forces and boil propan-1-ol.

M1: Identify hydrogen bonding as the strongest intermolecular force in propan-1-ol [1 mark]. M2: Identify permanent dipole-dipole forces as the strongest intermolecular force in propanone [1 mark]. M3: State that hydrogen bonds require more energy to break than permanent dipole-dipole forces [1 mark]. M4: Reference that both compounds have similar London forces because they have similar molecular masses [0.5 marks].

Using the formula: \(\text{Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of all reactants}} \times 100\). Here, the desired product is ethyl ethanoate (\(M_r = 88.0\)). The reactants are ethanoic acid and ethanol. Total mass of reactants = \(60.0 + 46.0 = 106.0 \text{ g mol}^{-1}\). \(\text{Atom Economy} = \frac{88.0}{106.0} \times 100 = 83.0188...\%\). To 3 significant figures, this is 83.0%.

M1: State or use the correct expression for percentage atom economy [1 mark]. M2: Calculate the total mass of reactants as 106.0 [1 mark]. M3: Set up the correct division \(\frac{88.0}{106.0} \times 100\) [1 mark]. M4: Provide the final answer rounded to 3 significant figures: 83.0% [0.5 marks].

Part (a):
Heating under reflux allows the organic reaction mixture to be heated to its boiling point for a prolonged period without the volatile reactants, products, or solvent evaporating into the atmosphere (1 mark).
The vapor rises, condenses in the vertical condenser, and drips back down into the reaction flask to continue reacting (1 mark).

Part (b):
1. Transfer the reaction mixture to a separating funnel, allow the layers to settle, and run off the organic layer (1-bromobutane is denser than water and will form the lower layer) (1 mark).
2. Return the organic layer to the separating funnel and wash it with aqueous sodium hydrogencarbonate (or sodium carbonate) to neutralize any remaining acid. Invert the separating funnel and open the tap periodically to release pressure from the carbon dioxide gas produced (1 mark).
3. Separate the layers again, run off the organic layer, and wash it with water to remove any remaining inorganic salts (1 mark).
4. Transfer the wet organic layer to a conical flask and add an anhydrous inorganic drying agent, such as anhydrous calcium chloride (or magnesium sulfate / sodium sulfate). Swirl the flask and leave it to stand until the liquid changes from cloudy to completely clear (1 mark).
5. Decant or filter the dry liquid into a clean distillation flask and perform a final distillation, collecting the fraction that boils at the boiling point of 1-bromobutane (approximately \(101-104^\circ\text{C}\)) (1 mark).

Part (c):
- Moles of butan-1-ol used = \(\frac{10.0}{74.0} = 0.1351\text{ mol}\) (0.5 marks).
- Theoretical mass of 1-bromobutane = \(0.1351\text{ mol} \times 137.0\text{ g mol}^{-1} = 18.51\text{ g}\) (0.5 marks).
- Percentage yield = \(\frac{12.5}{18.51} \times 100 = 67.53\% \approx 67.5\%\) (0.5 marks).

Part (a) [Max 2 marks]:
- 1 mark: Prevents the loss of volatile reactants, products, or solvents / allows prolonged heating without boiling dry.
- 1 mark: Vapors condense and return to the reaction flask.

Part (b) [Max 5 marks]:
- 1 mark: Use of a separating funnel to separate the aqueous and organic layers (identifying 1-bromobutane as the lower layer).
- 1 mark: Washing with aqueous sodium hydrogencarbonate (or sodium carbonate) to neutralize acid impurities AND releasing pressure/gas.
- 1 mark: Washing with water / separating layers.
- 1 mark: Adding an anhydrous inorganic drying agent (e.g., \(\text{CaCl}_2\), \(\text{MgSO}_4\), or \(\text{Na}_2\text{SO}_4\)) until the liquid is clear.
- 1 mark: Simple distillation to collect the fraction boiling at the boiling point of 1-bromobutane (\(101-104^\circ\text{C}\)).

Part (c) [Max 1.5 marks]:
- 0.5 marks: Calculates moles of butan-1-ol as \(0.135\text{ mol}\).
- 0.5 marks: Calculates theoretical yield of 1-bromobutane as \(18.5\text{ g}\).
- 0.5 marks: Calculates percentage yield as \(67.5\%\) (must be given to 3 significant figures; accept \(67.6\%\) if rounded differently at intermediate steps).

Part (a):
- Place equal volumes of 1-chlorobutane, 1-bromobutane, and 1-iodobutane into three separate test tubes in a water bath to maintain a constant temperature (1 mark).
- Add ethanol to each tube. Ethanol acts as a mutual solvent (co-solvent) to allow the halogenoalkanes (which are insoluble in water) and the aqueous silver nitrate to mix and form a single phase (1 mark).
- Add aqueous silver nitrate (where water acts as the nucleophile for hydrolysis, and silver ions react with halide ions) to each tube and start a timer. Measure the time taken for a precipitate to appear in each test tube to compare the relative rates (1 mark).

Part (b):
- The rate of hydrolysis increases in the order: 1-chlorobutane < 1-bromobutane < 1-iodobutane (1-iodobutane reacts the fastest / precipitate appears first, 1-chlorobutane reacts the slowest / precipitate appears last) (1 mark).
- The carbon-halogen bond strength/bond enthalpy decreases down the group (C-Cl is the strongest, C-I is the weakest) (1 mark).
- Bond enthalpy is the dominant factor determining the rate of nucleophilic substitution. Even though the C-Cl bond is the most polar and should attract nucleophiles most easily, it is too strong to break easily compared to the less polar but much weaker C-I bond (1 mark).

Part (c):
- (i) \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\) (1 mark for correct formulae and state symbols).
- (ii) The precipitate of silver bromide is cream in colour (0.5 marks).
- To confirm its identity, add dilute ammonia solution: the cream precipitate remains insoluble / does not dissolve (0.5 marks).
- Then add concentrated ammonia solution: the cream precipitate dissolves to form a colorless solution (0.5 marks).

Part (a) [Max 3 marks]:
- 1 mark: Keep temperature constant (e.g., water bath) AND use equal volumes/amounts of the halogenoalkanes.
- 1 mark: Use ethanol as a solvent to allow the halogenoalkanes and aqueous silver nitrate to mix/dissolve.
- 1 mark: Measure the time taken for a precipitate to appear.

Part (b) [Max 3 marks]:
- 1 mark: Rate trend: 1-iodobutane > 1-bromobutane > 1-chlorobutane (or equivalent wording indicating iodine is fastest, chlorine is slowest).
- 1 mark: Clarifies that bond enthalpy decreases down Group 7 (C-I is the weakest bond, C-Cl is the strongest bond).
- 1 mark: Explains that bond enthalpy (not bond polarity) is the dominant factor that determines the rate of reaction.

Part (c)(i) [Max 1 mark]:
- 1 mark: \(\text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq}) \rightarrow \text{AgBr}(\text{s})\) (state symbols must be included and correct; reject if incorrect ionic species are shown).

Part (c)(ii) [Max 1.5 marks]:
- 0.5 marks: Cream precipitate.
- 1 mark: Does not dissolve / is insoluble in dilute ammonia AND dissolves / is soluble in concentrated ammonia.