2025 Edexcel IAS-Level Chemistry (XCH11) Practice Paper with Answers

First, calculate the mass of water lost: \( 5.72 \text{ g} - 2.12 \text{ g} = 3.60 \text{ g} \). Next, determine the number of moles of anhydrous \( \text{Na}_2\text{CO}_3 \): \( n(\text{Na}_2\text{CO}_3) = \frac{2.12 \text{ g}}{106.0 \text{ g mol}^{-1}} = 0.020 \text{ mol} \). Determine the number of moles of water lost: \( n(\text{H}_2\text{O}) = \frac{3.60 \text{ g}}{18.0 \text{ g mol}^{-1}} = 0.20 \text{ mol} \). Find the simplest molar ratio: \( x = \frac{0.20 \text{ mol}}{0.020 \text{ mol}} = 10 \). Thus, the formula of the hydrated salt is \( \text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O} \).

1 mark: Correctly identifies x = 10 (Option D).

The successive ionization energies show a massive increase between the third and fourth ionization energies (from 2745 to 11577 kJ/mol). This indicates that the fourth electron is being removed from an inner quantum shell closer to the nucleus, which experiences a much stronger electrostatic attraction. Therefore, the element has three valence electrons and belongs to Group 3 (13).

1 mark: Correctly identifies Group 3 (13) (Option C).

The oxygen atom in \( \text{H}_3\text{O}^+ \) has 6 valence electrons, minus 1 for the positive charge, leaving 5. It forms 3 covalent bonds with hydrogen atoms, using 3 electrons, leaving one lone pair. With 3 bonding pairs and 1 lone pair, the electron-pair geometry is tetrahedral, resulting in a trigonal pyramidal molecular shape. Due to lone pair-bonding pair repulsion being greater than bonding pair-bonding pair repulsion, the bond angle is reduced from the tetrahedral angle of \( 109.5^\circ \) to approximately \( 107^\circ \).

1 mark: Correctly identifies trigonal pyramidal shape and 107 degree bond angle (Option B).

Option A is the initiation step. Option B is incorrect because hydrogen radicals do not form during this reaction; instead, \( \text{HCl} \) and \( \text{CH}_3^\bullet \) are produced. Option C is a correct propagation step as it starts with a radical (\( \text{CH}_3^\bullet \)) and produces a new radical (\( \text{Cl}^\bullet \)) alongside the stable organohalogen product. Option D represents a termination step where two radicals combine to form a neutral molecule.

1 mark: Correctly identifies the propagation step (Option C).

For \( E/Z \) isomerism, each carbon in the double bond must be attached to two different groups. Hex-3-ene and pent-2-ene both satisfy this. However, hex-3-ene is symmetrical: \( \text{CH}_3\text{CH}_2\text{CH}=\text{CH}\text{CH}_2\text{CH}_3 \). Adding \( \text{HBr} \) across this symmetrical double bond yields only 3-bromohexane as the sole product. Pent-2-ene is asymmetrical and would yield a mixture of 2-bromopentane and 3-bromopentane. 2-methylbut-2-ene and but-1-ene do not exhibit \( E/Z \) isomerism.

1 mark: Correctly identifies hex-3-ene (Option B).

Atom economy is defined as \( \frac{\text{Molar mass of desired product}}{\text{Total molar mass of all reactants}} \times 100\% \). Here, the desired product is ethanol. Total mass of desired product = \( 2 \times 46.0 = 92.0 \text{ g mol}^{-1} \). Total mass of reactants = \( 180.0 \text{ g mol}^{-1} \). Atom economy = \( \frac{92.0}{180.0} \times 100\% = 51.1\% \).

1 mark: Correctly calculates the percentage atom economy to be 51.1% (Option B).

The given electronic configuration has \( 2+2+6+2+6 = 18 \) electrons (the argon configuration).
- Sulfur (Z = 16) has 16 electrons; \( \text{S}^+ \) has 15 electrons.
- Calcium (Z = 20) has 20 electrons; \( \text{Ca}^+ \) has 19 electrons.
- Scandium (Z = 21) has 21 electrons; \( \text{Sc}^{3+} \) has 18 electrons, giving it the configuration \( 1s^2 2s^2 2p^6 3s^2 3p^6 \).
- Iron (Z = 26) has 26 electrons; \( \text{Fe}^{3+} \) has 23 electrons.

1 mark: Correctly identifies \( \text{Sc}^{3+} \) (Option C).

According to Fajans' rules, covalent character in ionic compounds increases with a highly polarising cation. A cation's polarising power increases with higher positive charge and smaller ionic radius (high charge density). Among the options, \( \text{Al}^{3+} \) has the highest positive charge (+3) and a small ionic radius, giving it the highest polarising power. This causes the greatest distortion of the chloride anion's electron cloud, resulting in the highest degree of covalent character in the metal-halogen bond.

1 mark: Correctly identifies \( \text{AlCl}_3 \) (Option C).

First, calculate the number of moles of \(\text{CO}_2\) gas produced: \(n(\text{CO}_2) = \frac{336}{24000} = 0.0140\text{ mol}\). Since the molar ratio of \(M\text{CO}_3 : \text{CO}_2\) is \(1:1\), the moles of \(M\text{CO}_3\) reacted is also \(0.0140\text{ mol}\). Next, calculate the molar mass of \(M\text{CO}_3\): \(M_r = \frac{1.40}{0.0140} = 100\text{ g mol}^{-1}\). Subtracting the mass of the carbonate group: \(A_r(M) = 100 - (12.0 + 3 \times 16.0) = 40.0\text{ g mol}^{-1}\). This corresponds to Calcium.

Award 1 mark for correct selection B.

The largest increase in ionization energy is between the third and fourth values (2745 to 11578), indicating that the element has three valence electrons and belongs to Group 13. Therefore, \(X\) forms a \(3+\) ion, and the formula of its chloride is \(X\text{Cl}_3\).

Award 1 mark for correct selection C.

\(\text{SO}_3\) has a trigonal planar geometry with no lone pairs on the sulfur atom, meaning its bond angles are exactly \(120^\circ\). \(\text{H}_2\text{O}\) is bent (\(104.5^\circ\)), \(\text{SF}_6\) is octahedral (\(90^\circ\)), and \(\text{NF}_3\) is trigonal pyramidal (\(107^\circ\)).

Award 1 mark for correct selection B.

In a propagation step, a free radical reacts with a stable molecule to form a new radical and a new stable molecule. The equation \(\text{C}_2\text{H}_6 + \text{Cl}^\bullet \rightarrow \text{C}_2\text{H}_5^\bullet + \text{HCl}\) meets this definition.

Award 1 mark for correct selection C.

For stereoisomerism to occur in an alkene, both carbon atoms involved in the double bond must be attached to two different groups. In pent-2-ene, carbon-2 is bonded to hydrogen and a methyl group, while carbon-3 is bonded to hydrogen and an ethyl group. Thus, it can form \(E\) and \(Z\) isomers.

Award 1 mark for correct selection C.

Total mass of reactants = \(M_r(\text{Fe}_2\text{O}_3) + 3 \times M_r(\text{CO}) = 159.6 + 3 \times 28.0 = 243.6\text{ g mol}^{-1}\). Mass of desired product = \(2 \times A_r(\text{Fe}) = 111.6\text{ g mol}^{-1}\). Atom economy = \(\frac{111.6}{243.6} \times 100\% = 45.8\%\).

Award 1 mark for correct selection B.

Bond polarity is determined by the difference in electronegativity between the bonded atoms. Electronegativity increases across Period 2 in the order: \(\text{C} < \text{N} < \text{O} < \text{F}\). Since Fluorine is the most electronegative element, the electronegativity difference is largest for the \(\text{C}-\text{F}\) bond, making it the most polar.

Award 1 mark for correct selection D.

An iron atom has the electronic configuration \([\text{Ar}] 3d^6 4s^2\). Transition metals lose electrons from their \(4s\) orbital before the \(3d\) orbital when forming ions. Thus, to form \(\text{Fe}^{3+}\), two electrons are removed from \(4s\) and one from \(3d\), leaving \([\text{Ar}] 3d^5\).

Award 1 mark for correct selection A.

The largest jump in successive ionisation energies occurs between the fifth and sixth ionisation energies (from \( 6274 \) to \( 21269 \text{ kJ mol}^{-1} \)). This shows that the sixth electron is removed from an inner quantum shell that is closer to the nucleus and experiences significantly less shielding. Therefore, the element has 5 valence electrons in its outer shell, placing it in Group 15 (Group 5). In Period 3, the Group 15 element is phosphorus.

1 mark: Correctly identifies the element as phosphorus (C) by locating the massive increase in energy between the 5th and 6th ionisation energies.

First, calculate the mass of anhydrous calcium sulfate:
\( \text{Mass of CaSO}_4 = 20.97 - 18.25 = 2.72\text{ g} \)

Next, calculate the mass of water lost:
\( \text{Mass of H}_2\text{O} = 21.69 - 20.97 = 0.72\text{ g} \)

Now, convert masses to moles:
\( \text{Moles of CaSO}_4 = \frac{2.72\text{ g}}{136.2\text{ g mol}^{-1}} = 0.020\text{ mol} \)
\( \text{Moles of H}_2\text{O} = \frac{0.72\text{ g}}{18.0\text{ g mol}^{-1}} = 0.040\text{ mol} \)

Determine the simplest molar ratio:
\( \text{Ratio } \frac{\text{H}_2\text{O}}{\text{CaSO}_4} = \frac{0.040}{0.020} = 2 \)

Therefore, \( x = 2 \).

1 mark: Correctly calculates \( x = 2 \) (C) from the mass data.

The arrangement of electron pairs around the central atom depends on the total number of bonding pairs and lone pairs:
- In \( \text{NH}_4^+ \), nitrogen has 4 bonding pairs and 0 lone pairs. Total = 4 (tetrahedral arrangement).
- In \( \text{H}_3\text{O}^+ \), oxygen has 3 bonding pairs and 1 lone pair. Total = 4 (tetrahedral arrangement).
- In \( \text{BF}_4^- \), boron has 4 bonding pairs and 0 lone pairs. Total = 4 (tetrahedral arrangement).
- In \( \text{SF}_4 \), sulfur has 6 valence electrons + 4 electrons from fluorine bonds = 10 electrons in outer shell, which equals 4 bonding pairs and 1 lone pair. Total = 5 electron pairs, which is based on a trigonal bipyramidal arrangement (resulting in a seesaw shape).

1 mark: Correctly identifies \( \text{SF}_4 \) (D) as the species with five electron pairs around the central atom rather than four.

Butane is formed during the chlorination of ethane when two ethyl radicals collide and combine. Since this step involves two free radicals reacting to form a stable molecule with no new radicals produced, it is classified as a termination step:

\( \text{CH}_3\text{CH}_2^\bullet + \text{CH}_3\text{CH}_2^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3 \)

- Option A is a propagation step.
- Option C is a propagation step.
- Option D is a termination step, but it produces chloroethane, not butane.

1 mark: Correctly selects equation B as the termination step that produces butane.

(a)(i) Mass of carbon = 4.40 * (12.0 / 44.0) = 1.20 g. Mass of hydrogen = 2.70 * (2.0 / 18.0) = 0.30 g. Mass of oxygen = 2.30 - 1.20 - 0.30 = 0.80 g. (a)(ii) Moles of C = 1.20 / 12.0 = 0.10 mol; Moles of H = 0.30 / 1.0 = 0.30 mol; Moles of O = 0.80 / 16.0 = 0.05 mol. Dividing by the smallest number of moles (0.05): C = 2, H = 6, O = 1. Empirical formula is C2H6O. (a)(iii) The empirical formula mass of C2H6O = 2(12.0) + 6(1.0) + 16.0 = 46.0 g mol^{-1}. Since the molar mass is 46.0 g mol^{-1}, the molecular formula is C2H6O. (b) C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l). (c) Moles of A = 11.5 / 46.0 = 0.25 mol. From the equation, 1 mole of A produces 2 moles of CO2. So, moles of CO2 = 2 * 0.25 = 0.50 mol. Volume of CO2 = 0.50 * 24.0 = 12.0 dm^3.

(a)(i) [3 marks]: 1 mark for mass of C (1.20 g); 1 mark for mass of H (0.30 g); 1 mark for mass of O (0.80 g). (a)(ii) [2 marks]: 1 mark for correct mole calculations of elements; 1 mark for the empirical formula C2H6O. (a)(iii) [1 mark]: 1 mark for deducing C2H6O. (b) [2 marks]: 1 mark for balanced equation; 1 mark for all correct state symbols. (c) [2 marks]: 1 mark for calculating moles of A as 0.25 mol and moles of CO2 as 0.50 mol; 1 mark for the final volume of 12.0 dm^3.

(a)(i) Relative isotopic mass is the mass of an atom of an isotope compared with one-twelfth of the mass of an atom of carbon-12. (a)(ii) Let the fractional abundance of \(^{63}\text{Cu}\) be \(x\). Then \((63 \times x) + (65 \times (1 - x)) = 63.55\). This gives \(63x + 65 - 65x = 63.55 \Rightarrow -2x = -1.45 \Rightarrow x = 0.725\). Thus, abundance of \(^{63}\text{Cu}\) is 72.5% and \(^{65}\text{Cu}\) is 27.5%. (b)(i) Copper is a transition metal with an anomalous configuration: \(\text{1s}^2\text{2s}^2\text{2p}^6\text{3s}^2\text{3p}^6\text{3d}^{10}\text{4s}^1\). (b)(ii) When \(\text{Cu}^+\) is formed, the outer \(\text{4s}\) electron is lost first: \(\text{1s}^2\text{2s}^2\text{2p}^6\text{3s}^2\text{3p}^6\text{3d}^{10}\). (c)(i) Element X belongs to Group 3 (or Group 13). There is a very large increase from the third to the fourth ionization energy (\(2745\) to \(11577\text{ kJ mol}^{-1}\)), which indicates the fourth electron is being removed from a shell closer to the nucleus (an inner shell). (c)(ii) Phosphorus has electronic configuration \([\text{Ne}]\text{3s}^2\text{3p}^3\) with three singly occupied \(\text{3p}\) orbitals. Sulfur has \([\text{Ne}]\text{3s}^2\text{3p}^4\), with one pair of electrons in a \(\text{3p}\) orbital. The mutual repulsion between the paired electrons in the sulfur orbital makes it easier to remove one of them, resulting in a lower first ionization energy.

(a)(i) [2 marks]: 1 mark for mass of an isotope's atom; 1 mark for compared to 1/12th of carbon-12. (a)(ii) [2 marks]: 1 mark for setting up equation; 1 mark for both abundances correct. (b)(i) [1 mark]: Correct configuration of Cu (must show 3d10 4s1, not 3d9 4s2). (b)(ii) [1 mark]: Correct configuration of Cu+ (must show 3d10). (c)(i) [2 marks]: 1 mark for identifying Group 3 (or 13); 1 mark for linking this to the large jump between 3rd and 4th ionization energies. (c)(ii) [2 marks]: 1 mark for identifying that sulfur has a paired electron in the 3p orbital; 1 mark for explaining that the repulsion between these paired electrons reduces the energy required to remove one.

(a)(i) The nitrogen atom in ammonia has three bonding pairs and one lone pair of electrons. These electron pairs repel each other to get as far apart as possible. Since lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion, the bonding pairs are pushed closer together. The shape is trigonal pyramidal and the bond angle is \(107^\circ\) (accept \(107.5^\circ\)). (a)(ii) A dative covalent (or coordinate) bond is formed. The lone pair of electrons on the nitrogen atom of \(\text{NH}_3\) is donated to the empty 1s orbital of the hydrogen ion, \(\text{H}^+\). (a)(iii) The \(\text{NH}_4^+\) ion has four bonding pairs of electrons and no lone pairs around the central nitrogen atom. To minimize repulsion, these four pairs arrange themselves symmetrically. The shape is tetrahedral and the bond angle is \(109.5^\circ\). (b) Oxygen in water has 2 bonding pairs and 2 lone pairs, whereas nitrogen in ammonia has 3 bonding pairs and 1 lone pair. Lone pairs exert greater repulsive forces than bonding pairs. The presence of two lone pairs in water results in greater repulsion on the bonding pairs, squeezing the \(\text{H}-\text{O}-\text{H}\) bond angle further (to \(104.5^\circ\)) than the \(\text{H}-\text{N}-\text{H}\) bond angle (which is \(107^\circ\)).

(a)(i) [3 marks]: 1 mark for stating 3 bonding pairs and 1 lone pair; 1 mark for explaining that electron pairs repel to minimize repulsion (with lone pair repelling more); 1 mark for trigonal pyramidal shape and \(107^\circ\) (or \(107.5^\circ\)). (a)(ii) [2 marks]: 1 mark for identifying dative covalent / coordinate bond; 1 mark for explaining that the lone pair on nitrogen is donated to \(\text{H}^+\). (a)(iii) [2 marks]: 1 mark for tetrahedral shape; 1 mark for \(109.5^\circ\) bond angle. (b) [3 marks]: 1 mark for stating that water has 2 lone pairs and ammonia has 1 lone pair; 1 mark for stating that lone pairs repel more than bonding pairs; 1 mark for stating that this extra repulsion in water squeezes the bond angle to a smaller value.

(a)(i) A hydrocarbon is a compound consisting of hydrogen and carbon atoms only. (a)(ii) The longest carbon chain has 4 carbons (butane), and there are methyl groups on carbon 2 and carbon 3, so the name is 2,3-dimethylbutane. (a)(iii) 2,2-dimethylbutane has a main chain of 4 carbons (butane) with two methyl groups attached to the second carbon. (b)(i) Ultraviolet light provides the activation energy needed to cause homolytic fission of the chlorine-chlorine (\(\text{Cl}-\text{Cl}\)) bond, generating chlorine radicals. (b)(ii) Initiation step: \(\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}^\bullet\). First propagation step: \(\text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl}\). Second propagation step: \(\text{CH}_3^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet\). (b)(iii) Two methyl radicals collide and combine in a termination step: \(\text{CH}_3^\bullet + \text{CH}_3^\bullet \rightarrow \text{C}_2\text{H}_6\). (b)(iv) Radical substitution is difficult to control. The chloromethane product can undergo further propagation steps with chlorine radicals, leading to a mixture of further substituted products (dichloromethane, trichloromethane, tetrachloromethane), which must be separated and lowers the yield of chloromethane.

(a)(i) [1 mark]: 1 mark for hydrogen and carbon only. (a)(ii) [1 mark]: 1 mark for 2,3-dimethylbutane. (a)(iii) [1 mark]: 1 mark for identifying 4-carbon chain with two methyl groups on the second carbon. (b)(i) [1 mark]: 1 mark for homolytic fission of \(\text{Cl}-\text{Cl}\) bond / radical generation. (b)(ii) [3 marks]: 1 mark for initiation equation; 1 mark for first propagation equation; 1 mark for second propagation equation. (b)(iii) [1 mark]: 1 mark for correct termination equation producing ethane. (b)(iv) [2 marks]: 1 mark for stating that further substitution occurs; 1 mark for stating that a mixture of products is formed which requires separation.

(a)(i) Stereoisomerism occurs when molecules have the same structural formula but different spatial arrangements of their atoms. (a)(ii) In *Z*-but-2-ene, the high-priority methyl groups (\(-\text{CH}_3\)) are on the same side of the carbon-carbon double bond (cis-like). In *E*-but-2-ene, the methyl groups are on opposite sides of the double bond (trans-like). (a)(iii) For a molecule to show stereoisomerism, there must be restricted rotation about the \(\text{C}=\text{C}\) double bond, and each carbon atom of the double bond must be attached to two different groups. In but-2-ene, each double-bonded carbon is attached to a hydrogen atom and a methyl group (different). In but-1-ene, the terminal carbon atom of the double bond is bonded to two identical hydrogen atoms, preventing stereoisomerism. (b)(i) The mechanism consists of: 1. Electrophilic attack: A curly arrow starts from the middle of the \(\text{C}=\text{C}\) double bond and points to the partially positive hydrogen atom (\(\delta+\)) of the \(\text{H}-\text{Br}\) molecule. Simultaneously, a curly arrow starts from the \(\text{H}-\text{Br}\) bond and points to the \(\text{Br}\) atom (\(\delta-\)). 2. Intermediate formation: This produces a secondary carbocation intermediate (\(\text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_3\)) and a bromide ion (\(\text{Br}^-\)). 3. Nucleophilic attack: A curly arrow starts from a lone pair of electrons on the \(\text{Br}^-\) ion and points to the positively charged carbon atom of the carbocation, forming 2-bromobutane. (b)(ii) The type of mechanism is electrophilic addition.

(a)(i) [2 marks]: 1 mark for same structural formula; 1 mark for different spatial arrangement of atoms. (a)(ii) [2 marks]: 1 mark for stating methyl groups on same side in Z; 1 mark for stating methyl groups on opposite sides in E. (a)(iii) [2 marks]: 1 mark for identifying restricted rotation around the \(\text{C}=\text{C}\) bond; 1 mark for explaining that but-1-ene has one carbon with two identical hydrogen atoms. (b)(i) [3 marks]: 1 mark for correct first step curly arrows and dipoles on \(\text{H}-\text{Br}\); 1 mark for correct structure of secondary carbocation intermediate and \(\text{Br}^-\); 1 mark for curly arrow from \(\text{Br}^-\) lone pair to the carbocation. (b)(ii) [1 mark]: 1 mark for electrophilic addition.

(a)(i) Relative formula masses: \(\text{CaCO}_3 = 100.1\), \(\text{HNO}_3 = 63.0\), \(\text{Ca(NO}_3)_2 = 164.1\). Total mass of reactants = \(100.1 + 2 \times 63.0 = 226.1\text{ g mol}^{-1}\). Atom economy = \(\frac{164.1}{226.1} \times 100\% = 72.58\% \approx 72.6\%\). (a)(ii) Moles of \(\text{CaCO}_3 = \frac{5.00}{100.1} = 0.0500\text{ mol}\). Since the molar ratio of \(\text{CaCO}_3\) to \(\text{Ca(NO}_3)_2\) is 1:1, the theoretical moles of calcium nitrate = 0.0500 mol. Theoretical mass of calcium nitrate = \(0.0500 \times 164.1 = 8.205\text{ g}\). Percentage yield = \(\frac{6.89}{8.205} \times 100\% = 83.97\% \approx 84.0\%\). (b)(i) Concordant titres are those within \(0.10\text{ cm}^3\) of each other. These are Titration 2 (\(21.80\text{ cm}^3\)), Titration 3 (\(21.90\text{ cm}^3\)), and Titration 4 (\(21.80\text{ cm}^3\)). Mean titre = \(\frac{21.80 + 21.90 + 21.80}{3} = 21.83\text{ cm}^3\) (or \(21.833\text{ cm}^3\)). (b)(ii) Number of moles of \(\text{HCl}\) used = \(\text{concentration} \times \text{volume} = 0.100 \times \frac{21.83}{1000} = 2.183 \times 10^{-3}\text{ mol}\). The reaction ratio of \(\text{NaOH}\) to \(\text{HCl}\) is 1:1, so moles of \(\text{NaOH} = 2.183 \times 10^{-3}\text{ mol}\). Concentration of \(\text{NaOH} = \frac{\text{moles}}{\text{volume}} = \frac{2.183 \times 10^{-3}}{0.0250} = 0.0873\text{ mol dm}^{-3}\).

(a)(i) [2 marks]: 1 mark for correct total mass of reactants (226.1); 1 mark for correct atom economy calculation (72.6%). (a)(ii) [3 marks]: 1 mark for calculating moles of calcium carbonate (0.0500 mol); 1 mark for calculating theoretical mass of calcium nitrate (8.21 g); 1 mark for correct percentage yield calculation (84.0%). (b)(i) [2 marks]: 1 mark for identifying concordant titres (2, 3, and 4) and excluding the rough titre; 1 mark for correct calculation of mean titre (21.83 cm3). (b)(ii) [3 marks]: 1 mark for calculating moles of HCl (2.18 x 10^-3 mol); 1 mark for matching moles of NaOH in 25 cm3; 1 mark for calculating correct concentration of NaOH (0.0873 mol dm^-3).

The equation for the formation of propane is:
\(3C(s) + 4H_2(g) \rightarrow C_3H_8(g)\)

Using Hess's Law with standard enthalpy changes of combustion:
\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)
\(\Delta_f H^\theta = [3 \times \Delta_c H^\theta(C) + 4 \times \Delta_c H^\theta(H_2)] - \Delta_c H^\theta(C_3H_8)\)
\(\Delta_f H^\theta = [3(-394) + 4(-286)] - (-2220)\)
\(\Delta_f H^\theta = [-1182 - 1144] + 2220 = -2326 + 2220 = -106\text{ kJ mol}^{-1}\)

Therefore, the correct option is A.

1 mark for the correct answer (A).

Hydrogen fluoride (\(HF\)) has hydrogen bonding between its molecules, which is the strongest intermolecular force, giving it the highest boiling temperature. For the other hydrogen halides (\(HCl\), \(HBr\), and \(HI\)), the only significant intermolecular forces are London forces and permanent dipole-dipole forces. As the size of the halogen atom increases down the group, the number of electrons increases, leading to stronger London forces. Therefore, the boiling temperatures of these three increase in the order \(HCl < HBr < HI\). Combining these factors gives the overall order of increasing boiling temperature: \(HCl < HBr < HI < HF\).

1 mark for the correct answer (A).

Thermal stability of Group 2 carbonates increases down the group. The cation size increases (\(Ba^{2+}\) is larger than \(Mg^{2+}\)) while the charge remains \(+2\), so the charge density of the cation decreases down the group. A lower charge density means the cation has less polarizing power, meaning it polarises the carbonate anion (specifically weakening the C-O covalent bond) to a lesser extent. Thus, more thermal energy is needed to decompose the carbonate.

1 mark for the correct answer (A).

Iodide ions are strong reducing agents and reduce concentrated sulfuric acid to sulfur dioxide (\(SO_2\)), sulfur (\(S\), a yellow solid), and hydrogen sulfide (\(H_2S\), a gas with a rotten egg smell), whilst being oxidized to iodine (\(I_2\), purple vapour/dark grey solid). Option A is correct because both sulfur and hydrogen sulfide are standard reduction products observed in this reaction.

1 mark for the correct answer (A).

The rate of hydrolysis depends on the bond enthalpy of the carbon-halogen bond. The C-I bond is the weakest bond (lowest bond enthalpy) among the three, so it breaks most easily, making 1-iodobutane the fastest to hydrolyse. The iodide ions (\(I^-\)) released react with \(Ag^+\) ions to form a yellow precipitate of silver iodide (\(AgI\)).

1 mark for the correct answer (A).

The infrared spectrum of \(Y\) shows a strong, broad absorption band at \(3000\text{ cm}^{-1}\) (characteristic of an O-H carboxylic acid bond) and a strong, sharp absorption band at \(1715\text{ cm}^{-1}\) (characteristic of a C=O bond). This indicates that \(Y\) is a carboxylic acid. Carboxylic acids are formed by the oxidation of primary alcohols under reflux. Therefore, \(X\) must be a primary alcohol. Among the choices, propan-1-ol is a primary alcohol, whereas propan-2-ol is a secondary alcohol (oxidizes to a ketone), propanone is a ketone (cannot be oxidized further), and methylpropan-2-ol is a tertiary alcohol (cannot be oxidized).

1 mark for the correct answer (A).

Adding a catalyst provides an alternative reaction pathway with a lower activation energy, so the activation energy, \(E_a\), decreases. However, a catalyst does not change the temperature of the system, meaning the kinetic energies of the molecules are unaffected, and the Maxwell-Boltzmann distribution curve remains in the same position (the peak position is unchanged).

1 mark for the correct answer (A).

The reaction is exothermic in the forward direction (\(\Delta H = -92\text{ kJ mol}^{-1}\)). According to Le Chatelier's principle, decreasing the temperature shifts the equilibrium in the exothermic (forward) direction to oppose the decrease in temperature, thereby increasing the yield of ammonia. Furthermore, for an exothermic reaction, decreasing temperature increases the value of the equilibrium constant, \(K_c\). Note that while increasing pressure increases the yield of ammonia, it has no effect on the value of \(K_c\).

1 mark for the correct answer (A).

A molecule has a non-zero dipole moment if its polar bonds are arranged asymmetrically so that their dipoles do not cancel out. In \(\text{CF}_4\) (tetrahedral), \(\text{BF}_3\) (trigonal planar), and \(\text{SF}_6\) (octahedral), the bonds are arranged symmetrically, so the individual bond dipoles cancel, resulting in a net dipole moment of zero. \(\text{SF}_4\) has a see-saw molecular geometry because the sulfur atom has four bonding pairs and one lone pair. This asymmetrical arrangement prevents the bond dipoles from canceling, making it polar.

1 mark: Correct option C selected.
0 marks: Any other option selected.

The reaction involves breaking one C-H bond and one Cl-Cl bond:
Energy required to break bonds = \(413 + 242 = +655\text{ kJ mol}^{-1}\).

The reaction involves forming one C-Cl bond and one H-Cl bond:
Energy released in forming bonds = \(346 + 432 = +778\text{ kJ mol}^{-1}\).

Enthalpy change = Energy of bonds broken - Energy of bonds formed = \(655 - 778 = -123\text{ kJ mol}^{-1}\).

1 mark: Correct option A selected.
0 marks: Any other option selected.

As you go down Group 2:
- The solubility of Group 2 hydroxides increases.
- The thermal stability of Group 2 nitrates and carbonates increases because the cation radius increases, reducing its charge density and polarizing effect on the anion.
- The solubility of Group 2 sulfates decreases.
- The first ionisation energy decreases due to increased atomic radius and shielding.

Therefore, statement C is the only correct statement.

1 mark: Correct option C selected.
0 marks: Any other option selected.

The rate of hydrolysis of halogenoalkanes depends on two factors:
1. The structure of the halogenoalkane: tertiary halogenoalkanes react the fastest via the \(\text{S}_\text{N}1\) mechanism due to the stability of the tertiary carbocation intermediate.
2. The identity of the halogen: iodoalkanes react faster than chloroalkanes because the C-I bond is much weaker than the C-Cl bond, requiring less energy to break.

Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts the fastest.

1 mark: Correct option D selected.
0 marks: Any other option selected.

According to Le Chatelier's principle:
1. Since the forward reaction is exothermic (\(\Delta H < 0\)), lowering the temperature shifts the equilibrium position to the right (the exothermic direction) to release heat, increasing the yield of \(\text{SO}_3\).
2. There are 3 moles of gaseous reactants and 2 moles of gaseous products. Increasing the pressure shifts the equilibrium position to the side with fewer gas moles (the right), increasing the yield of \(\text{SO}_3\).

Therefore, low temperature and high pressure will maximize the equilibrium yield.

1 mark: Correct option C selected.
0 marks: Any other option selected.

- The molecular formula is \(\text{C}_3\text{H}_6\text{O}\). Propan-1-ol has the formula \(\text{C}_3\text{H}_8\text{O}\), so option A is incorrect.
- The infrared spectrum shows a strong absorption at \(1720\text{ cm}^{-1}\), indicating a carbonyl group (\(\text{C=O}\)), and no broad peak in the range \(3200\text{--}3750\text{ cm}^{-1}\), showing that there is no alcohol hydroxyl group (\(\text{O-H}\)). This rules out prop-2-en-1-ol (which would have an \(\text{O-H}\) absorption), so option D is incorrect.
- The organic compound is oxidized by acidified potassium dichromate(VI) (indicated by the color change from orange to green). Aldehydes (such as propanal) are easily oxidized to carboxylic acids, whereas ketones (such as propanone) are resistant to oxidation.

Therefore, the compound must be propanal.

1 mark: Correct option B selected.
0 marks: Any other option selected.

Sodium iodide (\(\text{NaI}\)) is a strong reducing agent and reduces concentrated sulfuric acid to a mixture of products, including hydrogen sulfide, sulfur, and sulfur dioxide. During this redox reaction, iodide ions are oxidized to iodine (\(\text{I}_2\)), which is observed as a purple vapor.
- Sodium chloride and sodium fluoride are not strong enough reducing agents to reduce concentrated sulfuric acid; they only undergo acid-base reactions to form HF/HCl gases.
- Sodium bromide is oxidized to bromine, which is seen as a red-brown vapor.

1 mark: Correct option D selected.
0 marks: Any other option selected.

1. Calculate the total mass of the mixture:
\(m = 50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\).

2. Calculate the temperature change:
\(\Delta T = 25.0 - 18.5 = 6.5\text{ }^\circ\text{C}\).

3. Calculate the heat energy released:
\(q = m \cdot c \cdot \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 6.5\text{ }^\circ\text{C} = 2717\text{ J} = 2.717\text{ kJ}\).

4. Calculate the moles of water formed:
\(n(\text{H}_2\text{O}) = n(\text{HCl}) = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\).

5. Calculate the enthalpy change of neutralization:
\(\Delta H = -\frac{q}{n} = -\frac{2.717\text{ kJ}}{0.0500\text{ mol}} = -54.34\text{ kJ mol}^{-1}\).

To 3 significant figures, this is \(-54.3\text{ kJ mol}^{-1}\).

1 mark: Correct option A selected.
0 marks: Any other option selected.

First, calculate the heat energy absorbed by the water using the formula:
\( q = m c \Delta T \)
\( q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ }^\circ\text{C}^{-1} \times 18.8\text{ }^\circ\text{C} = 3929.2\text{ J} = 3.9292\text{ kJ} \)

Next, calculate the number of moles of alcohol \( \text{X} \) burned:
\( n = \frac{0.150\text{ g}}{60.0\text{ g mol}^{-1}} = 0.00250\text{ mol} \)

Finally, calculate the enthalpy of combustion:
\( \Delta_c H = -\frac{q}{n} = -\frac{3.9292\text{ kJ}}{0.00250\text{ mol}} = -1571.68\text{ kJ mol}^{-1} \approx -1570\text{ kJ mol}^{-1} \) (to 3 significant figures).

[1] Correct option selected with supporting calculations showing the correct heat energy absorbed, the moles of fuel burned, and the standard division to find enthalpy change.

Barium carbonate is more thermally stable than magnesium carbonate because the \( \text{Ba}^{2+} \) ion has a much larger ionic radius and consequently a lower charge density than the smaller \( \text{Mg}^{2+} \) ion. Because of this lower charge density, the \( \text{Ba}^{2+} \) ion has a weaker polarizing effect on the adjacent carbonate ion, keeping the C-O covalent bonds within the carbonate ion relatively strong. Therefore, more thermal energy is required to decompose barium carbonate.

[1] Correct option selected identifying the relationship between ionic radius, charge density, polarization of the carbonate ion, and thermal stability.

The rate of hydrolysis is determined by two main factors:
1. The strength of the carbon-halogen bond: C-I is the weakest bond (due to the large atomic radius of iodine and poor orbital overlap), making it much easier to break than C-Br or C-Cl bonds.
2. The class of halogenoalkane: Tertiary halogenoalkanes hydrolyze significantly faster than primary halogenoalkanes because they react via the \( \text{S}_\text{N}1 \) mechanism, which goes through a relatively stable tertiary carbocation intermediate.
Thus, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts the fastest to produce a precipitate of silver iodide.

[1] Correct option selected relating halogenoalkane hydrolysis rate to both bond enthalpy (C-I is weakest) and classification (tertiary reacts fastest via the \( \text{S}_\text{N}1 \) mechanism).

According to Le Chatelier's principle:
- The forward reaction is endothermic (\( \Delta H > 0 \)), so increasing the temperature shifts the equilibrium in the endothermic direction (to the right) to absorb excess thermal energy.
- The right-hand side has 2 moles of gas while the left-hand side has only 1 mole of gas. Decreasing the pressure shifts the equilibrium in the direction of more gas moles (to the right) to restore the pressure.
Therefore, both increasing temperature and decreasing pressure shift the position of equilibrium to the right.

[1] Correct option selected applying Le Chatelier's principle to both temperature (endothermic) and pressure (moles of gas) effects.

(a)(i) \(q = m c \Delta T\)
\(m = 150.0\text{ g}\)
\(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\)
\(\Delta T = 48.2 - 19.5 = 28.7\ ^\circ\text{C}\)
\(q = 150.0 \times 4.18 \times 28.7 = 17994.9\text{ J} = 18.0\text{ kJ}\) (or \(17.99\text{ kJ}\))

(ii) Molar mass of propan-1-ol (\(\text{C}_3\text{H}_8\text{O}\)):
\(M_r = (3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0\text{ g mol}^{-1}\)
\(n = \frac{0.820}{60.0} = 0.01367\text{ mol}\)
\(\Delta_c H = -\frac{q}{n} = -\frac{17.9949}{0.01367} = -1316.38\text{ kJ mol}^{-1}\)
To 3 significant figures: \(-1320\text{ kJ mol}^{-1}\)

(b) Experimental reasons:
1. Heat loss to the surroundings / calorimeter.
2. Incomplete combustion of propan-1-ol.
3. Evaporation of propan-1-ol from the wick after extinguishing the burner before weighing.
(Any two for 2 marks)

(c) Combustion cycle equation:
\(3\text{C(s)} + 4\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \to \text{C}_3\text{H}_7\text{OH(l)}\)
\(\Delta_f H^\ominus[\text{C}_3\text{H}_7\text{OH(l)}] = 3 \times \Delta_c H^\ominus[\text{C(graphite)}] + 4 \times \Delta_c H^\ominus[\text{H}_2\text{(g)}] - \Delta_c H^\ominus[\text{C}_3\text{H}_7\text{OH(l)}]\)
\(\Delta_f H^\ominus = [3 \times (-394) + 4 \times (-286)] - (-2021)\)
\(\Delta_f H^\ominus = [-1182 - 1144] + 2021\)
\(\Delta_f H^\ominus = -2326 + 2021 = -305\text{ kJ mol}^{-1}\)

(a)(i)
- M1: Calculation of temperature change \(\Delta T = 28.7\ ^\circ\text{C}\) or use of correct values in \(q = m c \Delta T\) (1)
- M2: Value of \(q = 18.0\text{ kJ}\) or \(17.99\text{ kJ}\) (1)

(ii)
- M1: Calculation of moles of propan-1-ol: \(n = 0.01367\text{ mol}\) (1)
- M2: Divide heat energy in kJ by moles: \(\frac{17.99}{0.01367} = 1316\text{ kJ mol}^{-1}\) (1)
- M3: Give answer with negative sign and to 3 significant figures: \(-1320\text{ kJ mol}^{-1}\) (1)
[Allow ECF from (a)(i)]

(b)
- M1: Heat lost to surroundings / copper calorimeter (1)
- M2: Incomplete combustion of propan-1-ol / evaporation of alcohol from the wick (1)

(c)
- M1: Correct Hess cycle or algebraic expression for \(\Delta_f H\) (1)
- M2: Correct substitution of values: \([3 \times (-394) + 4 \times (-286)] - (-2021)\) (1)
- M3: Correct evaluation: \(-305\text{ kJ mol}^{-1}\) (1)
[If final answer is +305, award 2 marks]

(a)(i) Misty fumes: hydrogen chloride / \(\text{HCl(g)}\) (1)
Equation: \(\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \to \text{NaHSO}_4\text{(s)} + \text{HCl(g)}\) (1)
(Alternative: \(2\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \to \text{Na}_2\text{SO}_4\text{(s)} + 2\text{HCl(g)}\))

(ii) No elements change their oxidation state during the reaction (or oxidation states of all species remain unchanged). (1)

(b)(i) Gas X: hydrogen sulfide / \(\text{H}_2\text{S}\) (1)
Oxidation state of sulfur: \(-2\) (or II-) (1)

(ii) \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \to \text{H}_2\text{S} + 4\text{H}_2\text{O}\) (2)
(Accept: \(\text{SO}_4^{2-} + 10\text{H}^+ + 8\text{e}^- \to \text{H}_2\text{S} + 4\text{H}_2\text{O}\))

(iii) Any two from:
- Purple vapour (iodine gas)
- Grey/black solid (iodine solid)
- Yellow solid (sulfur)
- Heat evolved / mixture gets hot
- Effervescence / bubbling

(iv) Iodide ions are stronger reducing agents than chloride ions (because they are larger and lose their outer electrons more easily).

(a)(i)
- M1: Identify misty fumes as hydrogen chloride / \(\text{HCl}\) (1)
- M2: Balanced equation: \(\text{NaCl} + \text{H}_2\text{SO}_4 \to \text{NaHSO}_4 + \text{HCl}\) (or forming \(\text{Na}_2\text{SO}_4\)) (1)
[State symbols are not required but must be correct if used]

(ii)
- M1: No change in oxidation numbers of any elements / S is always +6, H is always +1, Cl is always -1, etc. (1)

(b)(i)
- M1: Gas X is hydrogen sulfide / \(\text{H}_2\text{S}\) (1)
- M2: Oxidation state is \(-2\) (or \(-2\) written clearly, do not accept just '2-') (1)

(ii)
- M1: Correct species: \(\text{H}_2\text{SO}_4\) and \(\text{H}^+\) on left, \(\text{H}_2\text{S}\) and \(\text{H}_2\text{O}\) on right (1)
- M2: Fully balanced half-equation with electrons: \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \to \text{H}_2\text{S} + 4\text{H}_2\text{O}\) (1)

(iii)
- M1: Observation 1: Purple vapour / grey-black solid (1)
- M2: Observation 2: Yellow solid / heat evolved (1)
[Accept any two valid, distinct observations]

(iv)
- M1: Iodide ions are stronger reducing agents than chloride ions / iodide ions can reduce sulfur from +6 to -2 whereas chloride ions cannot reduce sulfuric acid (1)

(a)(i) Mass of Oxygen = \(100 - 60.0 - 13.3 = 26.7\%\)
Calculating molar ratios:
Carbon: \(\frac{60.0}{12.0} = 5.0\)
Hydrogen: \(\frac{13.3}{1.0} = 13.3\)
Oxygen: \(\frac{26.7}{16.0} = 1.669\)
Dividing by the smallest value (1.669):
Carbon: \(\frac{5.0}{1.669} = 3.0\)
Hydrogen: \(\frac{13.3}{1.669} = 8.0\)
Oxygen: \(\frac{1.669}{1.669} = 1.0\)
Empirical formula is \(\text{C}_3\text{H}_8\text{O}\). (2 marks)

(ii) The molar mass of the empirical formula \(\text{C}_3\text{H}_8\text{O}\) is \((3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0\text{ g mol}^{-1}\).
Since \(m/z\) of the molecular ion is 60, the molecular formula is the same as the empirical formula, which is \(\text{C}_3\text{H}_8\text{O}\). (1 mark)

(b)(i) \(\text{O-H}\) bond / alcohol group / hydroxyl group. (1 mark)

(ii) Reagents: Acidified potassium dichromate(VI) (or acidified sodium dichromate(VI)) and heat. (1 mark)
Observation: Orange solution turns green. (1 mark)
Reaction type: Oxidation. (1 mark)

(c)(i) Since compound Y can be oxidized to Z (which is an aldehyde because it reacts with Tollens' reagent), Y must be a primary alcohol. Therefore, Y is propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\). (1 mark)
Compound W is the product of complete reflux oxidation, which is propanoic acid, \(\text{CH}_3\text{CH}_2\text{COOH}\). (1 mark)

(ii) Sodium propanoate: \(\text{CH}_3\text{CH}_2\text{COONa}\) (or \(\text{CH}_3\text{CH}_2\text{COO}^-\text{Na}^+\)). (1 mark)

(a)(i)
- M1: Determine percentage of oxygen as \(26.7\%\) and divide all percentages by relative atomic masses (1)
- M2: Obtain ratio of \(3 : 8 : 1\) to show empirical formula is \(\text{C}_3\text{H}_8\text{O}\) (1)

(ii)
- M1: Relate empirical mass (60) to molecular ion peak (60) to deduce \(\text{C}_3\text{H}_8\text{O}\) (1)

(b)(i)
- M1: Alcohol O-H bond / hydroxyl group (1) [Reject carboxylic acid O-H]

(ii)
- M1: Add acidified potassium dichromate(VI) (or sodium dichromate(VI) / \(\text{H}^+\) and \(\text{Cr}_2\text{O}_7^{2-}\)) and warm/heat (1)
- M2: Solution changes from orange to green (1)
- M3: Reaction type is oxidation (1)

(c)(i)
- M1: Y is propan-1-ol / \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\) (1)
- M2: W is propanoic acid / \(\text{CH}_3\text{CH}_2\text{COOH}\) (1)
[Accept displayed, structural, or skeletal formulas]

(ii)
- M1: \(\text{CH}_3\text{CH}_2\text{COONa}\) / \(\text{CH}_3\text{CH}_2\text{COO}^-\text{Na}^+\) (1)

(a) Diagram details:
- y-axis labelled: Number of molecules / fraction of molecules
- x-axis labelled: Energy (or Kinetic Energy)
- The curve starts at the origin, rises to a single peak, and approaches the x-axis at high energy but never touches it.
- \(E_a\) is a vertical line on the right side of the peak, and the area to the right of this line is shaded.

(b) Temperature \(T_2\) curve:
- The curve has a lower peak shifted to the right.
- It starts at the origin, crosses the \(T_1\) curve once, and is higher than the \(T_1\) curve to the right of \(E_a\).
Explanation of rate increase:
- At higher temperature, the average kinetic energy of molecules increases, so a greater fraction/number of molecules have energy \(\ge E_a\) (shaded area is larger for \(T_2\)).
- This results in a higher frequency of successful collisions (more successful collisions per unit time).

(c)(i) Temperature increase:
- Since the forward reaction is exothermic (\(\Delta H = -88\text{ kJ mol}^{-1}\)), increasing temperature shifts the position of equilibrium to the left (the endothermic direction) to oppose the increase in temperature.
- The value of \(K_c\) decreases.

(ii) Catalyst addition:
- A catalyst increases the rates of both the forward and reverse reactions equally, so there is no change in the position of equilibrium.
- The value of \(K_c\) remains unchanged.

(a)
- M1: Correctly labelled axes: y-axis as 'Number of molecules' and x-axis as 'Energy' (1)
- M2: Correct curve shape starting at origin, having one peak, and asymptoting to the x-axis, with \(E_a\) marked on the x-axis (1)
- M3: Correctly shaded area to the right of \(E_a\) (1)

(b)
- M1: Curve for \(T_2\) drawn with peak lower and further to the right than the peak of \(T_1\), crossing the \(T_1\) curve only once (1)
- M2: Explains that at higher temperature a greater number / fraction of molecules have energy greater than or equal to the activation energy (represented by the larger shaded area under the \(T_2\) curve to the right of \(E_a\)) (1)
- M3: State that there are more frequent successful collisions / more successful collisions per unit time (1)

(c)(i)
- M1: Position of equilibrium shifts to the left because the forward reaction is exothermic / system moves in the endothermic direction to absorb heat (1)
- M2: Value of \(K_c\) decreases (1)

(ii)
- M1: Position of equilibrium is unchanged because the catalyst increases the rate of forward and reverse reactions equally (1)
- M2: Value of \(K_c\) remains unchanged (1)

(a) Equation: \((CH_3)_3COH + HCl \rightarrow (CH_3)_3CCl + H_2O\) (or molecular form: \(C_4H_{10}O + HCl \rightarrow C_4H_9Cl + H_2O\)). 2-methylpropan-2-ol is a tertiary alcohol, which reacts via a stable tertiary carbocation intermediate. Butan-1-ol is a primary alcohol, which does not form a stable carbocation and must react via a slower \(S_N2\) pathway with a higher activation energy at room temperature.

(b)(i) Reagent: Saturated \(NaHCO_3\) (sodium hydrogencarbonate) or \(Na_2CO_3\) (sodium carbonate) solution. Relieving pressure: Invert the separating funnel and open the tap/stopcock periodically to vent the gas.

(b)(ii) Drying agent: Anhydrous sodium sulfate (\(Na_2SO_4\)), anhydrous magnesium sulfate (\(MgSO_4\)), or anhydrous calcium chloride (\(CaCl_2\)). Indication of dryness: The liquid turns from cloudy to completely clear, and the drying agent flows freely as a powder (no longer clumping together).

(c)(i) Ethanol acts as a co-solvent (mutual solvent) to dissolve both the polar aqueous silver nitrate and the non-polar halogenoalkanes so they can mix in a single phase and react.

(c)(ii) Observation: A cream (or off-white) precipitate forms in Tube 2. Ionic equation: \(Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)\).

(c)(iii) Order: 2-iodo-2-methylpropane > 2-bromo-2-methylpropane > 2-chloro-2-methylpropane. Explanation: The rate of hydrolysis depends on the carbon-halogen bond strength (bond enthalpy). The \(C-I\) bond is the longest and weakest (lowest bond enthalpy), so it breaks most easily, leading to the fastest rate. The \(C-Cl\) bond is the shortest and strongest (highest bond enthalpy), so it breaks least easily, leading to the slowest rate.

(d)(i) Step 1: Curly arrow starts from the C-Cl bond to the Cl atom. Show partial positive charge (\(\delta+\)) on C and partial negative charge (\(\delta-\)) on Cl. This forms a stable tertiary carbocation, \((CH_3)_3C^+\), and a chloride ion, \(Cl^-\). Step 2: Curly arrow starts from a lone pair on the oxygen of a water molecule (\(H_2O\)) and points to the positive carbon of the carbocation. This forms the protonated intermediate \((CH_3)_3C-O^+H_2\). Step 3: A curly arrow from one of the O-H bonds to the positive oxygen atom to show loss of a proton, forming the final tertiary alcohol \((CH_3)_3COH\) and \(H^+\). (Alternatively, if using hydroxide ion \(OH^-\), show curly arrow from the lone pair of \(OH^-\) directly to the carbocation to form the alcohol in two steps).

(d)(ii) Only one reactant molecule (2-chloro-2-methylpropane) is involved in the slow, rate-determining step (the ionization step).

(e)(i) The skeletal structure of 3-chloro-3-methylhexane is a 6-carbon chain (hexane). Carbon-3 is bonded to a chlorine atom (represented by a line to Cl) and a methyl group (represented by a single line). Carbon-3 is marked with an asterisk (*).

(e)(ii) The carbocation intermediate formed in the first step is planar around the positively charged carbon atom. The nucleophile can attack this planar carbocation from either side (above or below the plane) with equal probability. This yields an equimolar (50:50) mixture of the two optical isomers (enantiomers), which cancel out each other's optical activity, resulting in an inactive racemic mixture.

(a) [Total: 3 marks]
• M1: Correct reactant and product formulas in equation: \((CH_3)_3COH + HCl \rightarrow (CH_3)_3CCl + H_2O\) (1)
• M2: Identifies 2-methylpropan-2-ol as a tertiary alcohol and butan-1-ol as a primary alcohol (1)
• M3: Explains that tertiary carbocations are much more stable than primary carbocations (or that tertiary alcohols undergo rapid \(S_N1\) reaction due to steric hindrance preventing \(S_N2\)) (1)

(b)(i) [Total: 2 marks]
• M1: Correct chemical formula of reagent (e.g. \(NaHCO_3\) / \(Na_2CO_3\)) (1)
• M2: Invert the funnel and open the stopcock / tap periodically to release carbon dioxide gas pressure (1)

(b)(ii) [Total: 2 marks]
• M1: Suitable anhydrous salt (e.g. anhydrous \(Na_2SO_4\) / \(MgSO_4\) / \(CaCl_2\)) (1)
• M2: The liquid turns from cloudy to clear OR the solid powder swirls freely without clumping (1)

(c)(i) [Total: 1 mark]
• M1: Acts as a mutual solvent / co-solvent to allow the immiscible layers to mix and react (1)

(c)(ii) [Total: 2 marks]
• M1: Cream precipitate (or off-white precipitate) (1)
• M2: Correct ionic equation with state symbols: \(Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)\) (1)

(c)(iii) [Total: 3 marks]
• M1: Correct order: 2-iodo-2-methylpropane > 2-bromo-2-methylpropane > 2-chloro-2-methylpropane (or Tube 3 > Tube 2 > Tube 1) (1)
• M2: States that reactivity/rate depends on the strength / bond enthalpy of the carbon-halogen bond (and not bond polarity) (1)
• M3: Explains that C-I bond has the lowest bond enthalpy (is weakest) and breaks most easily, while C-Cl bond has the highest bond enthalpy (is strongest) and breaks least easily (1)

(d)(i) [Total: 4 marks]
• M1: Correct dipole on C-Cl bond (\(\delta+\)/\(\delta-\)) and curly arrow from the C-Cl bond to the Cl atom (1)
• M2: Correct carbocation intermediate structural formula showing positive charge on the central carbon (1)
• M3: Curly arrow from the lone pair of the oxygen on \(H_2O\) (or \(OH^-\)) to the positive central carbon (1)
• M4: Correctly showing the subsequent deprotonation step of \((CH_3)_3C-O^+H_2\) to form \((CH_3)_3COH\) (or direct formation if \(OH^-\) used) (1)

(d)(ii) [Total: 1 mark]
• M1: Explanation that only one species / molecule is involved in the rate-determining (slow) step (1)

(e)(i) [Total: 1 mark]
• M1: Correct skeletal formula of 3-chloro-3-methylhexane with an asterisk (*) on C3 (1)

(e)(ii) [Total: 3 marks]
• M1: Explains that the intermediate carbocation is planar (around the positive C+) (1)
• M2: States that attack by the nucleophile can occur with equal probability from either side / above or below (1)
• M3: Results in an equimolar / 50:50 mixture of enantiomers, which cancels out optical activity (1)

(a) Ionic equation: \text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}

(b) Polystyrene is a good thermal insulator with low heat capacity, reducing heat transfer to the surroundings. Modifications to minimise heat loss: Add a lid; wrap the polystyrene cup in cotton wool or place it inside a larger beaker.

(c) Plot a graph of temperature (y-axis) against time (x-axis). Draw a horizontal line of best fit through the points from 0 to 3 minutes, and a straight line of best fit through the cooling points from 5 to 9 minutes. Extrapolate both lines to 4 minutes and measure the vertical temperature difference between the two lines at \(t = 4\text{ minutes}\).

(d) Heat energy released, \(q = m \cdot c \cdot \Delta T\)
Mass of solution, \(m = 50.0\text{ cm}^3 \times 1.00\text{ g cm}^{-3} = 50.0\text{ g}\)
\(q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 18.0\text{ K} = 3762\text{ J} = 3.762\text{ kJ}\)

Determine limiting reagent:
Moles of \(\text{CuSO}_4 = 0.500\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.0250\text{ mol}\)
Moles of \(\text{Zn} = \frac{2.00\text{ g}}{65.4\text{ g mol}^{-1}} = 0.0306\text{ mol}\)
Since moles of \(\text{Zn} > \text{moles of CuSO}_4\), zinc is in excess and \(\text{Cu}^{2+}\) is the limiting reagent (\(0.0250\text{ mol}\) reacts).

Enthalpy change \(\Delta H = -\frac{q}{n} = -\frac{3.762\text{ kJ}}{0.0250\text{ mol}} = -150.48\text{ kJ mol}^{-1}\)
To 3 significant figures: \(-150\text{ kJ mol}^{-1}\).

**Part (a): [1 Mark]**
- \(\text{Zn(s)} + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu(s)}\) (1) [including correct state symbols].

**Part (b): [3 Marks]**
- Polystyrene is a good thermal insulator / has a low heat capacity (1)
- Add a lid to the cup (1)
- Place the cup inside a beaker / wrap the cup in cotton wool (1)

**Part (c): [3 Marks]**
- Plot temperature against time (1)
- Draw a horizontal line of best fit through the points before adding zinc (0-3 mins), and a straight line of best fit through the points during cooling (5-9 mins) (1)
- Extrapolate both lines to 4 minutes and find the vertical temperature difference (1)

**Part (d): [5.5 Marks]**
- Calculation of heat transferred: \(q = 50.0 \times 4.18 \times 18.0 = 3762\text{ J}\) or \(3.762\text{ kJ}\) (1)
- Moles of \(\text{Cu}^{2+}\): \(0.500 \times 0.0500 = 0.0250\text{ mol}\) (1)
- Checking that zinc is in excess: \(\text{moles of Zn} = 2.00 / 65.4 = 0.0306\text{ mol}\) which is \(> 0.0250\text{ mol}\) (1)
- Division of \(q\) by moles: \(\frac{3.762}{0.0250} = 150.48\) (1)
- Final value to 3 s.f. and negative sign: \(-150\) (0.5)
- Correct units: \(\text{kJ mol}^{-1}\) (1)

(a) A brick-red flame indicates the presence of calcium ions, \(\text{Ca}^{2+}\).

(b) (i) Dilute nitric acid is added to react with and remove any carbonate (\(\text{CO}_3^{2-}\)) or hydroxide (\(\text{OH}^-\)) impurities, which would otherwise form a precipitate with silver ions (e.g. \(\text{Ag}_2\text{CO}_3\) or \(\text{AgOH}\)) and interfere with the test.
(ii) A cream precipitate indicates the presence of bromide ions, \(\text{Br}^-\).
- Addition of dilute ammonia: The cream precipitate does not dissolve / remains insoluble.
- Addition of concentrated ammonia: The cream precipitate dissolves to form a colourless solution.
(iii) The anhydrous formula is \(\text{CaBr}_2\).

(c) Molar mass of \(\text{AgBr} = 107.9 + 79.9 = 187.8\text{ g mol}^{-1}\)
Moles of \(\text{AgBr} = \frac{3.756\text{ g}}{187.8\text{ g mol}^{-1}} = 0.0200\text{ mol}\)

Since the formula is \(\text{CaBr}_2 \cdot n\text{H}_2\text{O}\), 1 mole of the salt yields 2 moles of \(\text{Br}^-\), and therefore 2 moles of \(\text{AgBr}\).
Moles of \(\text{CaBr}_2 \cdot n\text{H}_2\text{O} = \frac{0.0200\text{ mol}}{2} = 0.0100\text{ mol}\)

Molar mass of \(\text{CaBr}_2 \cdot n\text{H}_2\text{O} = \frac{2.36\text{ g}}{0.0100\text{ mol}} = 236\text{ g mol}^{-1}\)

Molar mass of anhydrous \(\text{CaBr}_2 = 40.1 + (2 \times 79.9) = 199.9\text{ g mol}^{-1}\)

Mass of water of crystallisation per mole of hydrated salt:
\(236 - 199.9 = 36.1\text{ g mol}^{-1}\)

Number of water molecules, \(n = \frac{36.1}{18.0} = 2.01 \approx 2\).
Thus, the value of \(n\) is 2.

**Part (a): [1 Mark]**
- Calcium / \(\text{Ca}^{2+}\) (1) [Reject: Ca, calcium metal]

**Part (b): [6 Marks]**
- (i) To react with/remove carbonate/hydroxide impurities (1) which would otherwise precipitate with silver ions / form \(\text{Ag}_2\text{CO}_3\) or \(\text{AgOH}\) (1).
- (ii) Bromide / \(\text{Br}^-\)\) (1).
- Dilute ammonia: precipitate remains/insoluble (1).
- Concentrated ammonia: precipitate dissolves / forms a colourless solution (1).
- (iii) \(\text{CaBr}_2\) (1).

**Part (c): [5.5 Marks]**
- Molar mass of \(\text{AgBr} = 187.8\text{ g mol}^{-1}\) and moles of \(\text{AgBr} = \frac{3.756}{187.8} = 0.0200\text{ mol}\) (1)
- Moles of \(\text{CaBr}_2 \cdot n\text{H}_2\text{O} = \frac{0.0200}{2} = 0.0100\text{mol}\) (1.5) [1 mark for dividing by 2, 0.5 mark for correct value]
- Molar mass of hydrated salt \(= \frac{2.36}{0.0100} = 236\text{ g mol}^{-1}\) (1)
- Molar mass of anhydrous \(\text{CaBr}_2 = 199.9\text{ g mol}^{-1}\) (1)
- Deduce \(n = 2\) by dividing remaining mass (36.1) by 18.0 (1)

(a) (i) \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\) (or \(2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HBr}\))
(ii) The reaction of concentrated sulfuric acid with water/sodium bromide is highly exothermic. Cooling prevents the mixture from boiling violently and minimises the oxidation of bromide ions to toxic bromine gas. Shaking ensures uniform temperature and thorough mixing of reactants.

(b) Reflux diagram:
- Round-bottomed or pear-shaped flask with heat source (e.g. heating mantle or water bath) shown (1)
- Vertical Liebig condenser attached directly to the flask (1)
- Correct direction of water flow (water in at the bottom jacket, water out at the top jacket) (1)
- The top of the condenser must remain open to the atmosphere (no stopper) (0.5)

(c) (i)
- Add the mixture to the separating funnel, insert the stopper, and allow the layers to separate (1).
- The organic layer (1-bromobutane) is more dense than water, so it forms the lower layer. (Alternatively, add a small volume of water; the layer that increases in volume is the aqueous layer) (1).
- Invert the separating funnel periodically and open the tap with the spout pointing away from people to release built-up pressure (1).
- Remove the stopper and run off the lower organic layer into a clean container, keeping the upper aqueous layer in the funnel (1).

(d) (i) Anhydrous calcium chloride (\(\text{CaCl}_2\)) / anhydrous sodium sulfate (\(\text{Na}_2\text{SO}_4\)) / anhydrous magnesium sulfate (\(\text{MgSO}_4\)).
(ii) The liquid changes from cloudy to clear/transparent.

**Part (a): [3 Marks]**
- (i) \(\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HBr}\) (1) [Accept ionic equation or \(2\text{NaBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HBr}\)]
- (ii) Exothermic reaction / releases a lot of heat (1) and cooling prevents boiling over / reduces side reactions / prevents oxidation of bromide to bromine gas (1).

**Part (b): [3.5 Marks]**
- Pear-shaped/round-bottom flask with heating source (1)
- Vertical condenser fitted into the flask (1)
- Water in at the bottom and out at the top of the condenser jacket (1)
- System open at the top (no stopper) (0.5)

**Part (c): [4 Marks]**
- Pour mixture into separating funnel and allow layers to separate (1)
- Identify organic layer as lower layer due to higher density / or by adding water to see which layer expands (1)
- Invert and open tap periodically to release pressure (1)
- Remove stopper and run off lower organic layer (1)

**Part (d): [2 Marks]**
- (i) Anhydrous calcium chloride / anhydrous sodium sulfate / anhydrous magnesium sulfate (1) [Reject: anhydrous cobalt chloride, anhydrous copper(II) sulfate]
- (ii) Changes from cloudy to clear/transparent (1) [Reject: changes to colourless]

(a)
- Dissolve the weighed solid in about \(100\text{ cm}^3\) of deionised water in a beaker, stirring with a glass rod.
- Transfer the solution quantitatively through a funnel into a \(250\text{ cm}^3\) volumetric flask.
- Rinse the beaker, glass rod, and funnel with deionised water several times, and transfer all washings to the volumetric flask.
- Add deionised water to the flask until the bottom of the meniscus is aligned with the graduation mark (at eye level). Use a dropping pipette for the final drops.
- Stopper the flask and invert it several times to ensure the solution is completely homogeneous.

(b) (i) Colourless to pale pink.
(ii) Concordant titres are within \(\pm 0.10\text{ cm}^3\) of each other. Titrations 2, 3, and 4 are concordant.
Mean titre \(= \frac{25.40 + 25.45 + 25.35}{3} = 25.40\text{ cm}^3\).

(c)
- Moles of \(\text{NaOH}\) in the mean titre \(= 0.100\text{ mol dm}^{-3} \times 0.02540\text{ dm}^3 = 0.00254\text{ mol}\).
- Moles of \(\text{H}_2\text{A}\) in \(25.0\text{ cm}^3\) portion \(= \frac{0.00254}{2} = 0.00127\text{ mol}\).
- Moles of \(\text{H}_2\text{A}\) in \(250.0\text{ cm}^3\) standard solution \(= 0.00127\text{ mol} \times 10 = 0.0127\text{ mol}\).
- Molar mass of \(\text{H}_2\text{A} = \frac{1.50\text{ g}}{0.0127\text{ mol}} = 118.11\text{ g mol}^{-1}\).
To 3 significant figures: \(118\text{ g mol}^{-1}\).

**Part (a): [4.5 Marks]**
- Dissolve the solid in a beaker with a volume of deionised water less than \(250\text{ cm}^3\) and stir with a glass rod (1)
- Transfer the solution quantitatively to a \(250\text{ cm}^3\) volumetric flask (1)
- Rinse the beaker, rod, and funnel and add washings to the flask (1)
- Fill with deionised water until the bottom of the meniscus is on the graduation mark (1)
- Stopper and invert the flask multiple times to mix (0.5)

**Part (b): [3 Marks]**
- (i) Colourless to pale pink (1) [Reject: clear to pink, colourless to red/purple]
- (ii) Concordant titres: Titrations 2, 3, and 4 (1). Mean titre \(= 25.40\text{ cm}^3\) (1).

**Part (c): [5 Marks]**
- Moles of \(\text{NaOH} = \frac{25.40}{1000} \times 0.100 = 0.00254\text{ mol}\) (1)
- Moles of \(\text{H}_2\text{A}\) in \(25.0\text{ cm}^3 = 0.00127\text{ mol}\) (1)
- Moles of \(\text{H}_2\text{A}\) in \(250.0\text{ cm}^3 = 0.0127\text{ mol}\) (1)
- Molar mass calculation: \(\frac{1.50}{0.0127} = 118.11\) (1)
- Final answer to 3 s.f. with unit: \(118\text{ g mol}^{-1}\) (or \(\text{g/mol}\)) (1)