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(a) Calculate empirical formula: Carbon: \(85.7 / 12.0 = 7.14\). Hydrogen: \(14.3 / 1.0 = 14.3\). Divide by smallest: \(7.14 / 7.14 = 1\) and \(14.3 / 7.14 = 2\). Thus, the empirical formula is \(\text{CH}_2\). (b) Use the ideal gas equation \(pV = nRT\). Convert units: \(p = 102 \times 10^3\text{ Pa}\), \(V = 573 \times 10^{-6}\text{ m}^3\), \(T = 310\text{ K}\). Rearrange to find moles: \(n = pV / RT = (102 \times 10^3 \times 573 \times 10^{-6}) / (8.31 \times 310) = 0.0227\text{ mol}\). Calculate relative molecular mass: \(M_r = \text{mass} / n = 1.27 / 0.0227 = 56.0\text{ g mol}^{-1}\). (c) Find empirical mass: \(\text{CH}_2 = 12.0 + 2.0 = 14.0\). Multiplier: \(56.0 / 14.0 = 4\). Molecular formula is \(\text{C}_4\text{H}_8\). Complete combustion equation: \(\text{C}_4\text{H}_8 + 6\text{O}_2 \rightarrow 4\text{CO}_2 + 4\text{H}_2\text{O}\).

(a) M1: correct use of Ar values to find mole ratio (1 mark). M2: empirical formula CH2 (1 mark). (b) M1: correct unit conversions for p and V (1 mark). M2: correct rearrangement of ideal gas equation (1 mark). M3: calculated moles n = 0.0227 mol (1 mark). M4: correct calculation of Mr = 56.0 g mol^-1 (1.6 marks). (c) M1: ratio of molecular mass to empirical mass shown (1 mark). M2: molecular formula C4H8 (1 mark). M3: balanced equation for complete combustion (2 marks).

(a) Phosphorus has the electron configuration \(1s^2 2s^2 2p^6 3s^2 3p^3\) with singly-occupied 3p orbitals. Sulfur has the configuration \(1s^2 2s^2 2p^6 3s^2 3p^4\), where one 3p orbital contains a pair of electrons. The repulsion between these paired electrons in sulfur makes the outermost electron easier to remove. (b) Element Y shows a massive increase between the 3rd and 4th ionization energies (from \(2745\) to \(11578\text{ kJ mol}^{-1}\)). This indicates that the fourth electron is removed from a inner shell closer to the nucleus, meaning Y has three outer electrons. Thus, the element in Period 3 is Aluminium (Al). The equation for the second ionization energy is: \(\text{Al}^+(g) \rightarrow \text{Al}^{2+}(g) + e^-\).

(a) M1: Phosphorus 3p orbitals are singly occupied, Sulfur has one paired 3p orbital (1 mark). M2: Paired electrons in S repel each other (1 mark). M3: Repulsion decreases energy required to remove the electron (2 marks). (b) M1: Identify Y as Aluminium / Al (1 mark). M2: Point out the huge jump between 3rd and 4th ionization energies (1.6 marks). M3: Explain that this jump corresponds to removing an electron from a core/inner shell (2 marks). (c) M1: Balanced chemical equation with Al^+ and Al^2+ (2 marks). M2: Inclusion of (g) state symbols on both Al ions (1 mark).

(a) Calculate heat energy transferred: \(q = m \cdot c \cdot \Delta T = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 9.5\text{ K} = 1985.5\text{ J} = 1.986\text{ kJ}\). Calculate moles of \(\text{CuSO}_4\): \(n = 3.99 / 159.6 = 0.0250\text{ mol}\). Enthalpy of solution: \(\Delta H_{\text{soln}} = -q / n = -1.9855 / 0.0250 = -79.42\text{ kJ mol}^{-1}\) (or \(-79.4\text{ kJ mol}^{-1}\)). (b) Addition of water to anhydrous copper(II) sulfate initiates both hydration and dissolution simultaneously; it is physically impossible to measure the heat of hydration without some copper(II) sulfate dissolving. (c) Using a Hess Cycle: \(\Delta H_{\text{hydration}} = \Delta H_{\text{soln}}\text{[CuSO}_4\text{(s)]} - \Delta H_{\text{soln}}\text{[CuSO}_4\cdot5\text{H}_2\text{O(s)]}\). \(\Delta H_{\text{hydration}} = -79.4 - (+11.5) = -90.9\text{ kJ mol}^{-1}\).

(a) M1: calculates heat energy q = 1985.5 J or 1.986 kJ (1.6 marks). M2: calculates moles of copper(II) sulfate = 0.0250 mol (1 mark). M3: calculates molar enthalpy with negative sign: -79.4 kJ mol^-1 (3 marks). (b) M1: explains that dissolving and hydration happen simultaneously and cannot be isolated (3 marks). (c) M1: sets up correct Hess Cycle relationship (1.5 marks). M2: calculates hydration enthalpy = -90.9 kJ mol^-1 (allow consequential error from a) (1.5 marks).

(a) For \(\text{PF}_5\): P is in Group 15 (5 outer electrons) and forms 5 covalent bonds with F. Total = 5 bonding pairs, 0 lone pairs. Shape is trigonal bipyramidal. Bond angles are \(90^\circ\) and \(120^\circ\). For \(\text{XeF}_4\): Xe is in Group 18 (8 outer electrons) and forms 4 covalent bonds with F. Total = 4 bonding pairs, 2 lone pairs. To minimize repulsion, lone pairs are positioned opposite each other. Shape is square planar. Bond angles are \(90^\circ\). (b) Oxygen has a high electronegativity due to its small size and high nuclear charge, creating a highly polar \(\text{O-H}\) bond. This allows water molecules to form strong intermolecular hydrogen bonds. Sulfur is less electronegative than oxygen, so the \(\text{S-H}\) bond is not polar enough to form hydrogen bonds. Instead, \(\text{H}_2\text{S}\) is held together by weaker dipole-dipole forces, which require less energy to break.

(a) PF5: M1: 5 bonding pairs and 0 lone pairs (1 mark). M2: trigonal bipyramidal shape (1 mark). M3: angles of 90 and 120 degrees (1 mark). XeF4: M1: 4 bonding pairs and 2 lone pairs (1 mark). M2: square planar shape (1 mark). M3: angles of 90 degrees (1 mark). (b) M1: oxygen is highly electronegative (1 mark). M2: water molecules form hydrogen bonds (1.6 marks). M3: sulfur is less electronegative, H2S only forms weaker dipole-dipole forces (2 marks). M4: more energy required to break hydrogen bonds (1 mark).

(a) The redox reaction between sodium bromide and concentrated sulfuric acid is: \(2\text{NaBr} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{SO}_2 + \text{Br}_2 + 2\text{H}_2\text{O}\). Sulfuric acid acts as an oxidizing agent, converting bromide ions to bromine gas while being reduced to sulfur dioxide. (b) Observations for NaI products: sulfur is a yellow solid/precipitate; sulfur dioxide is a colorless, choking/pungent gas; hydrogen sulfide is a gas with a distinct rotten egg smell. (c) Add aqueous silver nitrate: chloride gives a white precipitate, bromide gives a cream precipitate, and iodide gives a yellow precipitate. Upon addition of dilute ammonia, the white precipitate dissolves but cream and yellow do not. Upon addition of concentrated ammonia, the cream precipitate dissolves but the yellow precipitate remains insoluble.

(a) M1: correct products of NaBr redox reaction (1 mark). M2: balanced equation (1 mark). M3: role identified as oxidizing agent (2 marks). (b) M1: yellow solid for sulfur (1 mark). M2: choking gas for sulfur dioxide (1.6 marks). M3: rotten egg smell for hydrogen sulfide (1 mark). (c) M1: white, cream, yellow precipitate colors identified correctly (1.5 marks). M2: dilute ammonia dissolves white precipitate only (1.5 marks). M3: concentrated ammonia dissolves cream precipitate, yellow is insoluble in both (1 mark).

(a) The equilibrium constant expression is: \(K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]}\). Units: \(\frac{\text{mol dm}^{-3}}{\text{mol dm}^{-3} \times \text{mol dm}^{-3}} = \text{dm}^3\text{ mol}^{-1}\). (b) Initial moles: \(\text{PCl}_3 = 1.50\), \(\text{Cl}_2 = 1.50\), \(\text{PCl}_5 = 0.00\). Change: \(-0.45\) for reactants, \(+0.45\) for products. Equilibrium moles: \(\text{PCl}_3 = 1.05\), \(\text{Cl}_2 = 1.05\), \(\text{PCl}_5 = 0.45\). Divide by volume \(5.00\text{ dm}^3\) to get concentrations: \([\text{PCl}_3] = 0.21\text{ mol dm}^{-3}\), \([\text{Cl}_2] = 0.21\text{ mol dm}^{-3}\), \([\text{PCl}_5] = 0.09\text{ mol dm}^{-3}\). Substitute into expression: \(K_c = 0.09 / (0.21 \times 0.21) = 2.04\text{ dm}^3\text{ mol}^{-1}\). (c) The forward reaction is exothermic. According to Le Chatelier's principle, increasing the temperature shifts the equilibrium in the endothermic direction (to the left) to absorb the heat. Consequently, the concentrations of reactants increase, and the concentration of the product decreases, resulting in a decrease in the value of \(K_c\).

(a) M1: correct expression for Kc (1 mark). M2: correct units dm^3 mol^-1 (1 mark). (b) M1: calculates equilibrium moles of reactants (1.05 mol) (1.6 marks). M2: divides by volume to get concentrations (1 mark). M3: correct substitution into formula (1 mark). M4: correct final value of 2.04 (2 marks). (c) M1: states that forward reaction is exothermic / backward is endothermic (1 mark). M2: equilibrium shifts left (1 mark). M3: Kc decreases (2 marks).

(a)
1. Calculate heat energy transferred, \(q\):
\(q = m c \Delta T\)
\(q = 150.0 \times 4.18 \times (45.7 - 19.5)\)
\(q = 150.0 \times 4.18 \times 26.2 = 16427.4\ J = 16.43\ kJ\)

2. Calculate moles of propan-1-ol burned:
\(n = \frac{\text{mass}}{M_r} = \frac{0.650}{60.0} = 0.01083\ mol\)

3. Calculate enthalpy change, \(\Delta H_c\):
\(\Delta H_c = -\frac{16.4274}{0.01083} = -1516.8\ kJ\ mol^{-1}\)
Rounding to 3 sig figs: \(-1520\ kJ\ mol^{-1}\) (Accept \(-1516\) to \(-1520\)).

(b)
- Heat loss to the surroundings / copper beaker.
- Incomplete combustion of the propan-1-ol.
- Loss of fuel by evaporation from the wick of the burner.

(c)
Write the combustion equation:
\(C_3H_7OH(l) + 4.5\ O_2(g) \rightarrow 3\ CO_2(g) + 4\ H_2O(l)\)

Using Hess's Law:
\(\Delta_c H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})\)
\(-2021 = [3 \times (-393.5) + 4 \times (-285.8)] - [\Delta_f H[C_3H_7OH(l)] + 0]\)
\(-2021 = [-1180.5 - 1143.2] - \Delta_f H[C_3H_7OH(l)]\)
\(-2021 = -2323.7 - \Delta_f H[C_3H_7OH(l)]\)
\
\Delta_f H[C_3H_7OH(l)] = -2323.7 + 2021 = -302.7\ kJ\ mol^{-1}\) (Accept \(-303\ kJ\ mol^{-1}\))

(a) [4 marks]
- M1: Correct temperature change calculation (\(26.2\ K\)) and substitution into \(q = mc\Delta T\) (yielding \(16427.4\ J\) or \(16.43\ kJ\)). [1 mark]
- M2: Correct calculation of moles of propan-1-ol (\(0.01083\ mol\)). [1 mark]
- M3: Correct division of \(q\) by moles. [1 mark]
- M4: Correct sign and final answer to 3 significant figures (\(-1520\ kJ\ mol^{-1}\)). [1 mark]

(b) [2 marks]
- M1 & M2: Any two sensible reasons: Heat loss to surroundings [1 mark]; Incomplete combustion [1 mark]; Evaporation of water/alcohol [1 mark]. Reject 'experimental error' or 'not standard conditions'.

(c) [4 marks]
- M1: Correct stoichiometric expression for combustion reaction of propan-1-ol (\(3\ CO_2 + 4\ H_2O\)). [1 mark]
- M2: Correct calculation of sum of product formation enthalpies (\(-2323.7\ kJ\ mol^{-1}\)). [1 mark]
- M3: Correct algebraic setup representing Hess's Law cycle. [1 mark]
- M4: Correct final evaluation with units: \(-302.7\ kJ\ mol^{-1}\). [1 mark]

(a)
Mechanism Name: Nucleophilic substitution (or \(S_N1\)).

Mechanism Outline:
1. Curly arrow from the C-Cl bond to the Cl atom.
2. Formation of a stable tertiary carbocation intermediate, \((CH_3)_3C^+\), and a chloride ion, \(Cl^-\).
3. Curly arrow from a lone pair on the hydroxide ion, \(:OH^-\), to the positive carbon of the carbocation.
4. Structure of the product: 2-methylpropan-2-ol.

(b)
- 1-iodobutane reacts faster (produces a precipitate faster).
- The C-I bond is weaker than the C-Br bond (has lower bond enthalpy).
- The C-I bond breaks more easily, resulting in a lower activation energy for the reaction.

(c)
Equation:
\(CH_3CH_2CH_2CH_2Br + 2NH_3 \rightarrow CH_3CH_2CH_2CH_2NH_2 + NH_4Br\)
(Accept: \(C_4H_9Br + 2NH_3 \rightarrow C_4H_9NH_2 + NH_4Br\))

Name of organic product: Butan-1-amine (or 1-aminobutane or butylamine).

(a) [4 marks]
- M1: Name of mechanism: Nucleophilic substitution. [1 mark]
- M2: Correct arrow from C-Cl bond to Cl. [1 mark]
- M3: Correct structure of intermediate carbocation \((CH_3)_3C^+\). [1 mark]
- M4: Correct arrow from lone pair on \(OH^-\) to positive carbon. [1 mark]

(b) [3 marks]
- M1: 1-iodobutane reacts faster. [1 mark]
- M2: Carbon-iodine bond is weaker / has lower bond enthalpy than carbon-bromine bond. [1 mark]
- M3: Therefore, the C-I bond breaks more easily / has a lower activation energy. [1 mark]

(c) [3 marks]
- M1: Correct reactant structures and stoichiometric coefficients (2 ammonia molecules). [1 mark]
- M2: Correct products (primary amine and ammonium bromide). [1 mark]
- M3: Correct IUPAC name: Butan-1-amine (or 1-aminobutane). [1 mark]

(a)
Sketch features:
- x-axis labeled as 'Energy' (or \(E\)), y-axis labeled as 'Number of molecules' (or \(N\) / 'Fraction of molecules').
- Curves start at the origin \((0,0)\), rise to a peak, and taper off asymptotically without touching the x-axis at high energy.
- \(T_2\) curve has its peak lower and shifted to the right of the \(T_1\) peak, and is higher than the \(T_1\) curve at high energy values.
- The activation energy, \(E_a\), is marked as a vertical line on the x-axis.

Explanation:
- At the higher temperature \(T_2\), a larger fraction / proportion of molecules have energy greater than or equal to the activation energy (represented by the larger area under the \(T_2\) curve to the right of \(E_a\)).
- Therefore, there is a higher frequency of successful collisions (more successful collisions per unit time), which increases the reaction rate.

(b)
Activation energy is the minimum energy required for a collision between reacting particles to result in a chemical reaction.

(c)
- A catalyst provides an alternative reaction pathway.
- This alternative pathway has a lower activation energy (\(E_c < E_a\)).
- Consequently, a larger fraction of molecules have energy greater than or equal to this lower activation energy, leading to more successful collisions per unit time.

(a) [6 marks]
- M1: Correctly labeled axes (y: number/fraction of molecules, x: energy). [1 mark]
- M2: Correct curve shape for \(T_1\) starting at origin and not touching x-axis at high energy. [1 mark]
- M3: Correct curve shape for \(T_2\) (lower peak, shifted to the right, crossing \(T_1\) only once). [1 mark]
- M4: Correctly labeled activation energy \(E_a\) on the horizontal axis. [1 mark]
- M5: Stating that at \(T_2\), more molecules have energy \(\ge E_a\) / larger area under curve to the right of \(E_a\). [1 mark]
- M6: Concluding that this leads to a higher frequency of successful collisions / more successful collisions per unit time. [1 mark]

(b) [1 mark]
- M1: Minimum energy needed for a reaction to occur upon collision. [1 mark]

(c) [3 marks]
- M1: Provides an alternative reaction pathway. [1 mark]
- M2: The alternative pathway has a lower activation energy. [1 mark]
- M3: More molecules have energy \(\ge E_c\) leading to a higher frequency of successful collisions. [1 mark]

(a)
- Major product: 2-bromopropane
- Minor product: 1-bromopropane

(b)
Mechanism Steps:
1. The hydrogen atom in \(H^{\delta+} - Br^{\delta-}\) is attracted to the electron-rich double bond of propene.
2. Curly arrow starts from the double bond \(C=C\) and points to the hydrogen atom of \(H-Br\).
3. Curly arrow from the \(H-Br\) bond to the bromine atom.
4. This yields a secondary carbocation intermediate, \(CH_3C^+HCH_3\), and a bromide ion, \(:Br^-\) with a lone pair and a negative charge.
5. Curly arrow from the lone pair on the bromide ion to the positively charged carbon of the carbocation.
6. Formation of 2-bromopropane.

(c)
- The major product (2-bromopropane) is formed via a secondary carbocation intermediate (\(CH_3C^+HCH_3\)).
- The minor product is formed via a primary carbocation intermediate (\(CH_3CH_2C^+H_2\)).
- A secondary carbocation is more stable than a primary carbocation due to the electron-releasing inductive effect of two alkyl groups (instead of only one in the primary carbocation), reducing the positive charge density on the carbon.

(d)
Geometric / E-Z isomerism.

(a) [2 marks]
- M1: Major: 2-bromopropane. [1 mark]
- M2: Minor: 1-bromopropane. [1 mark]

(b) [4 marks]
- M1: Dipole shown correctly on \(H^{\delta+}-Br^{\delta-}\) and arrow from \(C=C\) double bond to \(H\). [1 mark]
- M2: Arrow from \(H-Br\) bond to \(Br\). [1 mark]
- M3: Correct structure of secondary carbocation intermediate with positive charge on the middle carbon. [1 mark]
- M4: Arrow from lone pair on \(:Br^-\) to positive carbon. [1 mark]

(c) [3 marks]
- M1: Identify major proceeds via secondary carbocation and minor via primary carbocation. [1 mark]
- M2: State that the secondary carbocation is more stable than the primary carbocation. [1 mark]
- M3: Explain stability in terms of the electron-releasing / inductive effect of two alkyl groups. [1 mark]

(d) [1 mark]
- M1: E-Z isomerism / geometric isomerism. [1 mark]

(a)
\(K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}\)

(b)
Units: \(mol^{-2}\ dm^6\) (or \(dm^6\ mol^{-2}\))

(c)
Set up an ICE (Initial, Change, Equilibrium) table:

Reaction: \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)

- Initial moles:
- \(N_2 = 2.00\ mol\)
- \(H_2 = 5.00\ mol\)
- \(NH_3 = 0.00\ mol\)

- At equilibrium, moles of \(NH_3 = 0.80\ mol\).
- Change in moles of \(NH_3 = +0.80\ mol\).
- Change in moles of \(N_2 = -\frac{0.80}{2} = -0.40\ mol\).
- Change in moles of \(H_2 = -3 \times 0.40 = -1.20\ mol\).

- Equilibrium moles:
- \(N_2 = 2.00 - 0.40 = 1.60\ mol\)
- \(H_2 = 5.00 - 1.20 = 3.80\ mol\)
- \(NH_3 = 0.80\ mol\)

- Equilibrium concentrations (divided by \(V = 10.0\ dm^3\)):
- \([N_2] = \frac{1.60}{10.0} = 0.160\ mol\ dm^{-3}\)
- \([H_2] = \frac{3.80}{10.0} = 0.380\ mol\ dm^{-3}\)
- \([NH_3] = \frac{0.80}{10.0} = 0.080\ mol\ dm^{-3}\)

- Calculate \(K_c\):
\(K_c = \frac{(0.080)^2}{(0.160) \times (0.380)^3}\)
\(K_c = \frac{0.0064}{0.160 \times 0.054872} = \frac{0.0064}{0.00877952} \approx 0.72897\)

Rounding to 3 significant figures: \(K_c = 0.729\).

(d)
- Value of \(K_c\) will decrease.
- By Le Chatelier's principle, an increase in temperature shifts the equilibrium in the endothermic direction (to the left) to absorb excess heat. This decreases the concentration of products (\(NH_3\)) and increases the concentration of reactants (\(N_2\) and \(H_2\)), resulting in a lower \(K_c\) value.

(a) [1 mark]
- M1: Correct expression for \(K_c\) (products over reactants with correct powers). [1 mark]

(b) [1 mark]
- M1: \(mol^{-2}\ dm^6\) or \(dm^6\ mol^{-2}\) (accept \(dm^6/mol^2\)). [1 mark]

(c) [6 marks]
- M1: Moles of \(N_2\) react = \(0.40\ mol\) and moles of \(H_2\) react = \(1.20\ mol\). [1 mark]
- M2: Equilibrium moles of \(N_2 = 1.60\ mol\) and \(H_2 = 3.80\ mol\). [1 mark]
- M3: Divide by volume (\(10.0\ dm^3\)) to find concentrations: \([N_2] = 0.160\), \([H_2] = 0.380\), \([NH_3] = 0.080\). [1 mark]
- M4: Correct substitution of concentrations into \(K_c\) expression. [1 mark]
- M5: Calculation of denominator \(0.00878\) and numerator \(0.0064\). [1 mark]
- M6: Final value of \(0.729\) (Accept \(0.73\)) to 3 significant figures. [1 mark]

(d) [2 marks]
- M1: State that \(K_c\) decreases. [1 mark]
- M2: Explain that increasing temperature shifts the position of equilibrium to the left (in the endothermic direction) to oppose the temperature rise. [1 mark]

(a)
- Equation: \(CH_2=CH_2 + H_2O \rightarrow CH_3CH_2OH\) (or \(C_2H_4 + H_2O \rightarrow C_2H_5OH\))
- Catalyst: Concentrated phosphoric acid (\(H_3PO_4\))
- Conditions: High temperature (approx. \(300\ ^\circ\text{C}\)) and high pressure (approx. \(60 - 70\ atm\)) / steam.

(b)
- Reagents: Ethanol and acidified potassium dichromate(VI) (\(H^+ / K_2Cr_2O_7\)).
- Apparatus Setup: Heated under reflux (to prevent volatile intermediate ethanal from escaping, ensuring complete oxidation to ethanoic acid).
- Observation: The reaction mixture changes color from orange to green.
- Safety: Use of an electric heating mantle / water bath (instead of a naked flame) because ethanol is highly flammable.

(c)
- Product Name: Propene
- Equation: \(CH_3CH(OH)CH_3 \rightarrow CH_3CH=CH_2 + H_2O\) (or \(C_3H_7OH \rightarrow C_3H_6 + H_2O\))
- Dehydrating Agent: Concentrated sulfuric acid (\(H_2SO_4\)) or concentrated phosphoric acid (\(H_3PO_4\)) or hot aluminium oxide catalyst (\(Al_2O_3\)).

(a) [3 marks]
- M1: Correct balanced equation (must show \(H_2O\) as a reactant). [1 mark]
- M2: Catalyst: Concentrated phosphoric acid / \(H_3PO_4\). [1 mark]
- M3: Conditions: \(300\ ^\circ\text{C}\) AND \(60-70\ atm\) pressure (or steam). [1 mark]

(b) [4 marks]
- M1: Acidified potassium dichromate(VI) / \(H^+\) and \(K_2Cr_2O_7\) (or sodium dichromate). [1 mark]
- M2: Heat under reflux. [1 mark]
- M3: Color change: Orange to green. [1 mark]
- M4: Safety point: Use of water bath / heating mantle due to flammability of ethanol. [1 mark]

(c) [3 marks]
- M1: Name: Propene. [1 mark]
- M2: Correct balanced equation (showing loss of \(H_2O\)). [1 mark]
- M3: Reagent: Concentrated sulfuric acid (or \(H_3PO_4\) or hot \(Al_2O_3\)). [1 mark]

(a)
- Test 1 (to identify the aldehyde, butanal):
- Reagent: Tollens' reagent (or Fehling's solution).
- Observations: Butanal forms a silver mirror (or red precipitate with Fehling's). Butan-1-ol and butanone show no visible change.
- Test 2 (to identify the alcohol, butan-1-ol from the remaining ketone, butanone):
- Reagent: Acidified potassium dichromate(VI) (\(H^+ / K_2Cr_2O_7\)).
- Observations: Butan-1-ol changes the solution color from orange to green (it is oxidized). Butanone shows no color change / remains orange.
- (Alternatively, sodium metal can be used: Butan-1-ol produces bubbles/effervescence of hydrogen gas; butanone shows no reaction).

(b)
1. Distinguishing Butan-1-ol from butanone:
- Butan-1-ol contains an alcohol \(O-H\) group which shows a broad, strong absorption peak in the range \(3230 - 3550\ cm^{-1}\).
- Butanone does not have this peak, but has a sharp \(C=O\) carbonyl peak in the range \(1680 - 1750\ cm^{-1}\) (which is absent in butan-1-ol).

2. Distinguishing Butanal from butanone:
- Butanal contains an aldehyde group with a unique \(C-H\) stretching absorption in the range \(2700 - 2900\ cm^{-1}\) (typically two weak-to-medium peaks near \(2720\) and \(2820\ cm^{-1}\)).
- Butanone (a ketone) does not have this specific aldehyde \(C-H\) absorption (though both contain a \(C=O\) peak in the range \(1680 - 1750\ cm^{-1}\)).

(a) [6 marks]
- M1: Test 1 reagent: Tollens' reagent / Fehling's solution. [1 mark]
- M2: Test 1 observation: Silver mirror with butanal AND no reaction for butanone and butan-1-ol. [1 mark]
- M3: Test 2 reagent: Acidified potassium dichromate(VI) (or sodium metal). [1 mark]
- M4: Test 2 observation: Orange to green with butan-1-ol AND no change with butanone (or effervescence with sodium for butan-1-ol). [1 mark]
- M5: A logical sequence described that successfully identifies all three substances. [1 mark]
- M6: Correct spelling and chemical terminology throughout. [1 mark]

(b) [4 marks]
- 1. Butan-1-ol vs butanone:
- M1: Butan-1-ol has a broad alcohol \(O-H\) peak at \(3230 - 3550\ cm^{-1}\). [1 mark]
- M2: This peak is absent in butanone (or butanone has a \(C=O\) peak at \(1680 - 1750\ cm^{-1}\) which is absent in butan-1-ol). [1 mark]
- 2. Butanal vs butanone:
- M3: Butanal has an aldehyde \(C-H\) absorption peak at \(2700 - 2900\ cm^{-1}\). [1 mark]
- M4: This peak is absent in butanone. [1 mark]

(a) Lattice enthalpy of dissociation is the enthalpy change when one mole of an ionic solid is completely dissociated into its constituent gaseous ions under standard conditions.

(b) According to the Born-Haber cycle:
\(\Delta H_f^\theta = \Delta H_{at}^\theta(\text{Ca}) + \text{1st IE}(\text{Ca}) + \text{2nd IE}(\text{Ca}) + 2 \times \Delta H_{at}^\theta(\text{Cl}) + 2 \times \text{EA}(\text{Cl}) - \Delta H_{latt\_diss}^\theta(\text{CaCl}_2)\)

Rearranging to solve for the lattice enthalpy of dissociation:
\(\Delta H_{latt\_diss}^\theta(\text{CaCl}_2) = \Delta H_{at}^\theta(\text{Ca}) + \text{1st IE}(\text{Ca}) + \text{2nd IE}(\text{Ca}) + 2 \times \Delta H_{at}^\theta(\text{Cl}) + 2 \times \text{EA}(\text{Cl}) - \Delta H_f^\theta\)

\(\Delta H_{latt\_diss}^\theta = 178 + 590 + 1145 + 2(122) + 2(-349) - (-796)\)
\(\Delta H_{latt\_diss}^\theta = 1913 + 244 - 698 + 796 = +2255 \text{ kJ mol}^{-1}\)

(c) Enthalpy of solution is given by:
\(\Delta H_{sol}^\theta = \Delta H_{latt\_diss}^\theta + \Delta H_{hyd}^\theta(\text{Ca}^{2+}) + 2 \times \Delta H_{hyd}^\theta(\text{Cl}^-)\)
\(\Delta H_{sol}^\theta = 2255 + (-1650) + 2(-364)\)
\(\Delta H_{sol}^\theta = 2255 - 1650 - 728 = -123 \text{ kJ mol}^{-1}\)

(a)
- M1: Enthalpy change when 1 mole of an ionic solid is separated/dissociated (1 mark)
- M2: Into its constituent gaseous ions under standard conditions (1 mark)

(b)
- M1: Correct expression for the Born-Haber cycle rearrangement (1 mark)
- M2: Account for doubling atomisation of chlorine (2 x 122) (1 mark)
- M3: Account for doubling electron affinity of chlorine (2 x -349) (1 mark)
- M4: Correct substitution of values (1 mark)
- M5: Correct answer: +2255 kJ mol^-1 (1 mark)

(c)
- M1: Correct expression: H_sol = H_latt_diss + H_hyd(Ca2+) + 2 x H_hyd(Cl-) (1 mark)
- M2: Substitution of values using student's answer from (b) (1 mark)
- M3: Correct final answer: -123 kJ mol^-1 (1 mark) (consequential marking allowed from part b)

(a)(i) \(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{1.45 \times 10^{-4}}{0.0500 \times 0.0800} = 0.03625 \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\) (or \(3.63 \times 10^{-2}\)).
Units: \(\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}\).

(ii) No effect. The rate constant \(k\) is independent of concentration and only changes with temperature.

(b) The Arrhenius equation in logarithmic form is \(\ln k = -\frac{E_a}{RT} + \ln A\). Therefore, \(\text{gradient} = -\frac{E_a}{R}\).
\(E_a = -\text{gradient} \times R = -(-6.25 \times 10^3) \times 8.31 = 51937.5 \text{ J mol}^{-1}\).
Converting to \(\text{kJ mol}^{-1}\): \(E_a = 51.9 \text{ kJ mol}^{-1}\) (accept 52).

(c) \(\text{Fe}^{2+}\) acts as a homogeneous catalyst.
Equations:
1) \(\text{S}_2\text{O}_8^{2-}(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) \rightarrow 2\text{SO}_4^{2-}(\text{aq}) + 2\text{Fe}^{3+}(\text{aq})\)
2) \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\)

(a)(i)
- M1: Rearranged expression or substitution of values (1 mark)
- M2: Correct calculation: 0.0363 (or 0.03625) (1 mark)
- M3: Correct units: dm^3 mol^-1 s^-1 (1 mark)

(a)(ii)
- M1: No effect (1 mark)
- M2: Because rate constant is only temperature-dependent / independent of concentration (1 mark)

(b)
- M1: Recalls/uses gradient = -Ea / R (1 mark)
- M2: Correct calculation in J: 51937.5 (1 mark)
- M3: Correct division by 1000 and conversion to 3sf: 51.9 kJ mol^-1 (1 mark)

(c)
- M1: State "homogeneous catalyst" (1 mark)
- M2: Both correct equations showing the regeneration of Fe2+ (1 mark)

(a)
- Equilibrium moles of \(\text{CO} = 1.50 - 0.60 = 0.90 \text{ mol}\)
- Equilibrium moles of \(\text{H}_2 = 3.00 - (2 \times 0.60) = 1.80 \text{ mol}\)

(b) Total moles at equilibrium = \(0.90 + 1.80 + 0.60 = 3.30 \text{ mol}\)
- Mole fraction of \(\text{CO} = \frac{0.90}{3.30} = 0.273\) (or \(\frac{3}{11}\))
- Mole fraction of \(\text{H}_2 = \frac{1.80}{3.30} = 0.545\) (or \(\frac{6}{11}\))
- Mole fraction of \(\text{CH}_3\text{OH} = \frac{0.60}{3.30} = 0.182\) (or \(\frac{2}{11}\))

(c) Partial pressure = Mole fraction \(\times\) Total pressure
- \(p(\text{CO}) = 0.2727 \times 240 = 65.5 \text{ kPa}\) (accept 65.45)
- \(p(\text{H}_2) = 0.5455 \times 240 = 130.9 \text{ kPa}\) (accept 131)
- \(p(\text{CH}_3\text{OH}) = 0.1818 \times 240 = 43.6 \text{ kPa}\) (accept 43.64)

(d) Expression:
\(K_p = \frac{p(\text{CH}_3\text{OH})}{p(\text{CO}) \times p(\text{H}_2)^2}\)

Calculation:
\(K_p = \frac{43.64}{65.45 \times (130.91)^2} = \frac{43.64}{1121643} = 3.89 \times 10^{-5} \text{ kPa}^{-2}\)
Units: \(\text{kPa}^{-2}\)

(a)
- M1: Equilibrium moles of CO = 0.90 mol (1 mark)
- M2: Equilibrium moles of H2 = 1.80 mol (1 mark)

(b)
- M1: Calculates total moles = 3.30 mol (1 mark)
- M2: Calculates all mole fractions correctly (1 mark)

(c)
- M1: Correct method for partial pressures (mole fraction x 240) (1 mark)
- M2: Correct values: p(CO) = 65.5, p(H2) = 130.9, p(CH3OH) = 43.6 kPa (1 mark)

(d)
- M1: Correct Kp expression with partial pressures, not concentration brackets (1 mark)
- M2: Correct substitution of partial pressures (1 mark)
- M3: Correct calculation: 3.89 x 10^-5 (1 mark)
- M4: Correct units: kPa^-2 (1 mark)

(a)(i) Equation:
\(\text{CH}_3\text{CH(OH)COOH} \rightleftharpoons \text{CH}_3\text{CH(OH)COO}^- + \text{H}^+\) (or with \(\text{H}_2\text{O}\) / \(\text{H}_3\text{O}^+\))
Expression:
\(K_a = \frac{[\text{H}^+][\text{CH}_3\text{CH(OH)COO}^-]}{[\text{CH}_3\text{CH(OH)COOH}]}\) (or \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\))

(ii) For a weak acid, \([\text{H}^+] \approx \sqrt{K_a \times [\text{HA}]}\)
\([\text{H}^+] = \sqrt{1.38 \times 10^{-4} \times 0.150} = \sqrt{2.07 \times 10^{-5}} = 4.550 \times 10^{-3} \text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(4.550 \times 10^{-3}) = 2.34\)

(b) Initial moles:
- \(n(\text{HA}) = \frac{50.0}{1000} \times 0.150 = 7.50 \times 10^{-3} \text{ mol}\)
- \(n(\text{OH}^-) = \frac{40.0}{1000} \times 0.100 = 4.00 \times 10^{-3} \text{ mol}\)

Upon neutralization:
\(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\)
- \(n(\text{A}^-)\text{ formed} = 4.00 \times 10^{-3} \text{ mol}\)
- \(n(\text{HA})\text{ remaining} = 7.50 \times 10^{-3} - 4.00 \times 10^{-3} = 3.50 \times 10^{-3} \text{ mol}\)

Using the buffer equation:
\([\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)} = 1.38 \times 10^{-4} \times \frac{3.50 \times 10^{-3}}{4.00 \times 10^{-3}} = 1.2075 \times 10^{-4} \text{ mol dm}^{-3}\)
\(\text{pH} = -\log_{10}(1.2075 \times 10^{-4}) = 3.92\)

(a)(i)
- M1: Correct balanced equation for dissociation (1 mark)
- M2: Correct Ka expression (1 mark)

(a)(ii)
- M1: Correct simplification: [H+] = sqrt(Ka x [HA]) (1 mark)
- M2: Correct [H+] calculation: 4.55 x 10^-3 mol dm^-3 (1 mark)
- M3: Correct pH value to 2 decimal places: 2.34 (1 mark)

(b)
- M1: Calculates initial moles of HA (7.50 x 10^-3) and OH- (4.00 x 10^-3) (1 mark)
- M2: Determines moles of A- formed (4.00 x 10^-3) (1 mark)
- M3: Determines moles of HA remaining (3.50 x 10^-3) (1 mark)
- M4: Correct calculation of [H+] (1.21 x 10^-4) (1 mark)
- M5: Correct pH calculation to 2 decimal places: 3.92 (1 mark)

(a) (i) \(\text{SO}_3(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_4(\text{aq})\)
(ii) \(\text{Na}_2\text{O}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow 2\text{NaOH}(\text{aq})\)
(iii) \(\text{P}_4\text{O}_{10}(\text{s}) + 12\text{NaOH}(\text{aq}) \rightarrow 4\text{Na}_3\text{PO}_4(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\)

(b) Silicon dioxide has a giant covalent / macromolecular structure. To melt it, many strong covalent bonds throughout the giant lattice must be broken, requiring a large amount of energy.
In contrast, phosphorus(V) oxide has a simple molecular structure with weak intermolecular van der Waals forces between individual molecules, which require much less energy to overcome.

(c) (i) \(\text{Al}_2\text{O}_3(\text{s}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Al}^{3+}(\text{aq}) + 3\text{H}_2\text{O}(\text{l})\)
(ii) \(\text{Al}_2\text{O}_3(\text{s}) + 2\text{OH}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow 2[\text{Al}(\text{OH})_4]^-(\text{aq})\) (or similar valid balanced ionic equations)

(a)(i)
- M1: Correct equation (1 mark)
(a)(ii)
- M1: Correct equation (1 mark)
(a)(iii)
- M1: Correct species (1 mark)
- M2: Correct balancing (1 mark)

(b)
- M1: SiO2 is macromolecular/giant covalent AND has strong covalent bonds that need to be broken (1 mark)
- M2: P4O10 is simple molecular (1 mark)
- M3: Weak van der Waals forces between molecules in P4O10 require less energy to overcome (1 mark)

(c)(i)
- M1: Correct ionic equation (1 mark)
(c)(ii)
- M1: Correct reactants and products (1 mark)
- M2: Correct balancing (1 mark)

(a) (i) A coordinate bond is a covalent bond in which both shared electrons originate from the same bonding atom / species / ligand.
(ii) A bidentate ligand is a species that donates two lone pairs of electrons from different atoms to a central metal ion to form two coordinate bonds.

(b) (i) Coordination number is 6 (since each of the three bidentate 'en' ligands forms two coordinate bonds).
(ii) The drawing must show two mirror-image isomers of an octahedral complex, with wedges and dashed lines clearly indicating the 3D octahedral geometry, and the bidentate ligands looping between adjacent cis positions.

(c) The colour changes from blue to yellow (or green-yellow).
The copper-containing product is the tetrahedral complex \([\text{CuCl}_4]^{2-}\).

(a)(i)
- M1: Covalent bond (1 mark)
- M2: Both electrons from the same atom / donor / ligand (1 mark)
(a)(ii)
- M1: Donates two lone pairs of electrons (1 mark)
- M2: To form two coordinate bonds to a metal ion (1 mark)

(b)(i)
- M1: 6 (1 mark)
(b)(ii)
- M1: Correct octahedral shape drawn with wedges/dashes (1 mark)
- M2: Correct non-superimposable mirror images showing 'en' loops in correct cis positions (1 mark)

(c)
- M1: Colour change: Blue to yellow / green-yellow (1 mark)
- M2: Product formula: [CuCl4]2- (1 mark) (reject without 2- charge)

(a)
- The \(\text{Fe}^{3+}\) ion is smaller and has a higher charge than the \(\text{Fe}^{2+}\) ion, meaning it has a much higher charge density.
- Thus, the \(\text{Fe}^{3+}\) ion is more strongly polarizing and pulls electron density from the \(\text{O}-\text{H}\) bonds of the water ligands more strongly.
- This weakens the \(\text{O}-\text{H}\) bonds in the water ligands more, making it easier for a hydrogen ion (\(\text{H}^+\)) to dissociate into the solution.

(b) (i)
Observation: A pale blue precipitate forms initially, which dissolves in excess ammonia to give a deep blue solution.
Equation:
\([\text{Cu(H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{NH}_3(\text{aq}) \rightarrow [\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)
(Accept a two-step representation via the copper hydroxide precipitate)

(ii)
Observation: A white precipitate forms and bubbles of gas (effervescence) are produced.
Equation:
\(2[\text{Al(H}_2\text{O})_6]^{3+}(\text{aq}) + 3\text{CO}_3^{2-}(\text{aq}) \rightarrow 2\text{Al(H}_2\text{O})_3(\text{OH})_3(\text{s}) + 3\text{CO}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l})\)

(a)
- M1: Fe3+ has a higher charge and smaller radius / higher charge density than Fe2+ (1 mark)
- M2: Fe3+ is more polarizing / polarizes water ligands more strongly (1 mark)
- M3: This weakens the O-H bond, making it easier to release H+ ions (1 mark)

(b)(i)
- M1: Blue precipitate dissolves to form deep blue solution (1 mark)
- M2: Correct balanced ionic equation for the ligand substitution (2 marks)

(b)(ii)
- M1: White precipitate (1 mark)
- M2: Bubbles / effervescence (1 mark)
- M3: Correct balanced equation (2 marks)

(a) \(\Delta H^\theta = \sum \Delta H_f^\theta(\text{products}) - \sum \Delta H_f^\theta(\text{reactants})\)
\(\Delta H^\theta = [0 + 2(-642)] - [-804 + 0] = -1284 - (-804) = -480 \text{ kJ mol}^{-1}\)

(b) \(\Delta S^\theta = \sum S^\theta(\text{products}) - \sum S^\theta(\text{reactants})\)
\(\Delta S^\theta = [31 + 2(90)] - [253 + 2(33)] = [211] - [319] = -108 \text{ J K}^{-1} \text{ mol}^{-1}\)

(c) \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\)
\(\Delta G^\theta = -480 - 298 \times (-108 \times 10^{-3}) = -480 - (-32.184) = -447.8 \text{ kJ mol}^{-1}\) (or \(-448 \text{ kJ mol}^{-1}\)).
Since \(\Delta G^\theta < 0\), the reaction is feasible at \(298 \text{ K}\).

(d) The reaction is feasible when \(\Delta G^\theta < 0\). It ceases to be feasible when \(\Delta G^\theta = 0\).
\(T = \frac{\Delta H^\theta}{\Delta S^\theta} = \frac{-480 \times 10^3 \text{ J mol}^{-1}}{-108 \text{ J K}^{-1} \text{ mol}^{-1}} = 4444.4 \text{ K}\) (accept \(4444 \text{ K}\) or \(4440 \text{ K}\)).
Because \(\Delta S^\theta\) is negative, the reaction is only feasible below this temperature, so it becomes non-feasible above \(4444 \text{ K}\).

(a)
- M1: Correct expression and doubling of MgCl2 (1 mark)
- M2: Correct calculation: -480 kJ mol^-1 (1 mark)

(b)
- M1: Correct expression and doubling of Mg/MgCl2 values (1 mark)
- M2: Correct calculation: -108 J K^-1 mol^-1 (1 mark)

(c)
- M1: Correct conversion of units (dividing dS by 1000) (1 mark)
- M2: Correct calculation of dG: -448 kJ mol^-1 (1 mark)
- M3: Correct conclusion that reaction is feasible because dG < 0 (1 mark)

(d)
- M1: Correctly states dG = 0 or T = dH / dS (1 mark)
- M2: Correct calculation of T: 4444 K (1 mark)
- M3: Clarifies that reaction is non-feasible ABOVE this temperature because dS is negative (1 mark)