2025 AQA IAL Chemistry (9620) 模擬試題連答案詳解

(a) The equation for electron impact ionisation is:
\(\text{Cu}(\text{g}) \rightarrow \text{Cu}^+(\text{g}) + \text{e}^-\)
(Accept: \(\text{Cu}(\text{g}) + \text{e}^- \rightarrow \text{Cu}^+(\text{g}) + 2\text{e}^-\))

(b) Isotopes have the same electron configuration / same number of electrons in their outer shell.

(c) The positive ions hit a negative detector plate and gain an electron, which generates an electric current. The size of the current is directly proportional to the abundance of that isotope.

(d) First, rearrange \(KE = \frac{1}{2}mv^2\) to find the velocity, \(v\):
\(v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2 \times 1.15 \times 10^{-13}}{1.05 \times 10^{-25}}} = 1.4800 \times 10^6\text{ m s}^{-1}\)
Then, calculate the time of flight, \(t\):
\(t = \frac{d}{v} = \frac{1.50}{1.4800 \times 10^6} = 1.01 \times 10^{-6}\text{ s}\)

(e) Since both ions are accelerated with the same kinetic energy, we can calculate \(t\) for \(^{65}\text{Cu}^+\) using:
\(v = \sqrt{\frac{2 \times 1.15 \times 10^{-13}}{1.08 \times 10^{-25}}} = 1.4593 \times 10^6\text{ m s}^{-1}\)
\(t = \frac{1.50}{1.4593 \times 10^6} = 1.03 \times 10^{-6}\text{ s}\)

(a)
- M1: Equation: \(\text{Cu}(\text{g}) \rightarrow \text{Cu}^+(\text{g}) + \text{e}^-\)
- M2: State symbols: \((\text{g})\) on both reactants and products (dependent on equation showing ionisation).

(b)
- M1: Same electron configuration / same number of electrons in outer shell.

(c)
- M1: Ions hit detector / plate and gain an electron / cause a current to flow.
- M2: Current size is proportional to abundance.

(d)
- M1: Rearrangement of equation to \(v = \sqrt{\frac{2KE}{m}}\) or \(t = d\sqrt{\frac{m}{2KE}}\).
- M2: Correct calculation of velocity \(v = 1.48 \times 10^6\text{ m s}^{-1}\).
- M3: Correct calculation of time \(t = 1.01 \times 10^{-6}\text{ s}\) (3 s.f. and standard form required).

(e)
- M1: Use of same kinetic energy or ratio method: \(\frac{t_1}{t_2} = \sqrt{\frac{m_1}{m_2}}\).
- M2: Correct calculation of velocity of \(^{65}\text{Cu}^+\) as \(1.46 \times 10^6\text{ m s}^{-1}\).
- M3: Correct calculation of time of flight \(t = 1.028 \times 10^{-6}\text{ s}\).
- M4: Correct rounding to 3 s.f. and standard form: \(1.03 \times 10^{-6}\text{ s}\).

(a) Moles of \(\text{HCl}\):
\(n(\text{HCl}) = C \times V = 0.125 \times \frac{22.40}{1000} = 2.80 \times 10^{-3}\text{ mol}\)

(b) Moles of \(\text{M}_2\text{CO}_3\) in \(25.0\text{ cm}^3\):
According to the stoichiometry, 1 mole of \(\text{M}_2\text{CO}_3\) reacts with 2 moles of \(\text{HCl}\).
\(n(\text{M}_2\text{CO}_3) = \frac{2.80 \times 10^{-3}}{2} = 1.40 \times 10^{-3}\text{ mol}\)

(c) Moles of \(\text{M}_2\text{CO}_3\) in \(250.0\text{ cm}^3\):
\(n(\text{total}) = 1.40 \times 10^{-3} \times 10 = 1.40 \times 10^{-2}\text{ mol}\)

(d) \(M_r\) of hydrated salt:
\(M_r = \frac{\text{mass}}{\text{moles}} = \frac{4.00}{1.40 \times 10^{-2}} = 286\) (to 3 s.f.)

(e)
(i) Mass of water of crystallisation:
\(\text{Mass of water} = 4.00 - 1.48 = 2.52\text{ g}\)

(ii) Moles of water:
\(n(\text{H}_2\text{O}) = \frac{2.52}{18.0} = 0.140\text{ mol}\)
Ratio of moles of water to moles of metal carbonate:
\(x = \frac{n(\text{H}_2\text{O})}{n(\text{M}_2\text{CO}_3)} = \frac{0.140}{1.40 \times 10^{-2}} = 10\)

(iii) Relative formula mass of anhydrous salt \(\text{M}_2\text{CO}_3\):
\(M_r(\text{M}_2\text{CO}_3) = \frac{1.48}{1.40 \times 10^{-2}} = 105.7\)
Now find the relative atomic mass of \(\text{M}\):
\(2 \times A_r(\text{M}) + 12.0 + 3(16.0) = 105.7\)
\(2 \times A_r(\text{M}) + 60.0 = 105.7\)
\(2 \times A_r(\text{M}) = 45.7\)
\(A_r(\text{M}) = 22.85\)
The alkali metal with \(A_r \approx 23.0\) is Sodium (\(\text{Na}\)).

(a)
- M1: \(2.80 \times 10^{-3}\text{ mol}\) (or 0.0028 mol)

(b)
- M1: \(1.40 \times 10^{-3}\text{ mol}\) (or 0.0014 mol) (Allow consequential error (conseq) from (a))

(c)
- M1: \(1.40 \times 10^{-2}\text{ mol}\) (or 0.0140 mol) (Allow conseq from (b))

(d)
- M1: Expression \(M_r = \frac{4.00}{\text{moles from (c)}}\)
- M2: \(286\) (accept 285.7)

(e)(i)
- M1: \(2.52\text{ g}\)

(e)(ii)
- M1: Moles of water = \(\frac{2.52}{18.0} = 0.140\text{ mol}\)
- M2: Ratio of moles of water to moles of carbonate (0.140 / 0.0140)
- M3: \(x = 10\)

(e)(iii)
- M1: calculation of \(M_r(\text{M}_2\text{CO}_3) = \frac{1.48}{0.0140} = 105.7\)
- M2: \(2 \times A_r(\text{M}) + 60.0 = 105.7 \implies A_r(\text{M}) = 22.85\)
- M3: Identity of \(\text{M}\) is Sodium / \(\text{Na}\) (conseq on calculated \(A_r\) close to 23.0)

(a)
(i) Drawings should show:
- \(\text{PCl}_3\): Pyramidal arrangement with three \(\text{P-Cl}\) bonds and one lone pair shown on the phosphorus atom.
- \(\text{PCl}_4^+\): Tetrahedral arrangement with four \(\text{P-Cl}\) bonds and positive charge indicated.

(ii)
- \(\text{PCl}_3\) Shape: Trigonal pyramidal / pyramidal. Bond angle: \(107^\circ\) (Accept any angle in the range \(100^\circ - 108^\circ\)).
- \(\text{PCl}_4^+\) Shape: Tetrahedral. Bond angle: \(109.5^\circ\).

(iii) Explanation:
- \(\text{PCl}_3\) has three bonding pairs and one lone pair of electrons around the central phosphorus atom, whereas \(\text{PCl}_4^+\) has four bonding pairs and zero lone pairs around phosphorus.
- Lone pairs repel other electron pairs more strongly than bonding pairs.
- This greater repulsion pushes the bonding pairs closer together, reducing the bond angle from the tetrahedral angle.

(b) \(\text{PCl}_5\) does not have a permanent dipole moment (it is non-polar).
- It has a symmetrical shape (trigonal bipyramidal).
- The individual \(\text{P-Cl}\) bond dipoles are equal and opposite, so they cancel out completely.

(a)(i)
- M1: Correct drawing of \(\text{PCl}_3\) showing 3D wedges/dashes and one lone pair on P.
- M2: Correct drawing of \(\text{PCl}_4^+\) showing tetrahedral arrangement with charge.

(a)(ii)
- M1: \(\text{PCl}_3\) shape: Pyramidal / Trigonal Pyramidal.
- M2: \(\text{PCl}_3\) bond angle: \(107^\circ\) (Accept range \(100^\circ - 108^\circ\)).
- M3: \(\text{PCl}_4^+\) shape: Tetrahedral.
- M4: \(\text{PCl}_4^+\) bond angle: \(109.5^\circ\).

(a)(iii)
- M1: State that \(\text{PCl}_3\) has 3 bonding pairs and 1 lone pair, and \(\text{PCl}_4^+\) has 4 bonding pairs.
- M2: State that lone pairs repel more than bonding pairs.
- M3: Conclude that the lone pair in \(\text{PCl}_3\) compresses the bond angles.

(b)
- M1: Non-polar / has no permanent dipole moment.
- M2: Symmetrical shape / trigonal bipyramidal.
- M3: Polar bonds / dipoles cancel out.

(a)
(i) Calculate temperature rise:
\(\Delta T = 31.2 - 19.5 = 11.7\text{ }^\circ\text{C}\) (or \(11.7\text{ K}\))
\(q = m \cdot c \cdot \Delta T = 50.0 \times 4.18 \times 11.7 = 2445.3\text{ J} = 2.45\text{ kJ}\) (to 3 s.f.)
(If mass of solution including solid is used, i.e. \(53.99\text{ g}\): \(q = 53.99 \times 4.18 \times 11.7 = 2640.4\text{ J} = 2.64\text{ kJ}\))

(ii) Moles of \(\text{CuSO}_4\):
\(n = \frac{3.99}{159.6} = 0.0250\text{ mol}\)

(iii) Enthalpy change of solution:
\(\Delta H_{\text{soln}} = -\frac{q}{n} = -\frac{2.4453}{0.0250} = -97.8\text{ kJ mol}^{-1}\)
(If using solution mass as \(53.99\text{ g}\): \(\Delta H_{\text{soln}} = -\frac{2.6404}{0.0250} = -105.6\text{ kJ mol}^{-1}\))

(b)
(i) Hess's cycle:
```
\Delta H_r
CuSO4(s) + 5H2O(l) -------------> CuSO4.5H2O(s)
\ /
\ /
\Delta H_soln(anhyd) \Delta H_soln(hyd)
\ /
v v
CuSO4(aq)
```

(ii) Calculation:
According to Hess's Law:
\(\Delta H_{\text{soln}}(\text{anhydrous}) = \Delta H_{\text{r}} + \Delta H_{\text{soln}}(\text{hydrated})\)
\(\Delta H_{\text{r}} = \Delta H_{\text{soln}}(\text{anhydrous}) - \Delta H_{\text{soln}}(\text{hydrated})\)
Using \(-97.8\text{ kJ mol}^{-1}\):
\(\Delta H_{\text{r}} = -97.8 - (+11.5) = -109.3\text{ kJ mol}^{-1}\)
(If using \(-105.6\text{ kJ mol}^{-1}\): \(\Delta H_{\text{r}} = -105.6 - 11.5 = -117.1\text{ kJ mol}^{-1}\))
(If using backup value \(-66.0\text{ kJ mol}^{-1}\): \(\Delta H_{\text{r}} = -66.0 - 11.5 = -77.5\text{ kJ mol}^{-1}\))

(a)(i)
- M1: Correct calculation of temperature rise (\(11.7\text{ K}\)).
- M2: Correct calculation of heat energy, \(q = 2.45\text{ kJ}\) (or \(2.64\text{ kJ}\) if mass \(53.99\text{ g}\) is used).

(a)(ii)
- M1: Correct calculation of moles: \(0.0250\text{ mol}\).

(a)(iii)
- M1: Dividing \(q\) by moles.
- M2: Minus sign included.
- M3: Correct value of \(-97.8\text{ kJ mol}^{-1}\) (or \(-106\text{ kJ mol}^{-1}\) if using \(53.99\text{ g}\)).

(b)(i)
- M1: Species cycle correct showing \(\text{CuSO}_4(\text{s}) + 5\text{H}_2\text{O}(\text{l})\), \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O}(\text{s})\) and \(\text{CuSO}_4(\text{aq})\).
- M2: Correct arrow directions: from anhydrous to hydrated, and both anhydrous and hydrated pointing down to aq.
- M3: Enthalpies labelled correctly on arrows.

(b)(ii)
- M1: Correct algebraic relationship: \(\Delta H_{\text{r}} = \Delta H_{\text{soln}}(\text{anhyd}) - \Delta H_{\text{soln}}(\text{hyd})\).
- M2: Correct substitution of values.
- M3: Final answer correct with sign and units: \(-109.3\text{ kJ mol}^{-1}\) (or \(-117.1\text{ kJ mol}^{-1}\) or \(-77.5\text{ kJ mol}^{-1}\) for backup value).

(a)
(i) \(\text{Cl}_2(\text{g}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{NaClO}(\text{aq}) + \text{H}_2\text{O}(\text{l})\)
(ii) Type: Disproportionation. Explanation: Chlorine is simultaneously oxidised and reduced. Its oxidation state changes from \(0\) in \(\text{Cl}_2\) to \(-1\) in \(\text{NaCl}\) (reduction) and to \(+1\) in \(\text{NaClO}\) (oxidation).

(b)
(i) Misty white fume: Hydrogen chloride / \(\text{HCl}\)
Equation: \(\text{KCl} + \text{H}_2\text{SO}_4 \rightarrow \text{KHSO}_4 + \text{HCl}\) (or \(2\text{KCl} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{HCl}\))
(ii) Dark grey solid: Iodine / \(\text{I}_2\)
Gas with bad egg smell: Hydrogen sulfide / \(\text{H}_2\text{S}\)
(iii) Half-equation:
\(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\)
(or \(\text{SO}_4^{2-} + 10\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\))

(c) Reducing ability increases down the group. As we go down the group, the halide ions become larger (ionic radius increases) and have more shielding. The outer electrons are further away and less strongly attracted by the nucleus, so they are lost more easily.

(a)(i)
- M1: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (Ignore states)

(a)(ii)
- M1: Disproportionation.
- M2: Cl oxidation state changes from 0 to -1 (reduced).
- M3: Cl oxidation state changes from 0 to +1 (oxidised).

(b)(i)
- M1: Hydrogen chloride / \(\text{HCl}\) (Reject hydrochloric acid).
- M2: \(\text{KCl} + \text{H}_2\text{SO}_4 \rightarrow \text{KHSO}_4 + \text{HCl}\) (or with \(\text{K}_2\text{SO}_4\)).

(b)(ii)
- M1: Iodine / \(\text{I}_2\) (Reject iodide).
- M2: Hydrogen sulfide / \(\text{H}_2\text{S}\).

(b)(iii)
- M1: Correct reactants and products: \(\text{H}_2\text{SO}_4\) and \(\text{H}_2\text{S}\).
- M2: Fully balanced: \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\).

(c)
- M1: Increases down the group as ionic size/radius increases / more shielding.
- M2: Outer electrons less strongly held by nucleus / easier to lose an electron.

(a) Reduction half-equation:
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)

(b) Oxidation half-equation:
\(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\)

(c) Multiply oxidation by 5 and add:
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}\)

(d)
(i) Hydrochloric acid contains chloride ions (\(\text{Cl}^-\)), which would be oxidised to chlorine (\(\text{Cl}_2\)) by the manganate(VII) ions. This would lead to an incorrectly large titre.
(ii) The end-point colour change is from colourless to a permanent pale pink.
(iii) Moles of \(\text{MnO}_4^-\):
\(n(\text{MnO}_4^-) = C \times V = 0.0100 \times \frac{15.20}{1000} = 1.52 \times 10^{-4}\text{ mol}\)
(iv) Moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\):
\(n(\text{Fe}^{2+}) = 5 \times 1.52 \times 10^{-4} = 7.60 \times 10^{-4}\text{ mol}\)
(v) Moles of \(\text{Fe}^{2+}\) in \(250.0\text{ cm}^3\):
\(n(\text{total}) = 7.60 \times 10^{-4} \times 10 = 7.60 \times 10^{-3}\text{ mol}\)
Mass of pure \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\):
\(m = 7.60 \times 10^{-3} \times 278.0 = 2.1128\text{ g} \approx 2.11\text{ g}\)
(vi) Percentage purity:
\(\text{Purity} = \frac{2.1128}{2.65} \times 100 = 79.73\% \approx 79.7\%\)

(a)
- M1: Correct species: \(\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\).
- M2: Balanced with \(5\text{e}^-\).

(b)
- M1: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\).

(c)
- M1: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}\).

(d)(i)
- M1: \(\text{Cl}^-\) from \(\text{HCl}\) is oxidised (to \(\text{Cl}_2\)) by manganate(VII).
- M2: Leading to a larger/incorrect volume of \(\text{MnO}_4^-\).

(d)(ii)
- M1: Colourless to (pale permanent) pink.

(d)(iii)
- M1: \(1.52 \times 10^{-4}\text{ mol}\) (or 0.000152 mol).

(d)(iv)
- M1: \(7.60 \times 10^{-4}\text{ mol}\) (or 0.00076 mol) (Allow conseq from (c) and d(iii)).

(d)(v)
- M1: Multiplies moles in \(25\text{ cm}^3\) by 10 to get \(7.60 \times 10^{-3}\text{ mol}\).
- M2: Correct calculation of mass = \(2.11\text{ g}\) (accept 2.1128 g).

(d)(vi)
- M1: \(79.7\%\) (Allow range 79.6% to 79.8%).

(a) The activation energy is the minimum energy [1 mark] that colliding particles must possess in order for a reaction to occur [1 mark].

(b) Maxwell-Boltzmann distribution:
- Axes labeled correctly: y-axis as number/fraction of molecules/particles, x-axis as energy [1 mark].
- Curve at \(T_1\) starts at origin, peaks, and decreases asymptotically toward the x-axis without touching it [1 mark].
- Curve at \(T_2\) has a lower peak [1 mark] and is shifted to the right (the peak is at higher energy) [1 mark].
- Explanation: At a higher temperature, a much larger fraction of molecules have energy greater than or equal to the activation energy (\(E \ge E_a\)), leading to a greater frequency of successful collisions per unit time [1 mark].

(c)
- A catalyst provides an alternative reaction pathway with a lower activation energy (\(E_c\)) [1 mark].
- On the distribution, this lower activation energy is located further to the left on the energy axis [1 mark].
- Consequently, a significantly larger fraction of molecules have energy greater than or equal to this lower activation energy (\(E \ge E_c\)), resulting in a higher rate of successful collisions [1 mark].

(a)
- M1: Minimum energy [1 mark]
- M2: required for colliding particles to react / for a reaction to occur [1 mark]

(b)
- M1: Correctly labeled axes (y = number/fraction of molecules, x = energy) [1 mark]
- M2: Curve at \(T_1\) starts at origin, has a peak, and does not touch the x-axis at high energy [1 mark]
- M3: Curve at \(T_2\) peak is lower than at \(T_1\) [1 mark]
- M4: Curve at \(T_2\) peak is shifted to the right of the \(T_1\) peak [1 mark]
- M5: Explanation that more molecules have energy \(\ge E_a\) at \(T_2\) [1 mark]

(c)
- M1: Catalyst provides alternative pathway with lower activation energy [1 mark]
- M2: This shifts the activation energy line to the left on the energy axis [1 mark]
- M3: Therefore, a greater fraction of molecules have sufficient energy to react [1 mark]

(a) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}(l) + 4.5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l)\) [1 mark]

(b) \(q = m c \Delta T\)
\(m = 150.0 \text{ g}\)
\(\Delta T = 52.4 - 18.2 = 34.2 \text{ K}\) (or °C)
\(q = 150.0 \times 4.18 \times 34.2 = 21443.4 \text{ J} = 21.44 \text{ kJ}\) [2 marks]

(c) \(\text{Moles} = \frac{\text{mass}}{M_r} = \frac{0.810}{60.0} = 0.0135 \text{ mol}\) [1 mark]

(d) \(\Delta H_c = -\frac{q}{\text{moles}} = -\frac{21.4434}{0.0135} = -1588.4 \text{ kJ mol}^{-1}\)
To 3 s.f.: \(-1590 \text{ kJ mol}^{-1}\) [3 marks]

(e) 1. Incomplete combustion of propan-1-ol (which forms carbon/soot or carbon monoxide instead of carbon dioxide) [1 mark].
2. Evaporation of the alcohol from the wick after weighing but before lighting / after extinguishing [1 mark].

(f) Use a copper calorimeter (or use wind shields, or place a lid on the beaker) [1 mark].

(a)
- M1: Correct balanced equation (allow fractional coefficients like 4.5 or 9/2; state symbols not required) [1 mark]

(b)
- M1: Correct calculation of temperature difference \(\Delta T = 34.2\) [1 mark]
- M2: Correct calculation of heat energy \(q = 21.4 \text{ kJ}\) (accept 21.44 kJ) [1 mark]

(c)
- M1: Correct calculation of moles = 0.0135 [1 mark]

(d)
- M1: Division of \(q\) by moles [1 mark]
- M2: Negative sign included [1 mark]
- M3: Correct value to 3 s.f. (\(-1590\)) [1 mark]
Note: Allow full error carried forward (ECF) from parts (b) and (c).

(e)
- M1: Incomplete combustion [1 mark]
- M2: Evaporation of alcohol from the wick [1 mark]

(f)
- M1: Wind shields / copper calorimeter / add a lid [1 mark]

(a) (i) Stereoisomers are compounds with the same structural formula but a different arrangement of atoms in space [1 mark].
(ii) The isomers of 3-methylpent-2-ene are:
- *E*-isomer: The higher priority group \(\text{-CH}_3\) on C2 and \(\text{-CH}_2\text{CH}_3\) on C3 are on opposite sides of the double bond [1 mark].
- *Z*-isomer: The higher priority group \(\text{-CH}_3\) on C2 and \(\text{-CH}_2\text{CH}_3\) on C3 are on the same side of the double bond [1 mark].
(iii) 3-methylpent-2-ene [1 mark].

(b) Mechanism:
- Dipole \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) shown on the \(\text{H-Br}\) molecule [1 mark].
- Curly arrow from the \(\text{C=C}\) double bond of *E*-3-methylpent-2-ene to the \(\text{H}^{\delta+}\) [1 mark].
- Curly arrow from the \(\text{H-Br}\) bond to the \(\text{Br}^{\delta-}\) atom [1 mark].
- Draw the stable tertiary carbocation intermediate: \(\text{CH}_3-\text{CH}_2-\text{C}^+(\text{CH}_3)-\text{CH}_2-\text{CH}_3\) [1 mark].
- Curly arrow from the lone pair on the \(\text{Br}^-\) ion to the positive carbon atom of the carbocation [1 mark]. (Max 4 marks total)

(c)
- The reaction to form **B** proceeds via a tertiary carbocation intermediate, whereas the reaction to form **C** proceeds via a secondary carbocation intermediate [1 mark].
- Tertiary carbocations are more stable than secondary carbocations due to the greater electron-releasing inductive effect of three alkyl groups compared to two [1 mark].

(a)(i)
- M1: Same structural formula, different spatial arrangement of atoms [1 mark]

(a)(ii)
- M1: Correct structure of *E*-3-methylpent-2-ene with stereochemistry clear [1 mark]
- M2: Correct structure of *Z*-3-methylpent-2-ene with stereochemistry clear [1 mark]

(a)(iii)
- M1: 3-methylpent-2-ene (accept 3-methyl-2-pentene) [1 mark]

(b)
- M1: Curly arrow from double bond to \(\text{H}\) and correct dipole on \(\text{H-Br}\) [1 mark]
- M2: Curly arrow from \(\text{H-Br}\) bond to \(\text{Br}\) [1 mark]
- M3: Correct structure of the tertiary carbocation intermediate [1 mark]
- M4: Curly arrow from lone pair of \(\text{Br}^-\) to positive carbon atom of carbocation [1 mark]

(c)
- M1: Mentions reaction goes via tertiary carbocation (for B) vs secondary carbocation (for C) [1 mark]
- M2: Mentions tertiary carbocations are more stable due to the electron-releasing inductive effect of alkyl groups [1 mark]

(a) (i) To provide energy to break the \(\text{Cl-Cl}\) bond homolytically / to generate chlorine radicals [1 mark].
(ii) \(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\) [1 mark].
(iii)
Propagation Step 1: \(\text{CH}_3\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2^\bullet + \text{HCl}\) [1 mark]
Propagation Step 2: \(\text{CH}_3\text{CH}_2^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{Cl}^\bullet\) [1 mark]
(iv) \(2\text{CH}_3\text{CH}_2^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\) (or \(\text{C}_4\text{H}_{10}\)) [1 mark].

(b) Further substitution reactions can occur [1 mark], leading to a mixture of multi-halogenated products (such as 1,1-dichloroethane or 1,2-dichloroethane) which are difficult to separate from chloroethane [1 mark].

(c) (i) \(\text{CF}_2\text{Cl}_2 \rightarrow \text{CF}_2\text{Cl}^\bullet + \text{Cl}^\bullet\) [1 mark].
(ii)
Step 1: \(\text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2\) [1 mark]
Step 2: \(\text{ClO}^\bullet + \text{O}_3 \rightarrow \text{Cl}^\bullet + 2\text{O}_2\) (or \(\text{ClO}^\bullet + \text{O} \rightarrow \text{Cl}^\bullet + \text{O}_2\)) [1 mark]

(a)(i)
- M1: To break the \(\text{Cl-Cl}\) covalent bond / initiate homolytic fission [1 mark]

(a)(ii)
- M1: \(\text{Cl}_2 \rightarrow 2\text{Cl}^\bullet\) (unpaired electron dots must be shown on radicals) [1 mark]

(a)(iii)
- M1: Correct propagation step 1: \(\text{CH}_3\text{CH}_3 + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{CH}_2^\bullet + \text{HCl}\) [1 mark]
- M2: Correct propagation step 2: \(\text{CH}_3\text{CH}_2^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Cl} + \text{Cl}^\bullet\) [1 mark]

(a)(iv)
- M1: \(2\text{CH}_3\text{CH}_2^\bullet \rightarrow \text{C}_4\text{H}_{10}\) [1 mark]

(b)
- M1: Further substitution occurs / chlorine radicals react with chloroethane [1 mark]
- M2: A complex mixture of products is formed / low yield of pure chloroethane [1 mark]

(c)(i)
- M1: \(\text{CF}_2\text{Cl}_2 \rightarrow \text{CF}_2\text{Cl}^\bullet + \text{Cl}^\bullet\) [1 mark]

(c)(ii)
- M1: \(\text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2\) [1 mark]
- M2: \(\text{ClO}^\bullet + \text{O}_3 \rightarrow \text{Cl}^\bullet + 2\text{O}_2\) [1 mark]

(a) \(K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^2}\) [1 mark]

(b) From the stoichiometry of the reaction, since 0.80 mol of **C** is produced:
- Change in **A** = \(-0.40 \text{ mol}\)
- Change in **B** = \(-0.80 \text{ mol}\)
Equilibrium amount of **A** = \(1.20 - 0.40 = 0.80 \text{ mol}\) [1 mark]
Equilibrium amount of **B** = \(2.00 - 0.80 = 1.20 \text{ mol}\) [1 mark]

(c) Calculate concentrations at equilibrium (volume = \(5.00 \text{ dm}^3\)):
\([\text{A}] = \frac{0.80}{5.00} = 0.16 \text{ mol dm}^{-3}\)
\([\text{B}] = \frac{1.20}{5.00} = 0.24 \text{ mol dm}^{-3}\)
\([\text{C}] = \frac{0.80}{5.00} = 0.16 \text{ mol dm}^{-3}\) [1 mark for all three]
Substitute these values into the \(K_c\) expression:
\(K_c = \frac{(0.16)^2}{0.16 \times (0.24)^2} = \frac{0.16}{(0.24)^2} = 2.7778\) [1 mark]
\(K_c = 2.78\) (to 3 s.f.) [1 mark]
Units: \(\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2} = \text{mol}^{-1} \text{ dm}^3\) (or \(\text{dm}^3 \text{ mol}^{-1}\)) [1 mark]

(d) (i) As the forward reaction is exothermic (\(\Delta H = -85 \text{ kJ mol}^{-1}\)), increasing the temperature shifts the equilibrium in the endothermic direction (to the left) [1 mark]. This decreases the equilibrium concentration of products and increases reactants, so the value of \(K_c\) decreases [1 mark].
(ii) No effect. A catalyst increases the rates of both the forward and reverse reactions equally [1 mark].

(a)
- M1: Correct expression for \(K_c\) with square brackets [1 mark]

(b)
- M1: Equilibrium moles of **A** = 0.80 mol [1 mark]
- M2: Equilibrium moles of **B** = 1.20 mol [1 mark]

(c)
- M1: Calculation of concentrations of **A**, **B**, and **C** by dividing by 5.00 [1 mark]
- M2: Substitution of correct values into the \(K_c\) expression [1 mark]
- M3: Correct calculation of numerical value as 2.78 (allow 2.8) [1 mark]
- M4: Correct units: \(\text{dm}^3 \text{ mol}^{-1}\) [1 mark]

(d)(i)
- M1: Shifts equilibrium to the left / in reverse direction because forward reaction is exothermic [1 mark]
- M2: \(K_c\) decreases [1 mark]

(d)(ii)
- M1: No change in value of \(K_c\) [1 mark]

(a) **X** (butan-1-ol) is a primary alcohol, **Y** (butan-2-ol) is a secondary alcohol, and **Z** (2-methylpropan-2-ol) is a tertiary alcohol [1 mark for all three correct].

(b) Test:
- Add acidified potassium dichromate(VI) to separate samples of each alcohol and warm/heat [1 mark].
- Alcohol **Z** (tertiary) does not react, and the solution remains orange [1 mark].
- Alcohols **X** (primary) and **Y** (secondary) both turn the solution from orange to green [1 mark].
- **X** is oxidized to butanal (or butanoic acid) and **Y** is oxidized to butanone [1 mark].

(c) Reagent: Concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) or concentrated phosphoric acid (\(\text{H}_3\text{PO}_4\)) [1 mark].
Conditions: Heat/high temperature (e.g., 170 °C or under reflux) [1 mark]. (Accept: pass vapour over hot aluminium oxide catalyst).

(d)
- Dehydration of butan-2-ol can eliminate a hydrogen atom from either C1 or C3, producing both but-1-ene and but-2-ene [1 mark].
- But-2-ene can exist as a pair of stereoisomers (*E*/*Z* isomers) because both carbons in the double bond are attached to two different groups [1 mark].
- Thus, the three isomeric alkenes formed are but-1-ene, *E*-but-2-ene, and *Z*-but-2-ene [1 mark].

(a)
- M1: Correct classification of all three alcohols (X = primary, Y = secondary, Z = tertiary) [1 mark]

(b)
- M1: Acidified potassium dichromate(VI) and heat/warm [1 mark]
- M2: Z (tertiary) remains orange [1 mark]
- M3: X and Y turn orange to green [1 mark]
- M4: Product of X is butanal/butanoic acid AND product of Y is butanone [1 mark]

(c)
- M1: Concentrated \(\text{H}_2\text{SO}_4\) / concentrated \(\text{H}_3\text{PO}_4\) [1 mark]
- M2: Heat / reflux / elevated temperature [1 mark]

(d)
- M1: Elimination of H from C1 or C3 gives but-1-ene or but-2-ene [1 mark]
- M2: But-2-ene exhibits stereoisomerism (*E*/*Z* or cis/trans) [1 mark]
- M3: Correctly identifies all three: but-1-ene, *E*-but-2-ene, and *Z*-but-2-ene [1 mark]

(a)
- Shape of \(\text{BF}_3\): trigonal planar [1 mark].
- Shape of \(\text{NF}_3\): trigonal pyramidal [1 mark].
- Explanation: Boron has 3 bonding pairs and 0 lone pairs around the central atom, which repel equally to find positions of maximum separation [1 mark]. Nitrogen has 3 bonding pairs and 1 lone pair of electrons around the central atom; since lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, the bonds are pushed closer together [1 mark].

(b)
- Fluorine is more electronegative than boron and nitrogen, meaning both molecules contain polar covalent bonds [1 mark].
- \(\text{BF}_3\) is highly symmetrical, so the individual bond dipoles cancel out, resulting in no net molecular dipole moment [1 mark].
- \(\text{NF}_3\) is asymmetrical (trigonal pyramidal shape / has a lone pair), meaning the individual bond dipoles do not cancel, giving a permanent net molecular dipole [1 mark].

(c)
- \(\text{HF}\) molecules are held together by hydrogen bonding [1 mark].
- \(\text{F}_2\) molecules are held together only by weak London / van der Waals / temporary induced dipole-dipole forces [1 mark].
- Hydrogen bonds are significantly stronger than van der Waals forces and require much more energy to overcome, resulting in a much higher boiling point [1 mark].

(a)
- M1: Shape of \(\text{BF}_3\) = trigonal planar [1 mark]
- M2: Shape of \(\text{NF}_3\) = trigonal pyramidal / pyramidal [1 mark]
- M3: B has 3 bonding pairs and 0 lone pairs [1 mark]
- M4: N has 3 bonding pairs and 1 lone pair (lone pair repels more than bonding pairs) [1 mark]

(b)
- M1: Polar bonds exist due to electronegativity difference [1 mark]
- M2: \(\text{BF}_3\) is symmetrical, so dipoles cancel [1 mark]
- M3: \(\text{NF}_3\) is asymmetrical / has a lone pair, so dipoles do not cancel [1 mark]

(c)
- M1: \(\text{HF}\) has hydrogen bonding [1 mark]
- M2: \(\text{F}_2\) has van der Waals / London / dispersion forces [1 mark]
- M3: Hydrogen bonds are stronger and require more energy to break than van der Waals forces [1 mark]

(a) Comparing Experiment 1 and Experiment 2: when \([\text{S}_2\text{O}_8^{2-}]\) doubles from \(0.0400\) to \(0.0800\text{ mol dm}^{-3}\) while \([\text{I}^-]\) remains constant at \(0.0800\text{ mol dm}^{-3}\), the initial rate doubles from \(1.16 \times 10^{-4}\) to \(2.32 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Thus, the order of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) is 1.

Comparing Experiment 2 and Experiment 3: when \([\text{I}^-]\) doubles from \(0.0800\) to \(0.1600\text{ mol dm}^{-3}\) while \([\text{S}_2\text{O}_8^{2-}]\) remains constant at \(0.0800\text{ mol dm}^{-3}\), the initial rate doubles from \(2.32 \times 10^{-4}\) to \(4.64 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}\). Thus, the order of reaction with respect to \(\text{I}^-\) is 1.

(b) The rate equation is: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\)

(c) Rearranging the rate equation to solve for \(k\) using Experiment 1 data:

\(k = \frac{\text{Rate}}{[\text{S}_2\text{O}_8^{2-}][\text{I}^-]} = \frac{1.16 \times 10^{-4}\text{ mol dm}^{-3}\text{ s}^{-1}}{(0.0400\text{ mol dm}^{-3}) \times (0.0800\text{ mol dm}^{-3})} = 3.625 \times 10^{-2}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

To 3 significant figures, \(k = 3.63 \times 10^{-2}\text{ dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

- **Order wrt \(\text{S}_2\text{O}_8^{2-}\)**: 1 mark for order 1 with explanation comparing Exp 1 & 2.
- **Order wrt \(\text{I}^-\)**: 1 mark for order 1 with explanation comparing Exp 2 & 3.
- **Rate equation**: 1 mark for correct rate equation: \(\text{Rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]\) (consecutive on orders).
- **Calculation of \(k\)**: 2 marks for correct calculation of value (3.63 x 10^-2 or 0.0363) (1 mark for working, 1 mark for final value to 3 sf).
- **Units**: 1 mark for correct units (\(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)).

(a) First, calculate the initial moles of propanoic acid and propanoate ions:
- \(n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.0500\text{ dm}^3 \times 0.250\text{ mol dm}^{-3} = 0.0125\text{ mol}\)
- \(n(\text{CH}_3\text{CH}_2\text{COO}^-) = 0.0500\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 0.00750\text{ mol}\)

Using the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_{\text{a}} + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = -\log_{10}(1.35 \times 10^{-5}) + \log_{10}\left(\frac{0.00750}{0.0125}\right)\)
\(\text{pH} = 4.870 - 0.222 = 4.648 \approx 4.65\)

(b) Moles of \(\text{NaOH}\) added:
\(n(\text{OH}^-) = 0.00200\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 2.00 \times 10^{-4}\text{ mol}\)

The added hydroxide ions react completely with propanoic acid:
\(\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}\)

New moles of propanoic acid (HA):
\(n_{\text{new}}(\text{HA}) = 0.0125 - 0.000200 = 0.0123\text{ mol}\)

New moles of propanoate ions (A\(^-\)):
\(n_{\text{new}}(\text{A}^-) = 0.00750 + 0.000200 = 0.00770\text{ mol}\)

Using the buffer equation again:
\(\text{pH} = 4.870 + \log_{10}\left(\frac{0.00770}{0.0123}\right) = 4.870 - 0.203 = 4.667 \approx 4.67\)

- **Part (a)**:
- 1 mark for finding initial moles of acid and salt.
- 1 mark for correct setup of pH expression or \([\text{H}^+]\).
- 1 mark for correct pH of 4.65 (accept 4.64–4.65).
- **Part (b)**:
- 1 mark for calculating moles of \(\text{OH}^-\).
- 1 mark for calculating new moles of acid (0.0123) and salt (0.00770).
- 1 mark for calculating final pH of 4.67 (accept 4.66–4.67).

(a) The pink solution of \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) turns dark blue when excess concentrated hydrochloric acid is added.

(b) The equation is: \([\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq}) + 4\text{Cl}^-(\text{aq}) \rightleftharpoons [\text{CoCl}_4]^{2-}(\text{aq}) + 6\text{H}_2\text{O}(\text{l})\)

(c) The starting complex, \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\), has an octahedral geometry. The product complex, \([\text{CoCl}_4]^{2-}\), has a tetrahedral geometry.

(d) The coordination number decreases from 6 to 4 because chloride ions are significantly larger than neutral water molecules. The increased steric hindrance (or repulsion) between the larger chloride ligands around the small central \(\text{Co}^{2+}\) ion prevents six chloride ions from coordinating, so only four can fit.

- **Part (a)**: 1 mark for color change from pink to blue.
- **Part (b)**: 2 marks for correct balanced ionic equation (1 mark for correct reactants/products, 1 mark for balancing with correct charges).
- **Part (c)**: 1 mark for identifying starting complex as octahedral and product complex as tetrahedral.
- **Part (d)**: 2 marks for explanation (1 mark for identifying chloride ligands are larger/have more steric hindrance, 1 mark for linking this to why coordination number decreases).

(a) Lattice enthalpy of formation is the enthalpy change when one mole of a solid ionic compound is formed from its constituent gaseous ions under standard conditions.

(b) According to Hess's Law, we can set up the Born-Haber cycle equation:
\(\Delta H_{\text{f}}(\text{MgO}) = \Delta H_{\text{at}}(\text{Mg}) + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + 0.5 \times E_{\text{bond}}(\text{O}_2) + \text{EA}_1(\text{O}) + \text{EA}_2(\text{O}) + \Delta H_{\text{L,form}}\)

Substitute the known values:
\(-602 = +148 + 738 + 1451 + (0.5 \times 496) + (-141) + 798 + \Delta H_{\text{L,form}}\)

Simplify the sum of the non-lattice terms:
\(-602 = 148 + 738 + 1451 + 248 - 141 + 798 + \Delta H_{\text{L,form}}\)
\(-602 = 3242 + \Delta H_{\text{L,form}}\)

Rearrange to solve for \(\Delta H_{\text{L,form}}\):
\(\Delta H_{\text{L,form}} = -602 - 3242 = -3844\text{ kJ mol}^{-1}\)

Thus, the lattice enthalpy of formation of magnesium oxide is \(-3844\text{ kJ mol}^{-1}\).

- **Part (a)**: 2 marks (1 mark for 'enthalpy change when 1 mole of a solid ionic lattice is formed', 1 mark for 'from its constituent gaseous ions').
- **Part (b)**: 4 marks (1 mark for showing correct use of half the bond enthalpy of oxygen, i.e., +248; 1 mark for correct cycle representation or algebraic formula; 1 mark for correct arithmetic process; 1 mark for final correct value of \(-3844\text{ kJ mol}^{-1}\) with correct sign and units).

(a) The standard IUPAC cell representation places the half-cell with the more negative electrode potential (the anode/oxidation half-cell) on the left:
\(\text{Cr}(\text{s}) | \text{Cr}^{3+}(\text{aq}) || \text{Fe}^{3+}(\text{aq}), \text{Fe}^{2+}(\text{aq}) | \text{Pt}(\text{s})\)

(b) Since the iron half-cell has a more positive standard electrode potential, it undergoes reduction, while the chromium half-cell undergoes oxidation.
Oxidation half-equation: \(\text{Cr}(\text{s}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{e}^-\)
Reduction half-equation: \(3\text{Fe}^{3+}(\text{aq}) + 3\text{e}^- \rightarrow 3\text{Fe}^{2+}(\text{aq})\)
Overall reaction:
\(\text{Cr}(\text{s}) + 3\text{Fe}^{3+}(\text{aq}) \rightarrow \text{Cr}^{3+}(\text{aq}) + 3\text{Fe}^{2+}(\text{aq})\)

(c) \(E_{\text{cell}}^\ominus = E_{\text{reduction}}^\ominus - E_{\text{oxidation}}^\ominus = +0.77 - (-0.74) = +1.51\text{ V}\)

(d) The platinum electrode is inert and provides a solid surface on which the electron transfer can occur, since both the reactant (\(\text{Fe}^{3+}\)) and product (\(\text{Fe}^{2+}\)) of the half-cell are in aqueous solution.

- **Part (a)**: 2 marks (1 mark for correct order with salt bridge, 1 mark for correct phases and including Pt(s) on the right-hand side).
- **Part (b)**: 2 marks (1 mark for correct species, 1 mark for correct balanced coefficients).
- **Part (c)**: 1 mark for \(+1.51\text{ V}\) (must have positive sign or unit).
- **Part (d)**: 1 mark for stating that it acts as an inert conductor / provides a surface for electron transfer.

(a)
1. **Reaction of sulfur(IV) oxide with water**:
\(\text{SO}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2\text{SO}_3(\text{aq})\)
(or \(\text{SO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{HSO}_3^-\)). This is a weak acid; the approximate pH is **1 to 3**.

2. **Reaction of phosphorus(V) oxide with water**:
\(\text{P}_4\text{O}_{10}(\text{s}) + 6\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}_3\text{PO}_4(\text{aq})\)
This is a strong acid; the approximate pH is **0 to 2**.

(b) Silicon dioxide, \(\text{SiO}_2\), is a giant covalent macromolecular structure with strong covalent bonds, making it insoluble in water. However, it behaves as an **acidic oxide**. It reacts with hot, concentrated strong bases like sodium hydroxide to form a soluble silicate salt and water.

Equation:
\(\text{SiO}_2(\text{s}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SiO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l})\)

- **Part (a)**:
- 1 mark for balanced equation for \(\text{SO}_2\).
- 1 mark for pH 1, 2, or 3.
- 1 mark for balanced equation for \(\text{P}_4\text{O}_{10}\).
- 1 mark for pH 0, 1, or 2.
- **Part (b)**:
- 1 mark for stating \(\text{SiO}_2\) is an acidic oxide.
- 1 mark for balanced equation.

(a) The \(\text{Al}^{3+}\) ion has a high positive charge and a small ionic radius, resulting in a very high charge density. This high charge density strongly polarizes the electron density in the \(\text{O-H}\) bonds of the coordinated water ligands. This weakens the \(\text{O-H}\) bonds, allowing a proton (\(\text{H}^+\)) to be easily lost to the solvent water molecules.

Equation:
\([\text{Al}(\text{H}_2\text{O})_6]^{3+}(\text{aq}) \rightleftharpoons [\text{Al}(\text{H}_2\text{O})_5(\text{OH})]^{2+}(\text{aq}) + \text{H}^+(\text{aq})\)
(or with water: \([\text{Al}(\text{H}_2\text{O})_6]^{3+} + \text{H}_2\text{O} \rightleftharpoons [\text{Al}(\text{H}_2\text{O})_5(\text{OH})]^{2+} + \text{H}_3\text{O}^+\))

(b)
- **When NaOH is added dropwise**:
A white precipitate of aluminium hydroxide forms.
\([\text{Al}(\text{H}_2\text{O})_6]^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Al}(\text{H}_2\text{O})_3(\text{OH})_3(\text{s}) + 3\text{H}_2\text{O}(\text{l})\)

- **When excess NaOH is added**:
The white precipitate dissolves to give a colorless solution.

\(\text{Al}(\text{H}_2\text{O})_3(\text{OH})_3(\text{s}) + \text{OH}^-(\text{aq}) \rightarrow [\text{Al}(\text{OH})_4]^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l})\)

- **Part (a)**:
- 1 mark for explanation involving high charge density of \(\text{Al}^{3+}\) and polarization of \(\text{O-H}\) bond.
- 1 mark for the correct deprotonation equation.
- **Part (b)**:
- 1 mark for observation of white precipitate and 1 mark for its dissolution to a colorless solution.
- 1 mark for the balanced equation of precipitate formation.
- 1 mark for the balanced equation of precipitate dissolution with excess hydroxide.

(a) The equilibrium constant expression is:
\(K_p = \frac{p(\text{NO}_2)^2}{p(\text{NO})^2 p(\text{O}_2)}\)

(b) Establish the moles at equilibrium:
* Initial moles: \(n(\text{NO}) = 2.00\text{ mol}\), \(n(\text{O}_2) = 1.00\text{ mol}\), \(n(\text{NO}_2) = 0\text{ mol}\).
* Change in moles: Since \(1.40\text{ mol}\) of \(\text{NO}_2\) is formed:
- \(\Delta n(\text{NO}_2) = +1.40\text{ mol}\)
- \(\Delta n(\text{NO}) = -1.40\text{ mol}\)
- \(\Delta n(\text{O}_2) = -0.70\text{ mol}\)
* Equilibrium moles:
- \(n_{\text{eq}}(\text{NO}) = 2.00 - 1.40 = 0.60\text{ mol}\)
- \(n_{\text{eq}}(\text{O}_2) = 1.00 - 0.70 = 0.30\text{ mol}\)
- \(n_{\text{eq}}(\text{NO}_2) = 1.40\text{ mol}\)

Total equilibrium moles:
\(n_{\text{total}} = 0.60 + 0.30 + 1.40 = 2.30\text{ mol}\)

Calculate partial pressures (\(p = \chi \times P_{\text{total}}\)) where \(P_{\text{total}} = 150\text{ kPa}\):
* \(p(\text{NO}) = \frac{0.60}{2.30} \times 150\text{ kPa} = 39.13\text{ kPa}\)
* \(p(\text{O}_2) = \frac{0.30}{2.30} \times 150\text{ kPa} = 19.57\text{ kPa}\)
* \(p(\text{NO}_2) = \frac{1.40}{2.30} \times 150\text{ kPa} = 91.30\text{ kPa}\)

(c) Calculate \(K_p\):
\(K_p = \frac{(91.30)^2}{(39.13)^2 \times 19.57} = \frac{8335.69}{1531.16 \times 19.57} = 0.2782 \approx 0.278\text{ kPa}^{-1}\)

Units:
\(\frac{\text{kPa}^2}{\text{kPa}^2 \times \text{kPa}} = \text{kPa}^{-1}\)

- **Part (a)**: 1 mark for the correct expression for \(K_p\).
- **Part (b)**: 3 marks:
- 1 mark for correct equilibrium moles (\(0.60\text{ mol NO}\), \(0.30\text{ mol O}_2\), and \(1.40\text{ mol NO}_2\)).
- 1 mark for total moles of \(2.30\text{ mol}\).
- 1 mark for correct partial pressures (allow ecf from incorrect moles, e.g., 39.1, 19.6, 91.3 kPa).
- **Part (c)**: 2 marks:
- 1 mark for correct calculation of \(K_p = 0.278\) (allow 0.28 or ecf from part b).
- 1 mark for correct units (\(\text{kPa}^{-1}\)).

**Part (a)**
The balanced chemical equation for the thermal decomposition of zinc carbonate with state symbols is:
\(\text{ZnCO}_3(\text{s}) \rightarrow \text{ZnO}(\text{s}) + \text{CO}_2(\text{g})\)

**Part (b)**
To find \(\Delta H^\ominus\):
\(\Delta H^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants})\)
\(\Delta H^\ominus = [(-350.5) + (-393.5)] - [-812.8]
\Delta H^\ominus = -744.0 + 812.8 = +68.8\text{ kJ mol}^{-1}\)

To find \(\Delta S^\ominus\):
\(\Delta S^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})\)
\(\Delta S^\ominus = [43.6 + 213.6] - 82.4
\Delta S^\ominus = 257.2 - 82.4 = +174.8\text{ J K}^{-1} \text{mol}^{-1}\)

**Part (c)**
A reaction is thermodynamically feasible when \(\Delta G^\ominus \le 0\).
\(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus = 0\) at the point of transition.
\(T = \frac{\Delta H^\ominus}{\Delta S^\ominus}\)
Note that \(\Delta H^\ominus\) must be converted to \(\text{J mol}^{-1}\) or \(\Delta S^\ominus\) to \(\text{kJ K}^{-1} \text{mol}^{-1}\):
\(T = \frac{68.8 \times 1000}{174.8} = 393.59\text{ K}\)
Thus, the minimum temperature is \(393.6\text{ K}\) (or \(394\text{ K}\) to 3 significant figures).

**Part (a): [1 mark]**
- 1 mark: \(\text{ZnCO}_3(\text{s}) \rightarrow \text{ZnO}(\text{s}) + \text{CO}_2(\text{g})\) (correct equation with correct state symbols).

**Part (b): [4 marks]**
- 1 mark: Correct expression or working for \(\Delta H^\ominus\).
- 1 mark: \(\Delta H^\ominus = +68.8\text{ kJ mol}^{-1}\) (accept without '+' sign but penalise incorrect sign).
- 1 mark: Correct expression or working for \(\Delta S^\ominus\).
- 1 mark: \(\Delta S^\ominus = +174.8\text{ J K}^{-1} \text{mol}^{-1}\).

**Part (c): [2 marks]**
- 1 mark: Correct formula \(T = \frac{\Delta H}{\Delta S}\) or correct substitution of values with unit conversion (e.g., \(68.8 \times 1000\)).
- 1 mark: Correct final temperature: \(393.6\text{ K}\) (accept range \(393\text{ K}\) to \(394\text{ K}\) depending on rounding).

**Part (a)(i)**
The ligand substitution reaction is:
\([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightarrow [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\)
The color of the starting octahedral complex \([\text{Co}(\text{H}_2\text{O})_6]^{2+}\) is pink, and the final tetrahedral complex \([\text{CoCl}_4]^{2-}\) is blue. Therefore, the color change is from pink to blue.
The coordination number decreases from 6 to 4 because the chloride ligands (\(\text{Cl}^-\)) are larger than water (\(\text{H}_2\text{O}\)) ligands, resulting in steric hindrance that prevents six chloride ligands from bonding around the cobalt central metal ion.

**Part (a)(ii)**
In the tetrahedral product \([\text{CoCl}_4]^{2-}\):
- The coordination number is 4.
- Let the oxidation state of cobalt be \(x\).
\(x + 4(-1) = -2 \Rightarrow x = +2\). Thus, the oxidation state of Co is +2.

**Part (b)**
The complex ion \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\) exhibits cis-trans (geometric) isomerism:
- In the **trans-isomer**, the two chloro (\(\text{Cl}^-\)) ligands are directly opposite to each other (angle of \(180^\circ\)), with the four ammine (\(\text{NH}_3\)) ligands in the equatorial plane.
- In the **cis-isomer**, the two chloro (\(\text{Cl}^-\)) ligands are adjacent to each other (angle of \(90^\circ\)).

**Part (a)(i): [3 marks]**
- 1 mark: Correct balanced equation \([\text{Co}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightarrow [\text{CoCl}_4]^{2-} + 6\text{H}_2\text{O}\).
- 1 mark: Correct color change (pink to blue).
- 1 mark: Explanation of change in coordination number (chloride ions are larger than water molecules, leading to steric hindrance / steric crowding).

**Part (a)(ii): [1 mark]**
- 1 mark: Coordination number = 4 AND oxidation state = +2 (both required).

**Part (b): [3 marks]**
- 1 mark: Correct 3D representation of the trans-isomer of \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\).
- 1 mark: Correct 3D representation of the cis-isomer of \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+\).
- 1 mark: Stating the type of isomerism is "cis-trans" or "geometric" isomerism.

**Part (a)**
First, calculate the initial moles of propanoic acid (\(\text{HA}\)) and propanoate ions (\(\text{A}^-\)):
\(n(\text{HA}) = \text{volume} \times \text{concentration} = 0.150\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.0300\text{ mol}\)
\(n(\text{A}^-) = \text{volume} \times \text{concentration} = 0.100\text{ dm}^3 \times 0.300\text{ mol dm}^{-3} = 0.0300\text{ mol}\)

Using the buffer equation:
\([\text{H}^+] = K_a \frac{[\text{HA}]}{[\text{A}^-]} = K_a \frac{n(\text{HA})}{n(\text{A}^-)}
[\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.0300}{0.0300} = 1.35 \times 10^{-5}\text{ mol dm}^{-3}\)

Calculate pH:
\(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.35 \times 10^{-5}) = 4.8697\)
Rounding to 2 decimal places gives \(\text{pH} = 4.87\).

**Part (b)**
When hydrochloric acid is added:
\(n(\text{H}^+\text{ added}) = 0.0100\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00100\text{ mol}\)

The added \(\text{H}^+\) reacts with the propanoate ions to form more propanoic acid:
\(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\)

The new moles of acid and salt are:
\(n(\text{HA})_{\text{new}} = 0.0300 + 0.00100 = 0.0310\text{ mol}
n(\text{A}^-)_{\text{new}} = 0.0300 - 0.00100 = 0.0290\text{ mol}\)

Using the buffer equation again:
\([\text{H}^+] = 1.35 \times 10^{-5} \times \frac{0.0310}{0.0290} = 1.4431 \times 10^{-5}\text{ mol dm}^{-3}\)

Calculate new pH:
\(\text{pH} = -\log_{10}(1.4431 \times 10^{-5}) = 4.8407\)
Rounding to 2 decimal places gives \(\text{pH} = 4.84\).

**Part (a): [3 marks]**
- 1 mark: Correct calculation of initial moles of \(\text{HA}\) and \(\text{A}^-\) (both are \(0.0300\text{ mol}\)).
- 1 mark: Correct expression for \([\text{H}^+]\) or pH formula.
- 1 mark: \(\text{pH} = 4.87\) (must be 2 decimal places).

**Part (b): [4 marks]**
- 1 mark: Correct calculation of moles of \(\text{H}^+\) added (\(0.00100\text{ mol}\)).
- 1 mark: Correct calculation of new moles of \(\text{HA}\) (\(0.0310\text{ mol}\)) and new moles of \(\text{A}^-\) (\(0.0290\text{ mol}\)).
- 1 mark: Correct substitution of new moles into the expression for \([\text{H}^+]\).
- 1 mark: \(\text{pH} = 4.84\) (must be 2 decimal places, accept error-carried-forward from incorrect initial moles of part a).

*(a) Calculation of Lattice Enthalpy of Dissociation:*

We apply Hess's Law using the Born-Haber cycle. The enthalpy of formation is equal to the sum of all the steps to form gaseous ions from standard elements, minus the lattice enthalpy of dissociation:

\(\Delta H_f = \Delta H_{\text{at}}(\text{Sr}) + \text{1st IE}(\text{Sr}) + \text{2nd IE}(\text{Sr}) + \Delta H_{\text{at}}(\text{O}) + \text{1st EA}(\text{O}) + \text{2nd EA}(\text{O}) - \Delta H_{\text{L, diss}}\)

Substitute the known values into the equation:

\(-592 = 164 + 550 + 1064 + 249 - 141 + 798 - \Delta H_{\text{L, diss}}\)

\(-592 = 2684 - \Delta H_{\text{L, diss}}\)

\(\Delta H_{\text{L, diss}} = 2684 + 592 = +3276 \text{ kJ mol}^{-1}\)

*(b) Explanation of the difference:*

- The perfect ionic model assumes that ions are perfectly spherical, point charges with purely electrostatic attraction.
- The iodide ion (\(\text{I}^-\)) has a very large ionic radius and a high polarisability.
- The strontium ion (\(\text{Sr}^{2+}\)) has a relatively high charge density and polarises the electron cloud of the iodide ion.
- This polarisation leads to partial covalent character in the bonding of strontium iodide, which strengthens the bonding, making the actual lattice enthalpy more endothermic than the theoretical model predicts.

**Part (a): 5 marks**
- M1: Correct expression or cycle structure showing the relationship of all terms (1 mark).
- M2: Substitution of correct values with correct signs (1 mark).
- M3: Calculation of sum of gaseous ion steps (\(2684 \text{ kJ mol}^{-1}\)) (1 mark).
- M4: Correct final calculation value (\(3276\)) (1 mark).
- M5: Correct positive sign and unit (\(\text{kJ mol}^{-1}\)) (1 mark).

**Part (b): 3.89 marks**
- M6: Mentions that the theoretical model assumes perfectly spherical ions / pure electrostatic attraction (1 mark).
- M7: Explains that the iodide ion is large and highly polarisable / polarised by the strontium cation (1 mark).
- M8: Explains that this causes covalent character (or partial covalent bonding) (1 mark).
- M9: States that covalent character makes the bonding stronger, requiring more energy to dissociate (0.89 marks).

*(a) Buffer pH Calculation:*

1. Calculate the initial amount of propanoic acid (\(\text{HA}\)):
\(n(\text{HA}) = \frac{25.0}{1000} \times 0.150 = 3.75 \times 10^{-3} \text{ mol}\)

2. Calculate the amount of sodium hydroxide (\(\text{OH}^-\)) added:
\(n(\text{OH}^-) = \frac{15.0}{1000} \times 0.100 = 1.50 \times 10^{-3} \text{ mol}\)

3. Propanoic acid reacts with hydroxide ions:
\(\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}\)

- Remaining \(n(\text{HA}) = 3.75 \times 10^{-3} - 1.50 \times 10^{-3} = 2.25 \times 10^{-3} \text{ mol}\)
- Formed \(n(\text{A}^-) = 1.50 \times 10^{-3} \text{ mol}\)

4. Calculate the hydrogen ion concentration \([\text{H}^+]\) using the buffer equation:
\([\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)}\)
\([\text{H}^+] = (1.35 \times 10^{-5}) \times \frac{2.25 \times 10^{-3}}{1.50 \times 10^{-3}} = 2.025 \times 10^{-5} \text{ mol dm}^{-3}\)

5. Calculate pH:
\(\text{pH} = -\log_{10}(2.025 \times 10^{-5}) = 4.6935 \approx 4.69\)

*(b) Buffer Action when Acid is Added:*
- When \(\text{H}^+\) ions are added from hydrochloric acid, they react with the conjugate base (propanoate ions, \(\text{CH}_3\text{CH}_2\text{COO}^-\)) present in the buffer.
- Equation: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\)
- The concentration of free hydrogen ions remains virtually unchanged, and therefore the pH remains constant.

**Part (a): 5 marks**
- M1: Calculates moles of propanoic acid correctly (1 mark).
- M2: Calculates moles of NaOH correctly (1 mark).
- M3: Calculates remaining moles of propanoic acid and moles of propanoate conjugate base (1 mark).
- M4: Correct substitution into \([\text{H}^+] = K_a \frac{[\text{HA}]}{[\text{A}^-]}\) or equivalent Henderson-Hasselbalch equation (1 mark).
- M5: Evaluates pH as 4.69 (must be to 2 decimal places to match standard pH rules) (1 mark).

**Part (b): 3.89 marks**
- M6: Identifies that added \(\text{H}^+\) reacts with propanoate ions (1 mark).
- M7: Writes a correct ionic equation: \(\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2\text{COOH}\) (state symbols not required) (1.89 marks).
- M8: Explains that the ratio of acid to salt / concentration of \(\text{H}^+\) remains almost constant (1 mark).

*(a) Deduction of orders of reaction:*
- **With respect to \(\text{A}\)**: Compare Experiment 1 and Experiment 2. Concentration of \(\text{B}\) and \(\text{C}\) are kept constant. Concentration of \(\text{A}\) doubles (\(0.10 \rightarrow 0.20 \text{ mol dm}^{-3}\)), while the rate increases by a factor of 4 (\(4.0 \times 10^{-4} \rightarrow 1.6 \times 10^{-3}\)). Since \(2^2 = 4\), the reaction is **second order** with respect to \(\text{A}\).
- **With respect to \(\text{B}\)**: Compare Experiment 1 and Experiment 3. Concentration of \(\text{A}\) and \(\text{C}\) are kept constant. Concentration of \(\text{B}\) doubles, while the initial rate remains unchanged (\(4.0 \times 10^{-4}\)). Therefore, the reaction is **zero order** with respect to \(\text{B}\).
- **With respect to \(\text{C}\)**: Compare Experiment 1 and Experiment 4. Concentration of \(\text{A}\) and \(\text{B}\) are kept constant. Concentration of \(\text{C}\) doubles (\(0.20 \rightarrow 0.40 \text{ mol dm}^{-3}\)), while the initial rate doubles (\(4.0 \times 10^{-4} \rightarrow 8.0 \times 10^{-4}\)). Therefore, the reaction is **first order** with respect to \(\text{C}\).

*(b) Rate equation and Calculation of \(k\):*
- The rate equation is:
$$\text{Rate} = k[\text{A}]^2[\text{C}]$$
- Rearranging for \(k\):
$$k = \frac{\text{Rate}}{[\text{A}]^2[\text{C}]}$$
- Using data from Experiment 1:
$$k = \frac{4.0 \times 10^{-4}}{(0.10)^2 \times (0.20)} = \frac{4.0 \times 10^{-4}}{0.01 \times 0.20} = \frac{4.0 \times 10^{-4}}{0.0020} = 0.20$$
- Units of \(k\):
$$\text{units} = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})} = \text{mol}^{-2} \text{dm}^6 \text{s}^{-1}$$

**Part (a): 4.89 marks**
- M1: Shows order with respect to A is 2 with reference to Exp 1 and 2 (1.63 marks).
- M2: Shows order with respect to B is 0 with reference to Exp 1 and 3 (1.63 marks).
- M3: Shows order with respect to C is 1 with reference to Exp 1 and 4 (1.63 marks).

**Part (b): 4 marks**
- M4: Correct rate equation matching the deduced orders: \(\text{Rate} = k[\text{A}]^2[\text{C}]\) (allow consequential on part a) (1 mark).
- M5: Correct rearrangement of rate equation and substitution of values (1 mark).
- M6: Correct value of \(k = 0.20\) (or \(2.0 \times 10^{-1}\)) (1 mark).
- M7: Correct units: \(\text{mol}^{-2} \text{dm}^6 \text{s}^{-1}\) (or \(\text{dm}^6 \text{mol}^{-2} \text{s}^{-1}\)) (1 mark).

*(a) Generation of electrophile:*

\(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\)
(Accept: \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\))

*(b) Mechanism:*
1. A curly arrow starts from the delocalised \(\pi\)-system of the methylbenzene ring (specifically from the 4-position, opposite the methyl group) pointing directly to the \(\text{N}\) in \(^{+}\text{NO}_2\).
2. Intermediate: Draw a benzene ring with a methyl group at position 1, a hydrogen atom and a nitro group (\(-\text{NO}_2\)) at position 4. The delocalised system is broken, represented by a horseshoe shape that is open towards carbon 4 and carries a positive charge inside the ring.
3. A curly arrow starts from the C-H bond at position 4 and points back into the delocalised ring to restore the aromatic system.
4. Products: 4-nitromethylbenzene and \(\text{H}^+\).

*(c) Explanation of relative rate:*
- The methyl group is an electron-donating group due to its positive inductive effect.
- This increases the electron density of the delocalised \(\pi\)-ring system compared to benzene.
- As a result, the ring in methylbenzene is more nucleophilic and polarises / attracts the electrophile (\(\text{NO}_2^+\)) more readily, lowering the activation energy for the reaction.

**Part (a): 2 marks**
- M1: Correct reactants (\(\text{HNO}_3 + \text{H}_2\text{SO}_4\)) and electrophile product (\(\text{NO}_2^+\)) (1 mark).
- M2: Fully balanced equation (1 mark).

**Part (b): 4 marks**
- M3: Curly arrow from the ring to the \(\text{NO}_2^+\) electrophile (1 mark).
- M4: Correct structure of the intermediate showing a broken ring with positive charge inside and both H and \(\text{NO}_2\) at position 4 (1 mark).
- M5: Curly arrow from the C-H bond to reform the aromatic delocalised ring (1 mark).
- M6: Correct final organic product and \(\text{H}^+\) shown (1 mark).

**Part (c): 2.89 marks**
- M7: Mentions the methyl group is electron-donating / positive inductive effect (1 mark).
- M8: Explains this increases electron density of the aromatic ring (1 mark).
- M9: Concludes that this attracts the electrophile more strongly / makes it more reactive (0.89 marks).

*(a) Mechanism:*
1. Nucleophilic attack: A curly arrow from the lone pair on the carbon of the cyanide ion (\(^{-}\text{:CN}\)) to the partially positive carbonyl carbon atom (\(\text{C}^{\delta+}\)) of butanone.
2. Heterolytic fission: A curly arrow from the \(\text{C}=\text{O}\) double bond to the oxygen atom (\(\text{O}^{\delta-}\)).
3. Intermediate formation: The intermediate is a tetrahedral alkoxide ion: \(\text{CH}_3\text{C}(\text{O}^-)(\text{CN})\text{CH}_2\text{CH}_3\).
4. Protonation: A curly arrow from the lone pair on the negative oxygen (\(\text{O}^-\)) of the intermediate to a hydrogen ion (\(\text{H}^+\)) from the acid (or water molecule) to form the final product: \(\text{CH}_3\text{C}(\text{OH})(\text{CN})\text{CH}_2\text{CH}_3\).

*(b) Explanation of optical inactivity:*
- The carbonyl group (\(\text{C}=\text{O}\)) of butanone is planar around the carbonyl carbon.
- The nucleophile (\(\text{CN}^-\)) can attack this flat, planar carbonyl group with equal probability from either the top side or the bottom side.
- This yields a racemic mixture (an equimolar mixture containing equal amounts of both enantiomers).
- Since the two enantiomers rotate plane-polarised light by equal amounts in opposite directions, the optical activities cancel out, rendering the mixture optically inactive.

**Part (a): 5 marks**
- M1: Show dipoles on the carbonyl group: \(\text{C}^{\delta+}=\text{O}^{\delta-}\) (1 mark).
- M2: Curly arrow from the lone pair on the carbon of \(^{-}\text{CN}\) to the carbonyl carbon (1 mark).
- M3: Curly arrow from the double bond of \(\text{C}=\text{O}\) to the oxygen atom (1 mark).
- M4: Correct structure of the tetrahedral intermediate with negative charge on oxygen (1 mark).
- M5: Curly arrow from the lone pair of the oxygen on intermediate to \(\text{H}^+\) to show protonation (1 mark).

**Part (b): 3.89 marks**
- M6: Mentions that the carbonyl carbon / group is planar (1 mark).
- M7: Explains that the nucleophile can attack with equal probability / chance from above or below (1 mark).
- M8: Explains that this results in an equimolar / equal mixture of two enantiomers (racemic mixture) (1 mark).
- M9: States that the opposite rotations of plane-polarised light cancel out (0.89 marks).

*(a) Repeating unit of poly(lactic acid):*
Lactic acid contains both an alcohol (\(-\text{OH}\)) and a carboxylic acid (\(-\text{COOH}\)) group. Under condensation polymerization, water is eliminated to form a polyester.
Repeating unit:

$$\text{-[O - CH(CH}_3\text{) - CO]-}$$

*(b) Repeating unit of Kevlar:*
Kevlar is formed from a diamine and a dicarboxylic acid, releasing water molecules.
Repeating unit:

$$\text{-[NH - C}_6\text{H}_4\text{ - NH - CO - C}_6\text{H}_4\text{ - CO]-}$$

*(c) Intermolecular forces and melting points:*
- Poly(lactic acid) contains ester links (\(-\text{COO}-\)). The main intermolecular forces between the polymer chains of PLA are permanent dipole-dipole forces and van der Waals forces.
- Kevlar contains amide links (\(-\text{CONH}-\)). The hydrogen atom on the highly electronegative nitrogen atom (\(\text{N}-\text{H}\)) can form **hydrogen bonds** with the lone pairs on the oxygen atom of the carbonyl group (\(\text{C}=\text{O}\)) of an adjacent chain.
- Hydrogen bonds are significantly stronger than permanent dipole-dipole forces.
- Consequently, a larger amount of thermal energy is required to overcome these strong hydrogen bonds in Kevlar, resulting in a much higher melting point.

**Part (a): 2 marks**
- M1: Correct ester linkage drawn (1 mark).
- M2: Correct connectivity of repeating unit showing open bonds on ends: \(\text{-[O-CH(CH}_3\text{)-CO]-}\) (1 mark).

**Part (b): 2.89 marks**
- M3: Correct structures of the aromatic rings (1 mark).
- M4: Correct amide linkage (\(-\text{NH}-\text{CO}-\)) shown (1 mark).
- M5: Correctly shows open bonds at the ends of the repeating unit: \(\text{-[NH-C}_6\text{H}_4\text{-NH-CO-C}_6\text{H}_4\text{-CO]-}\) (0.89 marks).

**Part (c): 4 marks**
- M6: Identifies that PLA has dipole-dipole forces (and van der Waals forces) between chains (1 mark).
- M7: Identifies that Kevlar has hydrogen bonding between chains (1 mark).
- M8: Explains hydrogen bonds form between the \(\text{N}-\text{H}\) group and the \(\text{C}=\text{O}\) group of adjacent chains (1 mark).
- M9: Concludes that hydrogen bonds are stronger than dipole-dipole forces and require more energy to break (1 mark).

*(a) Expression for \(K_p\) and Units:*

$$K_p = \frac{p_{\text{NO}}^2 \cdot p_{\text{Cl}_2}}{p_{\text{NOCl}}^2}$$

Units:
$$\text{Units} = \frac{(\text{kPa})^2 \cdot \text{kPa}}{(\text{kPa})^2} = \text{kPa}$$

*(b) Calculation of equilibrium moles and partial pressures:*
- Initial moles: \(n(\text{NOCl}) = 2.00 \text{ mol}\), \(n(\text{NO}) = 0 \text{ mol}\), \(n(\text{Cl}_2) = 0 \text{ mol}\).
- Moles of \(\text{NOCl}\) dissociated: \(25.0\% \text{ of } 2.00 = 0.50 \text{ mol}\).
- Equilibrium moles:
- \(n(\text{NOCl}) = 2.00 - 0.50 = 1.50 \text{ mol}\)
- \(n(\text{NO}) = 0.50 \text{ mol}\) (from \(1:1\) stoichiometry with dissociated \(\text{NOCl}\))
- \(n(\text{Cl}_2) = 0.25 \text{ mol}\) (from \(2:1\) stoichiometry with dissociated \(\text{NOCl}\))
- Total moles at equilibrium = \(1.50 + 0.50 + 0.25 = 2.25 \text{ mol}\).

- Partial pressures (\(p = \frac{\text{equilibrium moles}}{\text{total moles}} \times P_{\text{total}}\)):
- \(p_{\text{NOCl}} = \frac{1.50}{2.25} \times 150 \text{ kPa} = 100 \text{ kPa}\)
- \(p_{\text{NO}} = \frac{0.50}{2.25} \times 150 \text{ kPa} = 33.33 \text{ kPa}\)
- \(p_{\text{Cl}_2} = \frac{0.25}{2.25} \times 150 \text{ kPa} = 16.67 \text{ kPa}\)

*(c) Calculation of \(K_p\):*

$$K_p = \frac{(33.33)^2 \times 16.67}{100^2}$$
$$K_p = \frac{1111.1 \times 16.67}{10000} = 1.8518 \approx 1.85 \text{ kPa}$$

**Part (a): 2 marks**
- M1: Correct expression for \(K_p\) (1 mark).
- M2: Correct unit (\(\text{kPa}\)) (1 mark).

**Part (b): 4.89 marks**
- M3: Calculates equilibrium moles of all three gases: \(n(\text{NOCl}) = 1.50\), \(n(\text{NO}) = 0.50\), \(n(\text{Cl}_2) = 0.25\) (1.63 marks).
- M4: Correct total moles of \(2.25 \text{ mol}\) (1.63 marks).
- M5: Correct partial pressures: \(p_{\text{NOCl}} = 100 \text{ kPa}\), \(p_{\text{NO}} = 33.3 \text{ kPa}\), \(p_{\text{Cl}_2} = 16.7 \text{ kPa}\) (1.63 marks).

**Part (c): 2 marks**
- M6: Substitutes partial pressures into the correct \(K_p\) expression (1 mark).
- M7: Correct calculation of final value to 3 significant figures: \(1.85\) (1 mark).

*(a) Step-by-step Deduction of Structure:*

1. **IR Spectroscopy:**
- The sharp, strong absorption at \(1740 \text{ cm}^{-1}\) indicates the presence of a carbonyl group (\(\text{C}=\text{O}\)).
- The absence of a broad peak at \(3200-3600 \text{ cm}^{-1}\) indicates there is no alcohol (\(-\text{OH}\)) or carboxylic acid group.
- Combined with the molecular formula \(\text{C}_5\text{H}_{10}\text{O}_2\), this suggests that **X** is an ester.

2. **\(^1\text{H}\) NMR Analysis:**
- **Singlet at \(\delta = 3.67 \text{ ppm}\) (3H):** The chemical shift value is characteristic of protons adjacent to an ester oxygen (\(-\text{O}-\text{CH}_3\)). Because it is a singlet, there are no neighboring hydrogen atoms on the adjacent carbon (separated by the oxygen atom). This confirms a methyl ester group.
- **Triplet at \(\delta = 0.95 \text{ ppm}\) (3H):** This indicates a methyl group (\(-\text{CH}_3\)) adjacent to a \(-\text{CH}_2-\) group (split into a triplet by \(n + 1 = 2 + 1 = 3\)).
- **Sextet at \(\delta = 1.65 \text{ ppm}\) (2H):** This indicates a \(-\text{CH}_2-\) group coupled to 5 neighboring hydrogens (split into a sextet by \(n + 1 = 5 + 1 = 6\)). This implies it is flanked by a \(-\text{CH}_3\) group (3H) and a \(-\text{CH}_2-\) group (2H).
- **Triplet at \(\delta = 2.30 \text{ ppm}\) (2H):** This chemical shift is typical for protons on a carbon adjacent to a carbonyl group (\(-\text{CH}_2-\text{C}=\text{O}\)). It is split into a triplet, indicating it has a neighboring \(-\text{CH}_2-\) group.

3. **Putting the fragments together:**
- The alkyl chain must be: \(\text{CH}_3\text{CH}_2\text{CH}_2-\).
- This is attached to the ester carbonyl group: \(-\text{COOCH}_3\).
- Structure of **X** is **methyl butanoate** (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{COOCH}_3\)).

4. **\(^{13}\text{C}\) NMR Consistency:**
- The 5 peaks represent 5 unique carbon environments, which matches the 5 carbons of methyl butanoate. The peak at \(\delta = 174 \text{ ppm}\) corresponds to the ester carbonyl carbon.

**Marking Breakdown: 8.89 marks**
- M1: Deduces from IR (\(1740 \text{ cm}^{-1}\) peak and no \(3200-3600 \text{ cm}^{-1}\) broad peak) that an ester group is present (1.5 marks).
- M2: Assigns the singlet at \(3.67 \text{ ppm}\) to a \(-\text{O}-\text{CH}_3\) group (1.5 marks).
- M3: Explains the spin-spin coupling of the propyl chain (triplet-sextet-triplet pattern showing \(\text{CH}_3\text{-CH}_2\text{-CH}_2\text{-}\)) (2 marks).
- M4: Assigns the triplet at \(2.30 \text{ ppm}\) to the \(-\text{CH}_2\text{-}\) adjacent to the carbonyl (1.5 marks).
- M5: Relates the \(^{13}\text{C}\) NMR data (5 peaks including carbonyl carbon peak at 174 ppm) to the structure (1 mark).
- M6: Draws or names the correct final structure: **methyl butanoate** (1.39 marks).

(a)
- Comparing Experiments 1 and 2: \([B]\) is constant. \([A]\) doubles from \(0.100\) to \(0.200\text{ mol dm}^{-3}\), and the initial rate quadruples (increases by a factor of \(\frac{4.80 \times 10^{-4}}{1.20 \times 10^{-4}} = 4 = 2^2\)). Therefore, the reaction is second order with respect to A.
- Comparing Experiments 1 and 3: \([A]\) is constant. \([B]\) doubles from \(0.100\) to \(0.200\text{ mol dm}^{-3}\), and the initial rate remains unchanged. Therefore, the reaction is zero order with respect to B.

(b) Rate equation:
\(\text{Rate} = k[A]^2\)

(c) Calculation of \(k\):
Using data from Experiment 1:
\(k = \frac{\text{Rate}}{[A]^2} = \frac{1.20 \times 10^{-4}}{(0.100)^2} = 0.0120\) (or \(1.20 \times 10^{-2}\))

Units of \(k\):
\(\text{Units} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\)

(d) Two-step mechanism:
- Step 1 (Slow / Rate-determining step):
\(2A \rightarrow \text{Intermediate (I)}\)
- Step 2 (Fast):
\(\text{Intermediate (I)} + B \rightarrow C + D\)
- Justification: Step 1 must be the slow (rate-determining) step because only A is involved in this step, which matches the second-order dependence on A and zero-order dependence on B in the rate equation.

Part (a) [3 marks total]:
- 1 mark: Reasoning for A (doubling [A] quadruples the rate, hence second order with respect to A).
- 1 mark: Reasoning for B (doubling [B] does not change the rate, hence zero order with respect to B).
- 1 mark: Correctly stating both orders (order with respect to A = 2, order with respect to B = 0).

Part (b) [1 mark]:
- 1 mark: Correct rate equation: \(\text{Rate} = k[A]^2\) (Accept omission of state symbols; do not accept lowercase 'r' for rate if not defined, but allow minor notation variations. Must have 'k').

Part (c) [2 marks total]:
- 1 mark: Correct numerical value of \(k = 0.0120\) or \(1.20 \times 10^{-2}\) (Allow 2 or more significant figures).
- 1 mark: Correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\) (Accept \(\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}\)).

Part (d) [3 marks total]:
- 1 mark: Suitable Step 1 showing \(2A \rightarrow \text{intermediate}\) (or products including intermediate) AND labelled "slow" or "rate-determining step".
- 1 mark: Suitable Step 2 showing the intermediate reacting with B to produce \(C + D\) AND labelled "fast".
- 1 mark: Explanation that the rate-determining step must involve species matching the rate equation (only 2 molecules of A and 0 molecules of B).

(a) The half-equations are:\
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)\
\(\text{H}_2\text{C}_2\text{O}_4 \rightarrow 2\text{CO}_2 + 2\text{H}^+ + 2\text{e}^-\)\
Combining these gives:\
\(2\text{MnO}_4^- + 5\text{H}_2\text{C}_2\text{O}_4 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}\)\
\
(b) Autocatalysis is a reaction in which one of the products acts as a catalyst for the reaction. The catalyst in this reaction is manganese(II) ions, \(\text{Mn}^{2+}\).\
\
(c) The reaction is too slow at room temperature because the activation energy is very high (due to the collision of two negatively charged ions/molecules initially, or before the catalyst is formed). Heating provides the necessary activation energy to start the reaction.\
\
(d) Calculation:\
1. Moles of \(\text{MnO}_4^-\) used:\
\(n(\text{MnO}_4^-) = C \times V = 0.0200 \times \frac{23.80}{1000} = 4.76 \times 10^{-4}\text{ mol}\)\
\
2. Moles of \(\text{H}_2\text{C}_2\text{O}_4\) in the \(25.0\text{ cm}^3\) sample:\
Using the stoichiometry (ratio 5:2):\
\(n(\text{H}_2\text{C}_2\text{O}_4) = 4.76 \times 10^{-4} \times \frac{5}{2} = 1.19 \times 10^{-3}\text{ mol}\)\
\
3. Moles of \(\text{H}_2\text{C}_2\text{O}_4\) in the original \(250.0\text{ cm}^3\) volumetric flask:\
\(n(\text{total}) = 1.19 \times 10^{-3} \times 10 = 1.19 \times 10^{-2}\text{ mol}\)\
\
4. Mass of pure hydrated ethanedioic acid:\
\(M_r(\text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O}) = 2(1.0) + 2(12.0) + 4(16.0) + 2(18.0) = 126.0\text{ g mol}^{-1}\) (or using more precise values: 126.0)\
\(\text{Mass} = 1.19 \times 10^{-2}\text{ mol} \times 126.0\text{ g mol}^{-1} = 1.4994\text{ g}\)\
\
5. Percentage purity:\
\(\text{Percentage purity} = \frac{1.4994}{1.88} \times 100 = 79.755\\% \approx 79.8\\%\)

(a)\
- M1: For correct reactants and products: \(2\text{MnO}_4^- + 5\text{H}_2\text{C}_2\text{O}_4 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}\) [1 mark]\
- M2: For a fully balanced equation including charge and atoms [1 mark]\
\
(b)\
- M1: Autocatalysis is a process where a product of the reaction acts as a catalyst [1 mark]\
- M2: The catalyst is \(\text{Mn}^{2+}\) / manganese(II) ions [1 mark]\
\
(c)\
- M1: To provide the activation energy required because the reaction is slow at first / high activation energy [1 mark]\
\
(d)\
- M1: Calculates moles of \(\text{MnO}_4^-\) correctly as \(4.76 \times 10^{-4}\text{ mol}\) [1 mark]\
- M2: Calculates moles of \(\text{H}_2\text{C}_2\text{O}_4\) in \(25.0\text{ cm}^3\) correctly by multiplying by 2.5: \(1.19 \times 10^{-3}\text{ mol}\) [1 mark]\
- M3: Scales up to \(250.0\text{ cm}^3\) correctly: \(1.19 \times 10^{-2}\text{ mol}\) [1 mark]\
- M4: Calculates mass of pure hydrated ethanedioic acid: \(1.4994\text{ g}\) (accept \(1.50\text{ g}\) if rounded) [1 mark]\
- M5: Calculates percentage purity as \(79.8\\%\) (accept range 79.7% to 80.0% depending on intermediate rounding; must be 3 significant figures) [1 mark]

(a) Error 1 (Thermometer bulb too high): The thermometer will not be in contact with the vapour distilling over, resulting in a temperature reading that is lower than the actual boiling point of the distillate. \
Error 2 (Sealed system): Heating a closed system causes gas expansion, leading to a build-up of pressure and risking an explosion / apparatus shattering.\
\
(b) (i) Sodium hydrogencarbonate is added to neutralise any acid (phosphoric acid) impurities remaining in the crude product. \
Equation: \(\text{H}^+ + \text{HCO}_3^- \rightarrow \text{CO}_2 + \text{H}_2\text{O}\) or \(\text{H}_3\text{PO}_4 + 3\text{HCO}_3^- \rightarrow \text{PO}_4^{3-} + 3\text{CO}_2 + 3\text{H}_2\text{O}\)\
(ii) The neutralisation reaction produces carbon dioxide gas, which builds up pressure. Venting releases this gas safely.\
\
(c) Suitable anhydrous drying agent: anhydrous calcium chloride (\(\text{CaCl}_2\)) or anhydrous sodium sulfate (\(\text{Na}_2\text{SO}_4\)) or anhydrous magnesium sulfate (\(\text{MgSO}_4\)). (Do not accept silica gel or anhydrous cobalt chloride).\
How to know it is dry: The liquid changes from cloudy to clear/transparent, or the drying agent stops clumping together and forms a free-flowing powder when swirled.\
\
(d) Calculation:\
- Moles of cyclohexanol reacted: \(n = \frac{12.0}{100.0} = 0.120\text{ mol}\)\
- Theoretical yield of cyclohexene: \(0.120\text{ mol}\) (since the stoichiometry is 1:1)\
- Theoretical mass of cyclohexene: \(0.120 \times 82.0 = 9.84\text{ g}\)\
- Percentage yield: \(\frac{5.90}{9.84} \times 100 = 59.959\\% \approx 60.0\\%\)

(a)\
- M1: (Thermometer bulb too high): The boiling point recorded will be lower than actual / inaccurate [1 mark]\
- M2: (Sealed system): Explains that heating a closed system leads to pressure build-up / risk of explosion [1 mark]\
\
(b)\
- M1 (i): Neutralise the acid / phosphoric acid AND writes the correct ionic equation: \(\text{H}^+ + \text{HCO}_3^- \rightarrow \text{CO}_2 + \text{H}_2\text{O}\) [1 mark]\
- M2 (ii): To release the pressure built up by carbon dioxide gas [1 mark]\
\
(c)\
- M1: Identifies a correct drying agent: anhydrous calcium chloride / calcium sulfate / magnesium sulfate / sodium sulfate [1 mark]\
- M2: Liquid goes from cloudy to clear / drying agent is no longer clumping (free-flowing powder) [1 mark]\
\
(d)\
- M1: Calculates theoretical moles of reactant/product: \(0.120\text{ mol}\) [1 mark]\
- M2: Calculates theoretical mass: \(9.84\text{ g}\) [1 mark]\
- M3: Calculates percentage yield to 3 significant figures: \(60.0\\%\) [1 mark]

(a) Place the reaction mixture (beaker or flask) over a piece of paper with a black cross drawn on it. Measure the time taken, \(t\), from mixing the reactants until the cross is no longer visible when viewed from above due to the precipitation of colloidal sulfur.\
\
(b) Sulfur dioxide (\(\text{SO}_2\)) gas is toxic/irritating to the lungs. Precaution: Use a lid/stopper where possible, perform the reaction in a well-ventilated laboratory or fume cupboard, or use low concentrations to minimise gas release.\
\
(c) Calculation:\
- The Arrhenius equation in logarithmic form is: \(\ln k = -\frac{E_a}{RT} + \ln A\)\
- If we assume rate \(\propto k\), then \(\ln(\text{rate}) \approx -\frac{E_a}{RT} + \text{constant}\)\
- Plotting \(\ln\left(\frac{1}{t}\right)\) against \(\frac{1}{T}\) gives a gradient \(m = -\frac{E_a}{R}\)\
- \(-5640 = -\frac{E_a}{8.31}\) \
- \(E_a = 5640 \times 8.31 = 46868.4\text{ J mol}^{-1}\) \
- In \(\text{kJ mol}^{-1}\): \(E_a = 46.9\text{ kJ mol}^{-1}\) (to 3 significant figures).\
\
(d) The temperature should be controlled by placing the tubes/flasks containing the reactants in a water bath before mixing until they reach thermal equilibrium. The temperature must be monitored using a thermometer immediately before mixing and/or immediately after the cross disappears to find the mean temperature during the run.\
\
(e) The amount of sulfur precipitate required to obscure the cross is constant in each run. Therefore, the same amount of product is formed in each case, and rate is inversely proportional to time: \(\text{rate} \propto \frac{1}{t}\).

(a)\
- M1: Place the reaction vessel over a marked cross on paper [1 mark]\
- M2: Record the time taken for the cross to be completely obscured/disappear [1 mark]\
\
(b)\
- M1: Toxic / acidic / respiratory irritant gas (sulfur dioxide) [1 mark]\
- M2: Perform experiment in a fume cupboard / well-ventilated room [1 mark]\
\
(c)\
- M1: Connects gradient to the equation: \(\text{gradient} = -\frac{E_a}{R}\) [1 mark]\
- M2: Correct calculation of \(E_a\) in Joules: \(46868\text{ J mol}^{-1}\) (or \(46.9\text{ kJ mol}^{-1}\)) [1 mark]\
- M3: Correct conversion to \(\text{kJ mol}^{-1}\) and given to 3 significant figures: \(46.9\) [1 mark]\
\
(d)\
- M1: Use a water bath to heat / maintain the temperature of both solutions before mixing [1 mark]\
- M2: Use a thermometer to measure the temperature of the reaction mixture at the start and end of the reaction (and take an average) [1 mark]\
\
(e)\
- M1: A constant/fixed amount of sulfur is produced to obscure the cross [1 mark]

Moles of propanoic acid initially = \(0.0500 \times 0.100 = 0.00500\text{ mol}\). Moles of \(\text{NaOH}\) added = \(0.0250 \times 0.120 = 0.00300\text{ mol}\). The reaction is: \(\text{CH}_3\text{CH}_2\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{H}_2\text{O}\). Moles of propanoic acid remaining = \(0.00500 - 0.00300 = 0.00200\text{ mol}\). Moles of propanoate ions formed = \(0.00300\text{ mol}\). Using the \(K_{\text{a}}\) expression: \([\text{H}^+] = K_{\text{a}} \times \frac{[\text{acid}]}{[\text{salt}]} = 1.35 \times 10^{-5} \times \frac{0.00200}{0.00300} = 9.00 \times 10^{-6}\text{ mol dm}^{-3}\). \(\text{pH} = -\log_{10}(9.00 \times 10^{-6}) = 5.05\).

1 mark for correct selection of D (pH = 5.05).

From Experiment 1 and 2, doubling \([\text{NO}]\) with constant \([\text{H}_2]\) increases the rate by a factor of 4, so the order with respect to \(\text{NO}\) is 2. From Experiment 2 and 3, doubling \([\text{H}_2]\) with constant \([\text{NO}]\) doubles the rate, so the order with respect to \(\text{H}_2\) is 1. The rate equation is \(\text{Rate} = k[\text{NO}]^2[\text{H}_2]\). The overall order is 3. The units of \(k\) are: \(\frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{ dm}^6\text{ s}^{-1}\).

1 mark for correct selection of B.

The titration is self-indicating because the manganate(VII) ion, \(\text{MnO}_4^-\), is intense purple, while the product, \(\text{Mn}^{2+}\), is virtually colourless. The first permanent pale pink colour of excess \(\text{MnO}_4^-\) indicates the end-point. The reaction must be heated initially because it is slow at room temperature, and it is autocatalyzed by \(\text{Mn}^{2+}\) so the rate increases as the reaction proceeds. The stoichiometry is 2 moles of \(\text{MnO}_4^-\text{ to }5\text{ moles of }\text{C}_2\text{O}_4^{2-}\).

1 mark for correct selection of B.

The half-cell with the more positive potential (+0.77 V) undergoes reduction: \(\text{Fe}^{3+} + \text{e}^- \rightarrow \text{Fe}^{2+}\). The half-cell with the less positive potential (+0.15 V) undergoes oxidation: \(\text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2\text{e}^-\). Combining these and balancing the electrons gives: \(2\text{Fe}^{3+}(\text{aq}) + \text{Sn}^{2+}(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{Sn}^{4+}(\text{aq})\). The standard cell potential is \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = 0.77 - 0.15 = +0.62\text{ V}\).

1 mark for correct selection of A.

Testing with sodium carbonate solution causes effervescence (due to carbon dioxide release) only with hexanoic acid, identifying it first. Of the remaining three liquids, Fehling's solution reacts only with hexanal to form a brick-red precipitate, identifying it second. Finally, bromine water reacts only with the alkene hex-1-ene (decolourising the orange bromine water), identifying it third, and leaving hexan-1-ol identified by elimination as the unreacted liquid.

1 mark for correct selection of C.

According to the Born-Haber cycle for \(\text{MgCl}_2(\text{s})\): \(\Delta H^\ominus_{\text{f}} = \Delta H^\ominus_{\text{at}}(\text{Mg}) + \text{IE}_1(\text{Mg}) + \text{IE}_2(\text{Mg}) + 2 \times \Delta H^\ominus_{\text{at}}(\text{Cl}) + 2 \times \text{EA}_1(\text{Cl}) + \Delta H^\ominus_{\text{L,form}}\). Substituting the values: \(-642 = 148 + 738 + 1451 + 2(121) + 2(-349) + \Delta H^\ominus_{\text{L,form}}\). \(-642 = 1881 + \Delta H^\ominus_{\text{L,form}}\). Therefore, \(\Delta H^\ominus_{\text{L,form}} = -642 - 1881 = -2523\text{ kJ mol}^{-1}\).

1 mark for correct selection of B.

Chromium(III) ions exist as green-violet solutions in water. Upon adding dropwise NaOH, they form a green precipitate of chromium(III) hydroxide, \(\text{Cr(OH)}_3(\text{H}_2\text{O})_3\). Since this hydroxide is amphoteric, it reacts with excess hydroxide ions to form the soluble dark green complex \([\text{Cr(OH)}_6]^{3-}\). \(\text{Cr(OH)}_3\) does not dissolve in excess aqueous ammonia under standard conditions, which distinguishes it from other cations. This matches option B.

1 mark for correct selection of B.

Iodide ions, \(\text{I}^-\), are strong reducing agents and reduce concentrated sulfuric acid to sulfur dioxide (choking gas), sulfur (yellow solid), and hydrogen sulfide (rotten egg smell). The iodide ions themselves are oxidised to iodine, which is observed as a purple vapour and a grey-black solid. Bromide ions only reduce sulfuric acid to sulfur dioxide, and chloride ions are not oxidised by sulfuric acid at all.

1 mark for correct selection of B.

According to the Arrhenius equation, \(k = A e^{-E_a/RT}\). Taking natural logarithms gives: \(\ln(k) = \ln(A) - \frac{E_a}{R}\left(\frac{1}{T}\right)\). Since \(k\) is proportional to \(1/t\), we can substitute \(1/t\) for \(k\) (up to a constant multiplier), which yields: \(\ln(1/t) = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \text{constant}\). Thus, plotting \(\ln(1/t)\) on the y-axis against \(1/T\) on the x-axis gives a straight line with a gradient of \(-\frac{E_a}{R}\).

1 mark: Correctly identifies that plotting \(\ln(1/t)\) against \(1/T\) yields the required straight line.

The balanced redox equation is: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}\). Thus, \(1\text{ mol}\) of \(\text{Cr}_2\text{O}_7^{2-}\) reacts with \(6\text{ mol}\) of \(\text{Fe}^{2+}\). Moles of \(\text{Cr}_2\text{O}_7^{2-}\) used = \(0.0225\text{ dm}^3 \times 0.0160\text{ mol dm}^{-3} = 3.60 \times 10^{-4}\text{ mol}\). Moles of \(\text{Fe}^{2+}\) in the \(25.0\text{ cm}^3\) aliquot = \(6 \times 3.60 \times 10^{-4}\text{ mol} = 2.16 \times 10^{-3}\text{ mol}\). Moles of \(\text{Fe}^{2+}\) in the original \(250\text{ cm}^3\) solution = \(2.16 \times 10^{-3}\text{ mol} \times 10 = 2.16 \times 10^{-2}\text{ mol}\). Mass of iron in the alloy = \(2.16 \times 10^{-2}\text{ mol} \times 55.8\text{ g mol}^{-1} = 1.205\text{ g}\). Percentage by mass of iron = \(\left(\frac{1.205\text{ g}}{1.80\text{ g}}\right) \times 100\% = 67.0\%\).

1 mark: Correctly calculates the percentage by mass as 67.0%.

The standard electrode potential of the \(\text{Ag}^+/\text{Ag}\) half-cell is more positive than that of the \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell. Therefore, under standard conditions, \(\text{Ag}^+\text{(aq)}\) is reduced to \(\text{Ag(s)}\), and \(\text{Fe}^{2+}\text{(aq)}\) is oxidized to \(\text{Fe}^{3+}\text{(aq)}\). The spontaneous reaction is \(\text{Fe}^{2+}(\text{aq}) + \text{Ag}^+(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{Ag(s)}\). The EMF of this cell is \(+0.80 - 0.77 = +0.03\text{ V}\). Adding \(\text{NaCl(aq)}\) to the right-hand half-cell precipitates \(\text{Ag}^+\) as \(\text{AgCl(s)}\), reducing \([\text{Ag}^+]\), which shifts the reduction equilibrium to the left and decreases the EMF.

1 mark: Correctly identifies D as the only true statement.

Acid hydrolysis of an ester produces a carboxylic acid and an alcohol. Since \(\text{X}\) is oxidized to \(\text{Z}\), \(\text{X}\) must be the alcohol product. If \(\text{Z}\) does not react with sodium carbonate, it is not a carboxylic acid, meaning \(\text{X}\) must be a secondary alcohol (which oxidizes to a ketone) rather than a primary alcohol (which oxidizes to an aldehyde/carboxylic acid). The only secondary alcohol possible with a total carbon count of at most 4 in the ester is propan-2-ol. Therefore, the alcohol \(\text{X}\) is propan-2-ol, and the remaining acid portion must contain 1 carbon (methanoic acid) to give the formula \(\text{C}_4\text{H}_8\text{O}_2\). The ester is isopropyl methanoate (1-methylethyl methanoate).

1 mark: Correctly identifies the ester as isopropyl methanoate based on secondary alcohol oxidation.

The half-equivalence point occurs when exactly half the volume of \(\text{NaOH}\) required to reach the equivalence point has been added, which is \(12.5\text{ cm}^3\). At the half-equivalence point, the concentration of the weak acid \([\text{HA}]\) equals the concentration of its conjugate base \([\text{A}^-]\). According to the buffer equation, \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\), which simplifies to \(K_a = [\text{H}^+]\) (or \(\text{p}K_a = \text{pH}\)). Therefore, \(\text{p}K_a = 4.76\), and \(K_a = 10^{-4.76} = 1.74 \times 10^{-5}\text{ mol dm}^{-3}\).

1 mark: Correctly calculates \(K_a\) using the half-equivalence point relationship.

In TLC on silica gel, the stationary phase is highly polar (due to silanol groups). More polar molecules interact more strongly with the stationary phase, meaning they travel more slowly and have a lower \(R_f\) value. Non-polar molecules have weaker interactions with the polar stationary phase and are more soluble in the organic mobile phase, moving faster and having a larger \(R_f\) value. Lysine has a highly polar/charged side chain (hydrophilic), alanine has a small hydrophobic methyl group (intermediate), and leucine has a larger hydrophobic hydrocarbon side chain (least polar). Thus, the order of increasing \(R_f\) is Lysine < Alanine < Leucine.

1 mark: Correctly relates polarity of amino acid side chains to retention on silica gel.

The first electron affinity of oxygen (A) is exothermic because the electron is attracted to the neutral oxygen nucleus. The second electron affinity of oxygen (C) requires adding an electron to a negatively charged \(\text{O}^-\text{(g)}\) ion. This process involves overcoming strong electrostatic repulsion between the negative ion and the incoming electron, which requires an input of energy, making it endothermic. Lattice formation (B) and condensation (D) are both exothermic.

1 mark: Correctly identifies the second electron affinity of oxygen as endothermic.

The complex \([\text{Cu(NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\) is deep blue in color because it transmits blue light and absorbs its complementary color, which is orange-red. In colorimetry, the filter selected must be of the color that is absorbed most strongly by the sample (the complementary color of the solution). This maximizes the absorbance change per unit concentration, leading to the highest sensitivity. Therefore, an orange-red filter must be used.

1 mark: Correctly identifies that the complementary orange-red filter is chosen because the blue complex absorbs orange-red light.

Carbon dioxide is moderately soluble in water. When collecting gas over water, some of the carbon dioxide dissolves in the water, resulting in a recorded volume that is lower than the actual volume produced. Measuring mass loss on a balance avoids this source of error because it directly registers the mass of the escaping gas, regardless of its solubility properties.

1 mark for selecting A (recognising that solubility of carbon dioxide in water affects volume measurements).

The standard cell potential is calculated using \(E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}}\). The half-reaction with the more positive potential occurs as reduction (at the cathode): \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\) (\(E^\circ = +1.51\text{ V}\)). The other half-reaction occurs as oxidation (at the anode): \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-\). Therefore, \(E^\circ_{\text{cell}} = +1.51 - (+0.77) = +0.74\text{ V}\).

1 mark for the correct calculation of standard cell potential, option B.

The addition of excess ammonia to aqueous copper(II) ions yields the deep blue tetraamminediachorocopper(II) complex, \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\). To maximize sensitivity in colorimetry, a filter of the complementary color is used. Since the solution is deep blue, it absorbs orange/red light most strongly, so an orange/red filter should be selected.

1 mark for selecting B (correct complex formula and complementary filter combination).

In the dehydration of cyclohexanol, the crude cyclohexene is first obtained by distillation. To remove acidic impurities carried over from the acid catalyst, it is washed with sodium hydrogencarbonate solution in a separating funnel. The organic layer is then separated and dried using an anhydrous inorganic salt like calcium chloride to remove residual water. Finally, the pure product is collected via a second fractional distillation.

1 mark for option A, identifying the correct sequence of purification steps for a volatile liquid product.

Adding 12.5 cm3 of sodium hydroxide to 25.0 cm3 of propanoic acid of the same concentration reaches the half-neutralisation point. At this point, the concentration of the weak acid equals the concentration of its conjugate base (\([\text{HA}] = [\text{A}^-]\)). According to the Henderson-Hasselbalch equation, \(\text{pH} = \text{p}K_a + \log([\text{A}^-]/[\text{HA}])\), which simplifies to \(\text{pH} = \text{p}K_a\). Therefore, \(\text{pH} = -\log_{10}(1.35 \times 10^{-5}) = 4.87\).

1 mark for B (calculating pH = pKa at the half-neutralisation point).

For a reaction to be feasible, \(\Delta G \le 0\). At the threshold of feasibility, \(\Delta G = \Delta H - T\Delta S = 0\), which gives \(T = \frac{\Delta H}{\Delta S}\). Converting the enthalpy change to joules: \(\Delta H = 178000 \text{ J mol}^{-1}\). Thus, \(T = \frac{178000}{160} = 1112.5 \text{ K}\). Rounding to the nearest whole number gives 1113 K.

1 mark for C, correct calculation of temperature in Kelvin.

The silver mirror test confirms the presence of an aldehyde group, which also explains the C=O stretch at 1715 cm-1. The broad absorption at 3350 cm-1 indicates an O-H group (specifically an alcohol O-H, as carboxylic acid O-H typically centers lower, around 2500-3000 cm-1, and carboxylic acids do not reduce Tollens' reagent). 2-hydroxypropanal has both an alcohol O-H group and an aldehyde group, fitting all the observations.

1 mark for C (correct deduction of functional groups from chemical tests and IR data).

The rate of hydrolysis depends on the carbon-halogen bond strength. The C-I bond is the weakest (lowest bond enthalpy) and breaks most easily, so 1-iodobutane hydrolyses fastest. The iodide ions released react with silver ions to form a yellow precipitate of silver iodide (AgI).

1 mark for C (identifying the correct halogenoalkane and the correct precipitate color).

First, calculate the moles of the acid and base reactant:
\(n(\text{CH}_3\text{CH}_2\text{COOH}) = 0.0300\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 3.00 \times 10^{-3}\text{ mol}\)
\(n(\text{NaOH}) = 0.0200\text{ dm}^3 \times 0.150\text{ mol dm}^{-3} = 3.00 \times 10^{-3}\text{ mol}\)

Since the moles of propanoic acid and sodium hydroxide are equal, this is the equivalence point. All acid has been converted to sodium propanoate (\(\text{CH}_3\text{CH}_2\text{COONa}\)).

The total volume of the solution is:
\(20.0\text{ cm}^3 + 30.0\text{ cm}^3 = 50.0\text{ cm}^3 = 0.0500\text{ dm}^3\)

The concentration of propanoate ions, \([\text{A}^-]\), is:
\([\text{A}^-] = \frac{3.00 \times 10^{-3}\text{ mol}}{0.0500\text{ dm}^3} = 0.0600\text{ mol dm}^{-3}\)

For a solution containing only a weak conjugate base, we use the hydrolysis equation:
\([\text{H}^+] = \sqrt{\frac{K_w K_a}{[\text{A}^-]}}\)
\([\text{H}^+] = \sqrt{\frac{(1.00 \times 10^{-14}) \times (1.35 \times 10^{-5})}{0.0600}} = 1.50 \times 10^{-9}\text{ mol dm}^{-3}\)

\(\text{pH} = -\log_{10}(1.50 \times 10^{-9}) = 8.82\)

Award 1 mark for the correct option (C).

Incorrect pathways include:
- Option A (4.87): This is the \(\text{p}K_a\) of the acid, representing the pH at the half-equivalence point.
- Option B (5.18): This is the \(\text{pOH}\) of the equivalence solution instead of pH.
- Option D (9.13): Incorrect alkaline pH calculation.

The cell reaction is:
\(2\text{Cr}(\text{s}) + 6\text{H}^+(\text{aq}) \to 2\text{Cr}^{3+}(\text{aq}) + 3\text{H}_2(\text{g})\)

According to Le Chatelier's principle and the Nernst relationship:
- Diluting the \(\text{Cr}^{3+}(\text{aq})\) solution by adding water reduces the concentration of \(\text{Cr}^{3+}\).
- This shifts the oxidation-reduction equilibrium at the left-hand electrode to the left, making the electrode potential of the chromium half-cell more negative than its standard value (\(-0.74\text{ V}\)).
- Because \(E_{\text{cell}} = E_{\text{RHS}} - E_{\text{LHS}}\), reducing the potential of the left-hand electrode (making it more negative) results in a larger overall cell electromotive force.

Award 1 mark for the correct option (B).

Incorrect pathways include:
- Option A: Increasing \([\text{Cr}^{3+}]\) shifts the chromium potential to a less negative value, reducing the cell EMF.
- Option C: Adding sodium hydroxide decreases \([\text{H}^+]\) which decreases the potential of the standard hydrogen electrode, lowering the cell EMF.
- Option D: Increasing the pressure of \(\text{H}_2\) shifts the SHE potential to a more negative value, lowering the cell EMF.

The transition metal species is \([\text{Fe}(\text{H}_2\text{O})_6]^{2+}\):
- Adding a few drops of aqueous ammonia precipitates green iron(II) hydroxide, \([\text{Fe}(\text{H}_2\text{O})_4(\text{OH})_2]\).
- This precipitate is insoluble in excess ammonia because ammonia is not a sufficiently strong ligand to substitute hydroxide and form amine complexes with iron(II).
- Adding concentrated hydrochloric acid results in ligand substitution, forming the tetrahedral tetrachloroferrate(II) ion, \([\text{FeCl}_4]^{2-}\), which is a yellow-green solution.

Award 1 mark for the correct option (B).

Incorrect options:
- Option A: \([\text{Cu}(\text{H}_2\text{O})_6]^{2+}\) forms a blue precipitate that dissolves in excess ammonia to a deep blue solution.
- Option C: \([\text{Cr}(\text{H}_2\text{O})_6]^{3+}\) forms a green precipitate that dissolves in excess ammonia to a purple solution.
- Option D: \([\text{Al}(\text{H}_2\text{O})_6]^{3+}\) is not a transition metal ion and forms a white precipitate.

During the initial distillation, some of the acid catalyst (e.g., phosphoric acid or sulfuric acid) may vaporize or be carried over along with the cyclohexene and water. Washing the distillate mixture with aqueous sodium hydrogencarbonate (\(\text{NaHCO}_3\)) neutralizes these acid impurities, converting them into water-soluble salts that can be separated into the aqueous layer.

Award 1 mark for the correct option (B).

Reject options stating sodium hydrogencarbonate is a drying agent (which is calcium chloride or sodium sulfate), or that it separates cyclohexanol (which requires distillation due to boiling point differences).

A reaction becomes feasible when \(\Delta G^{\theta} \le 0\).
Using the Gibbs free energy relationship:
\(\Delta G^{\theta} = \Delta H^{\theta} - T\Delta S^{\theta}\)

To find the temperature where feasibility begins, set \(\Delta G^{\theta} = 0\):
\(T = \frac{\Delta H^{\theta}}{\Delta S^{\theta}}\)

Convert \(\Delta H^{\theta}\) from \(\text{kJ mol}^{-1}\) to \(\text{J mol}^{-1}\):
\(\Delta H^{\theta} = +178 \times 10^3\text{ J mol}^{-1}\)

Calculate \(T\):
\(T = \frac{178000}{161} = 1105.59\text{ K} \approx 1106\text{ K}\)

Award 1 mark for the correct option (C).

Incorrect pathways include:
- Option A: Forgetting to convert \(\text{kJ}\) to \(\text{J}\) (\(178 / 161 = 1.11\text{ K}\)).
- Option B: Random error value.
- Option D: Error in formula manipulation.

First, find the concentration of \(\text{Cu}^{2+}\) from the calibration equation:
\(c = \frac{A}{12.4} = \frac{0.310}{12.4} = 0.0250\text{ mol dm}^{-3}\)

Calculate the moles of \(\text{Cu}^{2+}\) in the volumetric flask volume of \(100\text{ cm}^3\) (\(0.100\text{ dm}^3\)):
\(n(\text{Cu}^{2+}) = 0.0250\text{ mol dm}^{-3} \times 0.100\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol}\)

Calculate the mass of copper:
\(m(\text{Cu}) = 2.50 \times 10^{-3}\text{ mol} \times 63.5\text{ g mol}^{-1} = 0.15875\text{ g}\)

Calculate the percentage by mass of copper in the brass alloy:
\(\%\text{ mass} = \frac{0.15875\text{ g}}{0.450\text{ g}} \times 100 = 35.28\% \approx 35.3\%\)

Award 1 mark for the correct option (B).

Incorrect pathways include:
- Option A: Error in mass scaling.
- Option D: Forgetting the dilution factor of 100 to 1000, leading to doubling error.