2025 Edexcel IAL Chemistry (YCH11) Practice Paper with Answers

The ground-state electronic configurations of the 3D transition metal ions are: \( \text{Fe}^{3+} \) has the configuration \( [\text{Ar}] 3\text{d}^5 \) with 5 unpaired electrons; \( \text{Co}^{2+} \) has the configuration \( [\text{Ar}] 3\text{d}^7 \) with 3 unpaired electrons; \( \text{Ni}^{2+} \) has the configuration \( [\text{Ar}] 3\text{d}^8 \) with 2 unpaired electrons; \( \text{Cu}^{2+} \) has the configuration \( [\text{Ar}] 3\text{d}^9 \) with 1 unpaired electron. Therefore, \( \text{Fe}^{3+} \) has the greatest number of unpaired d-electrons.

1 mark for identifying that \( \text{Fe}^{3+} \) has 5 unpaired electrons which is the greatest among the options.

First, calculate the amount of gas using the ideal gas equation: \( n = \frac{PV}{RT} = \frac{1.00 \times 10^5\text{ Pa} \times 59.0 \times 10^{-6}\text{ m}^3}{8.31\text{ J mol}^{-1}\text{ K}^{-1} \times 298\text{ K}} = 0.00238\text{ mol} \). The molar mass is \( M_r = \frac{\text{mass}}{n} = \frac{0.100\text{ g}}{0.00238\text{ mol}} = 42.0\text{ g mol}^{-1} \). The empirical formula ratio: \( \text{C} = \frac{85.7}{12.0} = 7.14 \), \( \text{H} = \frac{14.3}{1.0} = 14.3 \), which gives an empirical formula of \( \text{CH}_2 \) (formula mass = \( 14.0 \)). The molecular formula is therefore \( 3 \times \text{CH}_2 = \text{C}_3\text{H}_6 \).

1 mark for calculating the correct molar mass and deducing the molecular formula of \( \text{C}_3\text{H}_6 \).

Sulfur tetrafluoride (\( \text{SF}_4 \)) has 5 electron pairs around the central sulfur atom: 4 bonding pairs and 1 lone pair. This gives a trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position to minimize repulsion, resulting in a see-saw molecular shape.

1 mark for identifying \( \text{SF}_4 \) as having a see-saw molecular geometry.

Using Hess's law, the standard enthalpy change of formation is calculated from enthalpy changes of combustion: \( \Delta H_f^\ominus(\text{C}_3\text{H}_8\text{(g)}) = 3 \times \Delta H_c^\ominus(\text{C(s)}) + 4 \times \Delta H_c^\ominus(\text{H}_2\text{(g)}) - \Delta H_c^\ominus(\text{C}_3\text{H}_8\text{(g)}) \). Substituting the values: \( \Delta H_f^\ominus = 3(-394) + 4(-286) - (-2220) = -1182 - 1144 + 2220 = -106\text{ kJ mol}^{-1} \).

1 mark for the correct calculation showing \( -106\text{ kJ mol}^{-1} \).

In a propagation step, a free radical reacts with a stable molecule to form a new radical and a new stable molecule. The reaction of butane with a chlorine radical to produce a butyl radical and hydrogen chloride, \( \text{C}_4\text{H}_{10} + \text{Cl}^\bullet \rightarrow \text{C}_4\text{H}_9^\bullet + \text{HCl} \), is the correct propagation step. Option A is initiation, option B is chemically incorrect as H radicals are not formed, and option D is a termination step.

1 mark for identifying the correct propagation step equation.

An \( \text{S}_\text{N}1 \) mechanism involves the formation of a carbocation intermediate. Tertiary carbocations are highly stable due to the inductive electron-donating effect of three surrounding alkyl groups, making tertiary halogenoalkanes like 2-bromo-2-methylpropane react almost exclusively via \( \text{S}_\text{N}1 \). Primary halogenoalkanes like 1-chlorobutane and 1-bromobutane react predominantly via \( \text{S}_\text{N}2 \).

1 mark for identifying the tertiary halogenoalkane 2-bromo-2-methylpropane.

Since \( \text{en} \) is a bidentate ligand, three \( \text{en} \) ligands form a total of six coordinate bonds with the cobalt ion, resulting in a coordination number of 6 and an octahedral geometry. This non-symmetrical arrangement lacking a plane of symmetry results in optical isomerism.

1 mark for choosing the statement describing octahedral geometry and optical isomerism.

Initial moles of ethanoic acid (HA) = \( 0.0500\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.00500\text{ mol} \). Initial moles of NaOH (OH-) = \( 0.0500\text{ dm}^3 \times 0.0500\text{ mol dm}^{-3} = 0.00250\text{ mol} \). The NaOH reacts completely with the acid: \( \text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O} \). Moles of HA remaining = \( 0.00500 - 0.00250 = 0.00250\text{ mol} \). Moles of A- formed = \( 0.00250\text{ mol} \). Since \( [\text{HA}] = [\text{A}^-] \), using the Henderson-Hasselbalch equation: \( \text{pH} = \text{p}K_a = -\log_{10}(1.80 \times 10^{-5}) = 4.74 \).

1 mark for calculating the correct buffer pH of 4.74.

To find the empirical formula:
1. Calculate the number of moles of each element in \( 100 \text{ g} \) of the compound:
- Moles of \( \text{Fe} = \frac{69.94}{55.8} = 1.253 \text{ mol} \)
- Moles of \( \text{O} = \frac{30.06}{16.0} = 1.879 \text{ mol} \)

2. Divide by the smallest number of moles:
- Ratio of \( \text{Fe} = \frac{1.253}{1.253} = 1 \)
- Ratio of \( \text{O} = \frac{1.879}{1.253} = 1.50 \)

3. Multiply by 2 to obtain the simplest whole-number ratio:
- \( \text{Fe} = 2 \)
- \( \text{O} = 3 \)

Therefore, the empirical formula is \( \text{Fe}_2\text{O}_3 \).

[1] C - Correct empirical formula based on mole calculations.

Using Hess's Law, the enthalpy change of formation of ethane can be calculated using combustion data:
\( \Delta H_{\text{f}}^\ominus [\text{C}_2\text{H}_6\text{(g)}] = 2 \Delta H_{\text{c}}^\ominus [\text{C(graphite)}] + 3 \Delta H_{\text{c}}^\ominus [\text{H}_2\text{(g)}] - \Delta H_{\text{c}}^\ominus [\text{C}_2\text{H}_6\text{(g)}] \)

\( \Delta H_{\text{f}}^\ominus = 2(-393.5) + 3(-285.8) - (-1560.7) \)
\( \Delta H_{\text{f}}^\ominus = -787.0 - 857.4 + 1560.7 \)
\( \Delta H_{\text{f}}^\ominus = -83.7 \text{ kJ mol}^{-1} \)

[1] B - Correct standard enthalpy of formation using Hess's Law.

1. Comparing Exp 1 and Exp 2: When \( [\text{A}] \) is doubled while \( [\text{B}] \) remains constant, the rate quadruples (from \( 2.0 \times 10^{-4} \) to \( 8.0 \times 10^{-4} \)). This indicates the order with respect to \( \text{A} \) is 2.
2. Comparing Exp 1 and Exp 3: When \( [\text{B}] \) is doubled while \( [\text{A}] \) remains constant, the rate doubles (from \( 2.0 \times 10^{-4} \) to \( 4.0 \times 10^{-4} \)). This indicates the order with respect to \( \text{B} \) is 1.
3. The overall order of the reaction is \( 2 + 1 = 3 \).
4. The rate equation is: \( \text{Rate} = k [\text{A}]^2 [\text{B}] \).
5. Units of \( k = \frac{\text{Rate}}{[\text{A}]^2 [\text{B}]} = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6 \text{ mol}^{-2} \text{ s}^{-1} \).

[1] C - Correct identification of overall order and units of the rate constant.

The atomic number of Iron (\( \text{Fe} \)) is 26, so an iron atom has 26 electrons. Its electronic configuration is \( 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^6 4\text{s}^2 \), or \( [\text{Ar}] 3\text{d}^6 4\text{s}^2 \).
When forming the \( \text{Fe}^{3+} \) ion, three electrons are removed. Electrons are lost from the \( 4\text{s} \) subshell first, then from the \( 3\text{d} \) subshell. Removing two electrons from \( 4\text{s} \) and one from \( 3\text{d} \) yields \( [\text{Ar}] 3\text{d}^5 \).

[1] A - Correctly identifies the configuration as [Ar] 3d5.

- Reaction with 2,4-DNPH shows \( \text{Y} \) has a carbonyl group (aldehyde or ketone).
- No reaction with Tollens' reagent shows \( \text{Y} \) is a ketone.
- The triiodomethane test (reaction with alkaline iodine) shows \( \text{Y} \) is a methyl ketone, meaning it contains the \( \text{CH}_3\text{CO}- \) group.
- Optically active means \( \text{Y} \) has a chiral carbon atom.

Analyzing the options:
- \( \text{hexan-3-one} \) is not a methyl ketone and has no chiral carbon.
- \( \text{4-methylpentan-2-one} \) is a methyl ketone but contains no chiral carbon.
- \( \text{2-methylpentanal} \) is an aldehyde, so it would react with Tollens' reagent.
- \( \text{3-methylpentan-2-one} \), \( \text{CH}_3\text{COCH(CH}_3)\text{CH}_2\text{CH}_3 \), is a ketone, a methyl ketone, and has a chiral center at carbon-3 (bonded to \( -\text{H} \), \( -\text{CH}_3 \), \( -\text{CH}_2\text{CH}_3 \), and \( -\text{COCH}_3 \)).

[1] D - Correct deduction based on the chemical tests and stereochemical properties.

The buffer pH can be calculated using the Henderson-Hasselbalch equation:
\( \text{pH} = \text{p}K_{\text{a}} + \log_{10} \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \)

First, calculate the \( \text{p}K_{\text{a}} \) value:
\( \text{p}K_{\text{a}} = -\log_{10}(1.35 \times 10^{-5}) = 4.87 \)

Since equal volumes of acid and salt solutions are mixed, we can use the ratio of the number of moles directly:
- Moles of propanoate, \( \text{A}^- = 0.0500 \text{ dm}^3 \times 0.100 \text{ mol dm}^{-3} = 0.00500 \text{ mol} \)
- Moles of propanoic acid, \( \text{HA} = 0.0500 \text{ dm}^3 \times 0.200 \text{ mol dm}^{-3} = 0.0100 \text{ mol} \)

Now, substitute the values:
\( \text{pH} = 4.87 + \log_{10} \left( \frac{0.00500}{0.0100} \right) = 4.87 + \log_{10}(0.500) = 4.87 - 0.301 = 4.57 \)

[1] B - Correct buffer pH calculated using Henderson-Hasselbalch equation.

For a reaction to be thermodynamically feasible, the standard Gibbs free energy change must be less than or equal to zero (\( \Delta G^\ominus \le 0 \)).
\( \Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus \le 0 \)
Therefore, \( T \ge \frac{\Delta H^\ominus}{\Delta S^\ominus} \).

Converting \( \Delta H^\ominus \) into Joules:
\( \Delta H^\ominus = +145000 \text{ J mol}^{-1} \)

\( T \ge \frac{145000}{310} = 467.7 \text{ K} \approx 468 \text{ K} \).

Since both enthalpy and entropy changes are positive, the reaction is feasible at high temperatures, which corresponds to temperatures above \( 468 \text{ K} \).

[1] A - Correctly calculates the temperature boundary and determines that the reaction is feasible at high temperatures.

Basicity depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton:
1. Phenylamine is the weakest base because the lone pair on the nitrogen atom is partially delocalised into the benzene \( \pi \)-ring system, making it less available.
2. Ammonia is stronger than phenylamine because it has no delocalisation, but it is weaker than alkylamines.
3. Ethylamine is stronger than ammonia due to the positive inductive effect of the ethyl group, which pushes electron density towards the nitrogen atom, making the lone pair more available.
4. Diethylamine is a secondary aliphatic amine with two electron-releasing ethyl groups, making it the most basic of the group in aqueous solution.

Thus, the increasing order of basicity is: phenylamine < ammonia < ethylamine < diethylamine.

[1] C - Correctly identifies the order of basicity of the amines.

The largest jump in successive ionisation energies occurs between the third and fourth ionisation energies (from \( 2745 \) to \( 11577 \text{ kJ mol}^{-1} \)). This indicates that the fourth electron is being removed from an inner quantum shell closer to the nucleus. Therefore, the element has three valence electrons and is in Group 13 (Group 3 of Period 3, which is aluminium). It will most likely form a \( \text{X}^{3+} \) ion.

[1 mark] Correctly identifies that the large jump in IE occurs between the 3rd and 4th IEs, indicating 3 valence electrons, and selects option C.

1. Calculate moles of propane: \( n(\text{C}_3\text{H}_8) = \frac{4.40 \text{ g}}{44.0 \text{ g mol}^{-1}} = 0.100 \text{ mol} \).\
2. Write the balanced equation: \( \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \).\
3. Determine moles of oxygen required: \( n(\text{O}_2) = 5 \times 0.100 \text{ mol} = 0.500 \text{ mol} \).\
4. Use ideal gas equation: \( V = \frac{nRT}{P} = \frac{0.500 \times 8.31 \times 323}{100 \times 10^3} = 0.01342 \text{ m}^3 \).\
5. Convert to \( \text{dm}^3 \): \( 0.01342 \times 1000 = 13.4 \text{ dm}^3 \).

[1 mark] Correctly calculates moles of oxygen and uses the ideal gas equation to find the volume in dm3, matching option C.

Using the Born-Haber cycle equation: \( \Delta H_f^\ominus = \Delta H_{at}^\ominus[\text{Na}(s)] + E_{i1}[\text{Na}(g)] + \Delta H_{at}^\ominus[\frac{1}{2}\text{Cl}_2(g)] + E_{ea}[\text{Cl}(g)] + \Delta H_{lat}^\ominus[\text{NaCl}(s)] \). \
\( \Delta H_f^\ominus = (+107) + (+496) + (+121) + (-349) + (-787) = -412 \text{ kJ mol}^{-1} \).

[1 mark] Correctly substitutes all enthalpy changes with correct signs into the Born-Haber equation to calculate -412 kJ mol-1 (option A).

The amide ion, \( \text{NH}_2^- \), has a central nitrogen atom with 5 valence electrons, plus 1 electron from the negative charge, making 6. It forms 2 single bonds with hydrogen atoms, leaving 2 lone pairs. According to electron-pair repulsion theory, the 2 bonding pairs and 2 lone pairs repel each other to adopt a bent (non-linear) shape. The presence of two lone pairs reduces the tetrahedral bond angle from \( 109.5^\circ \) to approximately \( 104.5^\circ \).

[1 mark] Correctly determines the shape and bond angle of the amide ion based on electron pair repulsion, selecting option C.

The rate of hydrolysis of halogenoalkanes is determined by both the carbocation stability during the mechanism (tertiary halogenoalkanes react via the very fast \( \text{S}_\text{N}1 \) mechanism) and the strength of the C-Halogen bond (C-I is weaker than C-Cl). Therefore, the tertiary iodoalkane, 2-iodo-2-methylpropane, reacts fastest.

[1 mark] Deduces that tertiary iodoalkanes hydrolyse fastest due to carbocation stability and weak C-I bond, selecting option D.

Halving the volume of the reaction vessel doubles the concentration of all gaseous or aqueous species. If the initial concentrations are \( [\text{A}] \) and \( [\text{B}] \), the new concentrations are \( 2[\text{A}] \) and \( 2[\text{B}] \). Substituting these into the rate equation gives: \( \text{New rate} = k(2[\text{A}])(2[\text{B}])^2 = 8 \times k[\text{A}][\text{B}]^2 \). Thus, the rate increases by a factor of 8.

[1 mark] Realises that halving the volume doubles concentration and applies the rate equation exponents to find a factor of 8 (option C).

For a reaction to be feasible, \( E^\ominus_{\text{cell}} \) must be positive. \
For \( \text{Fe}^{3+} \) to oxidise \( \text{I}^- \): \( E^\ominus_{\text{cell}} = E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\ominus(\text{I}_2/\text{I}^-) = +0.77 - 0.54 = +0.23 \text{ V} \) (feasible).\
For \( \text{Fe}^{3+} \) to oxidise \( \text{Br}^- \): \( E^\ominus_{\text{cell}} = E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\ominus(\text{Br}_2/\text{Br}^-) = +0.77 - 1.09 = -0.32 \text{ V} \) (not feasible).\
Thus, \( \text{Fe}^{3+}(aq) \) can oxidise \( \text{I}^-(aq) \) but not \( \text{Br}^-(aq) \).

[1 mark] Correctly calculates cell potential values to determine thermodynamic feasibility, identifying option B as correct.

The ground-state d-electron configurations of the ions are:\
- \( \text{Fe}^{3+} \) is \( 3d^5 \) (5 unpaired d-electrons due to Hund's rule)\
- \( \text{Co}^{2+} \) is \( 3d^7 \) (3 unpaired d-electrons)\
- \( \text{Ni}^{2+} \) is \( 3d^8 \) (2 unpaired d-electrons)\
- \( \text{Cu}^{2+} \) is \( 3d^9 \) (1 unpaired d-electron)\
Thus, \( \text{Fe}^{3+} \) has the highest number of unpaired d-electrons.

[1 mark] Correctly identifies the d-orbital electronic configurations for each ion and applies Hund's rule to select option A.

First, find the empirical formula. \(n(\text{C}) = 40.0/12 = 3.33\), \(n(\text{H}) = 6.7/1.0 = 6.7\), \(n(\text{O}) = 53.3/16 = 3.33\). The ratio is 1 : 2 : 1, so the empirical formula is \(\text{CH}_2\text{O}\) (empirical formula mass = \(30\text{ g mol}^{-1}\)). Next, find the molar mass of \(X\) using the ideal gas equation: \(pV = nRT \implies n = \frac{pV}{RT} = \frac{101 \times 10^3\text{ Pa} \times 460 \times 10^{-6}\text{ m}^3}{8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 373\text{ K}} \approx 0.0150\text{ mol}\). Molar mass \(M = \frac{1.80\text{ g}}{0.0150\text{ mol}} = 120\text{ g mol}^{-1}\). Divide the molar mass by the empirical formula mass: \(120 / 30 = 4\). Therefore, the molecular formula is \(\text{C}_4\text{H}_8\text{O}_4\).

1 mark: Correctly identifies D as the molecular formula.

The largest increase in ionization energy occurs between the 5th and the 6th ionization energies (from 6274 to 21269 \(\text{kJ mol}^{-1}\)), indicating that the 6th electron is removed from an inner quantum shell. Thus, element \(Y\) has 5 valence electrons and is in Group 15 of the Periodic Table (phosphorus). Group 15 elements form hydrides with the general formula \(\text{YH}_3\).

1 mark: Correctly identifies B as the formula of the hydride.

In \(\text{SF}_6\), the shape is octahedral with bond angles of exactly \(90^\circ\). In \(\text{NH}_3\), the shape is trigonal pyramidal with a bond angle of \(107^\circ\). In \(\text{H}_2\text{O}\), the shape is non-linear (bent) with a bond angle of \(104.5^\circ\). In \(\text{BF}_3\), the shape is trigonal planar with bond angles of \(120^\circ\). Therefore, \(\text{SF}_6\) has the smallest bond angle.

1 mark: Correctly identifies C as the molecule with the smallest bond angle.

The equation for the standard enthalpy change of formation of propane is: \(3\text{C(s)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}\). Using Hess's Law with combustion data: \(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products}) = [3 \times (-394) + 4 \times (-286)] - [-2220] = [-1182 - 1144] + 2220 = -2326 + 2220 = -106\text{ kJ mol}^{-1}\).

1 mark: Correctly calculates \(-106\text{ kJ mol}^{-1}\) (option A).

As you descend Group 2, the cationic radius increases while the ionic charge stays at \(+2\). Therefore, the \(\text{Mg}^{2+}\) ion has a smaller ionic radius and a higher charge density than the \(\text{Ba}^{2+}\) ion. The higher charge density of the magnesium ion allows it to more strongly polarize the electron cloud of the carbonate ion, weakening the carbon-oxygen bonds within the carbonate group and decreasing the thermal stability, leading to decomposition at a lower temperature.

1 mark: Correctly identifies option A as the best explanation.

In an \(S_N1\) mechanism, the first, rate-determining step is the slow heterolytic fission of the carbon-halogen bond (C-Br) to form a tertiary carbocation intermediate. The nucleophile (hydroxide ion) then attacks the carbocation in a fast second step, meaning the concentration of hydroxide ions does not affect the rate of reaction.

1 mark: Correctly identifies C as the correct statement.

A reaction becomes thermodynamically feasible when the standard Gibbs free energy change, \(\Delta G^\theta\), is less than zero. Since \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta\), we set \(\Delta G^\theta < 0 \implies \Delta H^\theta < T\Delta S^\theta \implies T > \frac{\Delta H^\theta}{\Delta S^\theta}\). Converting units: \(T > \frac{178 \times 1000\text{ J mol}^{-1}}{160\text{ J K}^{-1}\text{ mol}^{-1}} = 1112.5\text{ K}\). Therefore, the reaction becomes feasible at temperatures above \(1112.5\text{ K}\), which rounds to \(1113\text{ K}\).

1 mark: Correctly calculates the feasibility temperature threshold as 1113 K (option A).

For a weak monobasic acid, \([H^+] \approx \sqrt{K_a \times [HA]}\). Substituting the given values: \([H^+] = \sqrt{1.80 \times 10^{-5}\text{ mol dm}^{-3} \times 0.0500\text{ mol dm}^{-3}} = \sqrt{9.00 \times 10^{-7}} = 9.487 \times 10^{-4}\text{ mol dm}^{-3}\). The pH is calculated as: \(\text{pH} = -\log_{10}[H^+] = -\log_{10}(9.487 \times 10^{-4}) \approx 3.02\).

1 mark: Correctly calculates the pH of the weak acid as 3.02 (option B).

Phosphorus has the electronic configuration \([\text{Ne}] 3\text{s}^2 3\text{p}^3\) with three singly-occupied \(3\text{p}\) orbitals. Sulfur has the configuration \([\text{Ne}] 3\text{s}^2 3\text{p}^4\), meaning one of its \(3\text{p}\) orbitals contains a pair of electrons. The mutual repulsion between these paired electrons (spin-pair repulsion) makes it easier to remove one of them, resulting in a lower first ionization energy.

1 mark for the correct option (C).

Using the ideal gas equation: \(PV = nRT\)
Convert temperature to Kelvin: \(T = 127 + 273 = 400\text{ K}\)
Convert volume to \(\text{m}^3\): \(V = 83.1 \times 10^{-6}\text{ m}^3\)
Calculate moles: \(n = \frac{PV}{RT} = \frac{1.00 \times 10^5 \times 83.1 \times 10^{-6}}{8.31 \times 400} = 0.00250\text{ mol}\)
Calculate molar mass: \(M = \frac{\text{mass}}{n} = \frac{0.120\text{ g}}{0.00250\text{ mol}} = 48.0\text{ g mol}^{-1}\)

1 mark for the correct option (B).

The equation for the formation of methane is:
\(\text{C(s, graphite)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}\)
Using the enthalpy of combustion cycle:
\(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)
\(\Delta_f H^\theta = [-393.5 + 2(-285.8)] - [-890.3]\)
\(\Delta_f H^\theta = [-393.5 - 571.6] + 890.3 = -965.1 + 890.3 = -74.8\text{ kJ mol}^{-1}\)

1 mark for the correct option (A).

The reaction proceeds via a carbocation intermediate. Electrophilic addition of \(\text{H}^+\) to 2-methylbut-2-ene can form either a secondary carbocation or a tertiary carbocation. The tertiary carbocation is more stable than the secondary carbocation because it has three electron-donating alkyl groups that help to disperse the positive charge (positive inductive effect).

1 mark for the correct option (A).

Propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) contains an \(-\text{OH}\) group and can form intermolecular hydrogen bonds. Hydrogen bonding is significantly stronger than the permanent dipole-dipole forces present in propanal and 1-fluoropropane, and the London dispersion forces present in butane. Therefore, more energy is required to separate the molecules of propan-1-ol, giving it the highest boiling temperature.

1 mark for the correct option (A).

The species being reduced is \(\text{Fe}^{3+}\) (acting as the cathode): \(E^\theta_{\text{red}} = +0.77\text{ V}\).
The species being oxidized is \(\text{I}^-\) (acting as the anode): \(E^\theta_{\text{ox}} = +0.54\text{ V}\).
\(E^\theta_{\text{cell}} = E^\theta_{\text{red}} - E^\theta_{\text{ox}} = 0.77 - 0.54 = +0.23\text{ V}\).
Since \(E^\theta_{\text{cell}} > 0\), the forward reaction is thermodynamically feasible under standard conditions.

1 mark for the correct option (B).

For a weak acid, \([\text{H}^+] \approx \sqrt{K_a \times c}\).
\([\text{H}^+] = \sqrt{(1.00 \times 10^{-5}) \times 0.100} = \sqrt{1.00 \times 10^{-6}} = 1.00 \times 10^{-3}\text{ mol dm}^{-3}\).
\(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.00 \times 10^{-3}) = 3.00\).

1 mark for the correct option (B).

Phenylamine is the weakest base because the lone pair of electrons on the nitrogen atom is delocalized into the \(\pi\)-system of the benzene ring, making it much less available to accept a proton. Ammonia is stronger than phenylamine but weaker than ethylamine. Ethylamine is the strongest base of the three because the ethyl group is electron-donating via the inductive effect, increasing the electron density on the nitrogen atom and making the lone pair more available to accept a proton.

1 mark for the correct option (B).

The successive ionization energies show a very large increase (jump) between the third and fourth ionization energies (from 2745 to 11577 \( \text{kJ mol}^{-1} \)). This indicates that the fourth electron is removed from an inner shell, which is closer to the nucleus and experiences much less shielding. Therefore, element X has three valence electrons and belongs to Group 13. The stable ion formed is \( \text{X}^{3+} \), which combines with chloride ions (\( \text{Cl}^- \)) to form the stable chloride \( \text{XCl}_3 \).

1 mark: Correctly identifies \( \text{XCl}_3 \) (option C).

According to Avogadro's Law, the volumes of gases are directly proportional to the amount (moles) of the gases. The ratio of volumes of \( \text{N}_x\text{O}_y : \text{N}_2 : \text{O}_2 \) is \( 50 : 50 : 25 \), which simplifies to a mole ratio of \( 2 : 2 : 1 \). The balanced equation for the decomposition is: \( 2\text{N}_x\text{O}_y(g) \rightarrow 2\text{N}_2(g) + 1\text{O}_2(g) \). Balancing the nitrogen atoms: \( 2x = 4 \Rightarrow x = 2 \). Balancing the oxygen atoms: \( 2y = 2 \Rightarrow y = 1 \). Therefore, the molecular formula of the oxide is \( \text{N}_2\text{O} \).

1 mark: Correctly identifies \( \text{N}_2\text{O} \) (option C).

The equation for the combustion of propan-2-ol is: \( \text{C}_3\text{H}_8\text{O}(l) + 4.5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(l) \). Using Hess's law: \( \Delta H_c^\ominus = \sum \Delta H_f^\ominus(\text{products}) - \sum \Delta H_f^\ominus(\text{reactants}) \). Note that the standard enthalpy of formation of \( \text{O}_2(g) \) is zero. \( \Delta H_c^\ominus = [3 \times (-394) + 4 \times (-286)] - [-318] = [-1182 - 1144] + 318 = -2326 + 318 = -2008\text{ kJ mol}^{-1} \).

1 mark: Correctly calculates \( -2008\text{ kJ mol}^{-1} \) (option A).

The rate of hydrolysis of halogenoalkanes depends on two factors: the strength of the C-X bond (bond enthalpy) and the mechanism of substitution. The C-I bond is much weaker than the C-Br and C-Cl bonds, meaning iodoalkanes react fastest. Furthermore, tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react much faster via the stable tertiary carbocation in the \( S_N1 \) mechanism than primary or secondary halogenoalkanes. Therefore, 2-iodo-2-methylpropane is the fastest to hydrolyse and form a precipitate.

1 mark: Correctly identifies 2-iodo-2-methylpropane (option D).

Let the initial rate be \( R_1 = k[\text{NO}]^2[\text{H}_2] \). If the concentration of \( \text{NO} \) is halved to \( 0.5[\text{NO}] \), and the concentration of \( \text{H}_2 \) is doubled to \( 2[\text{H}_2] \), the new rate \( R_2 \) is: \( R_2 = k(0.5[\text{NO}])^2(2[\text{H}_2]) = k(0.25[\text{NO}]^2)(2[\text{H}_2]) = 0.5 \times k[\text{NO}]^2[\text{H}_2] = 0.5 \times R_1 \). Thus, the initial rate of reaction is halved.

1 mark: Correctly identifies that the rate is halved (option B).

The standard enthalpy change of atomisation, \( \Delta H_{at}^\ominus \), is defined as the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state. The standard state of bromine is liquid bromine, \( \text{Br}_2(l) \). The equation representing the formation of one mole of gaseous bromine atoms from its standard state is therefore: \( \frac{1}{2}\text{Br}_2(l) \rightarrow \text{Br}(g) \).

1 mark: Correctly identifies the reaction (option A).

The atomic number of Cobalt (Co) is 27. The electronic configuration of a neutral Co atom in its ground state is \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^7 4s^2 \), or \( [\text{Ar}] 3d^7 4s^2 \). When transition metals form cations, the outer \( s \) electrons are lost first. Therefore, when Co forms a \( \text{Co}^{2+} \) ion, the two electrons in the \( 4s \) orbital are removed, leaving the ground state configuration as \( [\text{Ar}] 3d^7 4s^0 \).

1 mark: Correctly identifies the configuration as \( [\text{Ar}] 3d^7 4s^0 \) (option B).

For a weak monoprotic acid, the hydrogen ion concentration can be approximated using: \( [\text{H}^+] \approx \sqrt{K_a \times [\text{HA}]} \). Substituting the values: \( [\text{H}^+] = \sqrt{(1.00 \times 10^{-5}) \times 0.100} = \sqrt{1.00 \times 10^{-6}} = 1.00 \times 10^{-3}\text{ mol dm}^{-3} \). The pH is then: \( \text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.00 \times 10^{-3}) = 3.00 \).

1 mark: Correctly calculates the pH to be 3.00 (option B).

The volume of carbon dioxide produced is equal to the volume decrease when passed through potassium hydroxide: \(55 - 25 = 30\text{ cm}^3\). Since \(10\text{ cm}^3\) of the hydrocarbon was combusted, each molecule contains \(30 / 10 = 3\) carbon atoms. The volume of oxygen remaining is \(25\text{ cm}^3\), meaning the volume of oxygen reacted is \(70 - 25 = 45\text{ cm}^3\). The mole ratio of hydrocarbon to reacting oxygen is \(10 : 45\), which is \(1 : 4.5\). The equation for the complete combustion of a hydrocarbon \(\text{C}_3\text{H}_y\) is: \(\text{C}_3\text{H}_y + (3 + y/4)\text{O}_2 \rightarrow 3\text{CO}_2 + (y/2)\text{H}_2\text{O}\). Equating the oxygen coefficient to the volume ratio: \(3 + y/4 = 4.5 \Rightarrow y/4 = 1.5 \Rightarrow y = 6\). Thus, the molecular formula is \(\text{C}_3\text{H}_6\).

[1 mark] Correct answer is A. Deduce the volume of \(\text{CO}_2\) as \(30\text{ cm}^3\) to determine 3 carbon atoms. Deduce the volume of reacted \(\text{O}_2\) as \(45\text{ cm}^3\) to determine 6 hydrogen atoms.

The standard enthalpy of hydration is defined as the enthalpy change when one mole of gaseous ions is dissolved in water to form an infinitely dilute solution under standard conditions. This is represented by the equation: \(\text{Mg}^{2+}(g) + \text{aq} \rightarrow \text{Mg}^{2+}(\text{aq})\).

[1 mark] Correct answer is A. Identify the state symbols and charges corresponding to the standard definition of hydration enthalpy (gaseous ions to aqueous ions).

The successive ionization energies show a massive increase between the fifth and sixth ionization energies (from \(6274\) to \(21268\text{ kJ mol}^{-1}\)). This indicates that the sixth electron is removed from a principal quantum shell closer to the nucleus, meaning the element has 5 outer-shell (valence) electrons. In Period 3, the element in Group 15 with 5 valence electrons is phosphorus.

[1 mark] Correct answer is B. Locate the position of the major jump in successive ionization energies to determine the number of valence electrons (5), and match it with the correct Period 3 element.

Moles of \(\text{CH}_3\text{COOH} = 2.50 \times 10^{-3}\text{ mol}\). Moles of \(\text{NaOH} = 1.25 \times 10^{-3}\text{ mol}\). The sodium hydroxide reacts with half of the ethanoic acid to produce an equal number of moles of ethanoate ions (\(1.25 \times 10^{-3}\text{ mol}\)), leaving exactly half of the ethanoic acid unreacted (\(1.25 \times 10^{-3}\text{ mol}\)). This forms a buffer solution at its half-equivalence point where \([\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-]\). Under this condition, \(\text{pH} = \text{p}K_a = -\log_{10}(1.74 \times 10^{-5}) = 4.76\).

[1 mark] Correct answer is C. Identify that the mixture is at the half-equivalence point where pH equals pKa, and correctly evaluate the negative logarithm of Ka.

Comparing Experiments 1 and 2: when \([\text{B}]\) is held constant and \([\text{A}]\) is doubled, the initial rate increases by a factor of 4 (from \(1.2 \times 10^{-4}\) to \(4.8 \times 10^{-4}\)), which means the reaction is second order with respect to \(\text{A}\). Comparing Experiments 1 and 3: when \([\text{A}]\) is held constant and \([\text{B}]\) is doubled, the initial rate increases by a factor of 2 (from \(1.2 \times 10^{-4}\) to \(2.4 \times 10^{-4}\)), which means the reaction is first order with respect to \(\text{B}\). Therefore, the rate equation is \(\text{Rate} = k[\text{A}]^2[\text{B}]\).

[1 mark] Correct answer is B. Deduce the order of reaction with respect to A is 2, and the order of reaction with respect to B is 1, to formulate the correct rate law.

The atomic number of iron is 26, giving a neutral ground state configuration of \([\text{Ar}] 4\text{s}^2 3\text{d}^6\). When forming transition metal cations, the electrons from the \(4\text{s}\) subshell are lost before the \(3\text{d}\) electrons. Removing three electrons to form \(\text{Fe}^{3+}\) means removing both \(4\text{s}\) electrons and one \(3\text{d}\) electron, which leaves the ground state electronic configuration as \([\text{Ar}] 3\text{d}^5\).

[1 mark] Correct answer is A. Identify that 4s electrons are lost before 3d electrons when transition metal atoms form ions.

The sharp, strong peak at \(1715\text{ cm}^{-1}\) indicates a carbonyl group (\(\text{C}=\text{O}\)). The absence of any broad peak at \(3200\text{–}3600\text{ cm}^{-1}\) indicates there is no hydroxyl group (\(\text{O}-\text{H}\)), which rules out propan-2-ol and propanoic acid. Methoxyethane does not contain a carbonyl group and has a different molecular formula (\(\text{C}_3\text{H}_8\text{O}\)). Propanone is a ketone with molecular formula \(\text{C}_3\text{H}_6\text{O}\) and a carbonyl group, which perfectly fits the spectrum.

[1 mark] Correct answer is B. Use infrared data to identify functional groups present (C=O) and absent (O-H) and match with molecular formula.

For a reaction to be thermodynamically feasible, \(\Delta G^\ominus \le 0\). Using the relation \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\), we get \(\Delta H^\ominus - T\Delta S^\ominus \le 0\), which rearranges to \(T \ge \frac{\Delta H^\ominus}{\Delta S^\ominus}\). Converting \(\Delta H^\ominus\) to \(\text{J mol}^{-1}\) gives \(135000\text{ J mol}^{-1}\). Thus, \(T \ge \frac{135000}{320} = 421.875\text{ K}\). To the nearest Kelvin, this is \(422\text{ K}\).

[1 mark] Correct answer is A. Correctly convert enthalpy units, use the Gibbs free energy relationship, and find the minimum temperature.

The decrease in volume upon shaking with sodium hydroxide is due to the absorption of \( \text{CO}_2 \). Thus, the volume of \( \text{CO}_2 \) produced is \( 45.0 - 25.0 = 20.0\text{ cm}^3 \). Since \( 10.0\text{ cm}^3 \) of the compound was burned, each molecule of the compound must contain 2 carbon atoms (ratio of 1:2). The remaining \( 25.0\text{ cm}^3 \) of gas is the excess unreacted \( \text{O}_2 \). Thus, the volume of \( \text{O}_2 \) that reacted is \( 50.0 - 25.0 = 25.0\text{ cm}^3 \). This gives a reactant volume ratio of compound to \( \text{O}_2 \) of \( 1 : 2.5 \). Testing \( \text{C}_2\text{H}_4\text{O} \): \( \text{C}_2\text{H}_4\text{O}(g) + 2.5\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \), which fits the experimental stoichiometry perfectly.

1 mark: Correctly identifies C as the molecular formula by determining the volume of carbon dioxide produced and the volume of oxygen reacted.

A reaction is feasible when \( \Delta G^{\ominus} < 0 \). At the boundary temperature of feasibility, \( \Delta G^{\ominus} = 0 \), which gives \( \Delta H^{\ominus} - T\Delta S^{\ominus}_{\text{sys}} = 0 \). Rearranging for the entropy change of the system: \( \Delta S^{\ominus}_{\text{sys}} = \frac{\Delta H^{\ominus}}{T} \). Substituting the given values: \( \Delta S^{\ominus}_{\text{sys}} = \frac{-116 \times 10^3\text{ J mol}^{-1}}{542\text{ K}} = -214\text{ J K}^{-1}\text{ mol}^{-1} \). Since both enthalpy change and entropy change are negative, the reaction is feasible only at temperatures below this value.

1 mark: Correctly calculates the value and sign of the standard entropy change of the system.

There is a massive jump between the fifth and sixth ionization energies (from \( 6274 \) to \( 21269\text{ kJ mol}^{-1} \)), which indicates that the sixth electron is being removed from an inner quantum shell. This means element \( \text{X} \) has five valence electrons and belongs to Group 15 of the Periodic Table.

1 mark: Identifies that the jump in successive ionization energies indicates 5 valence electrons, corresponding to Group 15.

The rate of hydrolysis of halogenoalkanes is determined by the strength of the carbon-halogen bond and the mechanism. C-I bonds are weaker than C-Cl bonds, meaning iodoalkanes react much faster than chloroalkanes. Furthermore, tertiary halogenoalkanes (such as 2-iodo-2-methylpropane) react faster than primary ones because they undergo nucleophilic substitution via the \( \text{S}_{\text{N}}1 \) mechanism, forming a highly stable tertiary carbocation intermediate. Therefore, 2-iodo-2-methylpropane hydrolyses fastest.

1 mark: Correctly identifies 2-iodo-2-methylpropane as the halogenoalkane that hydrolyses fastest.

Since iodine does not appear in the rate equation, the reaction is zero-order with respect to iodine. Therefore, halving the concentration of iodine has no effect on the rate. The reaction is first-order with respect to both propanone and hydrogen ions. Doubling the concentration of propanone increases the rate by a factor of 2, and doubling the concentration of hydrogen ions increases the rate by another factor of 2. The combined effect is an increase in rate by a factor of \( 2 \times 2 = 4 \).

1 mark: Correctly determines that the rate increases by a factor of 4.

The reaction of butanone with hydrogen cyanide is a nucleophilic addition. The carbonyl carbon of the ketone is planar. The nucleophile (the cyanide ion, \( \text{CN}^- \)) can attack this planar carbon atom with equal probability from either above or below. This results in the formation of a racemic mixture containing equal amounts of both enantiomers, making the product mixture optically inactive.

1 mark: Correctly identifies that the planar carbonyl group is attacked with equal probability from either side, producing a racemic mixture.

Let's determine the d-orbital electronic configurations for each ion in its ground state: \( \text{Fe}^{3+} \) has the configuration \( [\text{Ar}] 3\text{d}^5 \), giving 5 unpaired electrons. \( \text{Co}^{2+} \) has the configuration \( [\text{Ar}] 3\text{d}^7 \), giving 3 unpaired electrons. \( \text{Cr}^{3+} \) has the configuration \( [\text{Ar}] 3\text{d}^3 \), giving 3 unpaired electrons. \( \text{Cu}^{2+} \) has the configuration \( [\text{Ar}] 3\text{d}^9 \), giving 1 unpaired electron. Therefore, \( \text{Fe}^{3+} \) has the greatest number of unpaired d-electrons.

1 mark: Correctly identifies Fe3+ as having the most unpaired d-electrons (5).

Concentrated sulfuric acid is a stronger acid than concentrated nitric acid. It protonates nitric acid, which then loses a water molecule to generate the active electrophile, the nitronium ion (\( \text{NO}_2^+ \)). Because the sulfuric acid is regenerated at the end of the electrophilic substitution process, it acts as an acid catalyst.

1 mark: Correctly identifies the role of sulfuric acid as an acid catalyst to generate the nitronium electrophile.

First, calculate the number of moles of \(\text{CO}_2\) gas produced:
\(n(\text{CO}_2) = \frac{576\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.024\text{ mol}\).

Since the decomposition reaction is:
\(\text{MCO}_3(s) \rightarrow \text{MO}(s) + \text{CO}_2(g)\)

The ratio of metal carbonate to carbon dioxide is 1:1, so \(n(\text{MCO}_3) = 0.024\text{ mol}\).

Next, calculate the molar mass of \(\text{MCO}_3\):
\(M_r(\text{MCO}_3) = \frac{2.40\text{ g}}{0.024\text{ mol}} = 100\text{ g mol}^{-1}\).

Subtract the relative formula mass of the carbonate ion (\(\text{CO}_3^{2-}\)):
\(M_r(\text{CO}_3^{2-}) = 12.0 + (3 \times 16.0) = 60.0\text{ g mol}^{-1}\).

\(A_r(\text{M}) = 100 - 60.0 = 40.0\text{ g mol}^{-1}\).

This corresponds to calcium (\(\text{Ca}\)).

1 mark for the correct option B.
- Award 1 mark for calculating correct molar mass of 100 g/mol and identifying Calcium.
- Reject other options.

The equation for the combustion of ethane is:
\(\text{C}_2\text{H}_6(g) + 3.5\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)\)

Using Hess's law:
\(\Delta H_c^{\ominus} = \sum \Delta H_f^{\ominus}(\text{products}) - \sum \Delta H_f^{\ominus}(\text{reactants})\)

\(\Delta H_c^{\ominus} = [2 \times (-393.5) + 3 \times (-285.8)] - [-84.7]\)
\(\Delta H_c^{\ominus} = [-787.0 - 857.4] + 84.7\)
\(\Delta H_c^{\ominus} = -1644.4 + 84.7 = -1559.7\text{ kJ mol}^{-1}\).

1 mark for the correct choice A.
- Reject B (did not add the enthalpy of formation of reactants).
- Reject C (used incorrect stoichiometric coefficients).
- Reject D (sign error).

Look for the largest relative jump between successive ionization energies. The values increase moderately from 578 to 1817 and 2745 \(\text{kJ mol}^{-1}\). The jump from the 3rd to the 4th ionization energy (2745 to 11577 \(\text{kJ mol}^{-1}\)) is extremely large. This indicates that the 4th electron is removed from a core shell, meaning there are 3 valence electrons. Therefore, element X belongs to Group 3 (Group 13).

1 mark for selecting C.

The strong, sharp peak at \(1715\text{ cm}^{-1}\) is characteristic of a carbonyl group (\(\text{C=O}\)). The absence of a broad band at \(3200\text{--}3750\text{ cm}^{-1}\) indicates that there is no alcohol (\(\text{O-H}\)) group present. Since the compound contains only three carbon atoms and one oxygen atom with one unit of unsaturation, it must be propanone. Propan-1-ol and propan-2-ol have \(\text{O-H}\) groups, and propanoic acid would show a very broad acid \(\text{O-H}\) absorption peak.

1 mark for identifying B as the correct molecule.

Compare Experiment 1 and 2: \([\text{B}]\) is kept constant while \([\text{A}]\) is doubled. The rate increases by a factor of \(\frac{8.0 \times 10^{-4}}{2.0 \times 10^{-4}} = 4\). Therefore, the order of reaction with respect to A is 2.

Compare Experiment 1 and 3: \([\text{A}]\) is kept constant while \([\text{B}]\) is doubled. The rate increases by a factor of \(\frac{4.0 \times 10^{-4}}{2.0 \times 10^{-4}} = 2\). Therefore, the order of reaction with respect to B is 1.

The rate equation is: \(\text{Rate} = k[\text{A}]^2[\text{B}]\).

1 mark for the correct option B.

- \(\text{SF}_4\) has 5 electron pairs around sulfur (4 bonding, 1 lone pair) and is seesaw shaped.
- \(\text{NH}_4^+\) has 4 electron pairs around nitrogen (4 bonding, 0 lone pairs) and is tetrahedral.
- \(\text{XeF}_4\) has 6 electron pairs around xenon (4 bonding, 2 lone pairs) and is square planar.
- \(\text{BF}_3\) has 3 electron pairs around boron (3 bonding, 0 lone pairs) and is trigonal planar.

1 mark for the correct choice B.

For an endothermic reaction, \(\Delta H > 0\). Since \(\Delta S_{\text{surroundings}} = -\frac{\Delta H}{T}\), \(\Delta S_{\text{surroundings}}\) must be negative (< 0).
For the reaction to be spontaneous, the total entropy change, \(\Delta S_{\text{total}}\), must be positive (> 0).
Since \(\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}}\), and \(\Delta S_{\text{surroundings}}\) is negative, \(\Delta S_{\text{system}}\) must be positive (> 0) and larger in magnitude than the surroundings' entropy change. Hence: \(\Delta S_{\text{system}} > 0\), \(\Delta S_{\text{surroundings}} < 0\), and \(\Delta S_{\text{total}} > 0\).

1 mark for the correct row choice A.

The addition of excess chloride ions from hydrochloric acid results in a ligand exchange reaction where six water ligands are replaced by four chloride ligands:

\([\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}\)

This reaction involves a decrease in the coordination number from 6 to 4, causing a transition in geometry from octahedral to tetrahedral, which changes the d-orbital splitting and shifts the color of the complex.

1 mark for the correct option B.

First, find the moles of \( \text{CO}_2 \) produced: \( \text{moles of CO}_2 = \frac{28.8\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 1.20 \times 10^{-3}\text{ mol} \). Since the reaction stoichiometry between \( \text{MCO}_3 \) and \( \text{CO}_2 \) is \( 1:1 \), the number of moles of \( \text{MCO}_3 \) is also \( 1.20 \times 10^{-3}\text{ mol} \). The molar mass of \( \text{MCO}_3 = \frac{0.120\text{ g}}{1.20 \times 10^{-3}\text{ mol}} = 100.0\text{ g mol}^{-1} \). Subtracting the molar mass of the carbonate group (\( 12.0 + 3 \times 16.0 = 60.0\text{ g mol}^{-1} \)) gives the molar mass of \( \text{M} \): \( 100.0 - 60.0 = 40.0\text{ g mol}^{-1} \). This atomic mass corresponds to calcium.

1 mark: Correctly identifies Calcium as the metal M by calculating the molar mass of the carbonate (100.0 g/mol) and subtracting the carbonate group mass to get 40.0 g/mol.

The successive ionization energies show a large increase between the third and fourth ionization energies (from 2745 to 11577 kJ/mol). This indicates that the fourth electron is being removed from an inner shell, meaning element X has three valence electrons and is in Group 13 (Aluminium). The outer electronic configuration of Aluminium is \( 3\text{s}^2 3\text{p}^1 \). Aluminium oxide is amphoteric, and its chloride is simple molecular.

1 mark: Correctly identifies that the large jump between the 3rd and 4th ionization energies indicates three valence electrons, corresponding to the \( 3\text{s}^2 3\text{p}^1 \) configuration.

Using Hess's Law: \( \Delta H^\ominus_{\text{f}}[\text{C}_3\text{H}_8(\text{g})] = 3 \times \Delta H^\ominus_{\text{c}}[\text{C}] + 4 \times \Delta H^\ominus_{\text{c}}[\text{H}_2] - \Delta H^\ominus_{\text{c}}[\text{C}_3\text{H}_8] \). Substituting the values: \( \Delta H^\ominus_{\text{f}} = [3 \times (-394) + 4 \times (-286)] - [-2220] = [-1182 - 1144] + 2220 = -2326 + 2220 = -106\text{ kJ mol}^{-1} \).

1 mark: Correctly calculates \( -106\text{ kJ mol}^{-1} \) using Hess's Law with correct signs and stoichiometry.

The rate of precipitate formation depends on the ease of carbon-halogen bond breaking (C-I is the weakest bond, so it breaks most rapidly) and the reaction mechanism (tertiary halogenoalkanes react much faster than primary or secondary ones via the stable tertiary carbocation intermediate in the \( \text{S}_{\text{N}}1 \) pathway). Therefore, 2-iodo-2-methylpropane (a tertiary iodoalkane) reacts the fastest.

1 mark: Correctly identifies 2-iodo-2-methylpropane as reacting fastest due to having the weakest C-I bond and undergoing rapid tertiary \( \text{S}_{\text{N}}1 \) hydrolysis.

The rate equation is \( \text{Rate} = k[\text{NO}]^2[\text{H}_2] \). If we double the concentration of both reactants, the new rate becomes: \( \text{Rate}' = k(2[\text{NO}])^2(2[\text{H}_2]) = 8 k[\text{NO}]^2[\text{H}_2] = 8 \times \text{Rate} \). Hence, the rate increases by a factor of eight.

1 mark: Correctly identifies that doubling both concentrations increases the rate by a factor of eight.

Mixing equal volumes (V) of \( 0.200\text{ mol dm}^{-3}\text{ HA} \) and \( 0.100\text{ mol dm}^{-3}\text{ NaOH} \) results in partial neutralisation. \( \text{Moles of HA} = 0.200 V \), \( \text{Moles of NaOH} = 0.100 V \). After reaction, remaining \( \text{HA} = 0.100 V \) and produced \( \text{A}^- = 0.100 V \). Since \( [\text{HA}] = [\text{A}^-] \), the buffer solution is at its half-neutralisation point where \( \text{pH} = \text{p}K_{\text{a}} = 4.76 \).

1 mark: Correctly identifies that the mixture forms a half-neutralised buffer where \( \text{pH} = \text{p}K_{\text{a}} = 4.76 \).

A positive 2,4-DNPH test indicates a carbonyl compound (aldehyde or ketone). A negative Tollens' or Fehling's test indicates that the compound is a ketone rather than an aldehyde. A positive triiodomethane (iodoform) test indicates the presence of a methyl ketone group, \( \text{CH}_3\text{C}=\text{O} \). The only four-carbon ketone is butan-2-one, which satisfies all these observations.

1 mark: Correctly identifies the compound as butan-2-one based on the functional group tests.

In the octahedral complex \( [\text{Ti}(\text{H}_2\text{O})_6]^{3+} \), the d-orbitals are split into two energy levels. Color arises when a d-electron in the lower energy level absorbs a photon of visible light (green-yellow region) and is promoted to a higher energy d-orbital (d-d transition). The complementary violet color is transmitted.

1 mark: Correctly identifies that the color arises from the promotion of a d-electron from a lower to a higher energy d-orbital upon absorbing visible light.

First, state the outer shell configurations: Phosphorus is \(3s^2 3p^3\) and Sulfur is \(3s^2 3p^4\). In phosphorus, the three 3p electrons occupy three separate 3p orbitals singly. In sulfur, there are four 3p electrons, meaning one of the 3p orbitals contains a pair of electrons with opposite spins. This pairing of electrons in the same orbital results in mutual electrostatic repulsion, which makes it easier to remove one of these paired electrons from sulfur compared to removing an unpaired electron from phosphorus, resulting in a lower first ionization energy.

M1: State or show that the outer configuration of P is \(3p^3\) and S is \(3p^4\) (1 mark). M2: State that the 3p orbitals in P are singly occupied or unpaired (1 mark). M3: State that in S, one 3p orbital contains a pair of electrons with opposite spins (1 mark). M4: Explain that mutual repulsion between these paired electrons in S makes the electron easier to remove, resulting in a lower first ionization energy (1 mark).

The feasibility boundary occurs when \(\Delta G = 0\). Since \(\Delta G = \Delta H - T\Delta S\), setting \(\Delta G = 0\) gives \(T = \frac{\Delta H}{\Delta S}\). First, convert \(\Delta H^{\ominus}\) to J: \(+178 \text{ kJ mol}^{-1} = +178000 \text{ J mol}^{-1}\). Next, calculate \(T\): \(T = \frac{178000}{160} = 1112.5 \text{ K}\) (accept \(1110 \text{ K}\) to 3 sig figs). Finally, state the assumption: It is assumed that both \(\Delta H^{\ominus}\) and \(\Delta S^{\ominus}_{\text{system}}\) remain constant and do not vary with temperature.

M1: Uses the feasibility condition \(\Delta G = 0\) or states the formula \(T = \frac{\Delta H}{\Delta S}\) (1 mark). M2: Correctly converts units of \(\Delta H\) or \(\Delta S\) (1 mark). M3: Calculates the temperature as \(1112.5 \text{ K}\) (accept \(1110 \text{ K}\) or \(1113 \text{ K}\)) (1 mark). M4: States that \(\Delta H^{\ominus}\) and \(\Delta S^{\ominus}\) are assumed to be independent of temperature (1 mark).

Under identical conditions, 1-iodobutane is hydrolyzed faster than 1-bromobutane. A yellow precipitate of silver iodide forms faster than a cream precipitate of silver bromide. Hydrolysis involves the breaking of the carbon-halogen bond (C-X). The rate of reaction depends on the strength of this bond. The bond enthalpy of the C-I bond is significantly lower than that of the C-Br bond because the larger iodine atom leads to poorer orbital overlap with carbon. Therefore, the C-I bond requires less energy to break, allowing the reaction to proceed faster.

M1: Identifies that 1-iodobutane hydrolyzes faster (or forms a precipitate quicker) than 1-bromobutane (1 mark). M2: States that the rate of hydrolysis depends on the carbon-halogen bond strength/enthalpy rather than bond polarity (1 mark). M3: States that the C-I bond is weaker than the C-Br bond or has a lower bond enthalpy (1 mark). M4: Explains that the C-I bond is weaker because the iodine atom is larger, resulting in less orbital overlap, so less energy is needed to break it (1 mark).

First, find the moles of ethanoic acid: \(n(CH_3COOH) = 0.100 \times \frac{50.0}{1000} = 5.00 \times 10^{-3} \text{ mol}\). Next, find the moles of ethanoate ions: \(n(CH_3COO^-) = 0.080 \times \frac{50.0}{1000} = 4.00 \times 10^{-3} \text{ mol}\). Use the buffer equation: \([H^+] = K_a \times \frac{[CH_3COOH]}{[CH_3COO^-]} = 1.7 \times 10^{-5} \times \frac{5.00 \times 10^{-3}}{4.00 \times 10^{-3}} = 2.125 \times 10^{-5} \text{ mol dm}^{-3}\). Calculate the pH: \(pH = -\log_{10}(2.125 \times 10^{-5}) = 4.6726\). Rounded to two decimal places, \(pH = 4.67\).

M1: Correctly calculates moles of ethanoic acid and ethanoate ions (1 mark). M2: Uses the correct equilibrium relationship: \([H^+] = K_a \times \frac{[acid]}{[conjugate\ base]}\) or equivalent (1 mark). M3: Computes \([H^+] = 2.125 \times 10^{-5} \text{ mol dm}^{-3}\) (1 mark). M4: Obtains the final pH of \(4.67\) to 2 decimal places (1 mark).

First, write the d-electron configurations: \(Cu^{2+}\) is \([Ar] 3d^9\) (partially filled d-subshell); \(Zn^{2+}\) is \([Ar] 3d^{10}\) (completely filled d-subshell). In the presence of water ligands, the five d-orbitals split into two sets of different energy levels. In \(Cu^{2+}\), an electron can absorb light from the visible region of the electromagnetic spectrum to transition from a lower-energy d-orbital to a higher-energy d-orbital (d-d transition). The non-absorbed frequencies are transmitted/reflected, producing the observed complementary blue color. Since all d-orbitals are completely filled in \(Zn^{2+}\), there are no empty states in the higher-energy d-orbitals, so d-d transitions cannot occur. No visible light is absorbed, making it colorless.

M1: Identifies that \(Cu^{2+}\) has an incomplete/partially filled d-subshell (\(3d^9\)) whereas \(Zn^{2+}\) has a complete d-subshell (\(3d^{10}\)) (1 mark). M2: States that ligands (water) cause the d-orbitals to split into different energy levels (1 mark). M3: Explains that \(Cu^{2+}\) absorbs a specific frequency of visible light during a d-d transition, and the complementary (blue) color is transmitted (1 mark). M4: Explains that \(Zn^{2+}\) is colorless because d-d transitions cannot occur since the d-orbitals are fully occupied (1 mark).

First, write the balanced equation for the formation of liquid ethanol: \(2C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)\). Next, construct the Hess's Law cycle using enthalpies of combustion: \(\Delta_f H^{\ominus} = \sum \Delta_c H^{\ominus}(\text{reactants}) - \sum \Delta_c H^{\ominus}(\text{products})\). Substitute the values: \(\Delta_f H^{\ominus} = [2 \times (-394) + 3 \times (-286)] - [-1367]\). Calculate the result: \(\Delta_f H^{\ominus} = [-788 - 858] + 1367 = -1646 + 1367 = -279 \text{ kJ mol}^{-1}\).

M1: Writes the balanced equation for the formation of ethanol: \(2C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)\) (1 mark). M2: Uses a valid Hess's cycle or algebraic formula: \(\Delta_f H = 2\Delta_c H(C) + 3\Delta_c H(H_2) - \Delta_c H(ethanol)\) (1 mark). M3: Correctly substitutes values: \([2 \times (-394) + 3 \times (-286)] - (-1367)\) (1 mark). M4: Obtains \(-279 \text{ kJ mol}^{-1}\) with correct sign and units (1 mark).

First, formation of the electrophile is shown by the equation: \(HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-\) (or \(HNO_3 + H_2SO_4 \rightarrow NO_2^+ + H_2O + HSO_4^-\)). Second, the mechanism involves a nucleophilic attack shown by a curly arrow from the ring (delocalized \(\pi\) system) to the \(N\) atom of the \(NO_2^+\) electrophile. Third, draw a cyclohexadienyl cation intermediate with a positive charge inside a horseshoe-shaped ring spanning five carbons, open towards the carbon bonded to both \(-H\) and \(-NO_2\). Finally, a curly arrow from the C-H bond back into the ring restores the aromatic pi-system, producing nitrobenzene and releasing \(H^+\).

M1: Correct equation for the generation of the \(NO_2^+\) electrophile (1 mark). M2: Correct curly arrow from the delocalized ring of benzene to the nitrogen of \(NO_2^+\) (1 mark). M3: Correct representation of the cationic intermediate with a positive charge and horseshoe shape open towards the carbon carrying the H and NO2 groups (1 mark). M4: Correct curly arrow from the C-H bond to the ring to reform the delocalized system (1 mark).

First, identify the rate-determining step: Step 2 is the slow step, meaning it determines the overall rate of the reaction. Therefore, the rate equation is \(\text{rate} = k_2[N_2O_2][O_2]\). Next, identify that \(N_2O_2\) is an intermediate and cannot be in the final rate expression, so its concentration must be expressed in terms of the reactants. Using the fast equilibrium in Step 1: \(K_c = \frac{[N_2O_2]}{[NO]^2}\), which gives \([N_2O_2] = K_c [NO]^2\). Finally, substitute \([N_2O_2]\) into the rate-determining step rate expression: \(\text{rate} = k_2 (K_c [NO]^2) [O_2] = k [NO]^2 [O_2]\) (where \(k = k_2 K_c\)). This matches the experimental rate equation.

M1: Identifies Step 2 as the rate-determining step and writes its rate equation: \(\text{rate} = k_2[N_2O_2][O_2]\) (1 mark). M2: Explains that \(N_2O_2\) is an intermediate and its concentration is governed by the fast equilibrium in Step 1 (1 mark). M3: Expresses the equilibrium relationship: \([N_2O_2] = K_c[NO]^2\) (1 mark). M4: Substitutes this into the rate-determining step rate equation to show that \(\text{rate} = k[NO]^2[O_2]\) (1 mark).

1. Calculate mass of water lost: 4.76 g - 2.60 g = 2.16 g. 2. Calculate moles of anhydrous CoCl2: Molar mass of CoCl2 = 58.9 + (2 * 35.5) = 129.9 g mol-1. Moles of CoCl2 = 2.60 / 129.9 = 0.0200 mol. 3. Calculate moles of H2O lost: Molar mass of H2O = 18.0 g mol-1. Moles of H2O = 2.16 / 18.0 = 0.120 mol. 4. Determine x: Ratio of H2O to CoCl2 = 0.120 / 0.0200 = 6.

M1: Mass of H2O lost = 2.16 g (1 mark). M2: Moles of CoCl2 = 0.0200 mol (1 mark). M3: Moles of H2O = 0.120 mol (1 mark). M4: Value of x = 6 (1 mark). Allow ecf throughout.

1. Calculate heat energy released, q: mass of solution = 50.0 + 50.0 = 100.0 g. delta T = 25.9 - 19.2 = 6.7 K. q = m * c * delta T = 100.0 * 4.18 * 6.7 = 2800.6 J = 2.8006 kJ. 2. Calculate moles of water formed: moles of HCl = 0.0500 mol; moles of NaOH = 0.0500 mol; moles of H2O formed = 0.0500 mol. 3. Calculate delta H: delta H = -q / n = -2.8006 kJ / 0.0500 mol = -56.012 kJ mol-1. 4. Rounding to 3 sig figs: -56.0 kJ mol-1.

M1: Calculates heat energy released q = 2800.6 J or 2.80 kJ (1 mark). M2: Moles of water formed = 0.0500 mol (1 mark). M3: Division of energy by moles (1 mark). M4: Correct final value of -56.0 with negative sign and to 3 significant figures (1 mark).

1. 2-bromo-2-methylpropane is a tertiary halogenoalkane. 2. Ionisation of the C-Br bond forms a tertiary carbocation intermediate, which is highly stable due to the positive inductive effect of the three electron-donating methyl groups. 3. The three bulky methyl groups also create significant steric hindrance around the central carbon atom, preventing the direct back-side attack of the hydroxide ion nucleophile that would be required for an SN2 mechanism.

M1: Identifies 2-bromo-2-methylpropane as a tertiary halogenoalkane (1 mark). M2: States that the tertiary carbocation intermediate is highly stable (1 mark). M3: Attributes carbocation stability to the positive inductive effect of three methyl groups (1 mark). M4: Explains that steric hindrance of bulky methyl groups blocks the SN2 pathway (1 mark).

1. Determine electron pairs: H2O has 2 bonding pairs and 2 lone pairs, whereas H3O+ has 3 bonding pairs and 1 lone pair. 2. Identify shapes: H2O is bent / V-shaped, and H3O+ is trigonal pyramidal. 3. Apply electron pair repulsion theory: Lone pairs repel other electron pairs more strongly than bonding pairs. 4. Because H2O has two lone pairs compared to only one lone pair in H3O+, the bonding pairs in H2O are repelled and pushed closer together, resulting in a smaller bond angle (104.5 degrees) than in H3O+ (107 degrees).

M1: States electron pairs for both: H2O has 2 BP and 2 LP, and H3O+ has 3 BP and 1 LP (1 mark). M2: States shapes: H2O is bent and H3O+ is trigonal pyramidal (1 mark). M3: States that lone pairs repel more than bonding pairs (1 mark). M4: Explains that the extra lone pair in H2O causes greater compression of the bond angle, making it smaller than in H3O+ (1 mark).

1. Find order with respect to A: comparing Experiment 1 and 2, [B] is constant, [A] doubles, and the initial rate increases by a factor of 4. Therefore, the order with respect to A is 2. 2. Find order with respect to B: comparing Experiment 1 and 3, [A] is constant, [B] doubles, and the initial rate increases by a factor of 2. Therefore, the order with respect to B is 1. 3. Write rate equation: Rate = k[A]^2[B]. 4. Calculate k: using Experiment 1, 2.0 x 10^-4 = k * (0.10)^2 * (0.10) => k = (2.0 x 10^-4) / 1.0 x 10^-3 = 0.20. 5. Units of k: rate / ([A]^2[B]) = (mol dm-3 s-1) / (mol dm-3)^3 = dm6 mol-2 s-1.

M1: Deduces order with respect to A is 2 and order with respect to B is 1 (1 mark). M2: Correct rate equation: Rate = k[A]^2[B] (1 mark). M3: Calculates k = 0.20 (1 mark). M4: Correct units of k: dm6 mol-2 s-1 (1 mark).

1. Write the balanced equation: [Cr(H2O)6]3+(aq) + EDTA4-(aq) -> [Cr(EDTA)]-(aq) + 6H2O(l). 2. Count the particles: on the reactant side, there are 2 free particles (the complex ion and the polydentate ligand); on the product side, there are 7 free particles (the new chromium complex and 6 released water molecules). 3. Explain the feasibility: the increase in the number of particles in solution creates a highly positive system entropy change (delta S is positive). Since delta G = delta H - T * delta S, this highly positive entropy change makes delta G negative, ensuring the reaction is thermodynamically feasible.

M1: Correct formulas for all species: [Cr(H2O)6]3+, EDTA4-, [Cr(EDTA)]-, and H2O (1 mark). M2: Correct balancing with 6H2O (1 mark). M3: States that 2 reactant particles form 7 product particles / the number of free particles increases (1 mark). M4: Explains that this leads to a large increase in system entropy / highly positive system entropy change (1 mark).

1. Calculate moles of HA: moles = (25.0 / 1000) * 0.200 = 0.00500 mol. 2. Calculate moles of A-: moles = (25.0 / 1000) * 0.100 = 0.00250 mol. 3. Use the buffer concentration equation: [H+] = Ka * ([HA] / [A-]). Since total volume is the same, we can use moles directly: [H+] = (1.80 x 10^-5) * (0.00500 / 0.00250) = 3.60 x 10^-5 mol dm-3. 4. Calculate pH: pH = -log10[H+] = -log10(3.60 x 10^-5) = 4.44.

M1: Moles of HA = 0.00500 mol and A- = 0.00250 mol (1 mark). M2: Uses correct buffer equation expression (1 mark). M3: Calculates [H+] = 3.60 x 10^-5 mol dm-3 (1 mark). M4: Calculates pH = 4.44 (must be to 2 decimal places) (1 mark).

1. The product of the reaction is optically inactive. 2. The carbonyl carbon in butanal has a trigonal planar geometry, which is flat. 3. The CN- nucleophile can attack this planar carbonyl carbon with equal probability from either above or below the plane. 4. This produces equal amounts of the two enantiomers (a racemic mixture). Since the enantiomers rotate plane-polarised light in equal and opposite directions, their optical rotations cancel out.

M1: States the product is optically inactive / a racemic mixture (1 mark). M2: States that the carbonyl carbon/group is planar (1 mark). M3: Explains that CN- attacks with equal probability from either above or below the plane (1 mark). M4: Explains that equal amounts of enantiomers are formed, which cancel each other's optical activity (1 mark).

1. Convert units to SI: V = 74.8 \(\times\) 10\(^{-6}\) m\(^{3}\), p = 101000 Pa, T = 373 K. 2. Use ideal gas equation: n = pV / RT = (101000 \(\times\) 74.8 \(\times\) 10\(^{-6}\)) / (8.31 \(\times\) 373) = 2.437 \(\times\) 10\(^{-3}\) mol. 3. Molar mass M = mass / n = 0.209 / (2.437 \(\times\) 10\(^{-3}\)) = 85.8 g mol\(^{-1}\). 4. Use general formula of alkanes C\(_n\)H\(_{2n+2}\): 12.0n + 2.0n + 2 = 85.8, which gives 14.0n = 83.8, so n = 6. Molecular formula is C\(_6\)H\(_{14}\).

M1: Convert volume to 74.8 \(\times\) 10\(^{-6}\) m\(^{3}\) and pressure to 101000 Pa. M2: Calculate amount of gas n = 2.44 \(\times\) 10\(^{-3}\) mol (accept 2.4 \(\times\) 10\(^{-3}\) mol). M3: Calculate molar mass M = 85.8 g mol\(^{-1}\) (allow 85.0 to 86.5 g mol\(^{-1}\)). M4: Deduce molecular formula as C\(_6\)H\(_{14}\) based on general formula C\(_n\)H\(_{2n+2}\).

Standard equation for enthalpy of formation: 3C(graphite) + 4H\(_2\)(g) + 0.5O\(_2\)(g) \(\rightarrow\) C\(_3\)H\(_7\)OH(l). According to Hess's Law: \(\Delta_f H^\theta\) = 3 \(\times\) \(\Delta_c H^\theta\)[C(graphite)] + 4 \(\times\) \(\Delta_c H^\theta\)[H\(_2\)(g)] - \(\Delta_c H^\theta\)[C\(_3\)H\(_7\)OH(l)]. \(\Delta_f H^\theta\) = 3(-394) + 4(-286) - (-2021) = -1182 - 1144 + 2021 = -305 kJ mol\(^{-1}\).

M1: Writes the balanced equation for the standard enthalpy of formation of propan-1-ol or constructs a valid Hess's Law cycle. M2: Multiplies carbon and hydrogen combustion values correctly by their coefficients: 3(-394) and 4(-286). M3: Subtracts the combustion enthalpy of propan-1-ol from the sum of reactant combustion values. M4: Obtains correct final answer of -305 kJ mol\(^{-1}\) with correct sign and units.

The huge jump in ionisation energy occurs between the 3rd and 4th ionisation energies (2745 to 11577 kJ mol\(^{-1}\)), indicating that the fourth electron is being removed from an inner shell which is closer to the nucleus and experiences less shielding. Thus, the element has 3 valence electrons and is Aluminium. Its electronic configuration is 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^1\).

M1: Identifies element Y as Aluminium (or Al). M2: Gives full electronic configuration: 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^1\) (accept noble gas core notation [Ne] 3s\(^2\) 3p\(^1\)). M3: Explains that the fourth electron is removed from an inner shell / 2p subshell / shell closer to the nucleus. M4: Mentions that this inner shell experiences significantly less shielding and a stronger electrostatic attraction to the nucleus.

1. Comparing Exp 1 and Exp 2: [B] is constant, [A] doubles, and rate increases by a factor of 4. Since 2\(^2\) = 4, the order with respect to A is 2. 2. Comparing Exp 2 and Exp 3: [A] is constant, [B] triples, and rate increases by a factor of 3. Since 3\(^1\) = 3, the order with respect to B is 1. 3. Rate equation: Rate = k[A]\(^2\)[B]. 4. Calculate k using Exp 1: k = Rate / ([A]\(^2\)[B]) = 1.2 \(\times\) 10\(^{-4}\) / (0.10\(^2\) \(\times\) 0.10) = 0.12 dm\(^6\) mol\(^{-2}\) s\(^{-1}\).

M1: Deduces order with respect to A is 2 AND order with respect to B is 1 with clear explanation from the experimental data. M2: Writes correct rate equation: Rate = k[A]\(^2\)[B] (consequential on orders). M3: Calculates k value as 0.12 (consequential on rate equation). M4: States correct units for k as dm\(^6\) mol\(^{-2}\) s\(^{-1}\) (consequential on rate equation).

2-bromo-2-methylpropane is a tertiary halogenoalkane. It undergoes hydrolysis via an S\(_N\)1 mechanism because it can lose a bromide ion to form a stable tertiary carbocation, which is stabilized by the positive inductive effect of three methyl groups. Furthermore, the bulky methyl groups cause steric hindrance, which prevents direct backside nucleophilic attack. In contrast, 1-bromobutane is a primary halogenoalkane. If it underwent S\(_N\)1 hydrolysis, it would form a highly unstable primary carbocation. Because there is minimal steric hindrance around the primary carbon, the hydroxide nucleophile can easily attack the carbon directly from the backside, leading to a single-step S\(_N\)2 mechanism.

M1: States that 2-bromo-2-methylpropane is a tertiary halogenoalkane and forms a stable tertiary carbocation intermediate (due to inductive effects of three methyl groups). M2: Explains that steric hindrance from bulky methyl groups in 2-bromo-2-methylpropane prevents direct backside attack, making S\(_N\)2 unfavourable. M3: States that 1-bromobutane is a primary halogenoalkane and would form a highly unstable primary carbocation intermediate, making S\(_N\)1 unfavourable. M4: Explains that 1-bromobutane has little steric hindrance, allowing the nucleophile to easily attack the central carbon directly from the backside in a single-step S\(_N\)2 process.

1. Standard entropy of system: \(\Delta S_{\text{system}}^\theta\) = 2 \(\times\) S\(^\theta\)[NH\(_3\)] - (S\(^\theta\)[N\(_2\)] + 3 \(\times\) S\(^\theta\)[H\(_2\)]) = 2(192.3) - [191.6 + 3(130.6)] = 384.6 - 583.4 = -198.8 J K\(^{-1}\) mol\(^{-1}\). 2. Standard Gibbs free energy change: \(\Delta G^\theta\) = \(\Delta H^\theta\) - T\(\Delta S^\theta\). Converting \(\Delta S^\theta\) to kJ K\(^{-1}\) mol\(^{-1}\): -0.1988 kJ K\(^{-1}\) mol\(^{-1}\). \(\Delta G^\theta\) = -92.2 - 298 \(\times\) (-0.1988) = -92.2 + 59.24 = -32.96 kJ mol\(^{-1}\). 3. Since \(\Delta G^\theta\) is negative, the reaction is feasible at 298 K.

M1: Correct calculation of \(\Delta S_{\text{system}}^\theta\) = -198.8 J K\(^{-1}\) mol\(^{-1}\) (with units). M2: Converts units of \(\Delta S^\theta\) or \(\Delta H^\theta\) correctly in the Gibbs free energy expression. M3: Correct calculation of \(\Delta G^\theta\) = -33.0 kJ mol\(^{-1}\) (accept range -32.9 to -33.0 kJ mol\(^{-1}\)). M4: States that the reaction is feasible because \(\Delta G^\theta\) is negative.

1. When ligands coordinate to the copper(II) ion, the five degenerate d-orbitals split in energy into two distinct levels. 2. Ground state d-electrons absorb a specific frequency of visible light to transition to a higher energy d-orbital (d-d transition). The remaining, unabsorbed wavelengths are transmitted/reflected and constitute the complementary color observed. 3. Ammonia is a stronger field ligand than water, causing a larger splitting of the d-orbitals, which increases the energy difference (\(\Delta E\)) between the levels. 4. A larger \(\Delta E\) means that the complex absorbs light of higher energy (shorter wavelength, e.g., yellow/orange instead of red), resulting in a complementary color shift to a darker royal blue.

M1: States that ligands cause the degenerate 3d orbitals to split into higher and lower energy levels. M2: Explains that d-electrons absorb visible light of a specific frequency/wavelength to promote to a higher energy d-orbital, and the complementary color is observed. M3: Explains that ammonia is a stronger ligand than water and causes a larger energy gap / d-orbital splitting (\(\Delta E\)). M4: Explains that a larger \(\Delta E\) shifts the absorption to higher energy / shorter wavelengths, leading to a complementary shift to a deeper royal blue.

1. The carbonyl group in butanone has a planar (trigonal planar) shape around the carbon-oxygen double bond. 2. The nucleophile (cyanide ion, CN\(^{-}\)) has an equal probability of attacking this planar carbonyl carbon from either side (above or below the plane). 3. This equal probability of attack leads to the formation of equal amounts of both enantiomers (a 50:50/equimolar or racemic mixture). 4. Because each enantiomer rotates the plane of polarised light by the same angle but in opposite directions, the optical rotations cancel each other out, making the product mixture optically inactive.

M1: Identifies that the carbonyl carbon/group in butanone is planar / trigonal planar. M2: Explains that the nucleophile (cyanide ion) can attack the planar carbon from either side (above or below) with equal probability. M3: Deduces that this equal probability produces a 1:1 / 50:50 mixture of the two enantiomers (a racemic mixture). M4: Explains that the optical rotations of the two enantiomers cancel out, resulting in zero net rotation of plane-polarised light.

The successive ionisation energies of element X show a very large jump between the 3rd and 4th ionisation energies (from 2745 to 11578 kJ/mol). This large increase indicates that the 4th electron is being removed from an inner quantum shell (the 2p subshell), which is closer to the nucleus and experiences significantly less shielding. This means that element X has three electrons in its outer shell, which corresponds to an element in Group 3. Since X is in Period 3, it must be Aluminium (Al) with the outer electronic configuration of \(3s^2 3p^1\).

M1: Identifies element X as Aluminium / Al. (1)
M2: Identifies the large jump between the 3rd and 4th ionisation energies. (1)
M3: Explains that the 4th electron is removed from an inner quantum shell / shell closer to the nucleus / shell experiencing less shielding. (1)
M4: Concludes X has three outer-shell electrons, placing it in Group 3. (1)

Volume of \(CO_2\) produced = decrease in volume upon treatment with NaOH = \(65.0 - 25.0 = 40.0\text{ cm}^3\). Since 10.0 cm\(^3\) of hydrocarbon produced 40.0 cm\(^3\) of \(CO_2\), the mole ratio is 1:4, which means \(x = 4\). Volume of unreacted \(O_2\) remaining = 25.0 cm\(^3\), so volume of \(O_2\) reacted = \(80.0 - 25.0 = 55.0\text{ cm}^3\). The general equation for combustion is \(C_xH_y + (x + y/4) O_2 \rightarrow x CO_2 + y/2 H_2O\). The mole ratio of hydrocarbon to reacted \(O_2\) is \(10.0 : 55.0 = 1 : 5.5\). Therefore, \(x + y/4 = 5.5\). Substituting \(x = 4\) gives \(4 + y/4 = 5.5\), which solves to \(y/4 = 1.5\), hence \(y = 6\). The molecular formula of the hydrocarbon is \(C_4H_6\).

M1: Determines the volume of \(CO_2\) produced is 40.0 cm\(^3\). (1)
M2: Deduces that \(x = 4\) from the volume of \(CO_2\). (1)
M3: Calculates the volume of \(O_2\) reacted is 55.0 cm\(^3\). (1)
M4: Uses the reacting ratio of 10.0 to 55.0 to show \(y = 6\), yielding \(C_4H_6\). (1)

The major product is 2-bromopropane. During the electrophilic addition of HBr to propene, the \(H^+\) electrophile adds to the double bond. This can form either a secondary carbocation intermediate (\(CH_3C^+HCH_3\)) or a primary carbocation intermediate (\(CH_3CH_2C^+H_2\)). The secondary carbocation is more stable than the primary carbocation due to the electron-donating inductive effect of two methyl groups, which disperse the positive charge more effectively than the single propyl/ethyl group on the primary carbocation. The reaction preferentially proceeds via this more stable carbocation intermediate, yielding 2-bromopropane as the major product.

M1: Identifies 2-bromopropane as the major product. (1)
M2: States that 2-bromopropane forms via a secondary carbocation and 1-bromopropane via a primary carbocation (or shows structures). (1)
M3: Explains that secondary carbocations are more stable than primary carbocations. (1)
M4: Attributes this stability to the electron-donating inductive effect of two alkyl groups. (1)

The equation for the standard enthalpy of formation of liquid ethanol is: \(2C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_2H_5OH(l)\). According to Hess's Law, the enthalpy of formation can be calculated from combustion enthalpies: \(\Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\). Therefore: \(\Delta_f H^\theta = [2 \times (-393.5) + 3 \times (-285.8)] - (-1367.3)\)
\(\Delta_f H^\theta = [-787.0 - 857.4] + 1367.3 = -1644.4 + 1367.3 = -277.1\text{ kJ/mol}\).

M1: States or uses Hess's law relation: \(\Delta_f H = 2\Delta_c H(C) + 3\Delta_c H(H_2) - \Delta_c H(\text{ethanol})\). (1)
M2: Multiplies C by 2 (-787.0) and \(H_2\) by 3 (-857.4). (1)
M3: Calculates sum of reactants as -1644.4 kJ/mol. (1)
M4: Obtains correct final answer of -277.1 kJ/mol with correct negative sign and units. (1)

The rate of hydrolysis increases in the order 1-chlorobutane < 1-bromobutane < 1-iodobutane. The C-Cl bond is the most polar because chlorine is the most electronegative halogen, which would theoretically attract nucleophiles more strongly. However, the C-I bond is the weakest (has the lowest bond enthalpy), while the C-Cl bond is the strongest. Because the carbon-halogen bond must break during the hydrolysis reaction, the bond enthalpy is the overriding factor. Thus, the weaker C-I bond breaks most easily, causing 1-iodobutane to react the fastest.

M1: States the rate of hydrolysis increases from 1-chlorobutane to 1-iodobutane (or 1-iodobutane is fastest). (1)
M2: Explains that the C-I bond is weaker / has lower bond enthalpy than C-Cl. (1)
M3: Explains that the C-Cl bond is more polar than the C-I bond. (1)
M4: Concludes that bond enthalpy (not polarity) is the dominant factor determining the rate of hydrolysis. (1)

For a reaction to be feasible, \(\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta \le 0\). At the boundary where it becomes feasible, \(\Delta G^\theta = 0\), so \(T = \Delta H^\theta / \Delta S^\theta\). Converting \(\Delta H^\theta\) to J/mol gives +178000 J/mol. Thus, \(T = 178000 / 161 = 1105.59\text{ K}\), which rounds to 1106 K (or 1110 K to 3 s.f.). The key assumption is that both \(\Delta H^\theta\) and \(\Delta S^\theta\) do not change with temperature.

M1: States that \(\Delta G \le 0\) for feasibility (or \(\Delta G = 0\) at the limit). (1)
M2: Converts units so that both enthalpy and entropy are in compatible energy units (e.g., \(\Delta H = 178000\text{ J/mol}\)). (1)
M3: Calculates T correctly as 1106 K (accept 1110 K or 1105.6 K). (1)
M4: States the assumption that \(\Delta H\) and \(\Delta S\) remain constant as temperature changes. (1)

The carbonyl carbon and its three attached atoms in propanal are arranged in a planar geometry. The cyanide ion (\(CN^-\)) nucleophile can attack this planar carbonyl carbon from either side (above or below the plane) with equal probability. This yields an equimolar (50:50) mixture of the two enantiomers (a racemic mixture). Since enantiomers rotate plane-polarized light in opposite directions by equal angles, the overall optical rotation of the mixture is zero, making it optically inactive.

M1: States that the carbonyl group / carbon-oxygen double bond in propanal is planar. (1)
M2: Explains that attack by the \(CN^-\)_ nucleophile can occur from above or below the plane with equal probability. (1)
M3: Identifies that this produces a racemic mixture / equimolar mixture of enantiomers. (1)
M4: Explains that the opposite rotations of the enantiomers cancel each other out, leading to optical inactivity. (1)

The copper(II) ion in \([Cu(H_2O)_6]^{2+}\) has the electronic configuration \([Ar] 3d^9\), indicating a partially filled 3d subshell. When water ligands coordinate, they split the 3d orbitals into two different energy levels. An electron can absorb a photon of visible light to be promoted from the lower to the higher energy level (a d-to-d transition). The unabsorbed wavelengths of visible light are transmitted, resulting in a blue color. In contrast, the scandium(III) ion in \([Sc(H_2O)_6]^{3+}\) has the electronic configuration \([Ar] 3d^0\). Since there are no electrons in the 3d orbitals, no d-to-d transitions can take place, meaning no visible light is absorbed, and the complex is colorless.

M1: Explains that water ligands split the d-orbitals into two distinct energy levels. (1)
M2: States that \(Cu^{2+}\) has a \(3d^9\) configuration (partially filled d-subshell) allowing d-to-d transition of electrons. (1)
M3: Explains that the absorption of visible light during this transition leaves the complementary blue light to be transmitted. (1)
M4: States that \(Sc^{3+}\) has a \(3d^0\) electronic configuration, meaning no d-d transitions are possible, so no visible light is absorbed. (1)

1. First, calculate the standard entropy change of the system (\(\Delta S^{\ominus}_{\text{system}}\)):
\(\Delta S^{\ominus}_{\text{system}} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})\)
\(\Delta S^{\ominus}_{\text{system}} = (192.8 + 186.9) - 94.6 = +285.1\text{ J K}^{-1}\text{ mol}^{-1}\) (or \(+0.2851\text{ kJ K}^{-1}\text{ mol}^{-1}\))

2. For the reaction to be feasible, the Gibbs free energy change must be less than or equal to zero (\(\Delta G \le 0\)). At the threshold of feasibility:
\(\Delta G = \Delta H - T\Delta S_{\text{system}} = 0\)

3. Rearrange the equation to solve for the minimum temperature \(T\):
\(T = \frac{\Delta H}{\Delta S_{\text{system}}}\)

4. Convert \(\Delta H\) to \(\text{J mol}^{-1}\) and substitute the values:
\(\Delta H = 176.0 \times 1000 = 176000\text{ J mol}^{-1}\)
\(T = \frac{176000}{285.1} = 617.3\text{ K}\) (or \(617\text{ K}\) to 3 significant figures)

**M1:** Calculates the standard entropy change of the system:
\(\Delta S^{\ominus}_{\text{system}} = +285.1\text{ J K}^{-1}\text{ mol}^{-1}\) (1)

**M2:** States the feasibility condition:
\(\Delta G \le 0\) (or \(\Delta H - T\Delta S = 0\) / \(\Delta S_{\text{total}} \ge 0\)) (1)

**M3:** Correctly converts units so that \(\Delta H\) and \(\Delta S\) are consistent (e.g. \(176000\text{ J mol}^{-1}\) or \(0.2851\text{ kJ K}^{-1}\text{ mol}^{-1}\)) and rearranges the formula to make \(T\) the subject (1)

**M4:** Calculates the final temperature correctly:
\(617\text{ K}\) or \(617.3\text{ K}\) (1)

*Note: Allow full error-carried-forward (ECF) from M1 to subsequent marks.*

1. Find equilibrium moles of each component:
- Equilibrium concentration of \(\text{C} = 0.200\text{ mol dm}^{-3}\)
- Moles of \(\text{C}\) at equilibrium \(= 0.200 \times 2.00 = 0.400\text{ mol}\)
- Moles of \(\text{A}\) reacted \(= \frac{1}{2} \times 0.400 = 0.200\text{ mol}\)
- Moles of \(\text{A}\) at equilibrium \(= 0.800 - 0.200 = 0.600\text{ mol}\)
- Moles of \(\text{B}\) reacted \(= 0.400\text{ mol}\)
- Moles of \(\text{B}\) at equilibrium \(= 0.600 - 0.400 = 0.200\text{ mol}\)

2. Find equilibrium concentrations:
- \([\text{A}] = \frac{0.600\text{ mol}}{2.00\text{ dm}^3} = 0.300\text{ mol dm}^{-3}\)
- \([\text{B}] = \frac{0.200\text{ mol}}{2.00\text{ dm}^3} = 0.100\text{ mol dm}^{-3}\)
- \([\text{C}] = 0.200\text{ mol dm}^{-3}\)

3. Write the \(K_c\) expression:
\(K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^2}\)

4. Calculate \(K_c\) value:
\(K_c = \frac{(0.200)^2}{(0.300)(0.100)^2} = \frac{0.0400}{0.00300} = 13.3\) (or \(13.33\))

5. Determine the units:
\(\text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{dm}^3\text{ mol}^{-1}\)

**M1:** Calculates correct equilibrium concentrations for both reactants:
\([\text{A}] = 0.300\text{ mol dm}^{-3}\) and \([\text{B}] = 0.100\text{ mol dm}^{-3}\) (1)
*(Accept equilibrium moles: \(n(\text{A}) = 0.600\text{ mol}\) and \(n(\text{B}) = 0.200\text{ mol}\))*

**M2:** Expresses the equilibrium constant formula correctly:
\(K_c = \frac{[\text{C}]^2}{[\text{A}][\text{B}]^2}\) (1)
*(Do not award if round brackets are used or if the expression is inverted)*

**M3:** Obtains the correct numerical value:
\(13.3\) / \(13.33\) / \(\frac{40}{3}\) (1)
*(Allow ECF from incorrect equilibrium concentrations in M1)*

**M4:** States the correct unit:
\(\text{dm}^3\text{ mol}^{-1}\) (or \(\text{mol}^{-1}\text{ dm}^3\)) (1)

1. Analysis of alcohol **Y**:
Since alcohol **Y** has the molecular formula \(\text{C}_4\text{H}_{10}\text{O}\) (from substitution of \(-\text{Cl}\) in \(\text{C}_4\text{H}_9\text{Cl}\) with \(-\text{OH}\)) and is resistant to oxidation, it must be a tertiary alcohol. The only tertiary alcohol with four carbon atoms is 2-methylpropan-2-ol.
Therefore, **Y** is 2-methylpropan-2-ol (or \((\text{CH}_3)_3\text{COH}\)).

2. Identification of halogenoalkane **W**:
Since **Y** is 2-methylpropan-2-ol, the starting halogenoalkane **W** must be 2-chloro-2-methylpropane (or \((\text{CH}_3)_3\text{CCl}\)).

3. Identification of alkene **X**:
Elimination of \(\text{HCl}\) from 2-chloro-2-methylpropane can only yield 2-methylpropene (or \((\text{CH}_3)_2\text{C}=\text{CH}_2\)) as the single alkene product.

4. Reaction mechanism:
The substitution of a tertiary halogenoalkane like 2-chloro-2-methylpropane with warm aqueous sodium hydroxide proceeds via an \(\text{S}_\text{N}1\) nucleophilic substitution mechanism.

**M1:** Identifies **W** as 2-chloro-2-methylpropane / \((\text{CH}_3)_3\text{CCl}\) (1)

**M2:** Identifies **X** as 2-methylpropene / methylpropene / \((\text{CH}_3)_2\text{C}=\text{CH}_2\) (1)

**M3:** Identifies **Y** as 2-methylpropan-2-ol / \((\text{CH}_3)_3\text{COH}\) (1)

**M4:** States the mechanism as \(\text{S}_\text{N}1\) (or nucleophilic substitution) (1)
*(Accept tertiary nucleophilic substitution; reject SN2)*

Mass of reaction mixture = \(50.0\text{ g} + 50.0\text{ g} = 100.0\text{ g}\)

Temperature change, \(\Delta T = 32.6 - 19.2 = 13.4\text{ }^\circ\text{C}\)

Heat exchanged, \(q = m c \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 13.4\text{ K} = 5601.2\text{ J} = 5.6012\text{ kJ}\)

Moles of acid/alkali reacting = \(c \times V = 2.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.100\text{ mol}\)

Enthalpy of neutralisation, \(\Delta H = -\frac{q}{n} = -\frac{5.6012\text{ kJ}}{0.100\text{ mol}} = -56.0\text{ kJ mol}^{-1}\) (to 3 sig figs).

Assumptions:
1. The density of the solutions is equal to that of pure water (\(1.00\text{ g cm}^{-3}\)).
2. The specific heat capacity of the mixture is equal to that of pure water (\(4.18\text{ J g}^{-1}\text{ K}^{-1}\)).

Modification to reduce heat loss: Place a lid on the polystyrene cup, or place the cup inside a larger beaker containing cotton wool as insulation.

1 mark for calculating heat exchanged \(q = 5.6012\text{ kJ}\).
1 mark for calculating moles of acid/alkali = \(0.100\text{ mol}\).
1 mark for correct \(\Delta H_{\text{neut}} = -56.0\text{ kJ mol}^{-1}\) (must include negative sign, 3 sig figs, and correct units).
1 mark each for any two valid assumptions (max 2 marks):
- Density of solutions is \(1.00\text{ g cm}^{-3}\)
- Specific heat capacity of solution is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\)
- No heat is lost to the surroundings / none absorbed by the calorimeter.
1 mark for suggesting a valid modification (e.g., adding a lid, using double cups, wrapping cup in cotton wool/insulation).

Method:
- Add equal volumes of ethanol to three separate test tubes.
- Add equal amounts (drops) of each halogenoalkane (1-chlorobutane, 1-bromobutane, and 1-iodobutane) to separate test tubes.
- Place the test tubes in a water bath at \(50\text{ }^\circ\text{C}\) to reach a constant temperature.
- Add aqueous silver nitrate, \(\text{AgNO}_3\text{(aq)}\), to each tube and start a stopwatch.
- Measure the time taken for a precipitate to form in each test tube.
- Ethanol is used as a co-solvent because halogenoalkanes are insoluble in water but soluble in ethanol, allowing them to mix with the aqueous silver nitrate.

Order of rate:
- 1-iodobutane is fastest, then 1-bromobutane, then 1-chlorobutane (slowest).

Explanation:
- The rate depends on the C-X bond strength (bond enthalpy).
- The C-I bond is the weakest (lowest bond enthalpy) and breaks most easily, while the C-Cl bond is the strongest (highest bond enthalpy) and requires the most energy to break.

1 mark for using silver nitrate solution to produce precipitates.
1 mark for using ethanol as a common solvent to allow mixing.
1 mark for measuring the time taken for precipitates to form (measuring rate).
1 mark for the correct order (1-iodobutane > 1-bromobutane > 1-chlorobutane).
1 mark for explaining that the rate depends on bond enthalpy / bond strength, not bond polarity.
1 mark for stating C-I bond is weakest/has lowest bond enthalpy, so it is cleaved fastest.

Part (i):
- The iodine produced (\(\text{I}_2\)) reacts immediately with the sodium thiosulfate (\(\text{S}_2\text{O}_3^{2-}\)) present in the mixture and is reduced back to iodide (\(\text{I}^-\)).
- Once all the sodium thiosulfate is completely reacted/consumed, any further iodine produced remains in solution and reacts with starch to form the blue-black complex.

Part (ii):
- From the stoichiometry: \(2\text{ S}_2\text{O}_3^{2-} \equiv 1\text{ I}_2 \equiv 1\text{ S}_2\text{O}_8^{2-}\).
- Therefore, the change in peroxodisulfate concentration, \(\Delta[\text{S}_2\text{O}_8^{2-}]\), when the blue-black color appears is:
\(\Delta[\text{S}_2\text{O}_8^{2-}] = \frac{1}{2} \times [\text{S}_2\text{O}_3^{2-}] = \frac{1}{2} \times 4.00 \times 10^{-3} = 2.00 \times 10^{-3}\text{ mol dm}^{-3}\).
- Initial rate of reaction with respect to \(\text{S}_2\text{O}_8^{2-}\) is:
\(\text{Rate} = \frac{\Delta[\text{S}_2\text{O}_8^{2-}]}{\Delta t} = \frac{2.00 \times 10^{-3}\text{ mol dm}^{-3}}{42\text{ s}} = 4.7619 \times 10^{-5} \approx 4.76 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\) (to 3 sig figs).

1 mark for explaining that iodine reacts with thiosulfate immediately as it forms.
1 mark for stating that the blue-black colour appears only after all thiosulfate is consumed.
1 mark for relating moles of thiosulfate to moles of iodine: \(2\text{S}_2\text{O}_3^{2-} : 1\text{I}_2\).
1 mark for relating iodine to peroxodisulfate: \(1\text{I}_2 : 1\text{S}_2\text{O}_8^{2-}\) (or directly \(2\text{S}_2\text{O}_3^{2-} : 1\text{S}_2\text{O}_8^{2-}\)).
1 mark for calculating \(\Delta[\text{S}_2\text{O}_8^{2-}] = 2.00 \times 10^{-3}\text{ mol dm}^{-3}\).
1 mark for calculating rate = \(4.76 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\) (allow 2 or 3 s.f.).

Part (i):
- At the half-neutralisation point, half of the weak acid \(\text{HA}\) has been converted to its conjugate base \(\text{A}^-\).
- Therefore, \([\text{HA}] = [\text{A}^-]\).
- The expression for \(K_a\) is: \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\).
- Since \([\text{HA}] = [\text{A}^-]\), these terms cancel, giving \(K_a = [\text{H}^+]\).
- Taking the negative logarithm of both sides gives \(\text{p}K_a = \text{pH}\).

Part (ii):
- Since \(\text{p}K_a = \text{pH} = 4.82\) at the half-neutralisation point:
- \(K_a = 10^{-\text{p}K_a} = 10^{-4.82} = 1.5136 \times 10^{-5} \approx 1.51 \times 10^{-5}\text{ mol dm}^{-3}\).

Part (iii):
- If the pH reading is consistently \(0.20\) units too high, the recorded pH at half-neutralisation would be \(5.02\).
- This means the experimental \(\text{p}K_a\) is recorded as \(5.02\), giving \(K_a = 10^{-5.02} = 9.55 \times 10^{-6}\text{ mol dm}^{-3}\).
- Therefore, the calculated value of \(K_a\) would be lower than the true value because a higher pH corresponds to a lower \([\text{H}^+]\) concentration.

1 mark for stating that at half-neutralisation, \([\text{HA}] = [\text{A}^-]\).
1 mark for showing how \(K_a = [\text{H}^+]\) leads to \(\text{p}K_a = \text{pH}\).
1 mark for calculating \(K_a = 1.51 \times 10^{-5}\text{ mol dm}^{-3}\) (allow \(1.5 \times 10^{-5}\)).
1 mark for stating that the recorded pH would be higher (5.02 instead of 4.82).
1 mark for concluding that the calculated \(K_a\) would be lower.
1 mark for explaining that a higher pH corresponds to a lower hydrogen ion concentration / weaker apparent acid strength.

Part (i):
- Colour change is from colourless to a permanent pale pink.

Part (ii):
- Moles of \(\text{MnO}_4^-\text{(aq)}\) in the mean titre:
\(n(\text{MnO}_4^-) = 0.0200\text{ mol dm}^{-3} \times \frac{23.40}{1000}\text{ dm}^3 = 4.68 \times 10^{-4}\text{ mol}\)
- From the balanced equation: \(n(\text{Fe}^{2+}) = 5 \times n(\text{MnO}_4^-)\)
\(n(\text{Fe}^{2+})\text{ in } 25.0\text{ cm}^3 = 5 \times 4.68 \times 10^{-4} = 2.34 \times 10^{-3}\text{ mol}\)
- Moles of \(\text{Fe}^{2+}\) in the total \(250.0\text{ cm}^3\) volumetric flask:
\(n(\text{Fe}^{2+})\text{ total} = 2.34 \times 10^{-3} \times 10 = 2.34 \times 10^{-2}\text{ mol}\)
- Since \(1\text{ mol}\) of \(\text{Fe}^{2+}\) comes from \(1\text{ mol}\) of \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\):
\(m(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}) = 2.34 \times 10^{-2}\text{ mol} \times 278.0\text{ g mol}^{-1} = 6.5052\text{ g}\)
- Percentage by mass of \(\text{FeSO}_4 \cdot 7\text{H}_2\text{O}\) in the sample:
\(\%\text{ mass} = \frac{6.5052\text{ g}}{6.85\text{ g}} \times 100 = 94.966\% \approx 95.0\%\).

1 mark for the correct end-point colour change: colourless to pale pink (reject pink to colourless).
1 mark for calculating moles of manganate(VII): \(4.68 \times 10^{-4}\text{ mol}\).
1 mark for multiplying by 5 to find moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\): \(2.34 \times 10^{-3}\text{ mol}\).
1 mark for scaling up to \(250\text{ cm}^3\): \(2.34 \times 10^{-2}\text{ mol}\).
1 mark for calculating mass of hydrated salt: \(6.51\text{ g}\).
1 mark for calculating percentage mass: \(95.0\%\) (accept \(94.9\%\) to \(95.0\%\)).

Part (i): Recrystallisation steps:
1. Dissolve the crude aspirin in the minimum volume of hot solvent (such as ethanol or water).
2. Filter the hot solution (using a preheated funnel and fluted filter paper) to remove any insoluble impurities.
3. Allow the filtrate to cool slowly to room temperature, and then place it in ice to recrystallise the aspirin.
4. Filter the crystals under reduced pressure using a Buchner funnel to separate the crystals from the soluble impurities (mother liquor).
5. Wash the crystals with a small volume of cold solvent.
6. Dry the purified crystals in a desiccator, warm oven, or between sheets of filter paper.

Part (ii): Verifying purity:
- Measure the melting temperature (melting point) of the dried crystals using a melting point apparatus.
- Compare the measured melting range with the literature value (aspirin melts at approximately \(136\text{ }^\circ\text{C}\)).
- A pure sample will have a sharp melting point that matches the literature value, whereas an impure sample will melt over a wider range and at a lower temperature than the literature value.

1 mark for dissolving crude solid in the minimum volume of hot solvent.
1 mark for hot filtration to remove insoluble impurities.
1 mark for cooling in ice to recrystallise and filtering under reduced pressure (using Buchner funnel).
1 mark for washing with cold solvent and drying.
1 mark for measuring the melting point / melting range.
1 mark for stating that a pure sample has a sharp melting point matching the literature value (or impure sample has a wider, lower melting range).

Part (i):
- To the unknown solution, add dilute nitric acid, \(\text{HNO}_3\text{(aq)}\), to remove any interfering carbonate impurities.
- Then, add aqueous silver nitrate, \(\text{AgNO}_3\text{(aq)}\).
- Note the colour of the precipitate formed:
- White precipitate indicates chloride ions, \(\text{Cl}^-\).
- Cream precipitate indicates bromide ions, \(\text{Br}^-\).
- Yellow precipitate indicates iodide ions, \(\text{I}^-\).
- Since these colours can be difficult to distinguish, add dilute aqueous ammonia, \(\text{NH}_3\text{(aq)}\):
- If the precipitate dissolves, the halide is chloride.
- If the precipitate does not dissolve, add concentrated aqueous ammonia:
- If the precipitate dissolves in concentrated ammonia, the halide is bromide.
- If the precipitate remains insoluble even in concentrated ammonia, the halide is iodide.

Part (ii):
- \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\)

1 mark for adding dilute nitric acid followed by silver nitrate solution.
1 mark for stating the correct initial precipitate colours (White = chloride, Cream = bromide, Yellow = iodide).
1 mark for testing with dilute aqueous ammonia (solubility of chloride).
1 mark for testing with concentrated aqueous ammonia (solubility of bromide vs insolubility of iodide).
1 mark for the correct ionic equation: \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\).
1 mark for correct state symbols in the ionic equation.

Part (i):
- The technique is fractional distillation.
- Labeled diagram should show:
- A round-bottom flask containing the mixture on a heat source.
- A fractionating column fitted vertically on the flask.
- A thermometer at the top of the column / entrance to the condenser to monitor the vapour temperature.
- A Liebig condenser connected to the column, angled downwards, with water entering at the bottom and exiting at the top.
- A receiver adapter and a collecting vessel.

Part (ii):
- The fractionating column provides a large surface area for repeated condensation and vaporisation cycles.
- As the vapour rises, it cools and condenses on the surface of the column packing.
- The lower boiling point component (hexane) vaporises more readily than the higher boiling point component (heptane), so the vapour becomes increasingly enriched in the more volatile component (hexane) as it moves up the column.
- Eventually, pure hexane vapour reaches the top of the column and enters the condenser first, while heptane remains in the flask or lower parts of the column.

1 mark for naming the technique: fractional distillation.
1 mark for a diagram showing the flask, fractionating column, and thermometer positioned correctly at the T-junction.
1 mark for a diagram showing the condenser with correct water flow (in at bottom, out at top) and a collecting vessel.
1 mark for explaining that the fractionating column provides a large surface area for repeated condensation and evaporation cycles.
1 mark for explaining that the more volatile component (hexane/lower boiling point) moves up the column faster.
1 mark for explaining that this results in a temperature gradient, allowing complete separation of the two liquids.

Mass of hydrated salt = \(21.67\text{ g} - 18.42\text{ g} = 3.25\text{ g}\). Mass of anhydrous salt = \(21.19\text{ g} - 18.42\text{ g} = 2.77\text{ g}\). Mass of water lost = \(21.67\text{ g} - 21.19\text{ g} = 0.48\text{ g}\). Moles of \(\text{BaCl}_2 = 2.77\text{ g} / 208.3\text{ g mol}^{-1} = 0.01330\text{ mol}\). Moles of \(\text{H}_2\text{O} = 0.48\text{ g} / 18.0\text{ g mol}^{-1} = 0.02667\text{ mol}\). Ratio \(x = 0.02667 / 0.01330 = 2.005\) which rounds to \(2.01\). If the crucible is not heated to constant mass, some water remains in the salt. This means the calculated mass of water lost is lower, which results in a lower calculated value of \(x\).

Award 1 mark for finding mass of water lost (0.48 g) and mass of anhydrous barium chloride (2.77 g). Award 1 mark for calculating moles of anhydrous barium chloride (0.01330 mol) and moles of water (0.02667 mol). Award 1 mark for calculating the ratio x = 2.01. Award 1 mark for expressing the answer to 2 decimal places. Award 1 mark for explaining that incomplete heating means less water is lost, making the calculated mass of water too low. Award 1 mark for stating this results in a lower calculated value of x.

Total mass of mixture, \(m = (50.0 + 50.0) \times 1.00 = 100.0\text{ g}\). Temperature change, \(\Delta T = 25.2 - 18.5 = 6.7\text{ K}\). Heat released, \(q = m \times c \times \Delta T = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 6.7\text{ K} = 2800.6\text{ J} = 2.8006\text{ kJ}\). Moles of \(\text{H}^+\) ions neutralized: \(n = 1.00\text{ mol dm}^{-3} \times 0.0500\text{ dm}^3 = 0.0500\text{ mol}\). \(\Delta H_{\text{neu}} = -q / n = -2.8006\text{ kJ} / 0.0500\text{ mol} = -56.0\text{ kJ mol}^{-1}\) (to 3 s.f.). Two systematic errors: 1. Heat lost to the surroundings or absorbed by the polystyrene cup. 2. Assuming the specific heat capacity and density of the salt solution are exactly the same as pure water.

Award 1 mark for calculating mass of solution (100.0 g) and temperature change (6.7 K). Award 1 mark for calculating heat energy released (q = 2.80 kJ). Award 1 mark for calculating moles of water formed or acid used (0.0500 mol). Award 1 mark for calculating the enthalpy change of neutralization as -56.0 kJ mol^-1 (including negative sign, correct unit, and 3 s.f.). Award 1 mark for identifying heat loss to surroundings/cup as a source of error. Award 1 mark for identifying assuming density and specific heat capacity are of pure water / neglecting heat capacity of calorimeter as a source of error.

The purification steps are: 1. Transfer the crude reaction mixture to a separating funnel. 2. Wash with water to remove most of the soluble inorganic impurities (such as sulfuric acid and sodium salts) and run off the lower organic layer (since 1-bromobutane is denser than water). 3. Wash the organic layer with aqueous sodium hydrogencarbonate solution to neutralize any remaining trace sulfuric acid. Invert the funnel and open the tap periodically to release the pressure built up by carbon dioxide gas. 4. Separate the organic layer and wash it with saturated sodium chloride solution (brine) to help separate the phases. 5. Transfer the lower organic layer to a clean conical flask and add an anhydrous inorganic salt (such as anhydrous calcium chloride or sodium sulfate) as a drying agent. Shake and leave until the liquid becomes completely clear. 6. Filter or decant the dry 1-bromobutane from the drying agent.

Award 1 mark for using a separating funnel and identifying that 1-bromobutane is the lower layer. Award 1 mark for washing with water to remove soluble acids or salts. Award 1 mark for washing with aqueous sodium hydrogencarbonate solution to neutralize remaining acid. Award 1 mark for releasing carbon dioxide pressure build-up by opening the tap. Award 1 mark for drying the organic layer using an anhydrous drying agent (e.g., anhydrous calcium chloride or sodium sulfate) until clear. Award 1 mark for decanting or filtering the liquid to remove the solid drying agent.

At the half-equivalence point, exactly half of the acid \(\text{HA}\) has been neutralized to form the conjugate base \(\text{A}^-\). Therefore, \([\text{HA}] = [\text{A}^-]\). The acid dissociation constant expression is: \(K_a = [\text{H}^+][\text{A}^-] / [\text{HA}]\). Since \([\text{HA}] = [\text{A}^-]\), these terms cancel, giving \(K_a = [\text{H}^+]\). Taking negative logarithms of both sides, \(\text{p}K_a = \text{pH}\). Given the \(\text{pH}\) at the half-equivalence point (\(11.20\text{ cm}^3\)) is \(4.76\), we have \(\text{p}K_a = 4.76\). Therefore, \(K_a = 10^{-4.76} = 1.74 \times 10^{-5}\text{ mol dm}^{-3}\).

Award 1 mark for explaining that at the half-equivalence point, [HA] = [A^-]. Award 1 mark for showing that Ka = [H^+] and therefore pH = pKa. Award 1 mark for stating that pKa = 4.76. Award 1 mark for calculating Ka = 1.74 * 10^-5. Award 1 mark for reporting the final answer to 3 significant figures. Award 1 mark for stating correct units as mol dm^-3.

(a) The ionic equation is: \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\). (b) Moles of thiosulfate ions used: \(n(\text{S}_2\text{O}_3^{2-}) = 0.0100\text{ mol dm}^{-3} \times 0.0050\text{ dm}^3 = 5.00 \times 10^{-5}\text{ mol}\). From stoichiometry, \(2\text{ mol}\) of \(\text{S}_2\text{O}_3^{2-}\) reacts with \(1\text{ mol}\) of \(\text{I}_2\), so \(n(\text{I}_2) = 2.50 \times 10^{-5}\text{ mol}\) has been produced when thiosulfate is consumed. The primary reaction is \(\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2\), so \(1\text{ mol}\) of \(\text{S}_2\text{O}_8^{2-}\) produces \(1\text{ mol}\) of \(\text{I}_2\). Thus, moles of peroxodisulfate reacted at color change = \(2.50 \times 10^{-5}\text{ mol}\). Change in peroxodisulfate concentration, \(\Delta [\text{S}_2\text{O}_8^{2-}] = 2.50 \times 10^{-5}\text{ mol} / 0.0500\text{ dm}^3 = 5.00 \times 10^{-4}\text{ mol dm}^{-3}\). Initial rate of reaction = \(\Delta [\text{S}_2\text{O}_8^{2-}] / \Delta t = 5.00 \times 10^{-4}\text{ mol dm}^{-3} / 42.5\text{ s} = 1.18 \times 10^{-5}\text{ mol dm}^{-3}\text{ s}^{-1}\).

Award 1 mark for the correct ionic equation for reaction of iodine with thiosulfate. Award 1 mark for calculating moles of thiosulfate as 5.00 * 10^-5 mol. Award 1 mark for deducing moles of peroxodisulfate reacted as 2.50 * 10^-5 mol. Award 1 mark for calculating change in concentration of peroxodisulfate as 5.00 * 10^-4 mol dm^-3. Award 1 mark for calculating the rate as 1.18 * 10^-5. Award 1 mark for writing the correct units as mol dm^-3 s^-1.

(a) The equation is: \(\text{CuSO}_4\cdot5\text{H}_2\text{O}(\text{aq}) + 4\text{NH}_3(\text{aq}) \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{SO}_4(\text{aq}) + 3\text{H}_2\text{O}(\text{l})\). (b) Molar mass of \(\text{CuSO}_4\cdot5\text{H}_2\text{O} = 249.6\text{ g mol}^{-1}\). Moles of starting material: \(n = 6.24\text{ g} / 249.6\text{ g mol}^{-1} = 0.0250\text{ mol}\). Molar mass of product \([\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{SO}_4 = 63.5 + 4(17.0) + 2(18.0) + 32.1 + 4(16.0) = 263.6\text{ g mol}^{-1}\). Theoretical yield: \(\text{mass} = 0.0250\text{ mol} \times 263.6\text{ g mol}^{-1} = 6.59\text{ g}\). Percentage yield: \(\text{yield} = (4.85\text{ g} / 6.59\text{ g}) \times 100\% = 73.6\%\) (to 3 s.f.).

Award 1 mark for correct formulas of reactants and products in equation. Award 1 mark for balancing equation. Award 1 mark for calculating moles of copper sulfate starting material as 0.0250 mol. Award 1 mark for calculating molar mass of product as 263.6 g mol^-1. Award 1 mark for calculating theoretical mass of product as 6.59 g. Award 1 mark for calculating percentage yield as 73.6% (to 3 s.f.).

(a) The ionic equation is: \(2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^-(\text{aq}) \rightarrow 2\text{CuI}(\text{s}) + \text{I}_2(\text{aq})\). (b) Moles of thiosulfate used: \(n(\text{S}_2\text{O}_3^{2-}) = 0.125\text{ mol dm}^{-3} \times 0.02140\text{ dm}^3 = 2.675 \times 10^{-3}\text{ mol}\). Stoichiometry of thiosulfate titration: \(\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}\), so \(n(\text{I}_2) = 0.5 \times n(\text{S}_2\text{O}_3^{2-}) = 1.3375 \times 10^{-3}\text{ mol}\). From copper reaction, \(2\text{ mol}\) of \(\text{Cu}^{2+}\) produces \(1\text{ mol}\) of \(\text{I}_2\), so \(n(\text{Cu}^{2+}) = 2 \times n(\text{I}_2) = n(\text{S}_2\text{O}_3^{2-}) = 2.675 \times 10^{-3}\text{ mol}\). Concentration of \(\text{Cu}^{2+}\) in sample: \([\text{Cu}^{2+}] = 2.675 \times 10^{-3}\text{ mol} / 0.02500\text{ dm}^3 = 0.107\text{ mol dm}^{-3}\).

Award 1 mark for correct reactants and products in equation. Award 1 mark for fully balanced equation. Award 1 mark for calculating moles of thiosulfate used. Award 1 mark for using stoichiometry to relate moles of thiosulfate to moles of copper(II) ions in a 1:1 ratio. Award 1 mark for calculating concentration of copper(II) ions as 0.107. Award 1 mark for writing correct units (mol dm^-3) and 3 s.f.

Dissolve the unknown sodium halide in distilled water. Add dilute nitric acid to remove any interfering carbonate impurities. Add silver nitrate solution. Observe the precipitate color: a white precipitate indicates chloride, cream indicates bromide, and yellow indicates iodide. Add dilute ammonia solution to the precipitates: if the white precipitate dissolves, it confirms chloride. If the precipitate does not dissolve in dilute ammonia, add concentrated ammonia: if the cream precipitate dissolves, it confirms bromide. If the yellow precipitate remains insoluble in both dilute and concentrated ammonia, it confirms iodide.

Award 1 mark for dissolving sample, adding dilute nitric acid followed by silver nitrate solution. Award 1 mark for stating correct precipitate observations: white for chloride, cream for bromide, yellow for iodide. Award 1 mark for stating that dilute ammonia is used to confirm chloride because silver chloride dissolves. Award 1 mark for stating that concentrated ammonia is used to confirm bromide because silver bromide dissolves only in concentrated ammonia. Award 1 mark for stating that silver iodide remains insoluble in both dilute and concentrated ammonia. Award 1 mark for presenting a logical, clear, and sequential laboratory procedure.

(a) 1. Calculate heat transferred, \(q\): \(m = 50.0 + 50.0 = 100.0\text{ g}\), \(\Delta T = 24.3 - 18.5 = 5.8\text{ °C}\), \(q = m c \Delta T = 100.0 \times 4.18 \times 5.8 = 2424.4\text{ J} = 2.4244\text{ kJ}\). 2. Calculate moles of water formed: \(n(\text{HA}) = 1.00 \times 0.0500 = 0.0500\text{ mol}\). Since the acid is monoprotic, \(n(\text{H}_2\text{O}) = 0.0500\text{ mol}\). 3. Calculate \(\Delta H_{\text{neut}}\): \(\Delta H_{\text{neut}} = -\frac{2.4244}{0.0500} = -48.5\text{ kJ mol}^{-1}\). (b) Since there are two temperature readings (initial and final), the total uncertainty is \(2 \times 0.1 = 0.2\text{ °C}\). Percentage uncertainty = \(\frac{0.2}{5.8} \times 100\% = 3.45\%\). (c) Place a lid on the polystyrene cup (or wrap the cup in cotton wool/insulating material).

M1: Correct calculation of heat transfer (2.42 kJ) [1 mark]. M2: Moles of acid/base calculated as 0.0500 mol [1 mark]. M3: Calculation of enthalpy change to 3 sig figs with a negative sign (-48.5 kJ mol^-1) [1 mark]. M4: Identification of total uncertainty in temperature change as 0.2 °C [1 mark]. M5: Percentage uncertainty of 3.45% (accept 3.4% or 3.5%) [1 mark]. M6: Suitable modification to reduce heat loss (e.g. use a lid) [1 mark].

(a) Mass of 2-methylpropan-2-ol used = \(10.0\text{ cm}^3 \times 0.789\text{ g cm}^{-3} = 7.89\text{ g}\). Moles of alcohol = \(7.89 / 74.12 = 0.10645\text{ mol}\). Theoretical mass of 2-chloro-2-methylpropane = \(0.10645\text{ mol} \times 92.57\text{ g mol}^{-1} = 9.854\text{ g}\). Percentage yield = \(\frac{6.12}{9.854} \times 100\% = 62.1\%\). (b) Aqueous sodium hydrogencarbonate is added to neutralise any remaining/excess hydrochloric acid. Ionic equation: \(\text{H}^+(aq) + \text{HCO}_3^-(aq) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(l)\). (c) Invert the separating funnel and open the tap regularly to release the pressure built up by carbon dioxide gas.

M1: Moles of reactant calculated as 0.106 mol [1 mark]. M2: Theoretical yield of product calculated as 9.85 g [1 mark]. M3: Percentage yield calculated as 62.1% (3 sig figs required) [1 mark]. M4: Purpose of NaHCO3 stated as neutralising excess HCl/acid [1 mark]. M5: Correct ionic equation (state symbols not required) [1 mark]. M6: Venting the funnel/releasing pressure described [1 mark].

1. Moles of \(\text{NaOH}\) in titration = \(0.120 \times \frac{23.50}{1000} = 2.82 \times 10^{-3}\text{ mol}\). 2. Moles of \(\text{HCl}\) in \(25.0\text{ cm}^3\) aliquot = \(2.82 \times 10^{-3}\text{ mol}\). 3. Moles of \(\text{HCl}\) remaining in \(250\text{ cm}^3\) flask = \(2.82 \times 10^{-3} \times 10 = 2.82 \times 10^{-2}\text{ mol}\). 4. Initial moles of \(\text{HCl}\) added = \(1.00 \times \frac{50.0}{1000} = 5.00 \times 10^{-2}\text{ mol}\). 5. Moles of \(\text{HCl}\) reacted with \(\text{CaCO}_3\) = \(5.00 \times 10^{-2} - 2.82 \times 10^{-2} = 2.18 \times 10^{-2}\text{ mol}\). 6. Moles of \(\text{CaCO}_3\) = \(\frac{2.18 \times 10^{-2}}{2} = 1.09 \times 10^{-2}\text{ mol}\). Mass of \(\text{CaCO}_3\) = \(1.09 \times 10^{-2} \times 100.1 = 1.091\text{ g}\). 7. Percentage purity = \(\frac{1.091}{1.50} \times 100\% = 72.7\%\) (or \(72.7\%\) using 100 for molar mass).

M1: Correct calculation of moles of NaOH (2.82 x 10^-3 mol) [1 mark]. M2: Multiplication by 10 to find unreacted HCl in 250 cm3 (2.82 x 10^-2 mol) [1 mark]. M3: Correct calculation of initial moles of HCl (5.00 x 10^-2 mol) [1 mark]. M4: Subtraction to find reacted moles of HCl (2.18 x 10^-2 mol) [1 mark]. M5: Divide by 2 to find moles of CaCO3 (1.09 x 10^-2 mol) [1 mark]. M6: Percentage purity calculation to 3 sig figs (72.7%) [1 mark].

(a) Ionic equation: \(\text{MnO}_4^-(aq) + 5\text{Fe}^{2+}(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 5\text{Fe}^{3+}(aq) + 4\text{H}_2\text{O}(l)\). (b) 1. Moles of \(\text{MnO}_4^-\): \(0.0150 \times \frac{18.25}{1000} = 2.7375 \times 10^{-4}\text{ mol}\). 2. Moles of \(\text{Fe}^{2+}\) in \(25.0\text{ cm}^3\): \(5 \times 2.7375 \times 10^{-4} = 1.36875 \times 10^{-3}\text{ mol}\). 3. Moles of \(\text{Fe}^{2+}\) in \(250\text{ cm}^3\): \(1.36875 \times 10^{-3} \times 10 = 1.36875 \times 10^{-2}\text{ mol}\). 4. Mass of Fe: \(1.36875 \times 10^{-2} \times 55.8 = 0.76376\text{ g}\). To 3 sig figs, \(0.764\text{ g}\). (c) The color change is from colourless to permanent pale pink.

M1: Correct ionic equation [1 mark]. M2: Moles of MnO4- calculated as 2.74 x 10^-4 mol [1 mark]. M3: Moles of Fe2+ in 25.0 cm3 aliquot calculated as 1.37 x 10^-3 mol [1 mark]. M4: Total moles of Fe2+ in 250 cm3 scaled to 1.37 x 10^-2 mol [1 mark]. M5: Mass of iron calculated to 3 sig figs (0.764 g) [1 mark]. M6: Correct end-point color change (colourless to permanent pale pink) [1 mark].