2025 Cambridge IAL Chemistry (9701) Practice Paper with Answers

Initially, aqueous copper(II) exists as the octahedral hexaaquacopper(II) complex, \([Cu(H_2O)_6]^{2+}\), which is pale blue. The coordination number of copper in this complex is 6.

When concentrated hydrochloric acid is added, ligand exchange occurs as the chloride ions displace the water ligands. This forms the tetrahedral tetrachlorocuprate(II) complex, \([CuCl_4]^{2-}\), which is yellow:

\([Cu(H_2O)_6]^{2+} + 4Cl^- \rightleftharpoons [CuCl_4]^{2-} + 6H_2O\)

The mixture appears yellow-green because of the equilibrium mixture of the blue and yellow complexes. In \([CuCl_4]^{2-}\), the coordination number of the copper ion is 4. Thus, the coordination number changes from 6 to 4.

- Option A is incorrect because the chloro complex formed is tetrahedral \([CuCl_4]^{2-}\), not octahedral.
- Option B is incorrect because diluting with water shifts the equilibrium back to the octahedral hexaaqua complex, \([Cu(H_2O)_6]^{2+}\).
- Option C is incorrect because no redox reaction occurs; copper remains in the +2 oxidation state.

[1 mark] Award for choosing option D.

Reject: A, B, C.

Using the Nernst equation for a system at 298 K:

\(E = E^\ominus + \frac{0.059}{z}\log\left(\frac{[\text{oxidised state}]}{[\text{reduced state}]}\right)\)

For the half-reaction \(Fe^{3+}(aq) + e^- \rightleftharpoons Fe^{2+}(aq)\), the number of electrons transferred, \(z\), is equal to 1. The oxidized state is \(Fe^{3+}(aq)\) and the reduced state is \(Fe^{2+}(aq)\).

Substitute the given values into the equation:

\(+0.711 = +0.770 + 0.059 \log\left(\frac{[Fe^{3+}]}{[Fe^{2+}]}\right)\)

\(-0.059 = 0.059 \log\left(\frac{[Fe^{3+}]}{[Fe^{2+}]}\right)\)

\(\log\left(\frac{[Fe^{3+}]}{[Fe^{2+}]}\right) = -1\)

\(\frac{[Fe^{3+}]}{[Fe^{2+}]} = 10^{-1} = 0.10\)

[1 mark] Award for choosing option A.

Reject: B, C, D.

Let us trace the reaction steps for methyl 3-chloropropanoate (Option D):

1. Heating with aqueous sodium hydroxide causes both ester hydrolysis and nucleophilic substitution of the chlorine atom:

\(Cl-CH_2-CH_2-COOCH_3 + 2NaOH \rightarrow HO-CH_2-CH_2-COONa + CH_3OH + NaCl\)

Thus, the product mixture Y contains sodium 3-hydroxypropanoate and methanol.

2. Acidification and oxidation with hot acidified \(KMnO_4\):

- The primary alcohol group of 3-hydroxypropanoic acid (\(HO-CH_2-CH_2-COOH\)) is oxidized to a carboxylic acid, giving propanedioic acid, \(HOOC-CH_2-COOH\).
- Methanol (\(CH_3OH\)) is oxidized to methanoic acid, which is readily oxidized further by hot acidified \(KMnO_4\) to form carbon dioxide, \(CO_2\), and water.

These products match the description perfectly.

- Option A: Ethyl chloroacetate would give glycolic acid and ethanol on hydrolysis. Oxidation would yield ethanedioic acid and ethanoic acid.
- Option B: Methyl 2-chloropropanoate would give lactic acid (a secondary alcohol) and methanol on hydrolysis. Oxidation of lactic acid would yield 2-oxopropanoic acid (pyruvic acid), not propanedioic acid.
- Option C: 4-chlorobutanoic acid would give 4-hydroxybutanoic acid on hydrolysis. Oxidation would yield butanedioic acid, with no carbon dioxide produced.

[1 mark] Award for choosing option D.

Reject: A, B, C.

1. Nitration of methylbenzene at 30 °C with concentrated \(HNO_3\) and concentrated \(H_2SO_4\) is an electrophilic aromatic substitution. Since the methyl group is ortho/para-directing, the major para-isomer Y is 1-methyl-4-nitrobenzene (4-nitrotoluene).

2. Reaction of 1-methyl-4-nitrobenzene with chlorine gas in the presence of UV light is a free-radical substitution. Under these conditions, substitution occurs selectively on the aliphatic alkyl side-chain rather than the aromatic ring.

Consequently, one hydrogen atom of the methyl group is substituted by chlorine, yielding 1-(chloromethyl)-4-nitrobenzene. If a halogen carrier (like \(AlCl_3\) or \(FeCl_3\)) in the dark were used instead of UV light, ring substitution would occur to yield 2-chloro-4-nitrotoluene.

[1 mark] Award for choosing option A.

Reject: B, C, D.

1. Calculate the initial moles of \(HCl\) added:

\(n(HCl)_{\text{initial}} = 50.0 \times 10^{-3}\text{ dm}^3 \times 0.200\text{ mol dm}^{-3} = 0.0100\text{ mol}\)

2. Calculate the moles of \(NaOH\) used to neutralize the excess \(HCl\):

\(n(NaOH) = 24.0 \times 10^{-3}\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 0.0024\text{ mol}\)

3. Since \(HCl\) and \(NaOH\) react in a 1:1 mole ratio, the excess \(HCl\) is 0.0024 mol. Calculate the moles of \(HCl\) that reacted with the metal carbonate:

\(n(HCl)_{\text{reacted}} = 0.0100 - 0.0024 = 0.0076\text{ mol}\)

4. Write the balanced equation for the reaction of \(MCO_3\) with \(HCl\):

\(MCO_3(s) + 2HCl(aq) \rightarrow MCl_2(aq) + CO_2(g) + H_2O(l)\)

This shows a 1:2 stoichiometric ratio between \(MCO_3\) and \(HCl\). Therefore, the moles of \(MCO_3\) are:

\(n(MCO_3) = \frac{0.0076}{2} = 0.0038\text{ mol}\)

5. Determine the relative formula mass (\(M_r\)) of \(MCO_3\):

\(M_r(MCO_3) = \frac{\text{mass}}{\text{moles}} = \frac{0.380\text{ g}}{0.0038\text{ mol}} = 100\)

6. Calculate the relative atomic mass (\(A_r\)) of \(M\):

\(M_r(MCO_3) = A_r(M) + A_r(C) + 3 \times A_r(O)\)

\(100 = A_r(M) + 12.0 + 48.0\)

\(A_r(M) = 100 - 60.0 = 40.0\)

Comparing this with the options, \(40.1\) corresponds to calcium (\(Ca\)).

[1 mark] Award for choosing option B.

Reject: A, C, D.

When the temperature of a gaseous mixture increases, the average kinetic energy of the molecules increases. This causes the Maxwell-Boltzmann distribution curve to change in the following ways:

- The curve shifts to the right (towards higher energy values).
- The peak of the curve flattens and becomes lower to keep the total area under the curve (representing the total number of molecules) constant.
- The activation energy, \(E_a\), remains unchanged because it is a constant property of the specific reaction pathway.
- Because the curve is shifted to the right, a larger fraction of the molecules now have kinetic energies greater than or equal to the activation energy, which leads to a higher frequency of successful collisions and a faster reaction rate.

Therefore, only Option B is correct.

[1 mark] Award for choosing option B.

Reject: A, C, D.

Using the ideal gas equation, \(pV = nRT = \frac{m}{M_r}RT\), we rearrange for relative molecular mass, \(M_r\):

\(M_r = \frac{mRT}{pV}\)

Convert all values to SI units:
- \(p = 100\text{ kPa} = 100 \times 10^3\text{ Pa} = 1.00 \times 10^5\text{ Pa}\)
- \(V = 83.1\text{ cm}^3 = 83.1 \times 10^{-6}\text{ m}^3\)
- \(T = 300\text{ K}\)
- \(m = 0.250\text{ g}\)
- \(R = 8.31\text{ J K}^{-1}\text{ mol}^{-1}\)

Calculate \(pV\):
\(pV = (1.00 \times 10^5\text{ Pa}) \times (83.1 \times 10^{-6}\text{ m}^3) = 8.31\text{ Pa m}^3\) (or J)

Substitute into the rearranged equation:
\(M_r = \frac{0.250 \times 8.31 \times 300}{8.31} = 0.250 \times 300 = 75.0\)

[1 mark] Award for choosing option B.

Reject: A, C, D.

Let us examine the trends of Group 2 elements and compounds down the group (from magnesium to barium):

- **Solubility of Group 2 hydroxides increases down the group.** This means barium hydroxide is more soluble than magnesium hydroxide. Thus, Option A is incorrect.
- **Solubility of Group 2 sulfates decreases down the group.** This means barium sulfate is less soluble than magnesium sulfate. Thus, Option B is incorrect.
- **Thermal stability of Group 2 carbonates increases down the group.** As you go down the group, the cationic radius increases, which decreases its charge density. The larger cation is less effective at polarizing the electron cloud of the carbonate ion (polarizing power decreases), making the C-O bonds in the carbonate ion harder to weaken. Consequently, a higher temperature is required for decomposition. Barium carbonate is therefore more stable and decomposes at a higher temperature than magnesium carbonate. Thus, Option C is correct.
- **Group 2 nitrates decompose on heating to form the metal oxide, nitrogen dioxide, and oxygen:**

\(2M(NO_3)_2(s) \rightarrow 2MO(s) + 4NO_2(g) + O_2(g)\)

They do not form the metal nitrite. Thus, Option D is incorrect.

[1 mark] Award for choosing option C.

Reject: A, B, D.

In a transition metal complex like \([Cu(H_2O)_6]^{2+}\), the presence of the water ligands splits the 3d orbitals into two different energy levels. d-electrons can absorb specific frequencies of visible light to be promoted from the lower energy level to the higher energy level. For the copper(II) complex, light in the red-orange region of the spectrum is absorbed, and the complementary color (blue) is transmitted, which is what we see.

1 mark for identifying the correct explanation of light absorption by d-d transitions and the transmission of the complementary color.

First, calculate the electrode potential of the \(Ce^{4+}/Ce^{3+}\) half-cell under the non-standard condition:
\(E = +1.61 + \frac{0.059}{1} \log \frac{0.010}{1.0} = 1.61 + 0.059(-2) = 1.61 - 0.118 = +1.492\text{ V}\).
The \(Fe^{3+}/Fe^{2+}\) half-cell remains under standard conditions, so its potential is \(+0.77\text{ V}\).
Because the potential of the cerium half-cell (\(+1.492\text{ V}\)) is more positive than that of the iron half-cell (\(+0.77\text{ V}\)), the cerium electrode acts as the cathode and the iron electrode as the anode.
\(E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 1.492 - 0.77 = +0.722\text{ V} \approx +0.72\text{ V}\).

1 mark for the correct application of the Nernst equation and calculation of the resulting cell potential.

Chlorobenzene (\(C_6H_5Cl\)) does not undergo hydrolysis because the C-Cl bond is very strong due to partial double-bond character from the overlap of the chlorine lone pair with the benzene pi-system; hence, no precipitate forms. Chloroethane (\(CH_3CH_2Cl\)) undergoes slow nucleophilic substitution at room temperature, yielding a slow-forming white precipitate of AgCl. Ethanoyl chloride (\(CH_3COCl\)) undergoes rapid hydrolysis at room temperature to release chloride ions, producing a white precipitate almost immediately.

1 mark for correctly matching the reactivity and observations of the three halogen compounds.

Under free-radical conditions (UV light and heat), substitution occurs on the alkyl side-chain of methylbenzene to produce (chloromethyl)benzene, \(C_6H_5CH_2Cl\) (Product X). Under electrophilic aromatic substitution conditions (halogen carrier \(AlCl_3\) in the dark), substitution occurs on the aromatic ring. Because the methyl group is ortho/para-directing (2,4-directing), a mixture of 2-chlorotoluene and 4-chlorotoluene is formed (Product Y).

1 mark for correctly identifying the structural products under the two distinct halogenation conditions.

First, calculate the mass of carbon and hydrogen in the sample:
- Mass of C = \(4.40 \times \frac{12.0}{44.0} = 1.20\text{ g}\)
- Mass of H = \(2.70 \times \frac{2.0}{18.0} = 0.30\text{ g}\)
- Mass of O = \(2.30 - (1.20 + 0.30) = 0.80\text{ g}\)

Next, convert to moles:
- Moles of C = \(1.20 / 12.0 = 0.10\text{ mol}\)
- Moles of H = \(0.30 / 1.0 = 0.30\text{ mol}\)
- Moles of O = \(0.80 / 16.0 = 0.05\text{ mol}\)

Divide by the smallest value (0.05):
- C = 2, H = 6, O = 1

The empirical formula is \(C_2H_6O\).

1 mark for the correct calculation of elements' masses, moles, and ratio leading to the empirical formula.

Increasing the temperature shifts the Maxwell-Boltzmann distribution curve so that the peak is lower and shifted to the right, representing a higher average kinetic energy of the molecules. Adding a catalyst provides an alternative reaction pathway with a lower activation energy, but it has no effect on the energy distribution of the molecules themselves.

1 mark for correctly describing both the shift of the curve with temperature and the lowering of the activation energy barrier with a catalyst.

A real gas behaves most ideally under high temperature and low pressure. At high temperature, the molecules have high kinetic energies, making intermolecular attractive forces negligible. At low pressure, the molecules are far apart, meaning the actual volume of the gas molecules is negligible compared to the volume of the container.

1 mark for identifying high temperature and low pressure along with the correct kinetic theory explanation.

Down Group 2, the size of the cation increases while keeping a constant \(2+\) charge. Consequently, the charge density of the cation decreases, meaning it has a lower polarizing effect on the nitrate anion. Because the nitrate ion's electron cloud is less distorted, more thermal energy is required to decompose the nitrate, making thermal stability increase down the group.

1 mark for identifying that stability increases and explaining it in terms of increasing cationic radius and decreasing polarization.

Ammonia (\(\text{NH}_3\)) is higher in the spectrochemical series than water (\(\text{H}_2\text{O}\)) and is a stronger field ligand. Consequently, it causes a larger splitting (\(\Delta E\)) of the d-orbitals in the octahedral complex. Since the energy of light absorbed is given by \(\Delta E = \frac{hc}{\lambda}\), an increase in \(\Delta E\) results in a decrease in the wavelength (\(\lambda\)) of the absorbed light.

Award 1 mark for the correct option B. Reject other options because ammonia is a stronger ligand (eliminating A) and both complexes have a coordination number of 6 (coordination geometry remains octahedral, eliminating C and D).

For the \(\text{Cu}^{2+}/\text{Cu}\) half-reaction, the number of electrons transferred, \(z\), is 2. The concentration \([\text{Cu}^{2+}(\text{aq})]\) is \(1.0 \times 10^{-3}\text{ mol dm}^{-3}\). Substituting these values into the Nernst equation:

\(E = +0.34 + \frac{0.059}{2}\log(1.0 \times 10^{-3})\)
\(E = +0.34 + 0.0295 \times (-3)\)
\(E = +0.34 - 0.0885 = +0.2515\text{ V}\)

This rounds to \(+0.25\text{ V}\).

Award 1 mark for the correct answer B.
- Option A is incorrect because it uses \(z=1\).
- Option C is standard potential.
- Option D is incorrect because the log term was added as positive instead of negative.

1. X reacts with \(\text{Na}_2\text{CO}_3\) to give \(\text{CO}_2\), showing that X has a carboxylic acid (\(-\text{COOH}\)) group.
2. Alkaline hydrolysis of the chlorine atom in X, followed by acidification, gives a hydroxycarboxylic acid Y, \(\text{C}_4\text{H}_8\text{O}_3\).
3. Since Y contains a chiral carbon, we can eliminate option C (\(\text{HOCH}_2\text{CH}_2\text{CH}_2\text{COOH}\)) and option D (\(\text{CH}_3\text{C(OH)(CH}_3)\text{COOH}\)) as neither has a chiral carbon.
4. Oxidation of Y with acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) yields Z. Z gives a positive iodoform test (yellow precipitate with \(\text{I}_2/\text{NaOH}\)), meaning Z must contain a methyl ketone (\(\text{CH}_3\text{CO}-\)) group.
5. If X is option A (\(\text{CH}_3\text{CH}_2\text{CHClCOOH}\)), then Y is \(\text{CH}_3\text{CH}_2\text{CH(OH)COOH}\) and Z is \(\text{CH}_3\text{CH}_2\text{COCOOH}\), which is not a methyl ketone.
6. If X is option B (\(\text{CH}_3\text{CHClCH}_2\text{COOH}\)), then Y is \(\text{CH}_3\text{CH(OH)CH}_2\text{COOH}\) and Z is \(\text{CH}_3\text{COCH}_2\text{COOH}\), which is a methyl ketone and will give a yellow precipitate of \(\text{CHI}_3\). Thus, B is correct.

Award 1 mark for the correct option B.
- Reject A: the oxidized product is not a methyl ketone.
- Reject C and D: the hydrolyzed products do not contain chiral carbon atoms.

1. Reaction 1 occurs via a free-radical substitution mechanism targeting the alkyl side-chain because of the high temperature and UV light. The methyl group (\(-\text{CH}_3\)) is chlorinated, forming (chloromethyl)benzene (\(\text{C}_6\text{H}_5\text{CH}_2\text{Cl}\)).
2. Reaction 2 is an electrophilic aromatic substitution. The anhydrous \(\text{AlCl}_3\) acts as a halogen carrier catalyst, generating the \(\text{Cl}^+\) electrophile that attacks the benzene ring. Since the methyl group is electron-donating and 2,4-directing, the chlorine atom is substituted onto the 2- and 4-positions of the ring, producing a mixture of 2-chlorotoluene and 4-chlorotoluene. Hence, option A is correct.

Award 1 mark for the correct option A.
- Reject B: it swaps the products of the two reactions.
- Reject C and D: because the methyl group is 2,4-directing, not 3-directing.

1. The mass loss during heating is due to the loss of \(\text{CO}_2(\text{g})\) and \(\text{H}_2\text{O}(\text{g})\).

\(\text{Mass loss} = 2.00\text{ g} - 1.38\text{ g} = 0.62\text{ g}\)

2. According to the balanced equation, \(2\text{ moles}\) of \(\text{NaHCO}_3\) (mass = \(2 \times 84.0 = 168.0\text{ g}\)) decompose to produce \(1\text{ mole}\) of \(\text{H}_2\text{O}\) and \(1\text{ mole}\) of \(\text{CO}_2\) (total mass of gases lost = \(18.0 + 44.0 = 62.0\text{ g}\)).

3. Set up the ratio to find the mass of \(\text{NaHCO}_3\) (let this be \(x\)) in the sample:

\(\frac{\text{mass of gases lost}}{\text{mass of }\text{NaHCO}_3\text{ decomposed}} = \frac{62.0}{168.0} = \frac{0.62}{x}\)

Solving for \(x\):
\(x = \frac{0.62 \times 168.0}{62.0} = 1.68\text{ g}\) of \(\text{NaHCO}_3\).

4. Calculate the percentage by mass:

\(\text{Percentage} = \frac{1.68\text{ g}}{2.00\text{ g}} \times 100\% = 84.0\%\).

Therefore, the correct option is D.

Award 1 mark for the correct option D.
- Reject A: this is the percentage of \(\text{Na}_2\text{CO}_3\) in the mixture.
- Reject B and C: these are incorrect values resulting from stoichiometry miscalculations (e.g. not accounting for the 2:1 molar ratio or using wrong molar masses).

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

2. Determine the order of reaction with respect to B: Compare Experiment 2 and 3. The concentration of A is kept constant while \([\text{B}]\) is tripled. The rate increases by a factor of \(\frac{2.4 \times 10^{-3}}{8.0 \times 10^{-4}} = 3 = 3^1\). Therefore, the reaction is first order with respect to B.

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

4. Determine the units of \(k\):
\(k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]}\)

\(\text{Units of } k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2(\text{mol dm}^{-3})} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol}^3\text{ dm}^{-9}} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

Therefore, the correct option is B.

Award 1 mark for the correct option B.
- Reject A: units of an overall second-order reaction.
- Reject C: units of an overall fourth-order reaction.
- Reject D: units of an overall first-order reaction.

1. Convert all values to SI units:
- Mass, \(m = 0.220\text{ g}\)
- Temperature, \(T = 127 + 273 = 400\text{ K}\)
- Pressure, \(p = 1.00 \times 10^5\text{ Pa}\)
- Volume, \(V = 83.1\text{ cm}^3 = 83.1 \times 10^{-6}\text{ m}^3\)

2. Use the ideal gas equation, \(pV = nRT = \frac{m}{M_{\text{r}}}RT\), to solve for \(M_{\text{r}}\):
\(M_{\text{r}} = \frac{mRT}{pV}\)

\(M_{\text{r}} = \frac{0.220 \times 8.31 \times 400}{1.00 \times 10^5 \times 83.1 \times 10^{-6}}\)

Notice that \(1.00 \times 10^5 \times 83.1 \times 10^{-6} = 8.31\). Substituting this in:
\(M_{\text{r}} = \frac{0.220 \times 8.31 \times 400}{8.31} = 0.220 \times 400 = 88.0\text{ g mol}^{-1}\)

3. Calculate the molar masses of the options:
- Propanone (\(\text{C}_3\text{H}_6\text{O}\)): \(3(12.0) + 6(1.0) + 16.0 = 58.0\text{ g mol}^{-1}\)
- Butan-1-ol (\(\text{C}_4\text{H}_{10}\text{O}\)): \(4(12.0) + 10(1.0) + 16.0 = 74.0\text{ g mol}^{-1}\)
- Ethyl ethanoate (\(\text{C}_4\text{H}_8\text{O}_2\)): \(4(12.0) + 8(1.0) + 2(16.0) = 88.0\text{ g mol}^{-1}\)
- Pentane (\(\text{C}_5\text{H}_{12}\)): \(5(12.0) + 12(1.0) = 72.0\text{ g mol}^{-1}\)

Therefore, the compound is ethyl ethanoate (option C).

Award 1 mark for the correct option C.
- Reject A, B, and D because their calculated molar masses do not match \(88.0\text{ g mol}^{-1}\).

1. The equation representing the standard enthalpy of formation of \(\text{MgF}_2(\text{s})\) is:
\(\text{Mg}(\text{s}) + \text{F}_2(\text{g}) \rightarrow \text{MgF}_2(\text{s})\)

2. The Born-Haber cycle relates this to the other enthalpy changes:
\(\Delta H_{\text{f}}^\theta = \Delta H_{\text{at}}^\theta[\text{Mg}] + 1\text{st } IE[\text{Mg}] + 2\text{nd } IE[\text{Mg}] + E(\text{F}-\text{F}) + 2 \times EA[\text{F}] + \Delta H_{\text{latt}}^\theta[\text{MgF}_2]\)

Note that \(1\text{ mole}\) of \(\text{F}_2(\text{g})\) is converted to \(2\text{ moles}\) of \(\text{F}(\text{g})\) atoms, which requires the dissociation of \(1\text{ mole}\) of \(\text{F}-\text{F}\) bonds (\(+158\text{ kJ mol}^{-1}\)). Therefore, we do not multiply the bond energy by 2.
However, since \(2\text{ moles}\) of gaseous fluorine atoms are converted to \(2\text{ moles}\) of fluoride ions, we must multiply the electron affinity by 2:
\(2 \times (-328) = -656\text{ kJ mol}^{-1}\).

3. Substitute the values into the equation:
\(-1123 = 148 + 738 + 1450 + 158 + 2(-328) + \Delta H_{\text{latt}}^\theta\)
\(-1123 = 2494 - 656 + \Delta H_{\text{latt}}^\theta\)
\(-1123 = 1838 + \Delta H_{\text{latt}}^\theta\)

\(\Delta H_{\text{latt}}^\theta = -1123 - 1838 = -2961\text{ kJ mol}^{-1}\).

Therefore, the correct option is C.

Award 1 mark for the correct option C.
- Reject A: calculated by forgetting to multiply the electron affinity of fluorine by 2.
- Reject B: calculated by multiplying the F-F bond energy by 2 instead of 1.

Since 2 moles of \(\text{AgCl}\) are precipitated per mole of complex, there must be 2 free chloride ions (\(\text{Cl}^-\)) outside the coordination sphere acting as counter-ions. This leaves 1 chloride ion and all 5 ammonia molecules coordinated directly to the cobalt central metal ion. Therefore, the formula of the complex is \([\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2\), and the complex cation is \([\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+}\).

1 mark: Correctly identify the chemical species that coordinate to the cobalt central ion and deduce that the remaining two chlorides are counter-ions, leading to option A.

To find the cell potential, we identify the oxidation and reduction half-reactions. \(\text{Fe}^{3+}\) is reduced to \(\text{Fe}^{2+}\) (\(E^\ominus = +0.77\text{ V}\)) and \(\text{I}^-\) is oxidized to \(\text{I}_2\) (\(E^\ominus = +0.54\text{ V}\)).

\(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V}\).
Since \(E^\ominus_{\text{cell}} > 0\), the forward reaction is thermodynamically feasible under standard conditions.

1 mark: Calculate the correct standard cell potential of +0.23 V and correctly state that it is feasible.

Acyl chlorides undergo nucleophilic substitution:
- Reaction with water yields a carboxylic acid (propanoic acid).
- Reaction with an alcohol yields an ester (ethanol gives ethyl propanoate).
- Reaction with ammonia yields an amide (propanamide).

1 mark: Correctly identify the products of nucleophilic substitution for all three reactions.

The active electrophile in the nitration of benzene is the nitronium ion, \(\text{NO}_2^+\). It is generated when concentrated sulfuric acid protonates nitric acid, which then loses a water molecule to form \(\text{NO}_2^+\). The overall equation for this electrophile generation step is: \(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\).

1 mark: Identify the correct equation demonstrating the acid-base interaction producing the nitronium electrophile.

1. The remaining gas after \(\text{NaOH}\) absorption is excess \(\text{O}_2 = 30\text{ cm}^3\).
2. The volume of \(\text{CO}_2\) produced is the volume decrease when passed through \(\text{NaOH}\), which is \(60 - 30 = 30\text{ cm}^3\).
3. The volume of oxygen that reacted is \(80 - 30 = 50\text{ cm}^3\).
4. Since \(10\text{ cm}^3\) of \(\text{C}_x\text{H}_y\) reacted, the mole ratio of \(\text{C}_x\text{H}_y : \text{O}_2 : \text{CO}_2\) is \(10 : 50 : 30 = 1 : 5 : 3\).
5. Thus, \(x = 3\) (since 1 mol of hydrocarbon gives 3 mol of \(\text{CO}_2\)).
6. Using the combustion equation: \(x + \frac{y}{4} = 5\). Substitute \(x = 3\): \(3 + \frac{y}{4} = 5 \implies \frac{y}{4} = 2 \implies y = 8\). The molecular formula is \(\text{C}_3\text{H}_8\).

1 mark: Deduce the volume of carbon dioxide and unreacted oxygen, set up the stoichiometry ratio, and solve for x and y to get C3H8.

1. Comparing Exp 1 and Exp 2: doubling \([\text{A}]\) doubles the rate. Order with respect to \(\text{A} = 1\).
2. Comparing Exp 1 and Exp 3: doubling \([\text{B}]\) quadruples the rate. Order with respect to \(\text{B} = 2\).
3. Comparing Exp 1 and Exp 4: doubling \([\text{C}]\) has no effect on the rate. Order with respect to \(\text{C} = 0\).
4. Rate equation: \(\text{Rate} = k[\text{A}][\text{B}]^2\). The overall order is \(1 + 2 = 3\).
5. Units of \(k = \frac{\text{Rate}}{[\text{A}][\text{B}]^2} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

1 mark: Deduce the individual orders of reaction (A = 1, B = 2, C = 0), calculate the overall order of 3, and find the corresponding correct units for k.

Using the ideal gas equation: \(pV = nRT = \frac{m}{M_r} RT\).
Rearranging for \(M_r\): \(M_r = \frac{mRT}{pV} = \frac{\rho RT}{p}\), where \(\rho\) is the density.
Converting units:
- \(T = 120 + 273 = 393\text{ K}\)
- \(p = 1.01 \times 10^5\text{ Pa}\)
- \(\rho = 2.60\text{ g dm}^{-3} = 2.60 \times 10^3\text{ g m}^{-3}\)

Substitute these into the equation:
\(M_r = \frac{2.60 \times 10^3 \times 8.31 \times 393}{1.01 \times 10^5} \approx 84.0\text{ g mol}^{-1}\).

1 mark: Correctly convert temperature to Kelvin, density to g/m^3 (or keep track of SI units correctly), and calculate the relative molecular mass of 84.

As we go down Group 2, the ionic radius of the metal cation increases, leading to a decrease in both lattice energy (less exothermic) and hydration enthalpy (less exothermic). However, because the sulfate ion is very large, the lattice energy of the sulfate decreases only slightly down the group. In contrast, the hydration enthalpy of the cation decreases much more rapidly because the charge density of the cation falls significantly. Consequently, the enthalpy of solution becomes more endothermic (or less exothermic) down the group, causing solubility to decrease.

1 mark: Identify that both lattice and hydration enthalpies decrease in magnitude (become less exothermic) but hydration enthalpy decreases more rapidly, explaining the solubility trend.

Using the Nernst equation:
\(E = E^\ominus + \frac{0.059}{z}\lg\left(\frac{[\text{Fe}^{3+}(\text{aq})]}{[\text{Fe}^{2+}(\text{aq}) ]}\right)\)

Substitute the given values:
- \(E^\ominus = +0.77\text{ V}\)
- \(z = 1\) (number of electrons transferred)
- \([\text{Fe}^{3+}] = 0.10\text{ mol dm}^{-3}\)
- \([\text{Fe}^{2+}] = 0.010\text{ mol dm}^{-3}\)

\(E = +0.77 + \frac{0.059}{1}\lg\left(\frac{0.10}{0.010}\right)\)
\(E = +0.77 + 0.059 \lg(10)\)

Since \(\lg(10) = 1\):
\(E = +0.77 + 0.059(1) = +0.829\text{ V} \approx +0.83\text{ V}\)

Therefore, option A is correct.

Award 1 mark for the correct calculation showing \(+0.83\text{ V}\).

Option B is correct because in acidic solution, the orange dichromate(VI) ions (\(\text{Cr}_2\text{O}_7^{2-}\)) are reduced by iron(II) ions (\(\text{Fe}^{2+}\)) to green chromium(III) ions (\(\text{Cr}^{3+}\)).

Option A is incorrect because the tetrachlorocuprate(II) ion, \([\text{CuCl}_4]^{2-}\), is tetrahedral rather than square planar.

Option C is incorrect because copper(I) ions are unstable in aqueous solution and readily undergo disproportionation into copper metal and copper(II) ions.

Option D is incorrect because manganese(VII) in \(\text{MnO}_4^-\right.\) is a powerful oxidizing agent and is easily reduced by sulfite ions.

Award 1 mark for identifying option B as the correct statement.

Option C is correct because the concentrated sulfuric acid acts as a catalyst by protonating the nitric acid, which subsequently loses a water molecule to generate the active electrophile, \(\text{NO}_2^+\).

Option A is incorrect because the electrophile is the nitronium ion, \(\text{NO}_2^+\), not the nitrite ion, \(\text{NO}_2^-\).

Option B is incorrect because the methyl group is an ortho/para-director (2,4-director), meaning the primary organic products formed are 2-nitromethylbenzene and 4-nitromethylbenzene, not 3-nitromethylbenzene.

Option D is incorrect because the mechanism is electrophilic substitution, which maintains the stable delocalised aromatic system in the product.

Award 1 mark for identifying option C as the correct statement.

Let us evaluate the options:
- For option A (\(\text{CH}_3\text{COOCH}_2\text{CH}_2\text{Cl}\)): Under reflux with excess \(\text{NaOH}(\text{aq})\), the ester linkage is hydrolyzed to produce ethanoate ions and 2-chloroethanol. The excess \(\text{NaOH}\) also nucleophically substitutes the chlorine atom in 2-chloroethanol to produce ethane-1,2-diol (\(\text{HOCH}_2\text{CH}_2\text{OH}\)). Upon acidification, the ethanoate ions are protonated to form ethanoic acid, and ethane-1,2-diol remains as the chlorine-free compound Y. Ethane-1,2-diol has 2 carbon atoms and is achiral (does not show optical isomerism). This matches all criteria.
- For option B (\(\text{ClCH}_2\text{COOCH}_2\text{CH}_3\)): Hydrolysis and substitution would yield hydroxyethanoic acid and ethanol, not ethanoic acid.
- For option C (\(\text{CH}_3\text{CH(Cl)COOCH}_3\)): Hydrolysis and substitution would yield 2-hydroxypropanoic acid (lactic acid) and methanol. Lactic acid contains 3 carbon atoms and is optically active.
- For option D (\(\text{ClCH}_2\text{CH}_2\text{COOCH}_3\)): Hydrolysis and substitution would yield 3-hydroxypropanoic acid (3 carbon atoms) and methanol (1 carbon atom).

Award 1 mark for identifying option A as the correct structure of X.

1. Find the mass of oxygen in the 4.05 g sample:
\(\text{Mass of O} = 4.05\text{ g} - 2.77\text{ g} = 1.28\text{ g}\)
(Percentage check: \(\frac{1.28}{4.05} \times 100\% = 31.6\%\), which is correct).

2. Calculate the number of moles of oxygen atoms:
\(n(\text{O}) = \frac{1.28\text{ g}}{16.0\text{ g mol}^{-1}} = 0.080\text{ mol}\)

3. Determine the relationship between the moles of oxygen and the metal \(\text{M}\):
- If the empirical formula is \(\text{M}_2\text{O}_3\), the mole ratio of \(\text{M} : \text{O}\) is \(2 : 3\).
Therefore, \(n(\text{M}) = \frac{2}{3} \times n(\text{O}) = \frac{2}{3} \times 0.080 = 0.0533\text{ mol}\).

4. Calculate the relative atomic mass (\(A_r\)) of the metal:
\(A_r(\text{M}) = \frac{\text{mass of M}}{n(\text{M})} = \frac{2.77\text{ g}}{0.0533\text{ mol}} = 51.9\text{ g mol}^{-1}\).

This relative atomic mass of \(51.9\) is consistent with chromium, which forms the stable oxide \(\text{Cr}_2\text{O}_3\). Thus, Option A is correct.

Award 1 mark for the correct calculation of relative atomic mass and empirical formula.

When the temperature of a gas is increased:
1. The molecules gain kinetic energy, shifting the Maxwell-Boltzmann distribution curve to the right (towards higher energies) and flattening it. Because the total area under the curve must remain constant (representing the total number of particles), the peak of the curve becomes lower.
2. The activation energy (\(E_a\)) is the minimum energy required for particles to react. It is a constant characteristic of the reaction pathway itself and does not change with temperature.

Therefore, the peak moves to the right and becomes lower, while the activation energy remains unchanged. This corresponds to Option B.

Award 1 mark for identifying option B.

Let us trace the two steps:
- **Initial State (1):**
\(P_1 = 1.00 \times 10^5\text{ Pa}\)
\(V_1 = 1.20 \times 10^{-3}\text{ m}^3\)
\(T_1 = 27 + 273 = 300\text{ K}\)

- **Step 1: Heating at constant pressure (\(P_2 = P_1\)) until volume doubles (\(V_2 = 2 V_1\)):**
Using Charles's Law: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)
\(T_2 = T_1 \times \left(\frac{V_2}{V_1}\right) = 300 \times 2 = 600\text{ K}\)

- **Step 2: Cooling at constant volume (\(V_3 = V_2 = 2.40 \times 10^{-3}\text{ m}^3\)) back to original temperature (\(T_3 = 300\text{ K}\)):**
Using Gay-Lussac's Law: \(\frac{P_2}{T_2} = \frac{P_3}{T_3}\)
\(P_3 = P_2 \times \left(\frac{T_3}{T_2}\right) = 1.00 \times 10^5 \times \left(\frac{300}{600}\right) = 5.00 \times 10^4\text{ Pa}\)

Therefore, the final pressure is \(5.00 \times 10^4\text{ Pa}\).

Award 1 mark for the correct final pressure calculation.

Option C is correct because:
1. The solubility of the Group 2 hydroxides increases down the group (from sparingly soluble magnesium hydroxide to highly soluble barium hydroxide).
2. The thermal stability of the Group 2 carbonates increases down the group. This is because the cationic radius increases from \(\text{Mg}^{2+}\) to \(\text{Ba}^{2+}\), which decreases its charge density and its polarizing power, meaning the carbonate anion is less distorted and thus more thermally stable.

Option A is incorrect because Group 2 sulfate solubility decreases down the group.
Option B is incorrect because the thermal stability of both Group 2 nitrates and carbonates increases down the group.
Option D is incorrect because the first ionisation energy of Group 2 elements decreases down the group.

Award 1 mark for identifying option C as the correct statement.

(a) The weighted average mass of atoms of an element compared to 1/12th of the mass of an atom of carbon-12.

(b)(i) To ensure that all of the water of crystallisation has been completely driven off / lost.
(ii) Mass of anhydrous \(BaCl_2 = 24.73 - 22.45 = 2.28\text{ g}\).
Mass of water lost = \(25.13 - 24.73 = 0.40\text{ g}\).
(iii) Molar mass of \(BaCl_2 = 137.3 + (2 \times 35.5) = 208.3\text{ g mol}^{-1}\).
Moles of \(BaCl_2 = 2.28 / 208.3 = 0.01095\text{ mol}\).
Moles of \(H_2O = 0.40 / 18.0 = 0.02222\text{ mol}\).
Ratio of \(H_2O : BaCl_2 = 0.02222 / 0.01095 = 2.03\).
Therefore, \(x = 2\) (to the nearest whole number).

(c)(i) \(Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)\)
(ii) Moles of \(BaCl_2 = 0.0250 \times 0.0500 = 0.00125\text{ mol}\).
Moles of \(Cl^- = 2 \times 0.00125 = 0.00250\text{ mol}\).
Moles of \(Ag^+ = 0.0350 \times 0.0800 = 0.00280\text{ mol}\).
Since \(Cl^-\) is the limiting reactant, moles of \(AgCl\) formed = \(0.00250\text{ mol}\).
Mass of \(AgCl = 0.00250\text{ mol} \times 143.4\text{ g mol}^{-1} = 0.359\text{ g}\) (or \(0.358\text{ g}\) using \(143.3\text{ g mol}^{-1}\)).

(d)(i) \(2Mg(NO_3)_2(s) \rightarrow 2MgO(s) + 4NO_2(g) + O_2(g)\)
(ii) Thermal stability increases down the group because the ionic radius of the Group 2 cation increases / charge density of the cation decreases down the group. This causes less polarisation / distortion of the nitrate electron cloud / nitrate ion, weakening the N-O bond less and requiring more thermal energy to break.

Award up to 15 marks as follows:

(a) [2 marks]
- 1 mark: Weighted average mass of atoms of an element.
- 1 mark: Compared to 1/12th of the mass of an atom of carbon-12.

(b) [5 marks]
- (i) 1 mark: To ensure all water of crystallisation has been removed.
- (ii) 1 mark: Correctly calculating both masses (anhydrous salt = 2.28 g and water = 0.40 g).
- (iii) 3 marks:
- 1 mark: Calculate moles of \(BaCl_2\) (0.01095 mol) AND moles of \(H_2O\) (0.02222 mol).
- 1 mark: Show the mole ratio (2.03).
- 1 mark: Correct integer value \(x = 2\).

(c) [5 marks]
- (i) 2 marks:
- 1 mark: Correct species \(Ag^+\), \(Cl^-\), \(AgCl\).
- 1 mark: Correct state symbols (aq, aq, s).
- (ii) 3 marks:
- 1 mark: Calculate moles of \(Cl^-\)\ (0.00250 mol) and moles of \(Ag^+\)\ (0.00280 mol).
- 1 mark: Identify \(Cl^-\) as the limiting reactant.
- 1 mark: Correct mass of \(AgCl\) (0.359 g, allow 0.358 g or 0.36 g).

(d) [3 marks]
- (i) 1 mark: Correctly balanced equation (including 2:2:4:1 stoichiometry or fractional coefficient for \(O_2\)).
- (ii) 2 marks:
- 1 mark: Cationic radius increases / cationic charge density decreases down the group.
- 1 mark: Nitrate ion is polarised/distorted less, requiring more thermal energy to decompose.

(a) The enthalpy change when 1 mole of a substance is burned completely in excess oxygen under standard conditions of 298 K and 101 kPa.

(b)(i) \(\Delta T = 45.8 - 19.5 = 26.3\text{ }^\circ\text{C}\) (or K).
\(q = m c \Delta T = 150.0 \times 4.18 \times 26.3 = 16490.1\text{ J}\) (or \(1.65 \times 10^4\text{ J}\)).
(ii) Mass of propan-1-ol burned = \(120.45 - 119.83 = 0.62\text{ g}\).
Molar mass of propan-1-ol, \(C_3H_8O = (3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0\text{ g mol}^{-1}\).
Moles of propan-1-ol = \(0.62 / 60.0 = 0.01033\text{ mol}\).
(iii) \(\Delta H_c = - (q / 1000) / n = - (16.4901) / 0.01033 = -1596\text{ kJ mol}^{-1}\) (or \(-1600\text{ kJ mol}^{-1}\) to 3 s.f.).
(iv) Any two of:
- Heat lost to the air/surroundings or copper calorimeter.
- Incomplete combustion of propan-1-ol (forming carbon monoxide or carbon soot).
- Evaporation of alcohol from the wick of the burner after extinguishing but before weighing.
- Non-standard conditions.

(c)(i) \(C_3H_7OH(l) + 4.5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)\)
(ii) Using Hess's Law:
\(\Delta H_c^\ominus = \sum \Delta H_f^\ominus(products) - \sum \Delta H_f^\ominus(reactants)\)
\(\Delta H_c^\ominus = [3 \times \Delta H_f^\ominus(CO_2) + 4 \times \Delta H_f^\ominus(H_2O)] - \Delta H_f^\ominus(C_3H_7OH)\)
\(-2021.0 = [3 \times (-393.5) + 4 \times (-285.8)] - \Delta H_f^\ominus(C_3H_7OH)\)
\(-2021.0 = [-1180.5 - 1143.2] - \Delta H_f^\ominus(C_3H_7OH)\)
\(-2021.0 = -2323.7 - \Delta H_f^\ominus(C_3H_7OH)\)
\(\Delta H_f^\ominus(C_3H_7OH) = -2323.7 - (-2021.0) = -302.7\text{ kJ mol}^{-1}\).

Award up to 15 marks as follows:

(a) [2 marks]
- 1 mark: Enthalpy change when 1 mole of a substance is burned completely in excess oxygen.
- 1 mark: Under standard conditions (298 K, 1 atm / 101 kPa).

(b) [8 marks]
- (i) 2 marks:
- 1 mark: Correct calculation of temperature rise (26.3 K).
- 1 mark: Correct calculation of \(q = 16490\text{ J}\) (or 16.5 kJ).
- (ii) 2 marks:
- 1 mark: Calculate mass of propan-1-ol burned = 0.62 g.
- 1 mark: Calculate moles of propan-1-ol = 0.0103 mol (or 0.01033 mol).
- (iii) 2 marks:
- 1 mark: Correct division of \(q\) by moles and converting to kJ.
- 1 mark: Correct sign (negative) and value \(-1596\) or \(-1600\text{ kJ mol}^{-1}\) (3 s.f.).
- (iv) 2 marks:
- 1 mark for each of any two correct reasons: heat loss to surroundings / incomplete combustion / evaporation of alcohol from the wick.

(c) [5 marks]
- (i) 1 mark: Correctly balanced equation with state symbols (allow fractional coefficients like 4.5 \(O_2\)).
- (ii) 4 marks:
- 1 mark: Constructing correct Hess's Law expression / cycle.
- 1 mark: Correct calculation of total products enthalpy of formation = \(-2323.7\text{ kJ mol}^{-1}\).
- 1 mark: Correct algebraic rearrangement.
- 1 mark: Correct final answer with sign and units: \(-302.7\text{ kJ mol}^{-1}\).

(a)(i) Any correct structural isomer without stereoisomerism, e.g., 2-methylbut-1-ene or 3-methylbut-1-ene or pent-1-ene.
(ii) 2-methylbut-2-ene does not show cis-trans isomerism. One of the carbon atoms in the double bond (C2) is bonded to two identical groups (two methyl groups), which means swapping their positions does not create a new isomer.

(b)(i) 2-bromo-2-methylbutane.
(ii) Mechanism:
- Electrophile \(H^{\delta+} - Br^{\delta-}\) approaches the double bond.
- Curly arrow from the double bond of \((CH_3)_2C=CHCH_3\) to the \(H\) of \(H-Br\).
- Curly arrow from the \(H-Br\) bond to \(Br\).
- Intermediate is a tertiary carbocation: \((CH_3)_2C^{+} - CH_2CH_3\) and a bromide ion, \(:Br^-\).
- Curly arrow from the lone pair on the bromide ion \(:Br^-\) to the carbon atom with the positive charge on the carbocation.
- Product is 2-bromo-2-methylbutane.
(iii) The major product is formed via the tertiary carbocation, \((CH_3)_2C^{+} - CH_2CH_3\), whereas the minor product is formed via the secondary carbocation, \((CH_3)_2CH - C^{+}HCH_3\).
Tertiary carbocations are more stable than secondary carbocations.
This is due to the greater positive inductive effect of three electron-releasing alkyl groups in the tertiary carbocation, which disperses the positive charge more effectively than the two alkyl groups in the secondary carbocation.

(c)(i) Propanone and ethanoic acid.
(ii) The purple solution decolourises (or turns from purple to colourless).
(iii) The product is 2-methylbutane-2,3-diol, which has the skeletal structure showing a 5-carbon chain with methyl at C2, and hydroxyl (-OH) groups at C2 and C3.

Award up to 15 marks as follows:

(a) [3 marks]
- (i) 1 mark: Correct IUPAC name of a valid isomer (e.g. 2-methylbut-1-ene or 3-methylbut-1-ene or pent-1-ene).
- (ii) 2 marks:
- 1 mark: Statement that it does not show cis-trans isomerism.
- 1 mark: Explanation that one carbon of the double bond (C2) is bonded to two identical methyl groups.

(b) [8 marks]
- (i) 1 mark: 2-bromo-2-methylbutane.
- (ii) 4 marks:
- 1 mark: Curly arrow from the double bond to the hydrogen of \(H-Br\), AND partial charges \(\delta+/\delta-\)\ on \(H-Br\) shown correctly.
- 1 mark: Curly arrow from the \(H-Br\) bond to \(Br\).
- 1 mark: Correct structure of the tertiary carbocation intermediate, including the positive charge on C2, AND \(Br^-\) shown with a lone pair.
- 1 mark: Curly arrow from the lone pair of \(Br^-\) to the positive carbon.
- (iii) 3 marks:
- 1 mark: Identify that major product forms via tertiary carbocation, minor via secondary carbocation.
- 1 mark: State that tertiary carbocations are more stable than secondary carbocations.
- 1 mark: Explain in terms of the positive inductive effect of the alkyl groups (3 alkyl groups vs 2 alkyl groups) which disperses the positive charge on the carbocation.

(c) [4 marks]
- (i) 2 marks: Propanone (1 mark) and ethanoic acid (1 mark).
- (ii) 1 mark: Purple to colourless.
- (iii) 1 mark: Correct skeletal structure of 2-methylbutane-2,3-diol.

(a)(i) Any two of:
- The intermolecular forces of attraction between molecules are negligible / zero.
- The volume occupied by the gas molecules themselves is negligible compared to the volume of the container.
- Collisions between molecules (and with the container walls) are perfectly elastic.
- Molecules are in continuous, random motion.
(ii) Conditions: High temperature and low pressure.
Explanation: At high temperature, the molecules have high kinetic energy, which overcomes the intermolecular forces of attraction. At low pressure, the molecules are far apart, meaning the volume occupied by the molecules is negligible compared to the total volume, and intermolecular forces are negligible.

(b)(i) \(PV = nRT\)
- \(P\): Pressure in Pascals (\(\text{Pa}\) or \(\text{N m}^{-2}\))
- \(V\): Volume in cubic metres (\(\text{m}^3\))
- \(n\): Amount of substance in moles (\(\text{mol}\))
- \(R\): Gas constant (\(\text{J K}^{-1}\text{ mol}^{-1}\))
- \(T\): Temperature in Kelvin (\(\text{K}\))
(ii) Convert units:
- \(T = 99 + 273 = 372\text{ K}\)
- \(V = 61.4\text{ cm}^3 = 61.4 \times 10^{-6}\text{ m}^3\)
Using \(PV = \frac{m}{M_r}RT\):
\(M_r = \frac{mRT}{PV} = \frac{0.150 \times 8.31 \times 372}{1.02 \times 10^5 \times 61.4 \times 10^{-6}}\)
\(M_r = \frac{463.698}{6.2628} = 74.049\)
\(M_r = 74.0\) (to 3 s.f.).

(c)(i) & (ii) Diagram details:
- Y-axis: Number of molecules / Fraction of molecules.
- X-axis: Molecular energy / Kinetic energy.
- Curve \(T_1\) starts at the origin, rises to a peak, and then decreases asymptotically to approach (but not touch) the x-axis.
- Activation energy, \(E_a\), marked on the x-axis, and the area under the curve to the right of \(E_a\) shaded.
- Curve \(T_2\) has a lower peak shifted to the right. It must cross the first curve once, and be higher than the first curve at higher energies.
(iii) Explanation: When temperature is increased, the average kinetic energy of the molecules increases. A significantly larger fraction of molecules have energy greater than or equal to the activation energy (represented by the larger shaded area under the curve at \(T_2\)). This leads to a greater frequency of successful/effective collisions, which increases the rate of reaction.

Award up to 15 marks as follows:

(a) [4 marks]
- (i) 2 marks: Any two of: intermolecular forces are negligible / molecular volume is negligible compared to container volume / collisions are perfectly elastic.
- (ii) 2 marks:
- 1 mark: High temperature and low pressure.
- 1 mark: Explanation: At high T, thermal energy overcomes intermolecular attractions; at low P, large distances make molecular volume and attraction negligible.

(b) [5 marks]
- (i) 2 marks: Correct equation \(PV = nRT\) with all five terms and units correctly identified (must include Pa, m\(^3\), mol, J K\(^{-1}\) mol\(^{-1}\), and K).
- (ii) 3 marks:
- 1 mark: Correct conversion of temperature (372 K) AND volume (\(61.4 \times 10^{-6}\text{ m}^3\)).
- 1 mark: Correct algebraic rearrangement of \(PV = nRT\) to solve for \(M_r\).
- 1 mark: Correct final calculation to 3 s.f. (74.0).

(c) [6 marks]
- (i) 3 marks:
- 1 mark: Axes labelled correctly (Y = number/fraction of molecules, X = energy/kinetic energy) AND curve starts at origin and goes to a peak and does not touch x-axis at high energy.
- 1 mark: \(E_a\) clearly marked on the x-axis.
- (ii) 1 mark: Curve for \(T_2\) drawn with peak lower and shifted to the right, crossing \(T_1\) once, and ending above \(T_1\).
- (iii) 2 marks:
- 1 mark: State that a greater fraction of molecules now have energy \(E \ge E_a\).
- 1 mark: State that this increases the frequency of successful/effective collisions (or more successful collisions per unit time).

Part (i): Typical times obtained might be: Run 1 (20 cm3 FA 1) = 25 s, giving a rate 1000/t = 40.0 s-1; Run 2 (15 cm3 FA 1) = 33 s, giving a rate = 30.3 s-1; Run 3 (10 cm3 FA 1) = 50 s, giving a rate = 20.0 s-1; Run 4 (5 cm3 FA 1) = 100 s, giving a rate = 10.0 s-1. Part (ii): A plot of 1000/t on the y-axis against the volume of FA 1 on the x-axis gives a straight line of best fit that passes through the origin (0,0). Part (iii): Since the graph of rate (1000/t) against volume of FA 1 (which is directly proportional to concentration) is a straight line through the origin, the rate is directly proportional to the concentration of iodide ions. Therefore, the reaction is first order with respect to iodide ions. Part (iv): Initial moles of iodide in Run 1 = Volume x Concentration = 0.0200 dm3 x 0.200 mol dm-3 = 4.00 x 10^-3 mol. Total volume of mixture = 50.0 cm3 = 0.0500 dm3. Concentration of iodide ions = 4.00 x 10^-3 mol / 0.0500 dm3 = 0.0800 mol dm-3. Part (v): A significant source of uncertainty is the human reaction time in starting/stopping the stopclock at the exact moment the color changes, or the subjective judgment of the blue color's intensity. An improvement would be to use a colorimeter connected to a data logger to measure the change in absorbance automatically.

M1 (Table presentation): Table has clear headings with units: volume of FA 1 / cm3, volume of water / cm3, time, t / s, and rate, 1000/t / s-1. [1] M2 (Data precision): All times recorded to the nearest second, and rates calculated correctly to 3 significant figures. [1] M3 (Graph axes): Graph plotted with rate (1000/t) on the y-axis and volume of FA 1 on the x-axis, with both axes clearly labelled with appropriate units. [1] M4 (Graph plotting): Linear scales used, points plotted accurately to within half a small square, and a straight line of best fit drawn passing through the origin. [1] M5 (Deduction of order): States that the rate is directly proportional to the volume of FA 1 because the line of best fit is straight and passes through the origin. [1] M6 (Order conclusion): Deduces that the reaction is first order with respect to iodide. [1] M7 (Calculation - moles of I-): Correctly calculates the moles of I- in Run 1 as 0.200 x (20/1000) = 4.00 x 10^-3 mol. [1] M8 (Calculation - concentration): Correctly divides by the total volume of 0.0500 dm3 to obtain 0.0800 mol dm-3 with appropriate units. [1] M9 (Uncertainty identified): Identifies human reaction time or difficulty in judging the precise end-point color change. [1] M10 (Improvement): Suggests using a colorimeter, light sensor, or comparing to a printed color reference standard. [1]

Part (i): Typical results show initial temperatures stable at 21.0 degrees C for 1, 2, and 3 minutes. At 4 minutes, FA 6 is added. Temperatures recorded are: 5 min = 34.2, 6 min = 34.0, 7 min = 33.8, 8 min = 33.6, 9 min = 33.4, 10 min = 33.2 degrees C. Part (ii): Extrapolating the cooling curve back to the 4th minute yields a theoretical maximum temperature of 34.6 degrees C. Part (iii): Delta T = 34.6 - 21.0 = 13.6 degrees C. Part (iv): Total volume of mixture = 25.0 + 25.0 = 50.0 cm3. Mass = 50.0 g. Heat released q = m x c x Delta T = 50.0 g x 4.18 J g-1 K-1 x 13.6 K = 2842.4 J (or 2.84 kJ). Part (v): Moles of HCl = Volume x Concentration = 0.0250 dm3 x 2.00 mol dm-3 = 0.0500 mol. (NaOH is in excess as 0.0250 x 2.20 = 0.0550 mol). Part (vi): Enthalpy change of neutralization = -q / (moles of HCl x 1000) = -2.8424 kJ / 0.0500 mol = -56.8 kJ mol-1. Part (vii): The mixture is an aqueous solution of sodium chloride, not pure water. The dissolved ions increase the density above 1.00 g cm-3 and lower the specific heat capacity below 4.18 J g-1 K-1. Using the values for pure water systematically overestimates or underestimates the heat energy.

M1 (Table): Clear columns for time / min and temperature / degrees C, with all temperatures recorded consistently to the nearest 0.1 or 0.5 degrees C. [1] M2 (Graph plotting): Appropriate scale used, all points correctly plotted, with temperature on the y-axis and time on the x-axis. [1] M3 (Extrapolation): Two straight lines of best fit drawn, with a clear vertical extrapolation at 4 minutes to determine Delta T. [1] M4 (Delta T): Accurate value of Delta T recorded (typically in the range of 13.0 to 14.0 degrees C). [1] M5 (Heat calculation): Correct substitution into q = m c Delta T using mass = 50.0 g and candidate's Delta T value. [1] M6 (Moles of HCl): Calculates moles of HCl correctly as 0.0250 x 2.00 = 0.0500 mol. [1] M7 (Enthalpy calculation): Division of q (in kJ) by 0.0500 mol to calculate Delta H. [1] M8 (Enthalpy sign and unit): Enthalpy change of neutralization reported with a negative sign and unit of kJ mol-1. [1] M9 (Systematic error): Explains that the reaction mixture contains ions (NaCl) which alters the density and specific heat capacity relative to pure water. [1]

### **Part (a): Expected Observations**
* **FA 1 (contains \(\text{Fe}^{2+}\), \(\text{NH}_4^+\), \(\text{SO}_4^{2-}\)):**
* *Test 1 (with NaOH):* Green precipitate is formed. It is insoluble in excess. On standing, the precipitate may turn brown at the surface due to oxidation to \(\text{Fe(OH)}_3\).
* *Test 2 (heating with NaOH):* A gas is evolved which turns damp red litmus paper blue, confirming ammonia, \(\text{NH}_3\).
* *Test 3 (with \(\text{NH}_3\)):* Green precipitate is formed, insoluble in excess.
* *Test 4 (with \(\text{Ba(NO}_3)_2\)):* White precipitate is formed (insoluble in dilute acid), confirming \(\text{SO}_4^{2-}\).
* *Test 5 (with \(\text{AgNO}_3\)):* No reaction / no precipitate.

* **FA 2 (contains \(\text{Zn}^{2+}\), \(\text{I}^-\)):**
* *Test 1 (with NaOH):* White precipitate is formed, which dissolves in excess to give a colorless solution.
* *Test 2 (heating with NaOH):* No gas evolved that turns red litmus paper blue.
* *Test 3 (with \(\text{NH}_3\)):* White precipitate is formed, which dissolves in excess to give a colorless solution.
* *Test 4 (with \(\text{Ba(NO}_3)_2\)):* No reaction / no precipitate.
* *Test 5 (with \(\text{AgNO}_3\)):* Yellow precipitate is formed, which is insoluble in aqueous ammonia.

---

### **Part (b): Identification and Equations**
1. **FA 1 Cations:** \(\text{Fe}^{2+}\) and \(\text{NH}_4^+\).
* *Evidence for \(\text{Fe}^{2+}\):* Green precipitate with NaOH and \(\text{NH}_3\) which is insoluble in excess.
* *Evidence for \(\text{NH}_4^+\):* Production of an alkaline gas (turns damp red litmus blue) upon heating with NaOH.
2. **FA 1 Anion:** \(\text{SO}_4^{2-}\).
* *Evidence:* White precipitate with acidified barium nitrate.
3. **FA 2 Cation:** \(\text{Zn}^{2+}\).
* *Evidence:* White precipitate with both NaOH and \(\text{NH}_3\), which is soluble in excess of both reagents to form a colorless solution.
4. **FA 2 Anion:** \(\text{I}^-\).
* *Evidence:* Yellow precipitate with acidified silver nitrate.
5. **Ionic Equation:**
\(\text{Zn}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Zn(OH)}_2\text{(s)}\)

---

### **Part (c): Confirmatory Test**
* **Reagent:** Add chlorine water (or aqueous bromine, or acidified hydrogen peroxide) to the solution of FA 2, followed by starch solution.
* **Observation:** The colorless solution turns blue-black (indicating the presence of free iodine, \(\text{I}_2\), produced by the oxidation of \(\text{I}^-\)).
* *Alternative:* Add cyclohexane after the oxidizing agent; the organic layer turns purple.

### **Part (a): Observations [7 marks]**
* **M1 [1 mark]:** Constructs a single, clearly organized table for all observations with appropriate headers.
* **M2 [1 mark]:** Correctly records for FA 1 and NaOH: green precipitate, insoluble in excess.
* **M3 [1 mark]:** Correctly records for FA 1 with NaOH and heating: gas evolved turns damp red litmus paper blue.
* **M4 [1 mark]:** Correctly records for FA 2 and NaOH: white precipitate, soluble in excess to form a colorless solution.
* **M5 [1 mark]:** Correctly records for FA 1 and FA 2 with \(\text{NH}_3\):
* FA 1: green precipitate, insoluble in excess.
* FA 2: white precipitate, soluble in excess to form a colorless solution.
* **M6 [1 mark]:** Correctly records for Test 4 (acidified barium nitrate):
* FA 1: white precipitate.
* FA 2: no precipitate / no reaction.
* **M7 [1 mark]:** Correctly records for Test 5 (acidified silver nitrate):
* FA 1: no precipitate / no reaction.
* FA 2: yellow precipitate.

### **Part (b): Identification and Equations [5 marks]**
* **M8 [1 mark]:** Identifies \(\text{Fe}^{2+}\) and \(\text{NH}_4^+\) as cations in FA 1 with correct supporting evidence (green precipitate insoluble in excess NaOH/\(\text{NH}_3\) and alkaline gas on heating with NaOH).
* **M9 [1 mark]:** Identifies \(\text{SO}_4^{2-}\) as the anion in FA 1 with correct supporting evidence (white precipitate with acidified barium nitrate).
* **M10 [1 mark]:** Identifies \(\text{Zn}^{2+}\) as the cation in FA 2 with correct supporting evidence (white precipitate soluble in excess NaOH and excess \(\text{NH}_3\)).
* **M11 [1 mark]:** Identifies \(\text{I}^-\) as the anion in FA 2 with correct supporting evidence (yellow precipitate with acidified silver nitrate).
* **M12 [1 mark]:** Writes the correct ionic equation: \(\text{Zn}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Zn(OH)}_2\text{(s)}\). Must include correct state symbols.

### **Part (c): Confirmatory Test [2 marks]**
* **M13 [1 mark]:** Identifies a suitable oxidizing agent to liberate iodine (e.g., chlorine water / aqueous bromine / acidified hydrogen peroxide) followed by an indicator (e.g., starch solution or cyclohexane).
* **M14 [1 mark]:** States the correct expected observation: solution turns blue-black (with starch) or organic layer turns purple (with cyclohexane).

(a) Cobalt has atomic number 27. The neutral atom configuration is \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^7 4\text{s}^2\). When forming the \(\text{Co}^{2+}\) ion, the two electrons are lost from the \(4\text{s}\) orbital first, giving: \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^7\).

(b)(i) Complex \(\mathbf{B}\) is formed by the oxidation of the hexamminecobalt(II) complex to hexamminecobalt(III) in the presence of excess ammonia. Its formula is \([\text{Co}(\text{NH}_3)_6]^{3+}\), and its charge is \(3+\).

(ii) The explanation of color in complexes:
- Ligands approach the central transition metal ion, and their lone pairs repel the electrons in the d-orbitals. This causes the five degenerate d-orbitals to split into two non-degenerate sets with an energy difference of \(\Delta E\).
- When white light passes through, electrons in the lower energy d-orbitals absorb a specific frequency of visible light to get promoted to a higher d-orbital (d-d transition), satisfying \(\Delta E = h\nu\).
- The complementary color of the absorbed light is transmitted or reflected.

(c)(i) The coordination number of cobalt is 6. Each of the two bidentate \(\text{en}\) ligands forms two coordinate bonds (totaling 4), and the two monodentate chloride ligands form one each (totaling 2), making a total of 6 coordinate bonds.

(ii) To have optical isomerism, the complex must exist as the cis-isomer. The trans-isomer is symmetrical and not chiral. The two optical isomers of cis-\([\text{Co}(\text{en})_2\text{Cl}_2]^+\) are non-superimposable mirror images of each other.

Part (a):
- [1 mark] for the correct electronic configuration of \(\text{Co}^{2+}\): \(1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^7\) (accept noble gas core \([\text{Ar}] 3\text{d}^7\)).

Part (b):
- [1 mark] for formula of B: \([\text{Co}(\text{NH}_3)_6]^{3+}\).
- [1 mark] for charge of B: \(3+\) (allow separately if formula is otherwise correct).
- [1 mark] for stating d-orbitals split into two energy levels due to ligands.
- [1 mark] for explaining that electrons absorb light of energy \(\Delta E\) to be promoted from lower to higher d-orbitals (d-d transition).
- [1 mark] for stating that the remaining/complementary wavelengths are transmitted/reflected.

Part (c):
- [1 mark] for coordination number of 6.
- [2 marks] for drawing correct, 3D octahedrally coordinated cis-isomers that are clearly non-superimposable mirror images (1 mark if structures are cis but mirror image is imperfect, 0 marks if trans-isomer is drawn).

(a) The iron half-reaction must proceed as oxidation because it has a less positive electrode potential:
\(5\text{Fe}^{2+}(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + 5\text{e}^-\)

Combining this with the reduction of permanganate gives the overall cell equation:
\(5\text{Fe}^{2+}(\text{aq}) + \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)

The standard cell potential is:
\(E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}} = +1.51 - (+0.77) = +0.74\text{ V}\).

(b)(i) The Nernst equation for this single-electron transfer process at \(298\text{ K}\) is:
\(E = E^\ominus + \frac{0.059}{z}\log\left(\frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]}\right)\)
Since \(z = 1\), it simplified to:
\(E = E^\ominus + 0.059\log\left(\frac{[\text{Fe}^{3+}]}{[\text{Fe}^{2+}]}\right)\)

(ii) Substituting the values into the equation:
\(E = +0.77 + 0.059\log\left(\frac{0.250}{0.015}\right)\)
\(E = +0.77 + 0.059\log(16.67)\)
\(E = +0.77 + 0.059(1.222)\)
\(E = +0.77 + 0.072 = +0.842\text{ V}\) (or \(+0.84\text{ V}\) to 2 decimal places).

(c) Increasing pH corresponds to a lower concentration of hydrogen ions, \([\text{H}^+]\). Looking at the half-cell equilibrium:
\(\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5\text{e}^- \rightleftharpoons \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})\)
A decrease in reactant concentration (\([\text{H}^+]\)) shifts the position of equilibrium to the left to favor the backward reaction. Consequently, the tendency to gain electrons decreases, which causes the electrode potential \(E\) of the half-cell to become less positive (decrease).

Part (a):
- [1 mark] for the correct overall balanced ionic equation (must include correct state symbols).
- [1 mark] for the calculation of cell potential: \(+0.74\text{ V}\).

Part (b):
- [1 mark] for the correct expression of the Nernst equation for the iron half-cell.
- [1 mark] for correctly substituting \([\text{Fe}^{3+}] = 0.250\) and \([\text{Fe}^{2+}] = 0.015\).
- [1 mark] for calculating \(E = +0.84\text{ V}\) (allow \(+0.842\text{ V}\)).

Part (c):
- [1 mark] for stating that increasing pH means \([\text{H}^+]\) decreases.
- [1 mark] for stating that the reduction equilibrium shifts to the left.
- [1 mark] for stating that the electrode potential decreases (becomes less positive).

(a)(i) Since compound \(\mathbf{X}\) is an ester and yields methanol on alkaline hydrolysis, it must be a methyl ester: \(\text{R-COOCH}_3\).
Since the formula is \(\text{C}_4\text{H}_7\text{ClO}_2\), subtracting the \(\text{-COOCH}_3\) group leaves a \(\text{-C}_2\text{H}_4\text{Cl}\) alkyl portion.
Therefore, the structural formula of \(\mathbf{X}\) is \(\text{ClCH}_2\text{CH}_2\text{COOCH}_3\) (methyl 3-chloropropanoate) or \(\text{CH}_3\text{CH(Cl)COOCH}_3\) (methyl 2-chloropropanoate).

(ii) Under reflux with excess \(\text{NaOH}\), both the ester functional group and the primary/secondary alkyl chloride functional group are hydrolyzed.
For \(\text{ClCH}_2\text{CH}_2\text{COOCH}_3\):
\(\text{ClCH}_2\text{CH}_2\text{COOCH}_3 + 2\text{NaOH} \rightarrow \text{HOCH}_2\text{CH}_2\text{COONa} + \text{CH}_3\text{OH} + \text{NaCl}\)
(The salt \(\mathbf{Y}\) is sodium 3-hydroxypropanoate, \(\text{HOCH}_2\text{CH}_2\text{COONa}\)).

(b) Relative rate of hydrolysis order:
Chlorobenzene < (Chloromethyl)benzene < Benzoyl chloride

Explanation:
- Chlorobenzene: The p-orbital containing a lone pair on the chlorine atom overlaps with the \(\pi\)-delocalised ring system of the benzene ring. This imparts partial double-bond character to the \(\text{C–Cl}\) bond, making it extremely strong and difficult to break. Additionally, the high electron density of the benzene ring repels nucleophiles.
- (Chloromethyl)benzene: The chlorine is attached to an \(\text{sp}^3\) hybridised carbon atom. The \(\text{C–Cl}\) bond is a normal single covalent bond, and the carbon is slightly positive (polarised), allowing substitution by nucleophiles at a moderate rate.
- Benzoyl chloride: The carbon of the carbonyl group is bonded to highly electronegative oxygen and chlorine atoms, leaving it extremely electron-deficient. This carbon is highly susceptible to nucleophilic attack, and the chlorine is a very good leaving group, resulting in rapid hydrolysis even with cold water.

Part (a):
- [1 mark] for identifying \(\mathbf{X}\) as a methyl ester with a chloro-substituted ethyl group: e.g. \(\text{ClCH}_2\text{CH}_2\text{COOCH}_3\) or \(\text{CH}_3\text{CH(Cl)COOCH}_3\).
- [1 mark] for balanced chemical equation with 2 moles of \(\text{NaOH}\) yielding the hydroxy salt, methanol, and sodium chloride.

Part (b):
- [1 mark] for correct order of reactivity: Chlorobenzene < (Chloromethyl)benzene < Benzoyl chloride.
- [1 mark] for explaining chlorobenzene has p-orbital overlap with the ring system, imparting partial double bond character to the \(\text{C-Cl}\) bond.
- [1 mark] for explaining (Chloromethyl)benzene has normal alkyl \(\text{C-Cl}\) single-bond behavior (no conjugation with the ring).
- [1 mark] for explaining benzoyl chloride has a highly electron-deficient carbonyl carbon, making it very susceptible to nucleophilic attack.

(a) Concentrated sulfuric acid acts as a catalyst and an acid (proton donor) to generate the electrophile, \(\text{NO}_2^+\) (nitronium ion).
Equation:
\(\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-\)
(Alternative: \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\))

(b) Mechanism:
1. A curly arrow must be drawn from the delocalised \(\pi\) cloud of the methylbenzene ring (specifically from the para position relative to the methyl group) to the electrophile, \(\text{NO}_2^+\).
2. This forms a positively charged arenium intermediate where carbon-4 is connected to both \(\text{-H}\) and \(\text{-NO}_2\). The ring carries a positive charge, delocalised over the remaining 5 carbons (draw a horse-shoe shape inside the ring, open towards carbon-4, with a '+' charge inside).
3. A curly arrow is drawn from the \(\text{C-H}\) bond at carbon-4 pointing back into the horseshoe of the ring to restore the fully delocalised aromatic \(\pi\) cloud.
4. A base, such as \(\text{HSO}_4^-\), abstracts the proton: \(\text{HSO}_4^- + \text{H}^+ \rightarrow \text{H}_2\text{SO}_4\), which regenerates the catalyst.

(c) The methyl group is an electron-donating alkyl group due to the positive inductive effect. This pushes electron density into the benzene ring, making the aromatic \(\pi\) cloud more electron-rich in methylbenzene compared to benzene. Thus, it attracts electrophiles more strongly and stabilizes the carbocation intermediate more effectively, resulting in a faster rate of reaction.

Part (a):
- [1 mark] for stating the role of \(\text{H}_2\text{SO}_4\) is a catalyst / proton donor / acid.
- [1 mark] for the correct chemical equation producing the \(\text{NO}_2^+\) electrophile.

Part (b):
- [1 mark] for curly arrow from the ring (at position 4) to the nitrogen atom of \(\text{NO}_2^+\).
- [1 mark] for drawing the correct intermediate structure showing both \(\text{-H}\) and \(\text{-NO}_2\) on C4 and a positive charge inside a broken ring.
- [1 mark] for curly arrow from the C-H bond to the ring system.
- [1 mark] for equation showing regeneration of the catalyst \(\text{H}_2\text{SO}_4\) (e.g. \(\text{HSO}_4^- + \text{H}^+ \rightarrow \text{H}_2\text{SO}_4\)).

Part (c):
- [1 mark] for identifying the methyl group as electron-donating due to the positive inductive effect.
- [1 mark] for explaining that this increases the electron density of the benzene ring (attracting electrophiles more easily).

(a)(i) Calculate the mass of each element in the sample:
- Mass of carbon: \(2.64\text{ g of CO}_2 \times \frac{12.0}{44.0} = 0.720\text{ g of C}\)
- Mass of hydrogen: \(1.08\text{ g of H}_2\text{O} \times \frac{2.0}{18.0} = 0.120\text{ g of H}\)
- Mass of oxygen: \(1.32\text{ g (total)} - 0.720\text{ g} - 0.120\text{ g} = 0.480\text{ g of O}\)

Convert these masses to moles:
- Moles of C: \(\frac{0.720}{12.0} = 0.060\text{ mol}\)
- Moles of H: \(\frac{0.120}{1.0} = 0.120\text{ mol}\)
- Moles of O: \(\frac{0.480}{16.0} = 0.030\text{ mol}\)

Divide each mole value by the smallest (0.030):
- Ratio C : H : O = \(2 : 4 : 1\).
Thus, the empirical formula of \(\mathbf{Z}\) is \(\text{C}_2\text{H}_4\text{O}\).

(ii) Use the ideal gas equation: \(pV = nRT = \frac{m}{M}RT\)
Rearranging for molar mass: \(M = \frac{mRT}{pV}\)
Convert units to standard SI units:
- \(T = 150 + 273.15 = 423.15\text{ K}\)
- \(V = 241\text{ cm}^3 = 2.41 \times 10^{-4}\text{ m}^3\)
- \(p = 1.01 \times 10^5\text{ Pa}\)
- \(m = 0.610\text{ g}\)

Substitute values into the formula:
\(M = \frac{0.610 \times 8.31 \times 423.15}{1.01 \times 10^5 \times 2.41 \times 10^{-4}}\)
\(M = \frac{2145}{24.34} = 88.1\text{ g mol}^{-1}\)

The empirical formula mass of \(\text{C}_2\text{H}_4\text{O}\) is \(2(12.0) + 4(1.0) + 16.0 = 44.0\text{ g mol}^{-1}\).
Since \(\frac{88.1}{44.0} \approx 2\), the molecular formula of \(\mathbf{Z}\) is twice the empirical formula: \(\text{C}_4\text{H}_8\text{O}_2\).

(b) The balanced equation for the complete combustion of \(\text{C}_4\text{H}_8\text{O}_2\) is:
\(\text{C}_4\text{H}_8\text{O}_2(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 4\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{g})\)

Part (a)(i):
- [1 mark] for calculating the masses or percentages of Carbon (0.72g), Hydrogen (0.12g), and Oxygen (0.48g).
- [1 mark] for converting masses to moles (0.06 mol C, 0.12 mol H, 0.03 mol O).
- [1 mark] for deducing empirical formula: \(\text{C}_2\text{H}_4\text{O}\).

Part (a)(ii):
- [1 mark] for correct temperature conversion (423 K) and volume conversion (\(2.41 \times 10^{-4}\text{ m}^3\)).
- [1 mark] for correct rearrangement of the ideal gas equation and substitution.
- [1 mark] for calculating molar mass: \(88\text{ g mol}^{-1}\) (accept \(88.0\) to \(88.1\)).
- [1 mark] for deducing molecular formula: \(\text{C}_4\text{H}_8\text{O}_2\).

Part (b):
- [1 mark] for balanced chemical equation (including state symbols; oxygen and products must be correct).

(a)(i) Order deduction:
- Propanone: Comparing Experiments 1 and 2, when \([\text{CH}_3\text{COCH}_3]\) is doubled (from 0.40 to 0.80) while keeping both \([\text{I}_2]\) and \([\text{H}^+]\) constant, the initial rate doubles (from \(1.2 \times 10^{-5}\) to \(2.4 \times 10^{-5}\)). Thus, the reaction is 1st order with respect to propanone.
- Iodine: Comparing Experiments 1 and 3, when \([\text{I}_2]\) is doubled (from 0.005 to 0.010) while keeping both \([\text{CH}_3\text{COCH}_3]\) and \([\text{H}^+]\) constant, the rate remains unchanged. Thus, the reaction is 0th order with respect to iodine.
- Hydrogen ions: Comparing Experiments 1 and 4, when \([\text{H}^+]\) is doubled (from 0.10 to 0.20) while keeping both \([\text{CH}_3\text{COCH}_3]\) and \([\text{I}_2]\) constant, the rate doubles (from \(1.2 \times 10^{-5}\) to \(2.4 \times 10^{-5}\)). Thus, the reaction is 1st order with respect to \([\text{H}^+]\).

(ii) Rate equation:
\(\text{Rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]\)

Using data from Experiment 1 to find \(k\):
\(1.2 \times 10^{-5} = k \times (0.40) \times (0.10)\)
\(1.2 \times 10^{-5} = 0.040 k\)
\(k = 3.0 \times 10^{-4}\)

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

(b) The Maxwell-Boltzmann distribution diagram should feature:
- Y-axis labeled 'Number of molecules' (or fraction of molecules) and X-axis labeled 'Energy'.
- Two curves: The curve for \(T_2\) should have a lower peak that is shifted to the right of the \(T_1\) peak, and its tail must remain higher at high energy values.
- A vertical line representing the activation energy, \(E_{\text{a}}\).
- Explaining: At \(T_2\) (higher temperature), a significantly larger fraction of molecules have energy equal to or greater than \(E_{\text{a}}\). This increases the frequency of successful collisions, leading to a faster rate of reaction.

Part (a)(i):
- [1 mark] for deducing 1st order for propanone with correct reasoning from Exp 1 & 2.
- [1 mark] for deducing 0th order for iodine with correct reasoning from Exp 1 & 3.
- [1 mark] for deducing 1st order for hydrogen ions with correct reasoning from Exp 1 & 4.

Part (a)(ii):
- [1 mark] for the correct rate equation: \(\text{Rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]\).
- [1 mark] for calculating \(k = 3.0 \times 10^{-4}\).
- [1 mark] for correct units: \(\text{dm}^3\text{ mol}^{-1}\text{ s}^{-1}\).

Part (b):
- [1 mark] for a well-labeled Maxwell-Boltzmann diagram showing both axes and two curves at different temperatures \(T_1\) and \(T_2\).
- [1 mark] for marking activation energy, \(E_{\text{a}}\), on the X-axis.
- [1 mark] for explaining that at \(T_2\), a greater area under the curve lies to the right of \(E_{\text{a}}\), meaning more molecules have energy \(\ge E_{\text{a}}\).

(a) The assumptions are:
1. There are no attractive or repulsive intermolecular forces between the gas molecules.
2. The actual volume occupied by the gas molecules themselves is negligible compared to the total volume of the container.

(b) Nitrogen, \(\text{N}_2\), is a non-polar molecule with only weak instantaneous dipole-induced dipole (London dispersion) forces between its molecules. On the other hand, ammonia, \(\text{NH}_3\), is a highly polar molecule with a lone pair on the nitrogen atom and strongly polar \(\text{N-H}\) bonds, allowing it to form strong hydrogen bonds between molecules. Stronger intermolecular forces lead to larger deviations from the ideal gas assumption of no intermolecular attractions.

(c)(i) Calculate the number of moles of each gas:
- \(n(\text{CO}) = \frac{14.0}{28.0} = 0.50\text{ mol}\)
- \(n(\text{O}_2) = \frac{16.0}{32.0} = 0.50\text{ mol}\)
- Total moles: \(n_{\text{total}} = 0.50 + 0.50 = 1.00\text{ mol}\)

Convert units to standard SI units:
- \(V = 5.00\text{ dm}^3 = 5.00 \times 10^{-3}\text{ m}^3\)
- \(T = 27.0 + 273.15 = 300.15\text{ K}\) (or \(300\text{ K}\))

Using \(pV = nRT\):
\(p_{\text{total}} = \frac{n_{\text{total}}RT}{V}\)
\(p_{\text{total}} = \frac{1.00 \times 8.31 \times 300.15}{5.00 \times 10^{-3}} = \frac{2494.2}{0.005} = 498849\text{ Pa} \approx 499\text{ kPa}\) (or \(498.6\text{ kPa}\) if \(T = 300\text{ K}\) is used).

(ii) The partial pressure of a gas is given by: \(p_i = x_i \times p_{\text{total}}\)
where \(x_i\) is the mole fraction.
\(x(\text{CO}) = \frac{0.50}{1.00} = 0.50\)
\(p(\text{CO}) = 0.50 \times 499\text{ kPa} = 249.5\text{ kPa}\) (or \(249.3\text{ kPa}\) if using \(498.6\text{ kPa}\)).

Part (a):
- [1 mark] for stating that ideal gases assume negligible volume of molecules.
- [1 mark] for stating that ideal gases assume zero intermolecular forces.

Part (b):
- [1 mark] for identifying that \(\text{N}_2\) has only London dispersion forces (van der Waals).
- [1 mark] for identifying that \(\text{NH}_3\) has hydrogen bonding.
- [1 mark] for explaining that stronger intermolecular forces lead to more significant deviations from ideality.

Part (c)(i):
- [1 mark] for calculating the correct number of moles of each gas and total moles (1.00 mol).
- [1 mark] for correct conversion of temperature to Kelvin (300 K) and volume to \(\text{m}^3\).
- [1 mark] for calculating total pressure \(499\text{ kPa}\) (accept range \(498\) to \(500\text{ kPa}\)).

Part (c)(ii):
- [1 mark] for finding the mole fraction of CO to be 0.50.
- [1 mark] for calculating the partial pressure of CO as \(249\text{ kPa}\) to \(250\text{ kPa}\) (must be consistent with the total pressure calculated in c(i)).

(a) Lattice energy is defined as the enthalpy change when one mole of an ionic crystalline solid is formed from its constituent gaseous ions under standard conditions.

(b) Set up the energy cycle for the formation of \(\text{CaCl}_2(\text{s})\):
\(\Delta H_{\text{f}}^\ominus = \Delta H_{\text{at}}^\ominus[\text{Ca}] + I_1[\text{Ca}] + I_2[\text{Ca}] + \Delta H_{\text{bond}}^\ominus[\text{Cl}_2] + 2 \times EA[\text{Cl}] + \Delta H_{\text{lattice}}\)

Substitute the given values into the cycle:
\(-796 = 178 + 590 + 1145 + 242 + 2(-348) + \Delta H_{\text{lattice}}\)
\(-796 = 2155 - 696 + \Delta H_{\text{lattice}}\)
\(-796 = 1459 + \Delta H_{\text{lattice}}\)
\(\Delta H_{\text{lattice}} = -796 - 1459 = -2255\text{ kJ mol}^{-1}\).

(c)(i) Use the relation: \(\Delta G_{\text{soln}}^\ominus = \Delta H_{\text{soln}}^\ominus - T\Delta S_{\text{soln}}^\ominus\)
Be careful with units: conversion of entropy from \(\text{J}\) to \(\text{kJ}\) is required.
- \(\Delta H_{\text{soln}}^\ominus = -83\text{ kJ mol}^{-1}\)
- \(T = 298\text{ K}\)
- \(\Delta S_{\text{soln}}^\ominus = -12.0 \times 10^{-3}\text{ kJ K}^{-1}\text{ mol}^{-1}\)

\(\Delta G_{\text{soln}}^\ominus = -83 - (298 \times (-0.0120))\)
\(\Delta G_{\text{soln}}^\ominus = -83 + 3.576 = -79.424\text{ kJ mol}^{-1}\) (or \(-79.4\text{ kJ mol}^{-1}\)).

(ii) Since \(\Delta S_{\text{soln}}^\ominus\) is negative, the term \(-T\Delta S_{\text{soln}}^\ominus\) is positive. As temperature increases, \(-T\Delta S_{\text{soln}}^\ominus\) becomes more positive, which makes the overall \(\Delta G_{\text{soln}}^\ominus\) less negative (or more positive). Therefore, the dissolving process becomes thermodynamically less feasible, and the solubility of calcium chloride decreases as temperature increases. (Alternatively, since the dissolution process is exothermic, according to Le Chatelier's principle, an increase in temperature shifts the equilibrium to favor the reactant solid, thus decreasing solubility).

Part (a):
- [1 mark] for defining lattice energy, mentioning '1 mole of solid ionic crystal' and 'gaseous ions' under standard conditions.

Part (b):
- [1 mark] for recognizing that the chlorine bond enthalpy equals the atomisation of 2 moles of chlorine atoms (no need to multiply 242 by 2).
- [1 mark] for multiplying the electron affinity of Cl by 2 (\(2 \times -348 = -696\)).
- [1 mark] for showing a correct Hess's law equation setup.
- [1 mark] for calculating the final lattice energy value: \(-2255\text{ kJ mol}^{-1}\) (deduct 1 mark if negative sign is omitted).

Part (c)(i):
- [1 mark] for converting entropy unit to \(\text{kJ K}^{-1}\text{ mol}^{-1}\).
- [1 mark] for correct substitution into Gibbs equation.
- [1 mark] for calculating \(\Delta G_{\text{soln}}^\ominus = -79.4\text{ kJ mol}^{-1}\).

Part (c)(ii):
- [1 mark] for stating that solubility decreases.
- [1 mark] for giving a valid explanation (either based on \(-T\Delta S\) making \(\Delta G\) less negative, OR based on Le Chatelier's principle and the exothermic enthalpy of solution).

(a) The standard electrode potential is defined as the electromotive force (e.m.f.) of a cell consisting of the half-cell of interest connected to a standard hydrogen electrode (S.H.E.) under standard conditions: temperature of \(298\text{ K}\), gas pressures of \(1\text{ atm}\) (or \(101\text{ kPa}\)), and solution concentrations of \(1.00\text{ mol dm}^{-3}\).

(b) (i) The half-cell with the more negative standard electrode potential is the stronger reducing agent and will undergo oxidation. Therefore, \(\text{Cr}^{2+}\) is oxidised to \(\text{Cr}^{3+}\), and \(\text{Ag}^+\) is reduced to \(\text{Ag}\).
Overall equation:
\(\text{Cr}^{2+}(\text{aq}) + \text{Ag}^+(\text{aq}) \rightarrow \text{Cr}^{3+}(\text{aq}) + \text{Ag}(\text{s})\)
(ii) \(E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = E^\ominus(\text{Ag}^+/\text{Ag}) - E^\ominus(\text{Cr}^{3+}/\text{Cr}^{2+}) = +0.80 - (-0.41) = +1.21\text{ V}\)

(c) According to the Nernst equation:
\(E = E^\ominus + \frac{0.059}{z} \lg \frac{[\text{oxidised}]}{[\text{reduced}]}\)
For the chromium half-cell, \(z = 1\) (transfer of 1 electron),
\([\text{oxidised}] = [\text{Cr}^{3+}(\text{aq})] = 0.015\text{ mol dm}^{-3}\),
\([\text{reduced}] = [\text{Cr}^{2+}(\text{aq})] = 0.45\text{ mol dm}^{-3}\).

\(E = -0.41 + \frac{0.059}{1} \lg \left( \frac{0.015}{0.45} \right)\)
\(E = -0.41 + 0.059 \lg(0.03333)\)
\(E = -0.41 + 0.059 \times (-1.477)\)
\(E = -0.41 - 0.087 = -0.497\text{ V}\) (or \(-0.50\text{ V}\))

(d) The relation between Gibbs free energy change and cell potential is:
\(\Delta G^\ominus = -n F E^\ominus_{\text{cell}}\)
Where \(n = 1\) mole of electrons transferred per mole of reaction as written,
\(F = 96500\text{ C mol}^{-1}\),
\(E^\ominus_{\text{cell}} = +1.21\text{ V}\).

\(\Delta G^\ominus = -1 \times 96500 \times 1.21 = -116765\text{ J mol}^{-1} = -116.8\text{ kJ mol}^{-1}\) (or \(-117\text{ kJ mol}^{-1}\))

(e) Decreasing \([\text{Ag}^+(\text{aq})]\) will shift the equilibrium position of the silver half-cell reaction, \(\text{Ag}^+(\text{aq}) + e^- \rightleftharpoons \text{Ag}(\text{s})\), to the left. This makes the electrode potential of the silver half-cell less positive (more negative). Since \(E_{\text{cell}} = E_{\text{reduction}} - E_{\text{oxidation}}\), a less positive \(E(\text{Ag}^+/\text{Ag})\) causes the overall cell potential, \(E_{\text{cell}}\), to decrease.

(a) [1] For stating e.m.f. relative to a standard hydrogen electrode (S.H.E.) / hydrogen electrode.
[1] For mentioning standard conditions: temperature of 298 K, pressure of 1 atm (or 101 kPa), and concentration of 1.00 mol dm^-3.

(b) (i) [1] For the correct balanced equation: Cr2+(aq) + Ag+(aq) -> Cr3+(aq) + Ag(s) (state symbols are not strictly required but species must be correct).
(ii) [1] For E⦵cell = +1.21 V.

(c) [1] For correct substitution of values into the Nernst equation (z = 1, [Cr3+]/[Cr2+] ratio = 0.015/0.45).
[1] For calculating the logarithmic term: lg(0.0333) = -1.48 (or -1.477).
[1] For the correct final answer: -0.50 V (accept -0.497 V to -0.50 V).

(d) [1] For correct substitution of values into standard relation: ΔG⦵ = -1 * 96500 * 1.21.
[1] For final value: -117 kJ mol^-1 (accept -116.8 to -117 kJ mol^-1, must include negative sign and correct units).

(e) [1] For stating that the cell potential (E_cell) decreases.
[1] For explaining that the equilibrium Ag+(aq) + e- <=> Ag(s) shifts left OR that the half-cell potential for Ag+/Ag becomes less positive.

**Part (a) Planning**

(i)
- Approximate mass of \(\text{Mg(OH)}_2\) in one tablet = \(1.20\text{ g} \times 0.30 = 0.36\text{ g}\).
- Moles of \(\text{Mg(OH)}_2 = \frac{0.36}{58.3} = 6.17 \times 10^{-3}\text{ mol}\) (or \(6.2 \times 10^{-3}\text{ mol}\)).

(ii)
- Initial moles of \(\text{HCl}\) added = \(25.0 \times 10^{-3}\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 0.0250\text{ mol}\).
- From equation \(\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}\), moles of \(\text{HCl}\) reacted = \(2 \times 6.17 \times 10^{-3}\text{ mol} = 0.01234\text{ mol}\).
- Moles of excess \(\text{HCl}\) in \(250.0\text{ cm}^3\) flask = \(0.0250 - 0.01234 = 0.01266\text{ mol}\).
- Moles of \(\text{HCl}\) in \(25.0\text{ cm}^3\) aliquot = \(0.01266 \times \frac{25.0}{250.0} = 1.266 \times 10^{-3}\text{ mol}\).
- Since \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\), moles of \(\text{NaOH}\) required = \(1.266 \times 10^{-3}\text{ mol}\).
- Volume of \(0.100\text{ mol dm}^{-3}\) \(\text{NaOH}\) required = \(\frac{1.266 \times 10^{-3}\text{ mol}}{0.100\text{ mol dm}^{-3}} = 0.01266\text{ dm}^3 = 12.66\text{ cm}^3\).
- This value is between \(10\text{ cm}^3\) and \(25\text{ cm}^3\), making it a suitable and highly reliable titration volume.

(iii)
1. Crush the tablet with a pestle and mortar to increase surface area, and weigh the crushed powder (or weigh by difference into a beaker).
2. Use a \(25.0\text{ cm}^3\) volumetric pipette to transfer exactly \(25.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) \(\text{HCl}\) to the beaker.
3. Stir with a glass rod (and optionally warm gently) to ensure complete reaction of all solid magnesium hydroxide.
4. Quantitatively transfer the contents of the beaker to a \(250.0\text{ cm}^3\) volumetric flask. Rinse the beaker and stirring rod multiple times with deionised water, adding all the washings to the flask.
5. Fill the flask with deionised water until the bottom of the meniscus is aligned with the graduation mark. Stopper the flask and invert several times to ensure a homogeneous solution.

(iv)
- Hazard: \(1.00\text{ mol dm}^{-3}\) \(\text{HCl}\) is an irritant.
- Safety precaution: Wear safety goggles to protect eyes, or wear protective gloves.

**Part (b) Analysis and Evaluation**

(i)
- Anomalous result: Titration 3 (Titre of \(11.80\text{ cm}^3\)).
- Possible experimental error: The student may have rinsed the conical flask with the acid solution instead of deionised water (introducing more acid, meaning less NaOH is needed to neutralise the aliquot), or they may have misread the burette scale.

(ii)
- Concordant titres are Titrations 1, 2, and 4 (which are within \(\pm 0.10\text{ cm}^3\) of each other).
- Mean titre = \(\frac{12.75 + 12.65 + 12.70}{3} = 12.70\text{ cm}^3\).

(iii)
- Moles of \(\text{NaOH}\) used = \(12.70 \times 10^{-3}\text{ dm}^3 \times 0.100\text{ mol dm}^{-3} = 1.27 \times 10^{-3}\text{ mol}\).
- Moles of excess \(\text{HCl}\) in \(25.0\text{ cm}^3\) aliquot = \(1.27 \times 10^{-3}\text{ mol}\).
- Moles of excess \(\text{HCl}\) in the total \(250.0\text{ cm}^3\) flask = \(1.27 \times 10^{-3} \times 10 = 1.27 \times 10^{-2}\text{ mol}\).
- Initial moles of \(\text{HCl}\) added = \(25.0 \times 10^{-3}\text{ dm}^3 \times 1.00\text{ mol dm}^{-3} = 2.50 \times 10^{-2}\text{ mol}\).
- Moles of \(\text{HCl}\) reacted with \(\text{Mg(OH)}_2\) = \((2.50 \times 10^{-2}) - (1.27 \times 10^{-2}) = 1.23 \times 10^{-2}\text{ mol}\).
- Moles of \(\text{Mg(OH)}_2\) in the tablet = \(\frac{1.23 \times 10^{-2}}{2} = 6.15 \times 10^{-3}\text{ mol}\).
- Mass of \(\text{Mg(OH)}_2 = 6.15 \times 10^{-3}\text{ mol} \times 58.3\text{ g mol}^{-1} = 0.3585\text{ g}\).
- Percentage by mass of \(\text{Mg(OH)}_2 = \frac{0.3585\text{ g}}{1.21\text{ g}} \times 100\% = 29.6\%\) (or \(29.63\%\)).

(iv)
1. Pipette percentage error = \(\frac{0.06}{25.0} \times 100\% = 0.24\%\).
2. Burette percentage error (two readings): Uncertainty of titre = \(2 \times 0.05 = 0.10\text{ cm}^3\).
Percentage error = \(\frac{0.10}{12.70} \times 100\% = 0.79\%\).

**Part (a) Planning [9 marks]**
- (i) [1 mark]: Correct calculation of moles of \(\text{Mg(OH)}_2\) as \(6.17 \times 10^{-3}\text{ mol}\) (accept \(6.2 \times 10^{-3}\text{ mol}\)).
- (ii) [3 marks total]:
- M1: Calculation of initial moles of \(\text{HCl}\) (\(0.0250\text{ mol}\)) and moles of reacted \(\text{HCl}\) (\(0.01234\text{ mol}\)).
- M2: Calculation of remaining excess moles in \(25.0\text{ cm}^3\) aliquot (\(1.266 \times 10^{-3}\text{ mol}\)).
- M3: Correct calculation of the titre volume as \(12.66\text{ cm}^3\) (or \(12.7\text{ cm}^3\)) and stating that it is a suitable volume because it is within \(10\text{ cm}^3\) to \(25\text{ cm}^3\).
- (iii) [4 marks total]:
- M1: Crush tablet and transfer to beaker (with mass weighed by difference).
- M2: Add exactly \(25.0\text{ cm}^3\) of \(1.00\text{ mol dm}^{-3}\) \(\text{HCl}\) using a volumetric pipette.
- M3: Stir/warm to ensure complete reaction of the solid.
- M4: Quantitative transfer (rinsing beaker and rod with deionised water into the flask) and make up to the graduation mark in a \(250.0\text{ cm}^3\) volumetric flask followed by thorough mixing.
- (iv) [1 mark]: Correctly identifies acid as an irritant and proposes wearing safety goggles/gloves.

**Part (b) Analysis and Evaluation [7 marks]**
- (i) [1 mark]: Identifies Titration 3 (\(11.80\text{ cm}^3\)) as anomalous AND gives a valid reason (e.g. conical flask rinsed with acid, misreading burette scale, or air bubble in tip).
- (ii) [1 mark]: Correctly calculates mean titre as \(12.70\text{ cm}^3\) using only concordant titres (1, 2, and 4).
- (iii) [3 marks total]:
- M1: Correct calculation of excess moles of \(\text{HCl}\) in the flask (\(1.27 \times 10^{-2}\text{ mol}\)).
- M2: Correct calculation of moles of reacted \(\text{HCl}\) (\(1.23 \times 10^{-2}\text{ mol}\)) and moles of \(\text{Mg(OH)}_2\) in the tablet (\(6.15 \times 10^{-3}\text{ mol}\)).
- M3: Correct calculation of percentage by mass as \(29.6\%\) (or \(29.63\%\)) to 3 significant figures.
- (iv) [2 marks total]:
- M1: Pipette error = \(0.24\%\).
- M2: Burette error = \(0.79\%\) (or \(0.787\%\)). Allow [1 mark] for \(0.39\%\) if they only used a single reading uncertainty of \(0.05\text{ cm}^3\) instead of \(0.10\text{ cm}^3\).

### **Part (a)**
First, convert the Celsius temperatures to Kelvin using \(T = \theta + 273\):
* Exp 1: \(20.0 + 273 = 293\text{ K}\)
* Exp 2: \(28.5 + 273 = 301.5\text{ K}\) (rounded to \(302\text{ K}\) if using whole numbers)
* Exp 3: \(37.0 + 273 = 310\text{ K}\)
* Exp 4: \(45.5 + 273 = 318.5\text{ K}\) (rounded to \(319\text{ K}\))
* Exp 5: \(54.0 + 273 = 327\text{ K}\)
* Exp 6: \(62.5 + 273 = 335.5\text{ K}\) (rounded to \(336\text{ K}\))

*(Note: If \(T = \theta + 273.1\) or \(T = \theta + 273.15\) is used, the Kelvin temperatures are 293.1, 301.6, 310.1, 318.6, 327.1, 335.6. The calculation below uses \(T = \theta + 273\))*

Next, calculate \(1/T \times 10^{-3} / \text{K}^{-1}\) to 3 significant figures:
* Exp 1: \(1/293 = 3.41 \times 10^{-3}\)
* Exp 2: \(1/302 = 3.31 \times 10^{-3}\)
* Exp 3: \(1/310 = 3.23 \times 10^{-3}\)
* Exp 4: \(1/319 = 3.13 \times 10^{-3}\)
* Exp 5: \(1/327 = 3.06 \times 10^{-3}\)
* Exp 6: \(1/336 = 2.98 \times 10^{-3}\)

Next, calculate \(\ln(1/t)\) to 2 decimal places:
* Exp 1: \(\ln(1/122) = -4.80\)
* Exp 2: \(\ln(1/65) = -4.17\)
* Exp 3: \(\ln(1/35) = -3.56\)
* Exp 4: \(\ln(1/30) = -3.40\)
* Exp 5: \(\ln(1/12) = -2.48\)
* Exp 6: \(\ln(1/7) = -1.95\)

### **Part (b)**
Plot a graph of \(\ln(1/t)\) (y-axis, values spanning \(-5.0\) to \(-1.5\)) against \(1/T \times 10^{-3} / \text{K}^{-1}\) (x-axis, values spanning \(2.9\) to \(3.5\)).
Points should be plotted accurately. A straight line of best fit must be drawn passing through points 1, 2, 3, 5, 6, clearly leaving point 4 (the anomalous point) significantly below the line.

### **Part (c)**
* **Anomalous point:** Experiment 4 (points plotted at \(3.13\), \(-3.40\)).
* **Reason:** The time taken was longer than expected (making \(\ln(1/t)\) more negative than the line of best fit), which suggests the rate of reaction was slower. This could be due to heat loss to the surroundings before mixing, meaning the actual temperature of the reaction was lower than \(45.5^\circ\text{C}\). Alternatively, the starch indicator or one of the reactants was added late, or the timer was started early.

### **Part (d)**
Choose two points on the line of best fit that are far apart. For example:
* Point A on line: \(x_1 = 3.40 \times 10^{-3}\), \(y_1 = -4.73\)
* Point B on line: \(x_2 = 2.98 \times 10^{-3}\), \(y_2 = -1.95\)

\[\text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1.95 - (-4.73)}{(2.98 - 3.40) \times 10^{-3}} = \frac{2.78}{-0.42 \times 10^{-3}} = -6619\text{ K}\]

*(Accept gradients in the range \(-6400\) to \(-6800\text{ K}\) depending on individual lines of best fit.)*

### **Part (e)**
From the Arrhenius equation: \(\text{gradient} = -\frac{E_a}{R}\)

\[E_a = -\text{gradient} \times R\]
\[E_a = -(-6619\text{ K}) \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} = 55004\text{ J mol}^{-1} = +55.0\text{ kJ mol}^{-1}\]

*(Using range \(-6400\) to \(-6800\text{ K}\) gives \(E_a\) in the range \(+53.2\) to \(+56.5\text{ kJ mol}^{-1}\). Answer must be positive, have 3 significant figures and include the unit \(\text{kJ mol}^{-1}\).)*

### **Part (f)**
* **Error:** The beaker of hot water cools down during the course of the experiment, meaning the temperature decreases during the run and the actual average reaction temperature is lower than the recorded starting temperature.
* **Improvement:** Use a thermostatically controlled water bath (or hot plate with feedback sensor) to maintain a constant temperature throughout each run.

**Part (a): [3 marks]**
* **1 Mark:** Correctly calculates and records all six temperature values in Kelvin (accept 293, 301.5/302, 310, 318.5/319, 327, 335.5/336).
* **1 Mark:** Correctly calculates and records all six \(1/T \times 10^{-3}\) values to 3 significant figures (3.41, 3.31, 3.23, 3.13, 3.06, 2.98).
* **1 Mark:** Correctly calculates and records all six \(\ln(1/t)\) values to 2 decimal places (-4.80, -4.17, -3.56, -3.40, -2.48, -1.95).

**Part (b): [3 marks]**
* **1 Mark:** Axes chosen with suitable linear scales (points must occupy more than half of the grid on both axes) and clearly labeled with units: y-axis label = \(\ln(1/t)\) (no unit), x-axis label = \(1/T \times 10^{-3} / \text{K}^{-1}\) (or \(1/T / 10^{-3}\ \text{K}^{-1}\)).
* **1 Mark:** Correct plotting of all 6 points (to within half a small square).
* **1 Mark:** Straight line of best fit drawn cleanly, omitting the anomalous point (Experiment 4).

**Part (c): [2 marks]**
* **1 Mark:** Identifies Experiment 4 as the anomalous point.
* **1 Mark:** Provides a logical chemical/experimental reason (e.g., reaction mixture cooled down before mixing, starch indicator was added late, or timer started too early / stopped too late).

**Part (d): [2 marks]**
* **1 Mark:** Shows coordinates of two points on the best-fit line (not data points from the table unless they lie precisely on the line) that are separated by at least half the length of the line drawn.
* **1 Mark:** Calculates the gradient correctly with negative sign and correct magnitude (typical range: -6400 to -6800).

**Part (e): [2 marks]**
* **1 Mark:** Correctly relates gradient to activation energy: \(E_a = -\text{gradient} \times 8.31\) (or uses shown rearrangement of Arrhenius equation).
* **1 Mark:** Obtains a positive value of \(E_a\) in \(\text{kJ mol}^{-1}\) to 3 significant figures (typical range: +53.2 to +56.5 \(\text{kJ mol}^{-1}\)) with correct unit.

**Part (f): [2 marks]**
* **1 Mark:** Identifies that the beaker of hot water cools down during the course of the experiment / heat is lost to the surroundings, meaning temperature is not kept constant.
* **1 Mark:** Suggests using a thermostatically controlled water bath (or insulated container / electric temperature controller).