2025 Cambridge IAL Chemistry (9701) 模擬試題連答案詳解

Shaking the gas mixture with NaOH absorbs \(CO_2\). Therefore, the volume decrease of 30 cm\(^3\) (from 50 cm\(^3\) to 20 cm\(^3\)) is the volume of \(CO_2\) produced. Since 10 cm\(^3\) of alkane produced 30 cm\(^3\) of \(CO_2\), the mole ratio of alkane to \(CO_2\) is 1 : 3. This means each molecule of alkane contains 3 carbon atoms (\(n = 3\)), giving the formula \(C_3H_8\). The remaining 20 cm\(^3\) of gas is the unreacted excess \(O_2\). The volume of \(O_2\) reacted is 70 - 20 = 50 cm\(^3\). This fits the stoichiometry of the equation: \(C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\) (ratio 10 : 50 : 30).

1 mark: Identify that 30 cm\(^3\) of CO\(_2\) is produced from 10 cm\(^3\) of alkane, yielding n = 3, which corresponds to C\(_3\)H\(_8\).

Bromide ions are strong enough reducing agents to reduce concentrated sulfuric acid to sulfur dioxide, \(SO_2\) (which turns acidified potassium dichromate(VI) paper from orange to green). The oxidation of bromide ions yields bromine, \(Br_2\) (a gaseous orange-brown element at the reaction temperature). The reaction also produces some hydrogen bromide, \(HBr\) (an acidic gas that fumes in moist air). Chloride and fluoride cannot reduce sulfuric acid, whereas iodide reduces it more aggressively to sulfur (solid) and hydrogen sulfide (bad egg smell gas).

1 mark: Correctly identify the halide as bromide based on the characteristic reduction product (SO\(_2\)) and elemental halogen (Br\(_2\)).

The original light blue solution contains the hexaaquacopper(II) ion, \([Cu(H_2O)_6]^{2+}\). When excess conc. \(HCl\) is added, ligand exchange occurs to form the tetrahedral tetrachlorocuprate(II) ion, \([CuCl_4]^{2-}\), which is yellow-green. When aqueous ammonia is added, ligand exchange forms the deep blue tetraamminedihydratocopper(II) ion, \([Cu(NH_3)_4(H_2O)_2]^{2+}\).

1 mark: Identify Cu\(^{2+}\)(aq) / [Cu(H\(_2\)O)\(_6\)]\(^{2+}\) as the unique species showing these characteristic color transitions.

Reaction of 2-methylbut-2-ene, \((CH_3)_2C=CHCH_3\), with cold, dilute \(KMnO_4\) results in mild oxidation to form the diol 2-methylbutane-2,3-diol, \((CH_3)_2C(OH)-CH(OH)CH_3\). Carbon-2 is a tertiary carbon bonded to an \(-OH\) group (tertiary alcohol), while Carbon-3 is a secondary carbon bonded to an \(-OH\) group (secondary alcohol). Reaction with hot, concentrated \(KMnO_4\) cleaves the \(C=C\) double bond to yield propanone (a ketone) from the \((CH_3)_2C=\) part, and ethanoic acid (a carboxylic acid) from the \(=CHCH_3\) part.

1 mark: Identify P as a diol with secondary and tertiary alcohol groups, and Q/R as a ketone and a carboxylic acid.

Let us calculate the oxidation state of the transition metals in each option: In \([Cr(H_2O)_6]Cl_3\), the complex ion has a charge of \(+3\), so \(Cr\) is in the \(+3\) oxidation state. In \(K_2MnO_4\), the two potassium ions have a combined charge of \(+2\) and the four oxides have a charge of \(-8\), so \(Mn\) is in the \(+6\) state. In \(NH_4VO_3\), the ammonium ion is \(NH_4^+\) (charge of \(+1\)) and the metavanadate ion is \(VO_3^-\) (charge of \(-1\)). In \(VO_3^-\), the three oxygen atoms have a total oxidation state of \(-6\), meaning \(V\) must be \(+5\). In \(Na_2Fe_2O_4\), sodium is \(+1\) and oxygen is \(-2\), giving \(Fe\) an oxidation state of \(+3\).

1 mark: Correctly calculate oxidation numbers to determine that V in NH\(_4\)VO\(_3\) is +5.

Let the initial rate be \(R_1 = k[A][B]^2\). If the concentration of \(A\) is doubled, the concentration becomes \(2[A]\). If the concentration of \(B\) is halved, the concentration becomes \(0.5[B]\). The new rate, \(R_2\), is given by:
\(R_2 = k(2[A])(0.5[B])^2 = k \times 2[A] \times 0.25[B]^2 = 0.5 k[A][B]^2 = 0.5 R_1\).
Thus, the initial rate of reaction halves (decreases by a factor of 0.5).

1 mark: Correctly scale the rate equation by 2\(^1\) for A and (0.5)\(^2\) for B, giving an overall factor of 0.5.

The equation for the standard enthalpy change of formation of propan-1-ol is:
\(3C(s) + 4H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_3H_7OH(l)\).
According to Hess's Law, using combustion data:
\(\Delta H_f^\ominus [C_3H_7OH(l)] = \sum \Delta H_c^\ominus[\text{reactants}] - \sum \Delta H_c^\ominus[\text{products}]\)
\(\Delta H_f^\ominus = 3 \times \Delta H_c^\ominus[C(s)] + 4 \times \Delta H_c^\ominus[H_2(g)] - \Delta H_c^\ominus[C_3H_7OH(l)]\)
\(\Delta H_f^\ominus = 3(-394) + 4(-286) - (-2021) = -1182 - 1144 + 2021 = -305\text{ kJ mol}^{-1}\).

1 mark: Formulate the correct Hess's Law expression using the stoichiometric coefficients (3 for C and 4 for H\(_2\)) and calculate -305 kJ mol\(^{-1}\).

Thermal stability of Group 2 carbonates increases down the group because the cationic radius increases and charge density decreases. Thus, the larger \(Ba^{2+}\) ion polarises the carbonate ion's electron cloud less strongly than the smaller \(Ca^{2+}\) ion, making \(BaCO_3\) more thermally stable than \(CaCO_3\).

The decomposition reaction is \(MCO_3(s) \rightarrow MO(s) + CO_2(g)\). The percentage mass loss is due to the loss of carbon dioxide gas: \(\%\text{ mass loss} = \frac{M_r(CO_2)}{M_r(MCO_3)} \times 100\%\).
For \(CaCO_3\) (\(M_r \approx 100.1\)): \(\%\text{ mass loss} = \frac{44.0}{100.1} \times 100\% = 44.0\%\).
For \(BaCO_3\) (\(M_r \approx 197.3\)): \(\%\text{ mass loss} = \frac{44.0}{197.3} \times 100\% = 22.3\%\).
Therefore, \(BaCO_3\) undergoes a smaller percentage mass loss than \(CaCO_3\).

1 mark: Identify that barium carbonate is more stable (due to lower polarising power of Ba\(^{2+}\)) and undergoes a lower percentage mass loss due to its higher molar mass.

First, calculate the number of moles of \(\text{CO}_2\) gas produced: \(n(\text{CO}_2) = \frac{240 \text{ cm}^3}{24000 \text{ cm}^3 \text{ mol}^{-1}} = 0.0100 \text{ mol}\). Next, write the balanced equation: \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2\). The stoichiometric ratio of \(\text{Na}_2\text{CO}_3\) to \(\text{CO}_2\) is 1:1, so \(n(\text{Na}_2\text{CO}_3) = 0.0100 \text{ mol}\). Calculate the mass of \(\text{Na}_2\text{CO}_3\): \(\text{mass} = 0.0100 \text{ mol} \times 106.0 \text{ g mol}^{-1} = 1.06 \text{ g}\). Finally, find the percentage by mass: \(\% \text{ by mass} = \frac{1.06 \text{ g}}{1.50 \text{ g}} \times 100\% = 70.7\%\).

1 mark for the correct calculation of percentage by mass. Allow 1 mark for correct intermediate steps (moles of CO2 and mass of sodium carbonate) leading to the correct option C.

Down Group 17, the halogens exist as diatomic covalent molecules. Volatility decreases (boiling points increase) down the group because the total number of electrons in the molecules increases, which increases the polarisability of the electron cloud. This leads to stronger instantaneous dipole-induced dipole (London dispersion) intermolecular forces, requiring more thermal energy to overcome.

1 mark for identifying that volatility decreases down the group and explaining this trend in terms of increasing strength of instantaneous dipole-induced dipole forces due to an increased number of electrons.

In the complex \([\text{Co}(\text{NH}_3)_6]\text{Cl}_3\), ammonia (\(\text{NH}_3\)) is a neutral ligand and there are three chloride counter-ions (\(3\text{Cl}^-\)), meaning the cobalt ion is in the +3 oxidation state (\(\text{Co}^{3+}\)). The atomic number of Cobalt is 27, so a neutral cobalt atom has the electronic configuration \([\text{Ar}] 3\text{d}^7 4\text{s}^2\). To form the \(\text{Co}^{3+}\) ion, three electrons are removed, starting with the outer 4s electrons first and then one 3d electron, resulting in \([\text{Ar}] 3\text{d}^6\).

1 mark for identifying the correct oxidation state of Co in the complex and deducing its corresponding d6 electronic configuration.

Hot, concentrated, acidified potassium manganate(VII) cleaves the carbon-carbon double bond in alkenes. Cyclohexene is a cyclic alkene with formula \(\text{C}_6\text{H}_{10}\). Cleavage of its single double bond opens the ring, converting both alkene carbons into carboxylic acid groups and forming hexanedioic acid (a dicarboxylic acid) as the sole organic product. 1-methylcyclopentene yields a keto-acid. Hexa-1,4-diene yields multiple organic fragments. 2,3-dimethylbut-2-ene yields a ketone.

1 mark for identifying cyclohexene as the only cyclic alkene of molecular formula C6H10 that produces a single dicarboxylic acid upon oxidative cleavage.

First, write the half-equations: \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\) and \(\text{Sn}^{2+} \rightarrow \text{Sn}^{4+} + 2\text{e}^-\). Combining these shows that 1 mole of dichromate reacts with 3 moles of tin(II). Calculate the moles of \(\text{Sn}^{2+}\): \(n(\text{Sn}^{2+}) = 0.0300 \text{ dm}^3 \times 0.0500 \text{ mol dm}^{-3} = 1.50 \times 10^{-3} \text{ mol}\). Calculate the moles of \(\text{Cr}_2\text{O}_7^{2-}\) required: \(1.50 \times 10^{-3} / 3 = 5.00 \times 10^{-4} \text{ mol}\). Finally, calculate the volume of dichromate: \(\text{Volume} = 5.00 \times 10^{-4} \text{ mol} / 0.0100 \text{ mol dm}^{-3} = 0.0500 \text{ dm}^3 = 50.0 \text{ cm}^3\).

1 mark for the correct stoichiometric calculation leading to the volume of 50.0 cm3.

Rearrange the rate equation to express the rate constant: \(k = \frac{\text{rate}}{[\text{A}]^2 [\text{B}]^{-1}}\). Substitute the units: \("[k] = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})^{-1}} = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{\text{mol dm}^{-3}} = \text{s}^{-1}"\).

1 mark for performing the correct dimensional analysis to determine the units as s^-1.

The equation for the combustion of propane is: \(\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}\). The enthalpy of combustion of C(graphite) is equal to the enthalpy of formation of \(\text{CO}_2\text{(g)}\), and the enthalpy of combustion of \(\text{H}_2\text{(g)}\) is equal to the enthalpy of formation of \(\text{H}_2\text{O(l)}\). Using Hess's Law: \(\Delta H^\theta_c = [3 \times \Delta H^\theta_f(\text{CO}_2) + 4 \times \Delta H^\theta_f(\text{H}_2\text{O})] - [\Delta H^\theta_f(\text{C}_3\text{H}_8)] = [3 \times (-393.5) + 4 \times (-285.8)] - [-103.8] = -2323.7 + 103.8 = -2219.9 \text{ kJ mol}^{-1}\), which is approximately \(-2220 \text{ kJ mol}^{-1}\).

1 mark for setting up the correct Hess's law equation and calculating the correct value of -2220 kJ mol-1.

Down Group 2, the size (ionic radius) of the \(M^{2+}\) cation increases. Because the ionic charge remains \(+2\), the charge density of the cation decreases. This causes the cation to have less polarising power, meaning it distorts the electron cloud of the nitrate anion less. This leaves the covalent bonds within the nitrate anion more intact, requiring higher temperatures (more thermal energy) to decompose the nitrate. Thus, thermal stability increases down the group.

1 mark for identifying that thermal stability increases down the group and explaining it via the decreasing polarising power of the larger cation.

Let the hydrocarbon be C\(_x\)H\(_y\). First, aqueous NaOH absorbs carbon dioxide, CO\(_2\). The decrease in volume upon passing through NaOH is 80 - 50 = 30 cm\(^3\). Thus, 30 cm\(^3\) of CO\(_2\) is produced. Since 10 cm\(^3\) of the hydrocarbon was used, x = 30 / 10 = 3. Next, the remaining 50 cm\(^3\) of gas is unreacted oxygen. The volume of oxygen reacted is 100 - 50 = 50 cm\(^3\). The equation for combustion is C\(_x\)H\(_y\) + (x + y/4)O\(_2\) \(\to\) xCO\(_2\) + (y/2)H\(_2\)O. The ratio of O\(_2\) reacted to hydrocarbon is 50 / 10 = 5. Therefore, x + y/4 = 5. Since x = 3, we have 3 + y/4 = 5, which gives y/4 = 2, so y = 8. The molecular formula is C\(_3\)H\(_8\).

1 mark: Correctly identifies the volume of CO\(_2\) as 30 cm\(^3\) to find x = 3, and the volume of O\(_2\) reacted as 50 cm\(^3\) to find y = 8, leading to option D.

Concentrated sulfuric acid acts as an acid with NaCl and NaF, producing misty fumes of HCl and HF, respectively, without redox occurring. With NaBr and NaI, the halide ions are oxidized because they are strong reducing agents. Iodide is strong enough to reduce sulfuric acid to sulfur (a yellow solid) and hydrogen sulfide, while iodide is oxidized to iodine (purple vapour). Thus, sulfuric acid acts as an oxidizing agent, making C the correct statement.

1 mark: Correctly identifies that concentrated sulfuric acid acts as an oxidizing agent in the reaction with sodium iodide, producing iodine and sulfur.

In the presence of ligands, the d-orbitals of a transition metal ion are split into two groups of different energy. When visible light falls on the complex, an electron absorbs a photon of energy corresponding to the difference in energy between the split d-orbitals (\(\Delta E = hf\)) and is promoted from the lower level to the higher level. The light that is not absorbed is transmitted or reflected, which we perceive as the complementary colour.

1 mark: Correctly identifies d-orbital splitting by ligands and the absorption of light during the promotion of an electron as the cause of colour.

Cold, dilute, alkaline KMnO\(_4\) oxidizes alkenes to diols by adding a hydroxyl group (-OH) to each carbon of the double bond. Pent-2-ene is CH\(_3\)CH=CHCH\(_2\)CH\(_3\). Reaction with cold KMnO\(_4\) produces pentane-2,3-diol, CH\(_3\)CH(OH)CH(OH)CH\(_2\)CH\(_3\). In this molecule, C2 is bonded to four different groups (-H, -OH, -CH\(_3\), -CH(OH)CH\(_2\)CH\(_3\)), and C3 is also bonded to four different groups (-H, -OH, -CH\(_2\)CH\(_3\), -CH(OH)CH\(_3\)). Both C2 and C3 are chiral, meaning the diol has two chiral carbon atoms. None of the other isomers form a diol with two chiral carbons.

1 mark: Deduces that pent-2-ene undergoes dihydroxylation to form pentane-2,3-diol, which has two chiral centres.

Let us calculate the change in oxidation state for each process: In VO\(_2^+\), V is +5; it is reduced to V\(^{2+}\) (+2), a change of 3. In Cr\(^{3+}\), Cr is +3; it is oxidized to Cr\(_2\)O\(_7^{2-}\) (+6), a change of 3. In MnO\(_4^-\), Mn is +7; it is reduced to Mn\(^{2+}\) (+2), a change of 5. In Fe\(^{2+}\), Fe is +2; it is oxidized to Fe\(^{3+}\) (+3), a change of 1. The greatest change in oxidation state occurs in process C (change of 5).

1 mark: Correctly calculates all oxidation state changes and identifies C as the largest change.

The original rate is given by r = k[A][B]\(^2\). When the concentration of A is doubled, [A]\(_{\text{new}}\) = 2[A]. When the concentration of B is halved, [B]\(_{\text{new}}\) = 0.5[B]. Substituting these into the rate equation gives: rate\(_{\text{new}}\) = k(2[A])(0.5[B])\(^2\) = k \(\times\) 2[A] \(\times\) 0.25[B]\(^2\) = 0.5 \(\times\) k[A][B]\(^2\) = 0.5r. Therefore, the rate is halved.

1 mark: Uses the rate equation to deduce that doubling [A] and halving [B] results in a rate that is 0.5 times the original.

Silicon has a giant molecular (covalent) structure with strong covalent bonds extending in three dimensions, which require a massive amount of energy to break, giving it the highest melting point of Period 3. Chlorine (Cl\(_2\)) is a diatomic molecule with more electrons than monatomic Argon (Ar), so chlorine has stronger London dispersion forces and a higher boiling point than argon. Aluminium has more delocalised electrons per atom than sodium, hence higher conductivity. First ionisation energy generally increases across the period with some drops, so it is not continuous.

1 mark: Recognises that silicon has the highest melting point due to its giant covalent lattice structure.

By Hess's law, the enthalpy change of formation of a compound can be calculated using enthalpies of combustion of the reactants and products: \(\Delta H_f^\ominus\)[CH\(_3\)OH(l)] = \(\Delta H_c^\ominus\)[C(s)] + 2 \(\times\) \(\Delta H_c^\ominus\)[H\(_2\)(g)] - \(\Delta H_c^\ominus\)[CH\(_3\)OH(l)]. Substituting the values: \(\Delta H_f^\ominus\) = -394 + 2(-286) - (-726) = -394 - 572 + 726 = -240 kJ mol\(^{-1}\).

1 mark: Correctly applies Hess's law using the combustion data and computes the result as -240 kJ mol\(^{-1}\).

1. Calculate the moles of \(\text{CO}_2\) produced: Moles = Volume / Molar volume = \(0.120\text{ dm}^3 / 24.0\text{ dm}^3\text{ mol}^{-1} = 0.00500\text{ mol}\). 2. The decomposition equation is: \(M\text{CO}_3(\text{s}) \rightarrow M\text{O}(\text{s}) + \text{CO}_2(\text{g})\). Therefore, the moles of \(M\text{CO}_3\) decomposed is also \(0.00500\text{ mol}\). 3. Calculate the molar mass of \(M\text{CO}_3\): \(M_r = \text{mass} / \text{moles} = 0.738\text{ g} / 0.00500\text{ mol} = 147.6\text{ g mol}^{-1}\). 4. Calculate the relative atomic mass of \(M\): \(A_r(M) = M_r(M\text{CO}_3) - M_r(\text{CO}_3^{2-}) = 147.6 - [12.0 + (3 \times 16.0)] = 147.6 - 60.0 = 87.6\). This relative atomic mass corresponds to Strontium (Sr).

Award 1 mark for the correct option C. Deduce moles of gas, calculate molecular mass of the metal carbonate, and identify the Group 2 metal by its atomic mass.

Using the Born-Haber cycle for \(\text{MgCl}_2(\text{s})\): \(\Delta H_f^\ominus = \Delta H_{at}^\ominus[\text{Mg}(\text{s})] + IE_1[\text{Mg}] + IE_2[\text{Mg}] + 2 \times \Delta H_{at}^\ominus[\text{Cl}_2(\text{g})] + 2 \times EA_1[\text{Cl}] + \Delta H_{latt}^\ominus[\text{MgCl}_2(\text{s})]\). Substituting the given values: \(\Delta H_f^\ominus = 148 + 738 + 1450 + 2(121) + 2(-349) + (-2526) = 148 + 738 + 1450 + 242 - 698 - 2526 = -646\text{ kJ mol}^{-1}\).

Award 1 mark for option B. Method: correctly apply stoichiometric coefficients (multiply chlorine atomisation and electron affinity by 2) and sum the terms according to Hess's law.

1. Determine the order of reaction: [A] is first order (doubling [A] doubles rate); [B] is second order (doubling [B] quadruples rate); [C] is zero order (doubling [C] has no effect). 2. Write the rate equation: Rate = \(k[\text{A}][\text{B}]^2\). 3. Calculate \(k\) using initial data: \(2.0 \times 10^{-4} = k(0.10)(0.10)^2 \Rightarrow 2.0 \times 10^{-4} = k(1.0 \times 10^{-3}) \Rightarrow k = 0.20\). 4. Units of \(k\) = \(\text{mol dm}^{-3}\text{ s}^{-1} / ((\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2) = \text{dm}^6\text{ mol}^{-2}\text{ s}^{-1}\).

Award 1 mark for option B. Deduce the order with respect to each reactant, set up the rate equation, calculate the numerical value of k, and determine the units of k.

1. Saturated compound \(X\) resisting oxidation with acidified potassium dichromate(VI) must be a tertiary alcohol. 2. Among the options, only 2-methylbutan-2-ol is a tertiary alcohol (3-methylbutan-2-ol is secondary; 2-methylbutan-1-ol and 2,2-dimethylpropan-1-ol are primary). 3. Dehydration of 2-methylbutan-2-ol yields exactly two structural isomers: 2-methylbut-1-ene and 2-methylbut-2-ene.

Award 1 mark for option A. Identify that resistance to oxidation confirms the tertiary alcohol structure, and verify the structural isomers formed upon dehydration.

1. The observation of purple fumes and a dark grey solid indicates the production of iodine, \(\text{I}_2\). Thus, the sodium halide must be sodium iodide, \(\text{NaI}\). 2. Concentrated sulfuric acid oxidizes the iodide ions (from hydrogen iodide) to iodine, which means the sulfuric acid acts as an oxidising agent.

Award 1 mark for option A. Identify the presence of iodine from the physical observations and deduce the redox role of sulfuric acid.

1. Establish the configuration of \(\text{Co}^{2+}\): \(\text{Co}\) is \([\text{Ar}] 3d^7 4s^2\), so \(\text{Co}^{2+}\) is \([\text{Ar}] 3d^7\). Following Hund's rule, a \(3d^7\) subshell contains 3 unpaired electrons. 2. Evaluate the options: \(\text{V}^{3+}\) is \([\text{Ar}] 3d^2\) (2 unpaired); \(\text{Cr}^{3+}\) is \([\text{Ar}] 3d^3\) (3 unpaired); \(\text{Fe}^{3+}\) is \([\text{Ar}] 3d^5\) (5 unpaired); \(\text{Ni}^{2+}\) is \([\text{Ar}] 3d^8\) (2 unpaired). Thus, \(\text{Cr}^{3+}\) also has exactly 3 unpaired d-electrons.

Award 1 mark for option B. Work out the d-electron configuration and apply Hund's rule to find the number of unpaired electrons in both species.

1. Thermal stability of Group 2 nitrates increases down the group because larger cations have lower charge densities and polarise the nitrate ion less. Since \(A\) decomposes at a lower temperature, metal \(A\) must be higher in the group than metal \(B\). 2. Solubility of Group 2 hydroxides increases down the group, so the pH of their saturated solutions increases down the group. Since the hydroxide of \(A\) has a lower pH, it is less soluble, confirming that \(A\) is higher in the group than \(B\). 3. Only option C (where \(A\) is Magnesium and \(B\) is Barium) satisfies this requirement.

Award 1 mark for option C. Connect thermal stability and hydroxide solubility trends to the vertical positions of the metals in Group 2.

1. The methyl group on methylbenzene is an electron-donating group, which activates the ring and directs incoming electrophiles to the 2- and 4-positions (2,4-directing). 2. The carboxyl group (-COOH) on benzoic acid is electron-withdrawing, deactivating the ring (requiring a higher reaction temperature of 60 °C) and directing substitution to the 3-position (3-directing). 3. Therefore, the major monosubstituted product Y is 3-nitrobenzoic acid.

Award 1 mark for option A. State correct directing effects for both activating (-CH3) and deactivating (-COOH) functional groups and identify the correct nitration product.

The decomposition equation for a metal(II) nitrate is:
\(2M(\text{NO}_3)_2(\text{s}) \rightarrow 2MO(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})\)

From the stoichiometry, 2 moles of \(M(\text{NO}_3)_2\) produce 5 moles of gas (4 moles of \(\text{NO}_2\) and 1 mole of \(\text{O}_2\)).

Moles of gas collected = \(\frac{600\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.025\text{ mol}\).

Moles of \(M(\text{NO}_3)_2\) decomposed = \(0.025\text{ mol} \times \frac{2}{5} = 0.010\text{ mol}\).

Molar mass of \(M(\text{NO}_3)_2 = \frac{1.88\text{ g}}{0.010\text{ mol}} = 188\text{ g mol}^{-1}\).

To find the relative atomic mass of metal \(M\):
\(A_r(M) + 2 \times [14.0 + (3 \times 16.0)] = 188\)
\(A_r(M) + 124.0 = 188\)
\(A_r(M) = 64.0\)

This is closest to the relative atomic mass of copper (63.5).

Award 1 mark for the correct answer (C).
- Method: Calculate moles of gas (0.025 mol), relate to moles of reactant using correct stoichiometry (1:2.5 ratio to find 0.010 mol of nitrate), determine molar mass of nitrate (188 g/mol) and relative atomic mass of M (~64), identifying it as Copper.

When solid sodium bromide, \(\text{NaBr}\), reacts with concentrated sulfuric acid, hydrogen bromide gas (\(\text{HBr}\)) is produced. This gas fumes in moist air.

When \(\text{HBr}\) dissolves in water, it forms hydrobromic acid. Reaction of this acidic solution with aqueous silver nitrate produces a cream precipitate of silver bromide, \(\text{AgBr}\).

Silver bromide (\(\text{AgBr}\)) is insoluble in dilute aqueous ammonia, but dissolves in concentrated aqueous ammonia.

Award 1 mark for identifying the correct halide ion based on physical characteristics and reactions of halide salts and silver halide precipitates.

When ligands coordinate to a transition metal ion, the electric field of the ligands splits the d-orbitals into two sets of non-degenerate energy levels. Electrons can transition between these lower and higher energy d-orbitals (d-d transitions) by absorbing specific wavelengths of light from the visible region. The colour observed is the complementary colour of the absorbed light.

Award 1 mark for identifying that ligand-induced d-orbital splitting and subsequent d-d electron absorption of visible light explains the color of complexes.

Hot, concentrated, acidified potassium manganate(VII) cleaves the double bond of an alkene.

Since propanone is a ketone, the carbon atoms on both sides of the double bond must be bonded to two other alkyl groups (i.e., they are disubstituted at each alkenyl carbon).

If the only organic product is propanone, the alkene must be symmetrical and yield two molecules of propanone upon cleavage.

Therefore, the original alkene must be 2,3-dimethylbut-2-ene:
\((\text{CH}_3)_2\text{C}=\text{C}(\text{CH}_3)_2 \rightarrow 2\text{ CH}_3\text{COCH}_3\).

Award 1 mark for analyzing the oxidative cleavage products to deduce the identity of the symmetrical alkene as 2,3-dimethylbut-2-ene.

Let's determine the oxidation states of vanadium in the species:
- In \(\text{VO}_2^+\): \(x + 2(-2) = +1 \Rightarrow x = +5\)
- In \(\text{VO}^{2+}\): \(x + (-2) = +2 \Rightarrow x = +4\)
- In \(\text{V}^{3+}\): oxidation state is \(+3\)
- In \(\text{V}^{2+}\): oxidation state is \(+2\)

Now look at the choices:
A) \(\text{VO}_2^+ \rightarrow \text{VO}^{2+}\): +5 to +4 (decrease of 1)
B) \(\text{VO}^{2+} \rightarrow \text{V}^{3+}\): +4 to +3 (decrease of 1)
C) \(\text{VO}^{2+} \rightarrow \text{V}^{2+}\): +4 to +2 (decrease of 2) - Correct!
D) \(\text{VO}_2^+ \rightarrow \text{V}^{2+}\): +5 to +2 (decrease of 3)

Award 1 mark for calculating the correct oxidation states for vanadium in each species and finding the transition corresponding to a decrease of exactly 2.

1. Determine the order with respect to each reactant:
- Comparing Expt 1 and Expt 2: \([A]\) is doubled while \([B]\) and \([C]\) remain constant. The rate increases by a factor of 4 (\(\frac{8.0 \times 10^{-4}}{2.0 \times 10^{-4}} = 4\)). Thus, the order with respect to \(A\) is 2.
- Comparing Expt 1 and Expt 3: \([B]\) is doubled while \([A]\) and \([C]\) remain constant. The rate does not change. Thus, the order with respect to \(B\) is 0.
- Comparing Expt 1 and Expt 4: \([C]\) is doubled while \([A]\) and \([B]\) remain constant. The rate doubles (\(\frac{4.0 \times 10^{-4}}{2.0 \times 10^{-4}} = 2\)). Thus, the order with respect to \(C\) is 1.

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

3. Determine the units of the rate constant \(k\):
\(k = \frac{\text{Rate}}{[A]^2[C]} = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{dm}^6\text{mol}^{-2}\text{ s}^{-1}\).

Award 1 mark for finding the orders of reaction for A, B, and C, formulating the correct rate equation, and correctly deducing the units of k.

Melting point depends on structure and bonding:
- Aluminium (metallic) has a high melting point, but it is lower than that of silicon.
- Silicon has a giant covalent macromolecular structure. To melt silicon, many strong, localized covalent bonds must be broken, which requires a huge amount of thermal energy.
- Phosphorus (\(P_4\)) and sulfur (\(S_8\)) are simple molecular elements with low melting points, as only weak London dispersion forces between molecules need to be overcome.

Award 1 mark for correctly identifying Silicon as having the highest melting point in Period 3 due to its giant covalent structure requiring high energy to break covalent bonds.

Calculate the energy required to break all reactant bonds:
- In 1 mole of \(\text{CH}_3\text{CH}_2\text{OH(g)}\):
- 5 \(\times C-H\) bonds = \(5 \times 413 = 2065\text{ kJ}\)
- 1 \(\times C-C\) bond = \(1 \times 347 = 347\text{ kJ}\)
- 1 \(\times C-O\) bond = \(1 \times 358 = 358\text{ kJ}\)
- 1 \(\times O-H\) bond = \(1 \times 464 = 464\text{ kJ}\)
- In 3 moles of \(\text{O}_2\):
- 3 \(\times O=O\) bonds = \(3 \times 498 = 1494\text{ kJ}\)
Total energy input (bonds broken) = \(2065 + 347 + 358 + 464 + 1494 = 4728\text{ kJ}\).

Calculate the energy released in forming product bonds:
- In 2 moles of \(\text{CO}_2\):
- 4 \(\times C=O\) bonds = \(4 \times 805 = 3220\text{ kJ}\)
- In 3 moles of \(\text{H}_2\text{O}\):
- 6 \(\times O-H\) bonds = \(6 \times 464 = 2784\text{ kJ}\)
Total energy output (bonds formed) = \(3220 + 2784 = 6004\text{ kJ}\).

Enthalpy change of combustion:
\(\Delta H = \text{Energy input} - \text{Energy output} = 4728 - 6004 = -1276\text{ kJ mol}^{-1}\).

Award 1 mark for the correct enthalpy change calculation (\(-1276\text{ kJ mol}^{-1}\)).
- Check bonds broken (4728) and bonds formed (6004).
- Correctly use the formula: Bonds Broken - Bonds Formed.

Detailed working:
(a) Balancing the atoms gives: \(\text{Cu}_2\text{CO}_3(\text{OH})_2(\text{s}) \rightarrow 2\text{CuO}(\text{s}) + \text{O}_2(\text{g}) + \text{H}_2\text{O}(\text{g})\).
(b)(i) \(n(\text{reactant}) = 4.42 / 221.0 = 0.0200\text{ mol}\). From stoichiometry, \(n(\text{CuO}) = 2 \times 0.0200 = 0.0400\text{ mol}\). \(m(\text{CuO}) = 0.0400 \times 79.5 = 3.18\text{ g}\).
(b)(ii) Total moles of gas = \(n(\text{CO}_2) + n(\text{H}_2\text{O}) = 0.0200 + 0.0200 = 0.0400\text{ mol}\). Volume = \(0.0400 \times 24.0 = 0.960\text{ dm}^3\).
(c)(i) \(\text{CO}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\).
(c)(ii) \(\text{I}^-\right.\) acts as a reducing agent because its oxidation number increases from -1 in \(\text{I}^-\right.\) to 0 in \(\text{I}_2\) (it reduces \(\text{Cu}^{2+}\) to \(\text{Cu}^{+}\)).
(c)(iii) \(n(\text{reactant}) = 0.5525 / 221.0 = 0.00250\text{ mol}\). Thus, \(n(\text{Cu}^{2+}) = 2 \times 0.00250 = 0.00500\text{ mol}\). From the equations, \(2\text{Cu}^{2+} \equiv \text{I}_2 \equiv 2\text{S}_2\text{O}_3^{2-}\), so mole ratio \(\text{Cu}^{2+} : \text{S}_2\text{O}_3^{2-} = 1 : 1\). Therefore, \(n(\text{S}_2\text{O}_3^{2-}) = 0.00500\text{ mol}\). \(V = n / C = 0.00500 / 0.100 = 0.0500\text{ dm}^3 = 50.0\text{ cm}^3\).

(a) 1 mark for correct formulas of products, 1 mark for correct balancing.
(b)(i) 1 mark for calculating 0.0200 mol of reactant and 0.0400 mol of CuO, 1 mark for mass 3.18 g (accept 3.2 g).
(b)(ii) 1 mark for identifying that 1 mol of reactant yields 2 mol of gas. 1 mark for total gas moles of 0.0400 mol. 1 mark for 0.960 dm3 (or 960 cm3).
(c)(i) 1 mark for correct species and balancing. 1 mark for correct state symbols.
(c)(ii) 1 mark for stating it is a reducing agent. 1 mark for explanation referring to oxidation states change from -1 to 0 (or copper being reduced from +2 to +1).
(c)(iii) 1 mark for calculating moles of copper reactant as 0.00250 mol and Cu2+ as 0.00500 mol. 1 mark for establishing the 1:1 mole ratio between Cu2+ and S2O32-. 1 mark for 0.00500 mol of S2O32-. 1 mark for volume 50.0 cm3.

Detailed working:
(a)(i) Volatility decreases down the group. This is because molecular size increases, meaning there are more electrons per molecule. Consequently, the temporary dipole-induced dipole (London dispersion) forces between the simple molecules become stronger, requiring more energy to overcome.
(a)(ii) \(3\text{Cl}_2 + 6\text{NaOH} \rightarrow 5\text{NaCl} + \text{NaClO}_3 + 3\text{H}_2\text{O}\). Chlorine is reduced to -1 in \(\text{NaCl}\) and oxidized to +5 in \(\text{NaClO}_3\).
(b)(i) \(\text{H}_2\text{SO}_4\) is a strong oxidizing agent that oxidizes \(\text{HBr}\) to \(\text{Br}_2\). Equation: \(2\text{HBr} + \text{H}_2\text{SO}_4 \rightarrow \text{Br}_2 + \text{SO}_2 + 2\text{H}_2\text{O}\).
(b)(ii) The atomic radius of the halogen increases down the group, causing the \(\text{H-X}\) bond length to increase. Longer bonds are weaker and have lower bond energies, so they require less thermal energy to break, decreasing thermal stability.
(c) Chloride ions form a white precipitate of \(\text{AgCl}\), which is soluble in dilute aqueous ammonia. Bromide ions form a cream precipitate of \(\text{AgBr}\), which is insoluble in dilute aqueous ammonia but soluble in concentrated ammonia.

(a)(i) 1 mark for stating volatility decreases. 1 mark for identifying instantaneous dipole-induced dipole forces between molecules. 1 mark for stating that forces increase with increasing number of electrons down the group.
(a)(ii) 1 mark for the correct balanced equation. 1 mark for identifying -1 oxidation state in NaCl. 1 mark for identifying +5 oxidation state in NaClO3.
(b)(i) 1 mark for stating H2SO4 is an oxidizing agent (or HBr is oxidized). 1 mark for correct species in equation: 2HBr + H2SO4 -> Br2 + SO2 + 2H2O. 1 mark for correct balancing.
(b)(ii) 1 mark for stating thermal stability decreases. 1 mark for stating that halogen atomic size increases down the group. 1 mark for explaining that the H-X bond length increases and therefore bond strength/energy decreases.
(c) 1 mark for chloride forms white ppt that dissolves in dilute NH3. 1 mark for bromide forms cream ppt. 1 mark for bromide ppt remains insoluble in dilute NH3.

Detailed working:
(a)(i) Hot, concentrated oxidation products propanone and ethanoic acid indicate the alkene double bond is between C2 and C3 of a 5-carbon skeleton, and that C2 has two methyl groups. Thus, Compound **A** is 2-methylbut-2-ene.
(a)(ii) It does not show geometrical isomerism because C2 is bonded to two identical groups (two methyl groups).
(b)(i) The purple color of the manganate(VII) solution is decolorized (becomes colorless).
(b)(ii) Cold dilute KMnO4 oxidizes the alkene to a diol. Compound **B** is 2-methylbutane-2,3-diol, structural formula: \((\text{CH}_3)_2\text{C}(\text{OH})\text{CH}(\text{OH})\text{CH}_3\).
(c)(i) Electrophilic addition of HBr yields the major product **C**, 2-bromo-2-methylbutane (from the more stable tertiary carbocation), and minor product **D**, 2-bromo-3-methylbutane (from the secondary carbocation).
(c)(ii) The mechanism proceeds via: 1) Attack of the pi electrons of the C=C double bond on the H delta+ of H-Br, with simultaneous cleavage of the H-Br bond, forming a tertiary carbocation intermediate \((\text{CH}_3)_2\text{C}^+-\text{CH}_2\text{CH}_3\) and a bromide ion. 2) Attack of the bromide ion's lone pair on the carbocation to form the product.
(c)(iii) The tertiary carbocation intermediate is more stable than the alternative secondary carbocation because the three electron-donating alkyl groups reduce the charge density on the positive carbon atom more effectively than two alkyl groups.

(a)(i) 1 mark for IUPAC name 2-methylbut-2-ene. 1 mark for correct skeletal structure.
(a)(ii) 1 mark for stating it does not exhibit geometrical isomerism. 1 mark for explanation (one of the C=C carbons is bonded to two identical methyl groups).
(b)(i) 1 mark for purple to colorless.
(b)(ii) 1 mark for correct structural formula of 2-methylbutane-2,3-diol. 1 mark for functional group class: diol (accept glycol or alcohol).
(c)(i) 1 mark for skeletal structure of 2-bromo-2-methylbutane (C). 1 mark for skeletal structure of 2-bromo-3-methylbutane (D).
(c)(ii) 1 mark for curly arrow from double bond to H and dipole charges on H-Br. 1 mark for curly arrow breaking H-Br bond. 1 mark for drawing the correct tertiary carbocation intermediate. 1 mark for curly arrow from lone pair on Br- to the C+ of the carbocation.
(c)(iii) 1 mark for stating C is formed via a tertiary carbocation and D via a secondary carbocation. 1 mark for explaining that tertiary carbocations are more stable due to the positive inductive effect of three alkyl groups dispersing the positive charge.

Detailed working:
(a) The standard enthalpy change of combustion, \(\Delta H_c^\ominus\), is the enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions of 100 kPa and 298 K.
(b)(i) \(q = m c \Delta T = 150.0 \times 4.18 \times (54.5 - 21.5) = 150.0 \times 4.18 \times 33.0 = 20691\text{ J} = 20.7\text{ kJ}\) (or 20.69 kJ).
(b)(ii) \(M_r(\text{C}_2\text{H}_5\text{OH}) = 46.0\text{ g mol}^{-1}\). \(n = 0.920 / 46.0 = 0.0200\text{ mol}\).
(b)(iii) \(\Delta H_c = -q / n = -20.691 / 0.0200 = -1034.55\text{ kJ mol}^{-1}\), which rounds to \(-1030\text{ kJ mol}^{-1}\) (3 sig fig) or \(-1035\text{ kJ mol}^{-1}\).
(b)(iv) Incomplete combustion of ethanol (forming soot or carbon monoxide), or evaporation of ethanol from the wick of the burner.
(c)(i) \(3\text{C}(\text{s, graphite}) + 4\text{H}_2(\text{g}) \rightarrow \text{C}_3\text{H}_8(\text{g})\) (accept \(\text{s}\) instead of \(\text{s, graphite}\)).
(c)(ii) Hess's Law states that the enthalpy change of a reaction is independent of the pathway or route taken, provided the initial and final states are the same.
(c)(iii) Combustion equation: \(\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})\).
\(\Delta H_c^\ominus = \sum \Delta H_f^\ominus[\text{products}] - \sum \Delta H_f^\ominus[\text{reactants}] = [3(-393.5) + 4(-285.8)] - [-103.8] = [-1180.5 - 1143.2] + 103.8 = -2323.7 + 103.8 = -2219.9\text{ kJ mol}^{-1}\) (rounds to \(-2220\text{ kJ mol}^{-1}\)).

(a) 1 mark for 'one mole of substance is completely burned in oxygen'. 1 mark for 'under standard conditions' (or 298 K and 100 kPa).
(b)(i) 1 mark for correct temperature difference of 33.0 K. 1 mark for correct calculation of 20.7 kJ (or 20.69 kJ).
(b)(ii) 1 mark for 0.0200 mol.
(b)(iii) 1 mark for value around 1030 to 1035 kJ mol-1. 1 mark for negative sign.
(b)(iv) 1 mark for incomplete combustion or evaporation of ethanol.
(c)(i) 1 mark for balanced equation. 1 mark for state symbols (s/s,graphite, g, g).
(c)(ii) 1 mark for 'enthalpy change is independent of the route'. 1 mark for 'provided the initial and final states are the same'.
(c)(iii) 1 mark for balanced combustion equation (can be implicit). 1 mark for Hess's Law expression: 3(-393.5) + 4(-285.8) - (-103.8). 1 mark for correct final answer: -2219.9 or -2220 kJ mol-1.

(a) Moles of \(\text{NaOH} = c \times V = 0.100\text{ mol dm}^{-3} \times \frac{26.00}{1000}\text{ dm}^3 = 2.60 \times 10^{-3}\text{ mol}\).

(b) Since \(\text{HCl}\) and \(\text{NaOH}\) react in a \(1:1\) ratio:
Moles of \(\text{HCl}\) in \(25.0\text{ cm}^3\) portion = \(2.60 \times 10^{-3}\text{ mol}\).

(c) Moles of excess \(\text{HCl}\) in \(250.0\text{ cm}^3 = 2.60 \times 10^{-3}\text{ mol} \times \frac{250.0}{25.0} = 0.0260\text{ mol}\).

(d) Initial moles of \(\text{HCl}\) added = \(1.00\text{ mol dm}^{-3} \times \frac{50.0}{1000}\text{ dm}^3 = 0.0500\text{ mol}\).

(e) Moles of \(\text{HCl}\) reacted = \(0.0500\text{ mol} - 0.0260\text{ mol} = 0.0240\text{ mol}\).

(f) From the stoichiometry, 1 mole of \(\text{CaCO}_3\) reacts with 2 moles of \(\text{HCl}\):
Moles of \(\text{CaCO}_3 = \frac{0.0240}{2} = 0.0120\text{ mol}\).
Mass of \(\text{CaCO}_3 = n \times M_r = 0.0120\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.2012\text{ g}\).

(g) Percentage by mass of \(\text{CaCO}_3 = \frac{1.2012\text{ g}}{1.50\text{ g}} \times 100\% = 80.08\% \approx 80.1\%\) (to 3 significant figures).

(h) If the sample was damp, the recorded mass of the sample (\(1.50\text{ g}\)) would include water, meaning there is less actual seashell solid weighed. This reduces the actual moles of \(\text{CaCO}_3\) present, which decreases the moles of acid reacted, resulting in a higher excess titre and a lower calculated mass of \(\text{CaCO}_3\). Thus, the calculated percentage purity would decrease.

1. Correct calculation of moles of \(\text{NaOH}\) [1 Mark]
2. Correct deduction of moles of excess acid in \(25.0\text{ cm}^3\) and scaling to \(250.0\text{ cm}^3\) [2 Marks]
3. Correct calculation of initial moles of acid [1 Mark]
4. Correct determination of moles of acid reacted with seashell [1 Mark]
5. Correct application of stoichiometry (dividing by 2) to find moles of \(\text{CaCO}_3\) [1 Mark]
6. Correct calculation of mass of \(\text{CaCO}_3\) and final percentage purity to 3 significant figures [2 Marks]
7. Correct explanation of the effect of dampness (water adds inactive mass, leading to a smaller mass of carbonate, less acid consumed, larger titre, and lower calculated percentage purity) [2 Marks]

(a) Initial temperature = \(21.5\ ^\circ\text{C}\).
Maximum temperature reached = \(29.5\ ^\circ\text{C}\).
Temperature change, \(\Delta T = 29.5 - 21.5 = +8.0\ ^\circ\text{C}\).

(b) \(q = m \cdot c \cdot \Delta T\)
\(m = 50.0\text{ g}\) (since \(50.0\text{ cm}^3\) has density \(1.00\text{ g cm}^{-3}\))
\(q = 50.0\text{ g} \times 4.18\text{ J g}^{-1}\ ^\circ\text{C}^{-1} \times 8.0\ ^\circ\text{C} = 1672\text{ J} = 1.672\text{ kJ}\).

(c) Moles of \(\text{Na}_2\text{CO}_3 = \frac{3.18\text{ g}}{106.0\text{ g mol}^{-1}} = 0.0300\text{ mol}\).

(d) \(\Delta H = -\frac{q}{\text{moles}} = -\frac{1.672\text{ kJ}}{0.0300\text{ mol}} = -55.73\text{ kJ mol}^{-1} \approx -55.7\text{ kJ mol}^{-1}\).

(e) Because two readings are used to find \(\Delta T\) (initial and final), the absolute uncertainty in \(\Delta T\) is \(2 \times 0.1 = \pm0.2\ ^\circ\text{C}\).
Percentage uncertainty = \(\frac{0.2}{8.0} \times 100\% = 2.5\%\).

(f) Improvement: Place a lid on the polystyrene cup or wrap the cup in cotton wool / lagging.
Effect: This minimizes heat loss to the surroundings, resulting in a larger temperature rise (\(\Delta T\)) and making the calculated enthalpy change of reaction more exothermic (more negative).

1. Correct values for initial temperature, maximum temperature, and \(\Delta T\) (+8.0 °C) [1 Mark]
2. Correct calculation of \(q\) (1670 J or 1.67 kJ) with correct units [2 Marks]
3. Correct calculation of moles of carbonate (0.0300 mol) [1 Mark]
4. Correct calculation of \(\Delta H\) with negative sign (\(-55.7\text{ kJ mol}^{-1}\)) [2 Marks]
5. Correct calculation of percentage uncertainty in temperature change (2.5%) accounting for two thermometer readings [2 Marks]
6. Identifies a suitable experimental modification to reduce heat loss (e.g. lid / lagging) [1 Mark]
7. Explains that reducing heat loss increases \(\Delta T\) and makes the calculated \(\Delta H\) more negative (more exothermic) [1 Mark]

(a) Concentrations of \(\text{Na}_2\text{S}_2\text{O}_3\):
- Exp 1: \(0.100\text{ mol dm}^{-3} \times \frac{40.0}{60.0} = 0.0667\text{ mol dm}^{-3}\)
- Exp 2: \(0.100\text{ mol dm}^{-3} \times \frac{30.0}{60.0} = 0.0500\text{ mol dm}^{-3}\)
- Exp 3: \(0.100\text{ mol dm}^{-3} \times \frac{20.0}{60.0} = 0.0333\text{ mol dm}^{-3}\)
- Exp 4: \(0.100\text{ mol dm}^{-3} \times \frac{10.0}{60.0} = 0.0167\text{ mol dm}^{-3}\)

(b) Rates (\(\frac{1}{t}\)):
- Exp 1: \(\frac{1}{25.0} = 0.0400\text{ s}^{-1}\)
- Exp 2: \(\frac{1}{33.3} = 0.0300\text{ s}^{-1}\)
- Exp 3: \(\frac{1}{50.0} = 0.0200\text{ s}^{-1}\)
- Exp 4: \(\frac{1}{100.0} = 0.0100\text{ s}^{-1}\)

(c) The graph would be a straight line passing through the origin, as \(\frac{1}{t}\) is directly proportional to concentration.

(d) The order of reaction is 1 (first-order). This is because doubling the concentration of sodium thiosulfate (e.g., from Exp 4 to Exp 3: \(0.0167\) to \(0.0333\text{ mol dm}^{-3}\)) doubles the rate of reaction (from \(0.0100\) to \(0.0200\text{ s}^{-1}\)).

(e) Keeping the total volume constant ensures that the concentration of the other reactant (\(\text{HCl}\)) is kept constant, so that any change in rate can be solely attributed to the change in sodium thiosulfate concentration.

(f) For Experiment 4, the reaction is much slower, so the end-point is less distinct because the sulfur precipitate forms very gradually. This increases human reaction-time uncertainty in determining exactly when the cross is completely obscured, leading to a higher human error. (Alternatively, the percentage error in stopwatch timing itself is lower because \(100\text{ s}\) is longer, but the visual judgment error is much larger.)

1. Correct calculations of final concentration of sodium thiosulfate for all 4 experiments [2 Marks]
2. Correct calculations of \(\frac{1}{t}\) to 3 significant figures [2 Marks]
3. Describes or sketches a straight line through the origin [1 Mark]
4. Correctly deduces first-order kinetics with a logical explanation based on proportional changes [2 Marks]
5. Explains the necessity of keeping total volume constant (to control the concentration of hydrochloric acid) [2 Marks]
6. Explains that slower reactions have a less distinct end-point where the cross fades gradually, leading to greater human observation error [1 Mark]

(a)
- Comparing Experiment 1 and Experiment 2: \([Y]\) and \([\text{H}^+]\) are constant. When \([X]\) doubles from 0.10 to 0.20 \(\text{mol dm}^{-3}\), the initial rate doubles from \(1.20 \times 10^{-4}\) to \(2.40 \times 10^{-4}\) \(\text{mol dm}^{-3}\text{ s}^{-1}\). Therefore, the order with respect to \(X\) is 1.
- Comparing Experiment 2 and Experiment 3: \([X]\) and \([\text{H}^+]\) are constant. When \([Y]\) doubles from 0.10 to 0.20 \(\text{mol dm}^{-3}\), the initial rate quadruples from \(2.40 \times 10^{-4}\) to \(9.60 \times 10^{-4}\) \(\text{mol dm}^{-3}\text{ s}^{-1}\). Therefore, the order with respect to \(Y\) is 2.
- Comparing Experiment 1 and Experiment 4: \([X]\) and \([Y]\) are constant. When \([\text{H}^+]\) doubles from 0.10 to 0.20 \(\text{mol dm}^{-3}\), the initial rate remains unchanged. Therefore, the order with respect to \([\text{H}^+]\) is 0.

(b) Rate = \(k[X][Y]^2\)

(c) Using the data from Experiment 1:
\[ 1.20 \times 10^{-4} = k(0.10)(0.10)^2 \]
\[ k = \frac{1.20 \times 10^{-4}}{1.0 \times 10^{-3}} = 0.12 \]
Units of \(k\): \(\text{dm}^6 \text{mol}^{-2} \text{s}^{-1}\)

(d) A possible mechanism must involve 1 molecule of \(X\) and 2 molecules of \(Y\) in the rate-determining step:
- Step 1 (Slow / Rate-determining step):
\[ X + 2Y \rightarrow \text{Intermediate} \]
- Step 2 (Fast):
\[ \text{Intermediate} + 2\text{H}^+ \rightarrow Z + 2\text{H}_2\text{O} \]

(e) An increase in temperature increases the average kinetic energy of the particles. Consequently, a significantly larger fraction of reactant molecules possess kinetic energy greater than or equal to the activation energy, \(E_a\). This leads to a higher frequency of successful, effective collisions, resulting in an exponential increase in the rate constant, \(k\).

(a) [3 marks]:
- 1 mark for order wrt X = 1 with a correct comparison explanation (Exp 1 and 2).
- 1 mark for order wrt Y = 2 with a correct comparison explanation (Exp 2 and 3).
- 1 mark for order wrt H+ = 0 with a correct comparison explanation (Exp 1 and 4).

(b) [1 mark]:
- 1 mark for correct rate equation: Rate = k[X][Y]^2 (allow error-carried-forward from part a).

(c) [3 marks]:
- 2 marks for calculation (1 mark for working, 1 mark for 0.12).
- 1 mark for correct units (dm6 mol-2 s-1).

(d) [2.5 marks]:
- 1.5 marks for proposing a plausible two-step mechanism where the slow step involves only 1 X and 2 Y.
- 1 mark for clearly labeling the slow step as the rate-determining step.

(e) [3 marks]:
- 1 mark for stating that increasing temperature increases the fraction of molecules with energy >= Ea.
- 1 mark for mentioning the frequency of successful collisions increases.
- 1 mark for stating that the rate constant, k, increases.

(a) Standard electrode potential, \(E^\ominus\), is the electromotive force (emf) of a half-cell coupled to a standard hydrogen electrode (SHE) measured under standard conditions of 298 K temperature, 1 atm gas pressure, and all aqueous solutions at \(1.0\text{ mol dm}^{-3}\).

(b) The experimental setup consists of two half-cells connected by a high-resistance voltmeter and a salt bridge:
- Standard Hydrogen Electrode (left): Platinum electrode in contact with \(1.0\text{ mol dm}^{-3}\ \text{H}^+(\text{aq})\) solution, with hydrogen gas bubbled over the electrode at 1 atm.
- \(\text{Fe}^{3+}/\text{Fe}^{2+}\) half-cell (right): Platinum electrode immersed in a solution containing both \(1.0\text{ mol dm}^{-3}\ \text{Fe}^{2+}(\text{aq})\) and \(1.0\text{ mol dm}^{-3}\ \text{Fe}^{3+}(\text{aq})\) ions.
- Connection: A high-resistance voltmeter connects the two platinum electrodes, and a salt bridge filled with saturated \(\text{KNO}_3\) solution links the two solutions.
- Conditions: The entire system is kept at a constant temperature of 298 K.

(c) For the reduction: \(\text{Cu}^{2+}(\text{aq}) + 2e^- \rightleftharpoons \text{Cu}(\text{s})\), the number of transferred electrons \(z = 2\).
Using the Nernst equation:
\[ E = +0.34 + \frac{0.059}{2}\log(0.015) \]
\[ E = +0.34 + 0.0295 \times (-1.8239) \]
\[ E = +0.34 - 0.0538 = +0.286\text{ V} \] (or \(+0.29\text{ V}\))

(d) The cell reaction is: \(2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})\).
\[ E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} \]
\[ E^\ominus_{\text{cell}} = E^\ominus(\text{Fe}^{3+}/\text{Fe}^{2+}) - E^\ominus(\text{I}_2/\text{I}^-) = +0.77\text{ V} - (+0.54\text{ V}) = +0.23\text{ V} \]
Since \(E^\ominus_{\text{cell}}\) is positive (\(E^\ominus_{\text{cell}} > 0\)), \(\text{Fe}^{3+}(\text{aq})\) can oxidize \(\text{I}^-(\text{aq})\) under standard conditions. The reaction is feasible.

(a) [2 marks]:
- 1 mark for voltage/emf measured relative to a standard hydrogen electrode.
- 1 mark for specifying standard conditions (298 K, 1 atm, 1.0 mol dm-3).

(b) [4.5 marks]:
- 1 mark for a correct, labeled diagram of the SHE (Pt electrode, H2 inlet, H+ solution).
- 1 mark for a correct, labeled diagram of the Fe3+/Fe2+ half-cell (Pt electrode, solution of Fe3+ and Fe2+).
- 1 mark for high-resistance voltmeter and salt bridge shown correctly connecting the half-cells.
- 1.5 marks for specifying concentrations of 1.0 mol dm-3 for H+, Fe2+, Fe3+, H2 pressure at 1 atm, and temperature at 298 K.

(c) [3 marks]:
- 1 mark for identifying z = 2.
- 1 mark for substitution and calculating log(0.015) = -1.82.
- 1 mark for the final answer of +0.286 V (or +0.29 V).

(d) [3 marks]:
- 1 mark for stating the correct half-equations and identifying the oxidation/reduction couples.
- 1 mark for calculating E_cell = +0.23 V.
- 1 mark for concluding that the reaction is feasible because E_cell is positive.

(a)
- Transition metals have partially filled d-orbitals.
- Ligands bond to the central metal ion using dative covalent bonds, which splits the degenerate 3d orbitals into two sets of different energy levels (separated by energy gap \(\Delta E\)).
- Electrons absorb a photon of light in the visible spectrum to undergo d-d transition (promotion from lower 3d orbital to higher 3d orbital).
- The complementary color of the light absorbed is observed.

(b)(i) Only the cis isomer exhibits optical isomerism as it is chiral and lacks a plane of symmetry. The 3D structures of the optical isomers are non-superimposable mirror images:
- Left diagram: Cobalt is central. Two Cl- ligands are placed on adjacent octahedral vertices (cis positions). Two 'en' loops connect the remaining positions.
- Right diagram: This is the exact non-superimposable mirror image of the left diagram.

(b)(ii) The trans isomer does not show optical isomerism. In the trans isomer, the two Cl- ligands are located on opposite octahedral vertices (180 degrees apart), giving the molecule a plane of symmetry which makes it achiral.

(c)(i) The ligand exchange reaction is:
\[ [\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{NH}_3 \rightleftharpoons [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{H}_2\text{O} \]
\[ K_{\text{stab}} = \frac{[[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}]}{[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}][\text{NH}_3]^4} \]

(c)(ii) Adding excess ammonia causes a ligand exchange where ammonia molecules replace water ligands. Because the value of \(K_{\text{stab}}\) is extremely large (\(1.2 \times 10^{13}\)), the position of equilibrium lies heavily to the right, converting the light-blue aqueous hexaaquacopper(II) complex completely into the deep-blue tetraaminediaquacopper(II) complex.

(a) [4 marks]:
- 1 mark for d-orbital splitting by ligands.
- 1 mark for d-d promotion / absorption of visible light.
- 1 mark for stating energy of photon absorbed matches the gap ΔE.
- 1 mark for mentioning the observed color is complementary to the absorbed color.

(b)(i) [3 marks]:
- 2 marks for drawing two 3D octahedral structures of the cis-isomer showing ethylenediamine as bidentate loops.
- 1 mark for showing them as non-superimposable mirror images.

(b)(ii) [1.5 marks]:
- 0.5 marks for identifying the trans-isomer.
- 1 mark for explaining that it contains a plane of symmetry / is achiral.

(c)(i) [2 marks]:
- 1 mark for correct numerator and denominator.
- 1 mark for omitting H2O concentration from the expression.

(c)(ii) [2 marks]:
- 1 mark for stating that ammonia replaces water because the copper-ammonia complex is much more stable (very high Kstab).
- 1 mark for describing the color change from light-blue solution to deep-blue solution.

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

(b) The Born-Haber cycle consists of the following steps:
- Starting elements: \(\text{Ca}(\text{s}) + \text{S}(\text{s})\)
- Direct path: Formation of \(\text{CaS}(\text{s})\) \((\Delta H^\ominus_f)\)
- Step 1: Atomisation of Ca: \(\text{Ca}(\text{s}) \rightarrow \text{Ca}(\text{g})\) \((\Delta H^\ominus_{at}[\text{Ca}])\)
- Step 2: 1st and 2nd Ionisation of Ca: \(\text{Ca}(\text{g}) \rightarrow \text{Ca}^{2+}(\text{g}) + 2e^-\rightleftharpoons\) \((\text{IE}_1 + \text{IE}_2)\)
- Step 3: Atomisation of S: \(\text{S}(\text{s}) \rightarrow \text{S}(\text{g})\) \((\Delta H^\ominus_{at}[\text{S}])\)
- Step 4: 1st Electron Affinity of S: \(\text{S}(\text{g}) + e^- \rightarrow \text{S}^-(\text{g})\) \((\text{EA}_1)\)
- Step 5: 2nd Electron Affinity of S: \(\text{S}^-(\text{g}) + e^- \rightarrow \text{S}^{2-}(\text{g})\) \((\text{EA}_2)\)
- Step 6: Lattice formation: \(\text{Ca}^{2+}(\text{g}) + \text{S}^{2-}(\text{g}) \rightarrow \text{CaS}(\text{s})\) \((\Delta H^\ominus_{\text{latt}})\)

(c) Using Hess's Law:
\[ \Delta H^\ominus_f = \Delta H^\ominus_{at}(\text{Ca}) + \text{IE}_1(\text{Ca}) + \text{IE}_2(\text{Ca}) + \Delta H^\ominus_{at}(\text{S}) + \text{EA}_1(\text{S}) + \text{EA}_2(\text{S}) + \Delta H^\ominus_{\text{latt}} \]
\[ -482 = 178 + 590 + 1145 + 279 + (-200) + \text{EA}_2(\text{S}) + (-3013) \]
\[ -482 = -1021 + \text{EA}_2(\text{S}) \]
\[ \text{EA}_2(\text{S}) = -482 + 1021 = +539\text{ kJ mol}^{-1} \]

(d) The calcium ion, \(\text{Ca}^{2+}\), has a higher ionic charge (\(+2\)) compared to the potassium ion, \(\text{K}^+\) (\(+1\)). Additionally, \(\text{Ca}^{2+}\) has a smaller ionic radius than \(\text{K}^+\). Consequently, the charge density of \(\text{Ca}^{2+}\) is much higher, which results in significantly stronger electrostatic attractions between \(\text{Ca}^{2+}\) and \(\text{S}^{2-}\) in the crystal lattice of \(\text{CaS}\), making the lattice energy much more exothermic.

(a) [2 marks]:
- 1 mark for the enthalpy change when 1 mole of solid ionic crystal is formed.
- 1 mark for stating that the starting reactants are gaseous ions.

(b) [4.5 marks]:
- 1.5 marks for all chemical formulas and states correct at each energy level.
- 1.5 marks for correct directions of all arrows in the cycle.
- 1.5 marks for correct labels of all enthalpy steps.

(c) [4 marks]:
- 1 mark for the correct algebraic expression of Hess's Law.
- 2 marks for correct calculation working.
- 1 mark for the final value of +539 kJ mol-1 with the correct sign.

(d) [2 marks]:
- 1 mark for stating that Ca2+ has a higher charge and smaller radius (higher charge density) than K+.
- 1 mark for concluding that this results in stronger ionic attractions/bonding in CaS.

(a) Nitric acid acts as a base and is protonated by sulfuric acid:
\[ \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + 2\text{HSO}_4^- \]
(Or simply: \(\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}\))

(b) Mechanism of mononation:
- Step 1: Curly arrow starting from the delocalised \(\pi\)-system of the methylbenzene ring to the nitrogen atom of the \(\text{NO}_2^+\) electrophile.
- Step 2: Draw the intermediate arenium carbocation, showing a partial ring with a positive charge in the center, open towards Carbon-4. Both hydrogen and the \(\text{NO}_2\) group must be shown bonded to Carbon-4.
- Step 3: A curly arrow from the C-H bond on Carbon-4 back into the ring, restoring the aromatic system, yielding 4-nitromethylbenzene and releasing \(\text{H}^+\).

(c) The methyl group is an electron-donating group through a positive inductive effect (\(+I\)). This electron donation increases the electron density on the benzene ring, specifically stabilizing the intermediate carbocations when the electrophile attacks positions 2 and 4.

(d)
- Reactivity order: Phenol > Methylbenzene > Benzene.
- Explanation:
- In phenol, the lone pair of electrons on the oxygen atom of the -OH group is delocalized into the \(\pi\)-system of the benzene ring. This significantly increases the electron density of the ring, making it much more reactive towards electrophiles.
- In methylbenzene, the methyl group is electron-donating via the positive inductive effect, which increases the ring's electron density relative to benzene, but to a lesser extent than the resonance donation of the oxygen lone pair in phenol.
- Benzene has no activating groups to increase its ring electron density, hence it is the least reactive.

(a) [2 marks]:
- 2 marks for a balanced equation showing the production of NO2+ (1 mark if minor balancing error but species are correct).

(b) [4.5 marks]:
- 1.5 marks for the curly arrow from the benzene ring to NO2+.
- 1.5 marks for the correct intermediate structure showing positive charge and both H and NO2 on C4.
- 1.5 marks for the curly arrow from the C-H bond back into the ring, and showing the correct products.

(c) [2 marks]:
- 1 mark for identifying the positive inductive effect (+I) of the methyl group.
- 1 mark for explaining that it stabilizes the positive carbocation intermediate at the ortho/para positions.

(d) [4 marks]:
- 1 mark for the correct order: Phenol > Methylbenzene > Benzene.
- 1 mark for explaining the high reactivity of phenol via delocalization of the oxygen lone pair.
- 1 mark for explaining the intermediate reactivity of methylbenzene via the inductive effect of the methyl group.
- 1 mark for stating that benzene lacks any activating group.

(a) \(\Delta H^\ominus = \sum \Delta H^\ominus_f(\text{products}) - \sum \Delta H^\ominus_f(\text{reactants})\)
\(\Delta H^\ominus = [(-602) + (-394)] - [-1096]\)
\(\Delta H^\ominus = -996 + 1096 = +100\text{ kJ mol}^{-1}\)

(b) \(\Delta S^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})\)
\(\Delta S^\ominus = [26.9 + 213.7] - [65.7]\)
\(\Delta S^\ominus = 240.6 - 65.7 = +174.9\text{ J K}^{-1}\text{ mol}^{-1}\)

(c) The sign of \(\Delta S^\ominus\) is positive, indicating an increase in entropy (disorder). This is because a solid reactant is decomposing to produce a gas (\(\text{CO}_2\)) alongside a solid product. Gas molecules have much higher translational energy and disorder than molecules in the solid state.

(d) \(\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus\)
At 298 K:
\(\Delta S^\ominus = +174.9\text{ J K}^{-1}\text{ mol}^{-1} = +0.1749\text{ kJ K}^{-1}\text{ mol}^{-1}\)
\(\Delta G^\ominus = 100 - (298 \times 0.1749)\)
\(\Delta G^\ominus = 100 - 52.12 = +47.88\text{ kJ mol}^{-1}\) (or \(+47.9\text{ kJ mol}^{-1}\))
Since \(\Delta G^\ominus\) is positive (\(\Delta G^\ominus > 0\)), the reaction is not feasible at 298 K.

(e) For the reaction to become feasible, \(\Delta G^\ominus < 0\).
\[ \Delta H^\ominus - T\Delta S^\ominus < 0 \]
\[ T > \frac{\Delta H^\ominus}{\Delta S^\ominus} \]
\[ T > \frac{100}{0.1749} = 571.75\text{ K} \]
Thus, the reaction becomes feasible above approximately 572 K.

(a) [2 marks]:
- 1 mark for correct working.
- 1 mark for correct value and units (+100 kJ mol-1).

(b) [2 marks]:
- 1 mark for correct working.
- 1 mark for correct value and units (+174.9 J K-1 mol-1).

(c) [1.5 marks]:
- 0.5 marks for identifying the state change from solid to gas.
- 1 mark for explaining that gas has much greater disorder/entropy.

(d) [4 marks]:
- 1 mark for correct formula.
- 1 mark for correct unit conversion of entropy change.
- 1 mark for correct calculation of +47.9 kJ mol-1.
- 1 mark for stating that the reaction is not feasible because ΔG is positive.

(e) [3 marks]:
- 1 mark for recognizing ΔG = 0 at the threshold.
- 1 mark for correct rearrangement and substitution.
- 1 mark for the correct temperature of 572 K.

(a) Reagents and conditions:
- Tin (\(\text{Sn}\)) and concentrated hydrochloric acid (\(\text{HCl}\)), heated under reflux.
- Followed by the addition of aqueous sodium hydroxide (\(\text{NaOH}\)) to liberate phenylamine from the phenylammonium ion.

(b)(i) Nitrous acid is unstable and is prepared in situ by reacting sodium nitrite (\(\text{NaNO}_2\)) with dilute hydrochloric acid (\(\text{HCl}\)) at a low temperature.

(b)(ii) Temperature range: \(0^\circ\text{C}\) to \(10^\circ\text{C}\).
Reasoning: Below \(0^\circ\text{C}\), the reaction rate is too slow. Above \(10^\circ\text{C}\), the benzenediazonium chloride is unstable and decomposes to form phenol and nitrogen gas.

(c) The equation for the coupling reaction is:
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_{10}\text{H}_7\text{O}^- \rightarrow \text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_{10}\text{H}_6\text{OH} + \text{Cl}^- \]
Structure of the azo dye: A benzene ring linked to position 1 of the naphthalen-2-ol group via a \(-\text{N}=\text{N}-\) azo group.

(d)
- Order of increasing basicity: phenylamine < ammonia < ethylamine.
- Explanation:
- Basicity depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton (\(\text{H}^+\)).
- In phenylamine, the nitrogen lone pair is delocalised into the \(\pi\)-system of the aromatic ring, making it significantly less available to accept a proton.
- In ethylamine, the ethyl group is electron-donating due to its positive inductive effect (\(+I\)), which increases the electron density on the nitrogen atom and makes the lone pair more available to accept a proton.
- Ammonia has no modifying groups, so its basicity lies in the middle.

(a) [2 marks]:
- 1 mark for Tin and concentrated HCl under reflux.
- 1 mark for adding aqueous sodium hydroxide.

(b)(i) [1 mark]:
- 1 mark for stating the reaction of sodium nitrite with hydrochloric acid.

(b)(ii) [2 marks]:
- 1 mark for specifying the temperature range of 0 to 10 °C.
- 1 mark for explaining that the diazonium salt is unstable and decomposes above 10 °C.

(c) [2.5 marks]:
- 1 mark for the correct equation.
- 1.5 marks for the correct structure of the azo dye showing coupling at Carbon-1 of naphthalene.

(d) [5 marks]:
- 1 mark for the correct order: phenylamine < ammonia < ethylamine.
- 1 mark for relating basicity to the availability of the nitrogen lone pair.
- 1 mark for explaining the weak basicity of phenylamine (delocalisation into the benzene ring).
- 1 mark for explaining the strong basicity of ethylamine (positive inductive effect of ethyl group).
- 1 mark for contrasting with ammonia as the baseline.

(a)(i) Equation:
\[ \text{CH}_3\text{COOH} + \text{PCl}_5 \rightarrow \text{CH}_3\text{COCl} + \text{POCl}_3 + \text{HCl} \]
Observation: Misty white fumes of \(\text{HCl}\) gas.

(a)(ii) Reagent: Thionyl chloride (or sulfur dichloride oxide), \(\text{SOCl}_2\).
Reason for preference: The by-products of the reaction (\(\text{SO}_2\) and \(\text{HCl}\)) are both gases, which escape the reaction mixture. This leaves the pure product, ethanoyl chloride, without the need for complex separation.

(b)(i) Equation:
\[ \text{CH}_3\text{COCl} + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{HCl} \]
Product name: Ethyl ethanoate.

(b)(ii) Equation:
\[ \text{CH}_3\text{COCl} + \text{CH}_3\text{CH}_2\text{NH}_2 \rightarrow \text{CH}_3\text{CONHCH}_2\text{CH}_3 + \text{HCl} \]
(Note: \(\text{CH}_3\text{COCl} + 2\text{CH}_3\text{CH}_2\text{NH}_2 \rightarrow \text{CH}_3\text{CONHCH}_2\text{CH}_3 + \text{CH}_3\text{CH}_2\text{NH}_3^+\text{Cl}^-\) is also accepted).
Product name: N-ethylethanamide.

(c)
- Relative ease of hydrolysis: Ethanoyl chloride > chloroethane > chlorobenzene.
- Explanation:
- In ethanoyl chloride, the carbonyl carbon is bonded to two highly electronegative atoms (oxygen and chlorine), making it highly electron-deficient (\(\delta+\)). It is highly susceptible to nucleophilic attack by water.
- In chloroethane, the carbon bonded to the chlorine atom is only moderately electron-deficient (\(\delta+\)), making nucleophilic attack much slower, requiring heating.
- In chlorobenzene, the lone pair of electrons on the chlorine atom is delocalised into the aromatic \(\pi\)-system. This gives the C-Cl bond a partial double-bond character, making it significantly stronger and extremely resistant to nucleophilic attack.

(a)(i) [2 marks]:
- 1 mark for the correct, balanced equation.
- 1 mark for stating the observation of misty/steamy white fumes.

(a)(ii) [2 marks]:
- 1 mark for naming SOCl2 (or thionyl chloride).
- 1 mark for explaining that gaseous by-products make purification easy.

(b)(i) [2.5 marks]:
- 1.5 marks for the correct equation.
- 1 mark for naming the product "ethyl ethanoate".

(b)(ii) [2.5 marks]:
- 1.5 marks for the correct equation.
- 1 mark for naming the product "N-ethylethanamide".

(c) [3.5 marks]:
- 1 mark for stating the correct order: Ethanoyl chloride > chloroethane > chlorobenzene.
- 1 mark for explaining the high reactivity of ethanoyl chloride due to the highly electrophilic carbon (bonded to O and Cl).
- 1 mark for explaining the inertness of chlorobenzene (delocalisation of Cl lone pair into the ring, strengthening the C-Cl bond).
- 0.5 marks for describing the normal single C-Cl polar bond reactivity of chloroethane.

(a) In a clock reaction, the concentration of iodine required to react with the fixed amount of thiosulfate and then trigger the blue starch-iodine endpoint is constant. Thus, the change in product concentration, \(\Delta [\text{I}_2]\), is a constant. Since \(\text{rate} = \frac{\Delta [\text{I}_2]}{\Delta t}\), the rate is directly proportional to \(\frac{1}{t}\). Reactant concentrations must remain virtually constant so that the measured rate accurately represents the *initial rate* before significant reactant consumption occurs.

(b) Calculations (using \(T(\text{K}) = T(^\circ\text{C}) + 273.15\) or \(T(^\circ\text{C}) + 273\)):
- **Exp 1**: \(T = 293\text{ K}\) (or \(293.2\text{ K}\)), \(\frac{1}{T} = 3.41 \times 10^{-3}\text{ K}^{-1}\), \(\ln(1/t) = -4.79\)
- **Exp 2**: \(T = 303\text{ K}\) (or \(303.2\text{ K}\)), \(\frac{1}{T} = 3.30 \times 10^{-3}\text{ K}^{-1}\), \(\ln(1/t) = -4.36\)
- **Exp 3**: \(T = 313\text{ K}\) (or \(313.2\text{ K}\)), \(\frac{1}{T} = 3.19 \times 10^{-3}\text{ K}^{-1}\), \(\ln(1/t) = -3.43\)
- **Exp 4**: \(T = 323\text{ K}\) (or \(323.2\text{ K}\)), \(\frac{1}{T} = 3.10 \times 10^{-3}\text{ K}^{-1}\), \(\ln(1/t) = -2.80\)
- **Exp 5**: \(T = 333\text{ K}\) (or \(333.2\text{ K}\)), \(\frac{1}{T} = 3.00 \times 10^{-3}\text{ K}^{-1}\), \(\ln(1/t) = -2.23\)
- **Exp 6**: \(T = 343\text{ K}\) (or \(343.2\text{ K}\)), \(\frac{1}{T} = 2.91 \times 10^{-3}\text{ K}^{-1}\) (or \(2.92 \times 10^{-3}\)), \(\ln(1/t) = -1.67\)

(c) From the Arrhenius equation: \(\text{Gradient} = -\frac{E_a}{R}\)
\[-6250 = -\frac{E_a}{8.31}\]
\[E_a = 6250 \times 8.31 = 51937.5\text{ J mol}^{-1} = +51.9\text{ kJ mol}^{-1}\text{ (or } +52.0\text{ kJ mol}^{-1}\text{)}\]

(d) Experiment 2 is anomalous. Based on the line of best fit of the other five points, the predicted value of \(\ln(1/t)\) at \(\frac{1}{T} = 3.30 \times 10^{-3}\text{ K}^{-1}\) should be approximately \(-4.10\). The measured value of \(\ln(1/t)\) is \(-4.36\). Since \(-4.36 < -4.10\), the rate (\(1/t\)) was lower than expected, which means the reaction was **slower** than expected (time taken was too long: 78.0 s instead of ~60 s). This could be caused by: the reaction mixture cooling down during the run because it was not insulated; incorrect reactant concentrations (e.g. dilution error); or an excess of sodium thiosulfate added. This point should be ignored when drawing the line of best fit.

(e) Improvement: Perform the reaction inside a thermostated water bath to maintain a constant temperature throughout each run. Direct heating using a Bunsen burner must be avoided because it causes uneven heating, resulting in temperature gradients within the flask, and cannot precisely maintain a constant target temperature.

Part (a) [2 Marks]:
- 1 Mark: For explaining that \(\Delta [\text{I}_2]\) is constant, meaning rate is directly proportional to \(1/t\).
- 1 Mark: For explaining that reactant concentrations must remain virtually constant to represent initial rate conditions.

Part (b) [3 Marks]:
- 1 Mark: All temperatures correctly converted to Kelvin (293, 303, 313, 323, 333, 343 or using .2 / .15).
- 1 Mark: All values of \(1/T\) correctly calculated to 3 significant figures with units \(\text{K}^{-1}\).
- 1 Mark: All values of \(\ln(1/t)\) correctly calculated to 3 significant figures (must be negative numbers).

Part (c) [4 Marks]:
- 1 Mark: Recalls relation \(\text{Gradient} = -E_a / R\).
- 1 Mark: Rearranges to solve for \(E_a\) and substitutes values correctly: \(E_a = 6250 \times 8.31\).
- 1 Mark: Obtains numerical value of 51.9 (or 52.0) with correct positive sign.
- 1 Mark: Correct units stated: \(\text{kJ mol}^{-1}\) (or \(\text{J mol}^{-1}\) with corresponding correct scale, i.e., \(+51900\text{ J mol}^{-1}\)).

Part (d) [3 Marks]:
- 1 Mark: Correctly identifies Experiment 2 as the anomaly.
- 1 Mark: States the reaction was slower than expected (or time taken was too long) and provides a valid explanation based on graph position (lies below the line).
- 1 Mark: Suggests a plausible source of error (e.g., cooling during the reaction, dilution, or excess thiosulfate) AND states the point should be excluded/ignored when drawing the best-fit line.

Part (e) [3 Marks]:
- 1 Mark: Suggests using a thermostated water bath to maintain constant temperature.
- 1 Mark: Identifies that direct heating causes uneven temperature distribution/gradients.
- 1 Mark: Identifies that direct heating lacks precise temperature control.

(a) Hess's Law Cycle:
Direct route:
\(\text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq) \xrightarrow{\Delta H_2} \text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l)\)

Indirect route:
1) Dissociation: \(\text{CH}_3\text{COOH}(aq) \xrightarrow{\Delta H_{\text{diss}}} \text{CH}_3\text{COO}^-(aq) + \text{H}^+(aq)\)
2) Neutralization of the resulting \(\text{H}^+(aq)\) with \(\text{OH}^-(aq)\): \(\text{H}^+(aq) + \text{OH}^-(aq) \xrightarrow{\Delta H_1} \text{H}_2\text{O}(l)\)

According to Hess's law:
\(\Delta H_2 = \Delta H_{\text{diss}} + \Delta H_1\)
Therefore:
\(\Delta H_{\text{diss}} = \Delta H_2 - \Delta H_1\)

(b) Since heat loss occurs immediately upon mixing, the temperature starts cooling before maximum temperature can be recorded directly. To correct for this, the linear cooling portion of the temperature-time curve (from 5.0 to 10.0 min) is extrapolated back to the time of mixing (4.0 min).
- **For Experiment A**:
The cooling rate between 5.0 and 10.0 min is constant:
\(\text{Cooling rate} = \frac{34.8 - 32.8}{10.0 - 5.0} = 0.40\ ^\circ\text{C min}^{-1}\).
Extrapolating back to 4.0 min (1.0 minute prior to the 5.0 min reading):
\(T_{\text{max, corrected}} = 34.8 + 0.40 = 35.2\ ^\circ\text{C}\).
Initial temperature = \(21.5\ ^\circ\text{C}\).
Corrected temperature rise, \(\Delta T_1 = 35.2 - 21.5 = 13.7\ ^\circ\text{C}\) (or \(13.7\text{ K}\)).

- **For Experiment B**:
The cooling rate between 5.0 and 10.0 min is constant:
\(\text{Cooling rate} = \frac{34.35 - 32.6}{10.0 - 5.0} = 0.35\ ^\circ\text{C min}^{-1}\).
Extrapolating back to 4.0 min:
\(T_{\text{max, corrected}} = 34.35 + 0.35 = 34.7\ ^\circ\text{C}\).
Initial temperature = \(21.5\ ^\circ\text{C}\).
Corrected temperature rise, \(\Delta T_2 = 34.7 - 21.5 = 13.2\ ^\circ\text{C}\) (or \(13.2\text{ K}\)).

(c) Mass of mixture, \(m = 50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ g}\) (since density = \(1.00\text{ g cm}^{-3}\)).
In both experiments, the limiting reactant is the acid (\(n_{\text{acid}} = 0.0500\text{ dm}^3 \times 2.00\text{ mol dm}^{-3} = 0.100\text{ mol}\); \(\text{NaOH}\) is in excess as \(n_{\text{NaOH}} = 0.1025\text{ mol}\)).

- **For Experiment A**:
\(q_1 = m c \Delta T_1 = 100.0 \times 4.18 \times 13.7 = 5726.6\text{ J} = 5.727\text{ kJ}\).
\(\Delta H_1 = -\frac{5.727\text{ kJ}}{0.100\text{ mol}} = -57.3\text{ kJ mol}^{-1}\).

- **For Experiment B**:
\(q_2 = m c \Delta T_2 = 100.0 \times 4.18 \times 13.2 = 5517.6\text{ J} = 5.518\text{ kJ}\).
\(\Delta H_2 = -\frac{5.518\text{ kJ}}{0.100\text{ mol}} = -55.2\text{ kJ mol}^{-1}\).

(d) \(\Delta H_{\text{diss}} = \Delta H_2 - \Delta H_1 = -55.2 - (-57.3) = +2.1\text{ kJ mol}^{-1}\).

(e) (i) A thermometer with graduations of \(0.1\ ^\circ\text{C}\) has an uncertainty of \(\pm 0.05\ ^\circ\text{C}\) per reading. Measuring \(\Delta T\) involves two temperature readings (initial and final).
\(\text{Total uncertainty} = 2 \times 0.05 = \pm 0.1\ ^\circ\text{C}\).
\(\text{Percentage error} = \frac{0.1}{13.7} \times 100\% = 0.730\%\) (or \(0.73\%\)).

(ii)
- **Apparatus modification**: Place a tight-fitting insulated lid on the polystyrene cup, or place the cup inside a larger beaker filled with cotton wool insulation to reduce convective/radiative heat loss.
- **Procedural modification**: Stir the mixture continuously and vigorously during the initial period after mixing to ensure rapid heat distribution, or measure the separate initial temperatures of both solutions to get an accurate starting average.

Part (a) [3 Marks]:
- 1 Mark: Correctly drawn Hess's Law cycle showing the reactant, product, and free ions species with clear arrows.
- 1 Mark: All three enthalpy changes labeled correctly in the cycle in accordance with the direction of the arrows.
- 1 Mark: Derives the correct equation: \(\Delta H_{\text{diss}} = \Delta H_2 - \Delta H_1\).

Part (b) [4 Marks]:
- 1 Mark: Explanation that the cooling curve after mixing (5.0 to 10.0 min) is extrapolated back to the time of mixing (4.0 min) to correct for heat loss.
- 1 Mark: Correct cooling rates calculated (\(0.40\ ^\circ\text{C min}^{-1}\) for A, \(0.35\ ^\circ\text{C min}^{-1}\) for B).
- 1 Mark: Extrapolated maximum temperatures are correct (\(35.2\ ^\circ\text{C}\) for A, \(34.7\ ^\circ\text{C}\) for B).
- 1 Mark: Obtains both temperature rises correctly: \(\Delta T_1 = 13.7\ ^\circ\text{C}\) (or K) and \(\Delta T_2 = 13.2\ ^\circ\text{C}\) (or K).

Part (c) [4 Marks]:
- 1 Mark: Correct mass of solution identified (100.0 g) and limiting reactant moles calculated (0.100 mol).
- 1 Mark: Calculated heat values correctly (\(q_1 = 5726.6\text{ J}\) and \(q_2 = 5517.6\text{ J}\), allow ECF from (b)).
- 1 Mark: Correct value and sign for \(\Delta H_1 = -57.3\text{ kJ mol}^{-1}\) (to 3 sig figs).
- 1 Mark: Correct value and sign for \(\Delta H_2 = -55.2\text{ kJ mol}^{-1}\) (to 3 sig figs).

Part (d) [1 Mark]:
- 1 Mark: Correctly calculates \(\Delta H_{\text{diss}} = +2.1\text{ kJ mol}^{-1}\) (must have "+" sign and correct units. Accept ECF from (c)).
- Reject: -2.1 or 2.1 without sign.

Part (e) [3 Marks]:
- 1 Mark: Shows total uncertainty is \(\pm 0.1\ ^\circ\text{C}\) and calculates percentage error as \(0.73\%\) (or \(0.730\%\)).
- 1 Mark: Suggests a valid apparatus modification (e.g., adding a lid to the cup or extra insulation/beaker wrapper).
- 1 Mark: Suggests a valid procedural modification (e.g., continuous stirring, or averaging initial separate temperatures of both solutions).