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(a) Electrospray ionisation:
- The sample is dissolved in a volatile polar solvent.
- It is injected through a fine hypodermic needle connected to a high-voltage power supply.
- The particles gain a proton (\( \text{H}^+ \)) from the solvent as they leave the needle, forming a fine mist of positive ions.
- Equation: \( M(g) + \text{H}^+(g) \rightarrow MH^+(g) \)

(b) Let the abundance of \( ^{25}\text{X} = y \% \).
Since the total abundance is 100%, the abundance of \( ^{26}\text{X} = 100 - 78.99 - y = 21.01 - y \% \).
\( A_r = \frac{(78.99 \times 24) + (y \times 25) + ((21.01 - y) \times 26)}{100} = 24.31 \)
\( 1895.76 + 25y + 546.26 - 26y = 2431 \)
\( 2442.02 - y = 2431 \)
\( y = 11.02 \)
Abundance of \( ^{25}\text{X} = 11.02\% \)
Abundance of \( ^{26}\text{X} = 21.01 - 11.02 = 9.99\% \)

(c) Iron has an atomic number of 26. The electron configuration of an Fe atom is \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2 \).
When forming an \( \text{Fe}^{2+} \) ion, the two \( 4s \) electrons are lost first.
Therefore, the electron configuration of \( \text{Fe}^{2+} \) is: \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 \).

(a) [Total: 4.66 marks]
- Dissolved in a volatile solvent and forced through a fine needle [1.00 mark]
- Connected to a high voltage supply [1.00 mark]
- Gains a proton / \( \text{H}^+ \) ion [1.00 mark]
- Equation: \( M(g) + \text{H}^+ \rightarrow MH^+(g) \) (accept without state symbols, but reject if charge is incorrect) [1.66 marks]

(b) [Total: 4.00 marks]
- Expression for sum of abundances equal to 100% (e.g., \( 21.01 - y \)) [1.00 mark]
- Correct setup of weighted average equation [1.00 mark]
- Correct rearrangement and calculation of \( y = 11.02\% \) [1.00 mark]
- Correct calculation of the second abundance as \( 9.99\% \) [1.00 mark]

(c) [Total: 3.00 marks]
- Identifies that \( 4s \) electrons are lost first [1.00 mark]
- Correct sub-shells shown up to \( 3p^6 \) [1.00 mark]
- Fully correct configuration: \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 \) [1.00 mark]

(a) Standard enthalpy of combustion is the enthalpy change when one mole of a substance is burned completely in excess oxygen under standard conditions (100 kPa and 298 K), with all reactants and products in their standard states.

(b) The equation for the formation of ethanoic acid is:
\( 2\text{C}(s) + 2\text{H}_2(g) + \text{O}_2(g) \rightarrow \text{CH}_3\text{COOH}(l) \)
Using Hess's Law:
\( \Delta_f H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products}) \)
\( \Delta_f H^\theta = [2 \times \Delta_c H^\theta(\text{C}) + 2 \times \Delta_c H^\theta(\text{H}_2)] - \Delta_c H^\theta(\text{CH}_3\text{COOH}) \)
\( \Delta_f H^\theta = [2(-394) + 2(-286)] - (-874) \)
\( \Delta_f H^\theta = [-788 - 572] + 874 \)
\( \Delta_f H^\theta = -1360 + 874 = -486\text{ kJ mol}^{-1} \).

(c) (i)
- Heat change, \( q = m c \Delta T \)
\( q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 36.0\text{ K} = 15048\text{ J} = 15.048\text{ kJ} \)
- Moles of ethanol burned:
\( n = \frac{0.920\text{ g}}{46.0\text{ g mol}^{-1}} = 0.0200\text{ mol} \)
- Enthalpy of combustion:
\( \Delta_c H = -\frac{15.048}{0.0200} = -752.4\text{ kJ mol}^{-1} \).

(ii) Reasons for the discrepancy:
- Heat loss to the surroundings / calorimeter.
- Incomplete combustion of ethanol (forming soot/CO instead of \( \text{CO}_2 \)).
- Evaporation of ethanol from the wick.

(a) [Total: 2.66 marks]
- Enthalpy change when 1 mole of substance is burned completely in oxygen [1.00 mark]
- Under standard conditions (298 K, 100 kPa) [1.00 mark]
- All reactants and products in standard states [0.66 marks]

(b) [Total: 4.00 marks]
- Writes correct equation or Hess's cycle for formation [1.00 mark]
- Calculates sum of reactants' combustion enthalpies: \( 2(-394) + 2(-286) = -1360\text{ kJ mol}^{-1} \) [1.00 mark]
- Correctly applies Hess's law equation: \( -1360 - (-874) \) [1.00 mark]
- Correct final answer: \( -486\text{ kJ mol}^{-1} \) (units required) [1.00 mark]

(c)(i) [Total: 3.00 marks]
- Correct calculation of heat energy: \( q = 15.05\text{ kJ} \) (or \( 15048\text{ J} \)) [1.00 mark]
- Correct calculation of moles: \( 0.0200\text{ mol} \) [1.00 mark]
- Correct calculation of experimental value: \( -752.4\text{ kJ mol}^{-1} \) (must have negative sign) [1.00 mark]

(c)(ii) [Total: 2.00 marks]
- Heat loss to surroundings [1.00 mark]
- Incomplete combustion of fuel / evaporation of fuel (any one) [1.00 mark]

(a)
- Moles of \( \text{NaOH} \) used in the titration:
\( n(\text{NaOH}) = 18.40 \times 10^{-3}\text{ dm}^3 \times 0.125\text{ mol dm}^{-3} = 2.30 \times 10^{-3}\text{ mol} \)
- From the 1:1 reaction ratio, moles of unreacted \( \text{HCl} \) in the \( 25.0\text{ cm}^3 \) aliquot = \( 2.30 \times 10^{-3}\text{ mol} \)
- Total moles of unreacted \( \text{HCl} \) in the \( 100.0\text{ cm}^3 \) volumetric flask:
\( n(\text{HCl})_{\text{unreacted, total}} = 2.30 \times 10^{-3}\text{ mol} \times \frac{100.0}{25.0} = 9.20 \times 10^{-3}\text{ mol} \)
- Initial moles of \( \text{HCl} \) added:
\( n(\text{HCl})_{\text{initial}} = 50.0 \times 10^{-3}\text{ dm}^3 \times 0.500\text{ mol dm}^{-3} = 2.50 \times 10^{-2}\text{ mol} = 0.0250\text{ mol} \)
- Moles of \( \text{HCl} \) that reacted with \( \text{CaCO}_3 \):
\( n(\text{HCl})_{\text{reacted}} = 0.02500 - 0.00920 = 0.01580\text{ mol} \).

(b)
- From the balanced equation:
The mole ratio of \( \text{CaCO}_3 \) to \( \text{HCl} \) is 1:2.
- Moles of \( \text{CaCO}_3 \) in sample:
\( n(\text{CaCO}_3) = \frac{0.01580}{2} = 7.90 \times 10^{-3}\text{ mol} \)
- Mass of pure \( \text{CaCO}_3 \):
\( m = 7.90 \times 10^{-3}\text{ mol} \times 100.1\text{ g mol}^{-1} = 0.79079\text{ g} \)
- Percentage purity of calcium carbonate:
\( \text{Percentage purity} = \frac{0.79079}{0.850} \times 100 = 93.03\% \approx 93.0\% \).

(c) The student should add distilled water dropwise near the graduation mark using a pipette, ensuring the bottom of the meniscus is aligned exactly on the mark at eye level.

(a) [Total: 5.66 marks]
- Calculates moles of \( \text{NaOH} \) correctly: \( 0.00230\text{ mol} \) [1.00 mark]
- Realises 1:1 ratio, moles of \( \text{HCl} \) in \( 25.0\text{ cm}^3 \) is \( 0.00230\text{ mol} \) [1.00 mark]
- Scales up to \( 100.0\text{ cm}^3 \) (multiplies by 4) to find total unreacted \( \text{HCl} = 0.00920\text{ mol} \) [1.00 mark]
- Calculates initial moles of \( \text{HCl} = 0.02500\text{ mol} \) [1.00 mark]
- Subtracts unreacted moles from initial to get reacted moles: \( 0.01580\text{ mol} \) [1.66 marks]

(b) [Total: 4.00 marks]
- Divides reacted moles of \( \text{HCl} \) by 2 to get moles of \( \text{CaCO}_3 = 0.00790\text{ mol} \) [1.00 mark]
- Calculates mass of \( \text{CaCO}_3 = 0.791\text{ g} \) [1.00 mark]
- Divides by sample mass (0.850 g) and multiplies by 100 [1.00 mark]
- Correct final percentage: \( 93.0\% \) (accept range 93.0 to 93.1) [1.00 mark]

(c) [Total: 2.00 marks]
- Use of a pipette/dropper to add water dropwise [1.00 mark]
- View at eye level and align bottom of meniscus with the graduation mark [1.00 mark]

(a) Trend in First Ionisation Energy across Period 3:
- General trend: increases from sodium to argon.
- Explanation of general trend: atomic radius decreases and nuclear charge increases (more protons), with similar shielding (electrons are in the same main energy level). Therefore, the electrostatic attraction between the nucleus and outer electron is stronger, requiring more energy to remove the electron.
- Deviation 1: Dip from Mg to Al.
- Mg has electron configuration \( 1s^2 2s^2 2p^6 3s^2 \), Al has \( 1s^2 2s^2 2p^6 3s^2 3p^1 \).
- The outer electron of Al is in a \( 3p \) sub-shell which is of higher energy and partially shielded by the \( 3s \) electrons, making it easier to remove.
- Deviation 2: Dip from P to S.
- P has configuration \( [Ne] 3s^2 3p^3 \) (singly occupied \( 3p \) orbitals), S has \( [Ne] 3s^2 3p^4 \) (one doubly occupied \( 3p \) orbital).
- The paired electrons in the same \( 3p \) orbital of sulfur experience mutual repulsion, making it easier to remove one of them.

(b) Sulfur trioxide reacts with water to form sulfuric acid:
\( \text{SO}_3(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_4(aq) \)
pH of resulting solution: \( 0 \), \( 1 \) or \( 2 \) (strongly acidic).

(c) Sodium oxide is a basic oxide, and phosphorus(V) oxide is an acidic oxide. They react to form sodium phosphate(V):
\( 6\text{Na}_2\text{O}(s) + \text{P}_4\text{O}_{10}(s) \rightarrow 4\text{Na}_3\text{PO}_4(s) \)
(accept \( 3\text{Na}_2\text{O} + \text{P}_2\text{O}_5 \rightarrow 2\text{Na}_3\text{PO}_4 \))

(a) [Total: 6.66 marks]
- General increase across Period 3 [1.00 mark]
- Explanation: increasing nuclear charge / atomic radius decreases with similar shielding [1.00 mark]
- Identifies dip at Al and explains using higher-energy \( 3p \) sub-shell / shielding by \( 3s \) [1.66 marks]
- Identifies dip at S and explains using repulsion of paired electrons in the same \( 3p \) orbital [1.00 mark]
- Mentions configuration of both pairs (Mg/Al and P/S) correctly [2.00 marks]

(b) [Total: 2.00 marks]
- Equation: \( \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 \) (ignore state symbols) [1.00 mark]
- pH: 0, 1 or 2 [1.00 mark]

(c) [Total: 3.00 marks]
- Correct formula of reactants: \( \text{Na}_2\text{O} \) and \( \text{P}_4\text{O}_{10} \) (or \( \text{P}_2\text{O}_5 \)) [1.00 mark]
- Correct formula of product: \( \text{Na}_3\text{PO}_4 \) [1.00 mark]
- Correctly balanced equation: \( 6\text{Na}_2\text{O} + \text{P}_4\text{O}_{10} \rightarrow 4\text{Na}_3\text{PO}_4 \) [1.00 mark]

(a) (i)
- Observation 1: Smell of bad eggs due to the reduction product hydrogen sulfide (\( \text{H}_2\text{S} \)).
- Observation 2: Yellow solid due to the reduction product sulfur (\( \text{S} \)).
- (Alternative): Purple fumes / black solid due to the oxidation product iodine (\( \text{I}_2 \)).

(ii) Reaction of \( \text{NaI} \) with \( \text{H}_2\text{SO}_4 \) to form \( \text{H}_2\text{S} \):
\( 8\text{NaI} + 9\text{H}_2\text{SO}_4 \rightarrow 8\text{NaHSO}_4 + \text{H}_2\text{S} + 4\text{I}_2 + 4\text{H}_2\text{O} \)
(Accept \( 8\text{NaI} + 5\text{H}_2\text{SO}_4 \rightarrow 4\text{Na}_2\text{SO}_4 + \text{H}_2\text{S} + 4\text{I}_2 + 4\text{H}_2\text{O} \))

(b) (i) The colorless solution turns orange or brown due to the formation of aqueous bromine (\( \text{Br}_2 \)).
(ii) Ionic equation:
\( \text{Cl}_2(aq) + 2\text{Br}^-(aq) \rightarrow 2\text{Cl}^-(aq) + \text{Br}_2(aq) \)

(c)
- Nitric acid is added to react with and remove any interfering anions, such as carbonate (\( \text{CO}_3^{2-} \)) or hydroxide (\( \text{OH}^- \)) ions, which would otherwise form precipitates with silver ions (e.g., \( \text{Ag}_2\text{CO}_3 \) or \( \text{AgOH} \)) and interfere with the test.
- Adding silver nitrate to aqueous chloride ions forms a white precipitate (\( \text{AgCl} \)).
- Upon adding dilute ammonia, the white precipitate dissolves to form a colorless solution.

(a)(i) [Total: 3.00 marks]
- Observation: Smell of bad eggs AND product: \( \text{H}_2\text{S} \) [1.00 mark]
- Observation: Yellow solid AND product: sulfur (\( \text{S} \)) [1.00 mark]
- (Accept): Choking gas AND product: sulfur dioxide (\( \text{SO}_2 \)) [1.00 mark]

(a)(ii) [Total: 2.66 marks]
- Correct species on both sides: \( \text{NaI} \) + \( \text{H}_2\text{SO}_4 \) \( \rightarrow \) \( \text{NaHSO}_4 \) (or \( \text{Na}_2\text{SO}_4 \)) + \( \text{H}_2\text{S} \) + \( \text{I}_2 \) + \( \text{H}_2\text{O} \) [1.00 mark]
- Fully balanced: either \( 8\text{NaI} + 9\text{H}_2\text{SO}_4 \rightarrow 8\text{NaHSO}_4 + \text{H}_2\text{S} + 4\text{I}_2 + 4\text{H}_2\text{O} \) OR \( 8\text{NaI} + 5\text{H}_2\text{SO}_4 \rightarrow 4\text{Na}_2\text{SO}_4 + \text{H}_2\text{S} + 4\text{I}_2 + 4\text{H}_2\text{O} \) [1.66 marks]

(b)(i) [Total: 1.00 mark]
- Solution turns orange/brown [1.00 mark]

(b)(ii) [Total: 2.00 marks]
- Correct reactants and products: \( \text{Cl}_2 + \text{Br}^- \rightarrow \text{Cl}^- + \text{Br}_2 \) [1.00 mark]
- Correctly balanced: \( \text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2 \) [1.00 mark]

(c) [Total: 3.00 marks]
- Nitric acid removes carbonate/hydroxide impurities which would also precipitate [1.00 mark]
- White precipitate formed with chloride [1.00 mark]
- Precipitate dissolves in dilute ammonia [1.00 mark]

(a)
(i) In \( \text{Cr}_2\text{O}_7^{2-} \), oxygen is \( -2 \).
\( 2(\text{Cr}) + 7(-2) = -2 \Rightarrow 2(\text{Cr}) - 14 = -2 \Rightarrow 2(\text{Cr}) = +12 \Rightarrow \text{Cr} = +6 \).
(ii) In \( \text{Na}\text{Cl}\text{O}_3 \), sodium is \( +1 \) and oxygen is \( -2 \).
\( +1 + \text{Cl} + 3(-2) = 0 \Rightarrow 1 + \text{Cl} - 6 = 0 \Rightarrow \text{Cl} = +5 \).
(iii) In \( \text{S}_2\text{O}_3^{2-} \), oxygen is \( -2 \).
\( 2(\text{S}) + 3(-2) = -2 \Rightarrow 2(\text{S}) - 6 = -2 \Rightarrow 2(\text{S}) = +4 \Rightarrow \text{S} = +2 \).

(b)
- Oxidation of thiosulfate ions half-equation:
\( 2\text{S}_2\text{O}_3^{2-}(aq) \rightarrow \text{S}_4\text{O}_6^{2-}(aq) + 2e^- \)
- Reduction of iodine half-equation:
\( \text{I}_2(aq) + 2e^- \rightarrow 2\text{I}^-(aq) \)
- Overall balanced equation (electrons cancel directly):
\( 2\text{S}_2\text{O}_3^{2-}(aq) + \text{I}_2(aq) \rightarrow \text{S}_4\text{O}_6^{2-}(aq) + 2\text{I}^-(aq) \).

(c)
(i) Half-equation for reduction of dichromate(VI):
- Balance chromium atoms: \( \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} \)
- Balance oxygen atoms using water: \( \text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)
- Balance hydrogen atoms using \( \text{H}^+ \): \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)
- Balance charge using electrons: Left side charge \( = +12 \), Right side charge \( = +6 \). Add \( 6e^- \) to left:
\( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \).

(ii) Oxidation of Fe(II) half-equation:
\( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \)
Multiply this half-equation by 6 to match electron count, then add to the dichromate half-equation:
\( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O} \).

(a) [Total: 3.00 marks]
- (i) +6 (accept 6) [1.00 mark]
- (ii) +5 (accept 5) [1.00 mark]
- (iii) +2 (accept 2) [1.00 mark]

(b) [Total: 3.66 marks]
- Thiosulfate half-equation: \( 2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} + 2e^- \) [1.66 marks]
- Overall equation: \( 2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^- \) [2.00 marks]

(c)(i) [Total: 2.00 marks]
- Correct species: \( \text{Cr}_2\text{O}_7^{2-} \), \( \text{H}^+ \), \( \text{Cr}^{3+} \), \( \text{H}_2\text{O} \) [1.00 mark]
- Correct balancing and electrons: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \) [1.00 mark]

(c)(ii) [Total: 3.00 marks]
- Correct Fe(II) oxidation half-equation or recognition of 6:1 ratio [1.00 mark]
- Correctly combines and balances equations [1.00 mark]
- Final correct ionic equation: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O} \) [1.00 mark]

The mechanism proceeds via a nucleophilic substitution (\(\text{S}_\text{N}2\)) pathway:
1. The \(\text{C}-\text{Br}\) bond is polar. The carbon atom is electron-deficient (\(\text{C}^{\delta+}\)) and the bromine atom is electronegative (\(\text{Br}^{\delta-}\)).
2. The hydroxide ion (\(\text{OH}^-\)) acts as a nucleophile. A curly arrow starts from the lone pair on the oxygen of the hydroxide ion and points directly to the \(\text{C}^{\delta+}\) atom.
3. Simultaneously, the \(\text{C}-\text{Br}\) bond breaks heterolytically. A curly arrow starts from the middle of the \(\text{C}-\text{Br}\) bond and points to the bromine atom.
4. This yields propan-1-ol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) and a bromide ion (\(\text{Br}^-\)).

Definition of nucleophile: An electron-pair donor.

M1: Showing the correct dipoles on the reactant: \(\text{C}^{\delta+}\) on carbon and \(\text{Br}^{\delta-}\) on bromine. (1 mark)
M2: Drawing a curly arrow starting from a lone pair on the oxygen atom of the hydroxide ion (\(\text{OH}^-\)) to the carbon atom of the \(\text{C}-\text{Br}\) bond. (1 mark)
M3: Drawing a curly arrow from the \(\text{C}-\text{Br}\) bond to the bromine atom. (1 mark)
M4: Correct structure of the product propan-1-ol and the leaving group \(\text{Br}^-\). (1 mark)
M5: Correct definition of a nucleophile as an electron-pair donor. (1 mark)
M6: Correctly identifying that the oxygen atom donates its lone pair to form a new covalent bond with the electron-deficient carbon. (1 mark)

The mechanism proceeds via elimination (\(\text{E}2\)):
1. The hydroxide ion (\(\text{OH}^-\)) acts as a base and attacks a hydrogen atom on an adjacent carbon (carbon-3 of 2-bromobutane).
2. A curly arrow starts from a lone pair on the oxygen of the hydroxide ion to the hydrogen atom on C3.
3. A second curly arrow starts from the C3-H bond and moves to the C2-C3 bond to form the double bond.
4. A third curly arrow starts from the C-Br bond and moves to the bromine atom, releasing the bromide ion.
5. This yields but-2-ene, water, and bromide ion.

Role of hydroxide ion: It acts as a base (proton acceptor).
Role of ethanol solvent: Ethanol acts as a co-solvent to dissolve both the inorganic potassium hydroxide and the organic halogenoalkane, allowing them to mix and react.

M1: Drawing a curly arrow from a lone pair on the oxygen of \(\text{OH}^-\) to a hydrogen atom on a carbon adjacent to the \(\text{C}-\text{Br}\) carbon (specifically C3 or C1). (1 mark)
M2: Drawing a curly arrow from the adjacent \(\text{C}-\text{H}\) bond to the carbon-carbon single bond (forming the \(\text{C}=\text{C}\) double bond). (1 mark)
M3: Drawing a curly arrow from the \(\text{C}-\text{Br}\) bond to the bromine atom. (1 mark)
M4: Drawing the correct structure of the product but-2-ene. (1 mark)
M5: Correctly stating that the hydroxide ion acts as a base / proton acceptor. (1 mark)
M6: Correctly explaining that ethanol is used to dissolve both reactants because halogenoalkanes are insoluble in water alone. (1 mark)

The mechanism proceeds via electrophilic addition:
1. The hydrogen bromide molecule is polar (\(\text{H}^{\delta+}-\text{Br}^{\delta-}\)).
2. A curly arrow starts from the \(\text{C}=\text{C}\) double bond of propene and points to the hydrogen atom of \(\text{HBr}\).
3. A curly arrow starts from the \(\text{H}-\text{Br}\) bond and points to the bromine atom.
4. This forms a secondary carbocation intermediate: \(\text{CH}_3-\text{C}^+\text{H}-\text{CH}_3\).
5. A curly arrow starts from a lone pair on the bromide ion (\(\text{Br}^-\)) and points to the positively charged carbon atom.
6. This yields the major product, 2-bromopropane.

Explanation of product ratio: The major product is formed via a secondary carbocation intermediate, which is more stable than the primary carbocation intermediate (leading to the minor product). This is because the secondary carbocation has two electron-releasing alkyl groups that stabilize the positive charge by the inductive effect.

M1: Drawing a curly arrow from the \(\text{C}=\text{C}\) double bond of propene to the H atom of \(\text{H}-\text{Br}\). (1 mark)
M2: Showing dipoles on \(\text{H}^{\delta+}-\text{Br}^{\delta-}\) and a curly arrow from the \(\text{H}-\text{Br}\) bond to the Br atom. (1 mark)
M3: Drawing the structure of the correct secondary carbocation intermediate: \(\text{CH}_3\text{C}^+\text{HCH}_3\). (1 mark)
M4: Drawing a curly arrow from a lone pair on the bromide ion to the positive carbon of the carbocation. (1 mark)
M5: Identifying that the intermediate is a secondary carbocation and stating that it is more stable than a primary carbocation. (1 mark)
M6: Explaining that the stability is due to the electron-releasing/inductive effect of the two methyl/alkyl groups. (1 mark)

The mechanism is an electrophilic addition:
1. Sulfuric acid is polar: \(\text{H}^{\delta+}-\text{OSO}_3\text{H}^{\delta-}\).
2. A curly arrow starts from the \(\text{C}=\text{C}\) double bond of but-1-ene and points to the acidic hydrogen atom of \(\text{H}_2\text{SO}_4\).
3. A curly arrow starts from the polar \(\text{H}-\text{O}\) bond in sulfuric acid and points to the oxygen atom, forming a hydrogen sulfate ion (\(\text{HSO}_4^-\)).
4. This step forms a secondary carbocation intermediate: \(\text{CH}_3\text{CH}_2\text{C}^+\text{HCH}_3\).
5. A curly arrow starts from a lone pair on an oxygen of the \(\text{HSO}_4^-\)

M1: Correct curly arrow from the double bond of but-1-ene to the hydrogen atom of sulfuric acid. (1 mark)
M2: Correct curly arrow from the \(\text{H}-\text{O}\) bond to the oxygen of the sulfuric acid molecule. (1 mark)
M3: Correct structure of the secondary carbocation intermediate (\(\text{CH}_3\text{CH}_2\text{C}^+\text{HCH}_3\)). (1 mark)
M4: Correct curly arrow from a lone pair on the oxygen of \(\text{HSO}_4^-\)

The mechanism is nucleophilic substitution:
1. A curly arrow starts from the lone pair on the nitrogen atom of the ammonia molecule (\(\text{NH}_3\)) and points to the carbon atom bonded to bromine in bromoethane.
2. A curly arrow starts from the \(\text{C}-\text{Br}\) bond and points to the bromine atom, producing a bromide ion.
3. This forms the intermediate ethylammonium ion, \(\text{CH}_3\text{CH}_2\text{N}^+\text{H}_3\).
4. A second ammonia molecule acts as a base. A curly arrow starts from its lone pair and points to one of the hydrogens on the nitrogen of the ethylammonium ion.
5. A curly arrow starts from the adjacent \(\text{N}-\text{H}\) bond and points to the nitrogen atom, neutralizing the positive charge.
6. This yields the primary amine ethylamine (\(\text{CH}_3\text{CH}_2\text{NH}_2\)) and an ammonium ion.

Explanation: Excess ammonia is used to ensure that the primary amine product (ethylamine) is not further alkylated by remaining bromoethane to form secondary, tertiary, or quaternary ammonium salts.

M1: Correct curly arrow from the lone pair on \(\text{NH}_3\) to the carbon atom of the \(\text{C}-\text{Br}\) bond. (1 mark)
M2: Correct curly arrow from the \(\text{C}-\text{Br}\) bond to the bromine atom. (1 mark)
M3: Correct structure of the intermediate ethylammonium ion (including the positive charge on nitrogen). (1 mark)
M4: Correct curly arrow from the lone pair of a second ammonia molecule to an H atom of the intermediate. (1 mark)
M5: Correct curly arrow from the intermediate's \(\text{N}-\text{H}\) bond to the positively charged nitrogen atom. (1 mark)
M6: Explanation that excess ammonia suppresses further substitution reactions, maximizing the yield of the primary amine. (1 mark)

The reaction proceeds via electrophilic addition:
1. As the non-polar \(\text{Br}_2\) molecule approaches the electron-dense \(\pi\)-bond of cyclohexene, the electrons in the \(\text{Br}-\text{Br}\) bond are repelled away from the double bond towards the more distant bromine atom. This induces a dipole: \(\text{Br}^{\delta+}-\text{Br}^{\delta-}\).
2. A curly arrow starts from the \(\text{C}=\text{C}\) double bond of cyclohexene and points to the closer, electropositive bromine atom (\(\text{Br}^{\delta+}\)).
3. A curly arrow starts from the \(\text{Br}-\text{Br}\) bond and points to the further bromine atom (\(\text{Br}^{\delta-}\)).
4. This forms a carbocation intermediate (or cyclic bromonium intermediate) and a bromide ion.
5. A curly arrow starts from a lone pair on the bromide ion and points to the carbocation carbon.
6. This yields 1,2-dibromocyclohexane.

M1: Correct curly arrow from the double bond of cyclohexene to the nearer bromine atom. (1 mark)
M2: Showing induced dipoles \(\text{Br}^{\delta+}-\text{Br}^{\delta-}\) and a correct curly arrow from the bromine-bromine single bond to the outer bromine atom. (1 mark)
M3: Drawing a correct intermediate structure (either carbocation or cyclic bromonium ion). (1 mark)
M4: Correct curly arrow from the lone pair of the bromide ion to the positive carbon. (1 mark)
M5: Structure of the final product: 1,2-dibromocyclohexane. (1 mark)
M6: Explanation of polarization: the high electron density in the \(\pi\)-bond of cyclohexene repels the electron cloud of the bromine molecule, inducing a temporary/induced dipole. (1 mark)

The reaction is free-radical substitution:

1. Initiation step:
\(\text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}^\bullet\)

2. Propagation steps:
Step 1: \(\text{C}_4\text{H}_{10} + \text{Cl}^\bullet \rightarrow \text{C}_4\text{H}_9^\bullet + \text{HCl}\)
Step 2: \(\text{C}_4\text{H}_9^\bullet + \text{Cl}_2 \rightarrow \text{C}_4\text{H}_9\text{Cl} + \text{Cl}^\bullet\)

3. Termination step yielding \(\text{C}_8\text{H}_{18}\):
\(2\text{C}_4\text{H}_9^\bullet \rightarrow \text{C}_8\text{H}_{18}\)

Explanation of chain reaction: The chlorine radical (\(\text{Cl}^\bullet\)) consumed in the first propagation step is regenerated in the second propagation step. This allows the cycle to continue repeatedly without requiring continuous initiation.

M1: Correct equation for initiation showing homolytic fission of chlorine with radical dots on chlorine. (1 mark)
M2: Correct equation for the first propagation step showing the abstraction of hydrogen by the chlorine radical to form a butyl radical. (1 mark)
M3: Correct equation for the second propagation step showing the butyl radical reacting with molecular chlorine to form 2-chlorobutane and regenerating the chlorine radical. (1 mark)
M4: Correct equation for the termination step showing the dimerization of two butyl radicals to form octane (\(\text{C}_8\text{H}_{18}\)). (1 mark)
M5: Explanation that the propagation steps form a chain reaction because a radical reactant is regenerated as a product, sustaining the process. (1 mark)
M6: Correct inclusion of radical dots (\(^\bullet\)) on all radical species in the equations. (1 mark)

The mechanism proceeds via acid-catalysed elimination:
1. Protonation of the hydroxyl group: A curly arrow starts from a lone pair on the oxygen atom of butan-2-ol to the \(\text{H}^+\) ion (from phosphoric acid).
2. This forms a protonated intermediate: \(\text{CH}_3\text{CH}_2\text{CH}(\text{O}^+\text{H}_2)\text{CH}_3\).
3. Leaving group loss: A curly arrow starts from the \(\text{C}-\text{O}\) bond and points to the oxygen atom, releasing a water molecule.
4. This forms a secondary carbocation intermediate: \(\text{CH}_3\text{CH}_2\text{C}^+\text{HCH}_3\).
5. Proton loss: A curly arrow starts from a \(\text{C}-\text{H}\) bond on C3 and moves to the carbon-carbon single bond (C2-C3) to form the double bond, releasing an \(\text{H}^+\).
6. This yields trans-but-2-ene.

Explanation of stereoisomerism: cis-but-2-ene and trans-but-2-ene are stereoisomers because:
1. There is restricted rotation around the carbon-carbon double bond (\(\text{C}=\text{C}\)) due to the presence of the \(\pi\)-bond.
2. Each carbon atom involved in the double bond is bonded to two different groups (a methyl group, \(-\text{CH}_3\), and a hydrogen atom, \(-\text{H}\)).

M1: Correct curly arrow from the lone pair on the oxygen atom of butan-2-ol to an \(\text{H}^+\). (1 mark)
M2: Correct structure of the protonated alcohol intermediate. (1 mark)
M3: Correct curly arrow showing the loss of \(\text{H}_2\text{O}\) to form the carbocation. (1 mark)
M4: Correct structure of the secondary carbocation and curly arrow showing loss of proton from C3. (1 mark)
M5: Correctly identifying that restricted rotation about the \(\text{C}=\text{C}\) double bond is necessary for stereoisomerism. (1 mark)
M6: Explaining that each carbon of the double bond must be attached to two different groups (here, \(-\text{H}\) and \(-\text{CH}_3\)). (1 mark)

(a)(i) Nucleophilic substitution.
(a)(ii) 1. The C-Br bond is polar due to the electronegativity difference, with a partial positive charge on the carbon (\(\text{C}^{\delta+}\)) and a partial negative charge on the bromine (\(\text{Br}^{\delta-}\)). 2. A lone pair of electrons on the oxygen of the hydroxide ion (\(\text{OH}^-\)) is donated to the electron-deficient carbon atom (\(\text{C}^{\delta+}\)), forming a C-O bond. 3. The C-Br bond breaks heterolytically, with both electrons from the bond moving to the bromine atom to form a bromide ion (\(\text{Br}^-\)).
(b)(i) Base (or proton acceptor).
(b)(ii) Product: propene. Reaction type: elimination.
(b)(iii) Use hot, concentrated ethanolic potassium hydroxide (or higher temperature and ethanol solvent instead of water).

(a)(i) Nucleophilic substitution [1 mark] - reject 'nucleophilic' alone or 'substitution' alone.
(a)(ii) Mark 1: Mention of \(\text{C}^{\delta+}\) and \(\text{Br}^{\delta-}\) polar bond [1 mark]. Mark 2: Attack of the lone pair of electrons from the oxygen of the \(\text{OH}^-\) ion onto the \(\text{C}^{\delta+}\) carbon atom [1 mark]. Mark 3: Heterolytic fission of the C-Br bond, with the electron pair moving from the bond to the bromine atom [1 mark].
(b)(i) Base / proton acceptor [1 mark] - reject nucleophile.
(b)(ii) Propene [1 mark]. Elimination [1 mark].
(b)(iii) Ethanol/ethanolic solvent and/or high temperature/hot/reflux [1 mark].

(a) The ionic equation representing neutralisation is:
\(\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})\)

(b) Heat energy released, \(q = m \cdot c \cdot \Delta T\)
Total volume of mixture = \(50.0\text{ cm}^3 + 50.0\text{ cm}^3 = 100.0\text{ cm}^3\)
Mass of mixture, \(m = 100.0\text{ g}\) (since density = \(1.00\text{ g cm}^{-3}\))
\(q = 100.0\text{ g} \times 4.18\text{ J g}^{-1}\text{ K}^{-1} \times 5.40\text{ K} = 2257.2\text{ J} = 2.26\text{ kJ}\)

(c) Moles of \(\text{H}^+\) ions added = \(2 \times \text{moles of H}_2\text{SO}_4 = 2 \times (0.0500\text{ dm}^3 \times 0.400\text{ mol dm}^{-3}) = 0.0400\text{ mol}\)
Moles of \(\text{OH}^-\) ions added = \(0.0500\text{ dm}^3 \times 0.900\text{ mol dm}^{-3} = 0.0450\text{ mol}\)
Since \(\text{H}^+\) is the limiting reagent, \(0.0400\text{ mol}\) of water (\(\text{H}_2\text{O}\)) is formed.
\(\Delta H_{\text{neut}} = -\frac{q}{\text{moles of } \text{H}_2\text{O} \text{ formed}} = -\frac{2.2572\text{ kJ}}{0.0400\text{ mol}} = -56.43\text{ kJ mol}^{-1}\)
To 3 significant figures, \(\Delta H_{\text{neut}} = -56.4\text{ kJ mol}^{-1}\).

(a)
- \(\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}\) (1 mark)
- Correct state symbols: \(\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})\) (1 mark)

(b)
- Use of \(m = 100\text{ g}\) in \(q = mc\Delta T\) (1 mark)
- Correct calculation: \(2.26\text{ kJ}\) (or \(2.257\text{ kJ}\)) (1 mark)

(c)
- Moles of \(\text{H}^+\) calculated as \(0.0400\text{ mol}\) (1 mark)
- Moles of \(\text{OH}^-\) calculated as \(0.0450\text{ mol}\) and identification of \(\text{H}^+\) as limiting reagent (1 mark)
- Correct calculation value: \(56.4\) (1 mark)
- Correct negative sign and units \(\text{kJ mol}^{-1}\) (1 mark)

(a) (i) The functional group present is a carboxylic acid (\(-\text{COOH}\)) (1). The very broad absorption between \(2500\text{--}3000\text{ cm}^{-1}\) is characteristic of the \(\text{O--H}\) stretch in carboxylic acids (1). The strong absorption at \(1715\text{ cm}^{-1}\) is due to the \(\text{C=O}\) carbonyl bond stretch (1).

(ii) There are two carboxylic acid isomers with the molecular formula \(\text{C}_4\text{H}_8\text{O}_2\):
1. Butanoic acid: A straight chain of four carbons with a terminal carboxylic acid group.
Skeletal structure: Draw a zig-zag chain of 4 carbons ending in \(=\text{O}\) and \(-\text{OH}\).
2. 2-Methylpropanoic acid: A branched three-carbon chain with a methyl group at position 2 and a terminal carboxylic acid group.
Skeletal structure: Draw a branched propanoic acid with a methyl group on carbon 2.

(b) The gas produced is carbon dioxide, \(\text{CO}_2\) (1).
Test: Bubble the gas through limewater (aqueous calcium hydroxide) (1).
Positive result: The limewater turns cloudy/milky (1).

(a) (i)
- Functional group: Carboxylic acid (1 mark)
- \(\text{O--H}\) stretch in carboxylic acid: \(2500\text{--}3000\text{ cm}^{-1}\) (1 mark)
- \(\text{C=O}\) stretch: \(1715\text{ cm}^{-1}\) (or range \(1680\text{--}1750\text{ cm}^{-1}\)) (1 mark)

(a) (ii)
- Correct skeletal structure of butanoic acid (1.5 marks)
- Correct skeletal structure of 2-methylpropanoic acid (1.5 marks)
(Allow 1 mark overall if structures are drawn correctly but not as skeletal structures)

(b)
- Gas: Carbon dioxide / \(\text{CO}_2\) (1 mark)
- Test: Bubble through limewater AND turns cloudy/milky (1 mark) [Reject other tests such as lit splint]

(a) Silicon has a giant covalent macromolecular structure (1). It contains many strong covalent bonds between silicon atoms throughout the structure, which require a large amount of energy to break (1). Sodium has a giant metallic structure with metallic bonding. The electrostatic forces of attraction between \(\text{Na}^+\) ions and the delocalised electrons are weaker and require less energy to overcome than the covalent bonds in silicon (1).

(b) Both sulfur and chlorine have simple molecular structures (1). Sulfur exists as \(\text{S}_8\) molecules, whereas chlorine exists as \(\text{Cl}_2\) molecules. The strength of the van der Waals forces between molecules depends on the size of the electron cloud (1). \(\text{S}_8\) molecules are larger and have more electrons than \(\text{Cl}_2\) molecules, resulting in stronger van der Waals forces between \(\text{S}_8\) molecules which require more energy to overcome (1).

(c) Reaction of sodium with excess oxygen:
\(2\text{Na(s)} + \text{O}_2\text{(g)} \rightarrow \text{Na}_2\text{O}_2\text{(s)}
(Alternatively, \)4\text{Na(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{Na}_2\text{O(s)}\)) (1).
Observation: burns with a bright yellow / orange flame OR forms a white solid/smoke (1).

(a)
- Silicon has a giant covalent / macromolecular structure AND sodium has a giant metallic structure (1 mark)
- Silicon requires many strong covalent bonds to be broken (1 mark)
- Breaking covalent bonds in silicon requires much more energy than overcoming metallic bonding in sodium (1 mark)

(b)
- Both have simple molecular structures with intermolecular van der Waals forces (1 mark)
- Sulfur exists as \(\text{S}_8\) and chlorine as \(\text{Cl}_2\) (or sulfur has a larger molecular size/more electrons) (1 mark)
- Stronger van der Waals forces between sulfur molecules require more energy to overcome (1 mark)

(c)
- Balanced equation: \(2\text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O}_2\) OR \(4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}\) (state symbols not required) (1 mark)
- Observation: Yellow/orange flame OR white solid/smoke (1 mark)

(a) (i) \(\text{NaCl(s)} + \text{H}_2\text{SO}_4\text{(l)} \rightarrow \text{NaHSO}_4\text{(s)} + \text{HCl(g)}\)
(Accept: \(2\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{HCl}\))
(ii) Sulfuric acid acts as an acid / proton donor.

(b) (i) Yellow solid: Sulfur (\(\text{S}\)).
Gas with bad-egg smell: Hydrogen sulfide (\(\text{H}_2\text{S}\)).
(ii) Reduction half-equation:
\(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\)
(Accept: \(\text{SO}_4^{2-} + 10\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\))
(iii) Sulfuric acid acts as an oxidising agent (1). Iodide ions are larger than chloride ions, meaning their outer electrons are further from the nucleus and more shielded, so they are more easily lost / iodide is a stronger reducing agent than chloride (1).

(a) (i)
- \(\text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl}\) (or molecular/ionic equivalents) (1 mark)

(a) (ii)
- Acid / proton donor (1 mark) [Reject: oxidising agent]

(b) (i)
- Yellow solid: Sulfur / \(\text{S}\) (or \(\text{S}_8\)) (1 mark)
- Bad-egg smell gas: Hydrogen sulfide / \(\text{H}_2\text{S}\) (1 mark)

(b) (ii)
- Reactants and products correct: \(\text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}\) (1 mark)
- Balanced with \(8\text{H}^+\), \(8\text{e}^-\), and \(4\text{H}_2\text{O}\) (1 mark)

(b) (iii)
- Role: Oxidising agent (1 mark)
- Reason: Iodide ion is a stronger reducing agent than chloride ion / outer electron is further from nucleus and easier to lose (1 mark)

(a) The standard enthalpy of combustion is the enthalpy change when:
- 1 mole of a substance (1)
- is burned completely in excess oxygen (1)
- under standard conditions (298 K and 100 kPa), with all reactants and products in their standard states (1).

(b) We construct a Hess's law cycle where both sides are combusted to their standard combustion products (\(\text{N}_2(\text{g}) + 3\text{H}_2\text{O}(\text{l})\)).
By Hess's Law:
\(\Delta_{\text{r}}H^\theta = \sum \Delta_{\text{c}}H^\theta(\text{reactants}) - \sum \Delta_{\text{c}}H^\theta(\text{products})\)

Reactants: \(2\text{NH}_3(\text{g})\)
\(\sum \Delta_{\text{c}}H^\theta(\text{reactants}) = 2 \times (-383\text{ kJ mol}^{-1}) = -766\text{ kJ mol}^{-1}\)

Products: \(\text{N}_2\text{H}_4(\text{l}) + \text{H}_2(\text{g})\)
\(\sum \Delta_{\text{c}}H^\theta(\text{products}) = (-622) + (-286) = -908\text{ kJ mol}^{-1}\)

\(\Delta_{\text{r}}H^\theta = -766 - (-908) = -766 + 908 = +142\text{ kJ mol}^{-1}\).

(a)
- Enthalpy change when one mole of a substance (1 mark)
- Reacts/burns completely in oxygen (1 mark)
- Under standard conditions (with all reactants/products in standard states) (1 mark)

(b)
- Hess's cycle or algebraic formula relating \(\Delta_{\text{c}}H\) values (1 mark)
- Calculation of reactant combustion energy: \(2 \times (-383) = -766\text{ kJ}\) (1 mark)
- Calculation of product combustion energy: \(-622 + (-286) = -908\text{ kJ}\) (1 mark)
- Correct subtraction: \(-766 - (-908)\) (1 mark)
- Correct answer with positive sign and units: \(+142\text{ kJ mol}^{-1}\) (1 mark)

(a) \(K_{\text{c}} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}\)

(b)
- Initial moles: \(\text{H}_2 = 1.50\text{ mol}\), \(\text{I}_2 = 1.50\text{ mol}\), \(\text{HI} = 0\text{ mol}\).
- Equilibrium moles of \(\text{I}_2 = 0.360\text{ mol}\).
- Change in moles of \(\text{I}_2 = 0.360 - 1.50 = -1.14\text{ mol}\).
- Since the stoichiometry is \(1:1\), change in moles of \(\text{H}_2 = -1.14\text{ mol}\).
- Equilibrium moles of \(\text{H}_2 = 1.50 - 1.14 = 0.360\text{ mol}\).
- Since \(1\text{ mol}\) of \(\text{I}_2\) reacts to form \(2\text{ mol}\) of \(\text{HI}\), change in moles of \(\text{HI} = +2 \times 1.14 = +2.28\text{ mol}\).
- Equilibrium moles of \(\text{HI} = 2.28\text{ mol}\).

(c) Substituting concentration terms into \(K_{\text{c}}\):
\(K_{\text{c}} = \frac{\left(\frac{2.28}{V}\right)^2}{\left(\frac{0.360}{V}\right) \times \left(\frac{0.360}{V}\right)} = \frac{(2.28)^2}{(0.360)^2} = \frac{5.1984}{0.1296} \approx 40.1\)

Why volume is not needed:
The total number of moles on the left-hand side (\(1 + 1 = 2\)) is equal to the total number of moles on the right-hand side (\(2\)). Thus, the volume terms \(V^2\) in the numerator and denominator cancel out completely.

(a)
- \(K_{\text{c}} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}\) (square brackets required, reject round brackets) (1 mark)

(b)
- Deduces equilibrium moles of \(\text{H}_2 = 0.360\text{ mol}\) (1 mark)
- Deduces change in moles for \(\text{HI} = 2.28\text{ mol}\) (2 marks: 1 mark for correct mole ratio change \(+2 \times 1.14\), 1 mark for final value)

(c)
- Correct substitution of values into \(K_{\text{c}}\) expression (1 mark)
- Calculates \(K_{\text{c}} = 40.1\) (accept \(40\) or \(40.11\)) (1 mark)
- Explanation: Equal number of moles on both sides of the reaction equation (1 mark)
- Explains that the volume term, \(V\), cancels out in the expression (1 mark)

(a) A yellow precipitate of sulfur (\(\text{S}\)) forms (1), which causes the solution to become cloudy / opaque (1).

(b) The reaction is stopped when a fixed / constant amount of sulfur precipitate has formed to obscure the cross (1). Since rate is defined as \(\frac{\Delta \text{concentration}}{\Delta t}\), and the change in concentration of sulfur is constant in each experiment, the rate is proportional to \(\frac{1}{t}\) (1).

(c) (i) Rate at \(30.0\text{ }^\circ\text{C} = \frac{1}{28.0\text{ s}} = 0.0357\text{ s}^{-1}\) (1 mark for \(0.0357\), 1 mark for unit \(\text{s}^{-1}\)).

(ii) Increasing the temperature increases the kinetic energy of all molecules. A small increase in temperature results in a very large increase in the fraction of molecules with energy greater than or equal to the activation energy (\(E \ge E_{\text{a}}\)) (1). Therefore, a much larger proportion of collisions are successful in leading to a reaction (1).

(a)
- Formation of a yellow precipitate / solid (1 mark)
- Sulfur (\(\text{S}\)) is insoluble in water / causes the mixture to go cloudy (1 mark)

(b)
- A constant/fixed amount of sulfur is produced before the cross is obscured (1 mark)
- Since change in concentration of product is constant, rate \(\propto \frac{1}{t}\) (1 mark)

(c) (i)
- Value: \(0.0357\) (accept \(3.57 \times 10^{-2}\)) (1 mark)
- Unit: \(\text{s}^{-1}\) (1 mark)

(c) (ii)
- Many more molecules / a much greater fraction of molecules have energy \(\ge E_{\text{a}}\) (1 mark)
- A significantly higher proportion / fraction of collisions are successful (1 mark) [Reject: just 'more successful collisions']

(a) Gallium atoms are vaporised and high-energy electrons are fired at them from an electron gun (1). This knocks off one electron from each atom, creating a positive ion (1).
Equation: \(\text{Ga}(\text{g}) \rightarrow \text{Ga}^+(\text{g}) + \text{e}^-\) (1)

(b) Velocity, \(v = \frac{d}{t}\), where \(d\) is the distance/length of the flight tube and \(t\) is the time of flight.
Substitute \(v\) into the kinetic energy equation:
\(KE = \frac{1}{2} m \left(\frac{d}{t}\right)^2\) (1)
Rearranging for \(t\):
\(t^2 = \frac{m d^2}{2KE} \implies t = d \sqrt{\frac{m}{2KE}}\)
Since \(d\) and \(KE\) are constant for all ions in the spectrometer, \(t\) is directly proportional to \(\sqrt{m}\) (1).

(c) Because \(t \propto \sqrt{m}\), we can use the ratio:
\(\frac{t(^{71}\text{Ga}^+)}{t(^{69}\text{Ga}^+)} = \sqrt{\frac{m(^{71}\text{Ga}^+)}{m(^{69}\text{Ga}^+)}}\) (1)
Substituting the given values:
\(t(^{71}\text{Ga}^+) = 2.12 \times 10^{-5}\text{ s} \times \sqrt{\frac{70.9}{68.9}}\) (1)
\(t(^{71}\text{Ga}^+) = 2.12 \times 10^{-5} \times \sqrt{1.0290275}\)
\(t(^{71}\text{Ga}^+) = 2.12 \times 10^{-5} \times 1.01441 \approx 2.15 \times 10^{-5}\text{ s}\) (1)

(a)
- High-energy electrons fired from an electron gun (1 mark)
- Knock off an electron to form a \(1+\) / positive ion (1 mark)
- Equation: \(\text{Ga(g)} \rightarrow \text{Ga}^+\text{(g)} + \text{e}^-\) (must include state symbols) (1 mark)

(b)
- Substitute \(v = \frac{d}{t}\) into \(KE = \frac{1}{2}mv^2\) (1 mark)
- Rearrange to show \(t = d\sqrt{\frac{m}{2KE}}\) and state that \(d\) and \(KE\) are constant, hence \(t \propto \sqrt{m}\) (1 mark)

(c)
- Use of \(\frac{t_1}{\sqrt{m_1}} = \frac{t_2}{\sqrt{m_2}}\) or calculation of constant \(k = 2.554 \times 10^{-6}\) (1 mark)
- Correct substitution: \(t_2 = 2.12 \times 10^{-5} \times \sqrt{\frac{70.9}{68.9}}\) (1 mark)
- Correct answer: \(2.15 \times 10^{-5}\text{ s}\) (accept \(2.15 \times 10^{-5}\) to \(2.151 \times 10^{-5}\)) (1 mark)

**Part (a)**
The standard enthalpy of combustion is the enthalpy change when one mole of a substance is burned completely in excess oxygen under standard conditions (100 kPa and 298 K), with all reactants and products in their standard states.

**Part (b)**
1. Calculate heat energy transferred to the water:
\(q = m c \Delta T\)
\(q = 150.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times (54.6 - 20.2) \text{ K}\)
\(q = 150.0 \times 4.18 \times 34.4 = 21568.8 \text{ J} = 21.57 \text{ kJ}\)

2. Calculate the amount in moles of propan-1-ol burned:
\(n = \frac{\text{mass}}{M_r} = \frac{0.900 \text{ g}}{60.0 \text{ g mol}^{-1}} = 0.0150 \text{ mol}\)

3. Calculate the enthalpy of combustion:
\(\Delta_c H = -\frac{q}{n} = -\frac{21.5688 \text{ kJ}}{0.0150 \text{ mol}} = -1437.92 \text{ kJ mol}^{-1}\)
To 3 significant figures, this is \(-1440 \text{ kJ mol}^{-1}\) (accept values in the range \(-1438\) to \(-1440 \text{ kJ mol}^{-1}\)).

**Part (c)**
The equation for the formation of liquid propan-1-ol is:
\(3\text{C(s)} + 4\text{H}_2\text{(g)} + 0.5\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_7\text{OH(l)}\)

Using a Hess cycle based on enthalpies of combustion:
\(\Delta_f H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})\)
\(\Delta_f H^\ominus = [3\Delta_c H^\ominus(\text{C}) + 4\Delta_c H^\ominus(\text{H}_2)] - \Delta_c H^\ominus(\text{propan-1-ol})\)
\(\Delta_f H^\ominus = [3(-394) + 4(-286)] - \Delta_c H^\ominus(\text{propan-1-ol})\)
\(\Delta_f H^\ominus = -2326 - \Delta_c H^\ominus(\text{propan-1-ol})\)

- If using the calculated experimental value of \(-1440 \text{ kJ mol}^{-1}\):
\(\Delta_f H^\ominus = -2326 - (-1440) = -886 \text{ kJ mol}^{-1}\) (or \(-888 \text{ kJ mol}^{-1}\) if using \(-1438\))

- If using the literature value of \(-2021 \text{ kJ mol}^{-1}\):
\(\Delta_f H^\ominus = -2326 - (-2021) = -305 \text{ kJ mol}^{-1}\)

**Part (a)** [2 marks]
* **M1**: Enthalpy change when 1 mole of a substance is burned completely in oxygen / burned in excess oxygen. [1 mark]
* **M2**: Under standard conditions (100 kPa and 298 K) with all reactants and products in their standard states. [1 mark]

**Part (b)** [3 marks]
* **M1**: Correctly calculates heat energy absorbed: \(q = 150.0 \times 4.18 \times 34.4 = 21568.8 \text{ J}\) (or \(21.57 \text{ kJ}\)). [1 mark]
* **M2**: Correctly calculates moles of propan-1-ol: \(0.900 / 60.0 = 0.0150 \text{ mol}\). [1 mark]
* **M3**: Correctly calculates enthalpy change with negative sign to 3 sig figs: \(-1440 \text{ kJ mol}^{-1}\) (allow range \(-1438\) to \(-1440\)). [1 mark]

**Part (c)** [3 marks]
* **M1**: Uses correct application of Hess's Law: \(\Delta_f H^\ominus = \sum\Delta_c H^\ominus(\text{reactants}) - \Delta_c H^\ominus(\text{product})\). [1 mark]
* **M2**: Calculates the combustion of reactants sum: \(3(-394) + 4(-286) = -2326 \text{ kJ mol}^{-1}\). [1 mark]
* **M3**: Correctly calculates final enthalpy of formation including correct sign and units: \(-886 \text{ kJ mol}^{-1}\) (or \(-888\) from \(-1438\)), OR \(-305 \text{ kJ mol}^{-1}\) using the literature value. [1 mark]

**Part (a)**
1. Calculate the mass of carbon present in the carbon dioxide:
\(\text{Mass of C} = 3.52 \text{ g} \times \frac{12.0}{44.0} = 0.960 \text{ g}\)
\(\text{Moles of C} = \frac{0.960 \text{ g}}{12.0 \text{ g mol}^{-1}} = 0.0800 \text{ mol}\)

2. Calculate the mass of hydrogen present in the water:
\(\text{Mass of H} = 1.80 \text{ g} \times \frac{2.0}{18.0} = 0.200 \text{ g}\)
\(\text{Moles of H} = \frac{0.200 \text{ g}}{1.0 \text{ g mol}^{-1}} = 0.200 \text{ mol}\)

3. Calculate the mass of oxygen by subtraction from the initial mass of compound X:
\(\text{Mass of O} = 1.48 \text{ g} - (0.960 \text{ g} + 0.200 \text{ g}) = 0.320 \text{ g}\)
\(\text{Moles of O} = \frac{0.320 \text{ g}}{16.0 \text{ g mol}^{-1}} = 0.0200 \text{ mol}\)

4. Determine the simplest whole-number ratio:
\(\text{Ratio of C} : \text{H} : \text{O} = \frac{0.0800}{0.0200} : \frac{0.200}{0.0200} : \frac{0.0200}{0.0200} = 4 : 10 : 1\)
The empirical formula is **\(\text{C}_4\text{H}_{10}\text{O}\)**.

**Part (b)**
(i) The formula mass of the empirical formula \(\text{C}_4\text{H}_{10}\text{O}\) is \((4 \times 12.0) + (10 \times 1.0) + 16.0 = 74.0\). Since the molecular ion peak in the mass spectrum matches this mass (\(m/z = 74.0\)), the molecular formula is also **\(\text{C}_4\text{H}_{10}\text{O}\)**.
(ii) The peak at \(m/z = 75\) is called the **\(M+1\) peak**. It arises because a small fraction of the carbon atoms in the molecule are the heavier carbon-13 (\(^{13}\text{C}\)) isotope.

**Part (c)**
The very broad peak at 3230–3550 cm\(^{-1}\) is characteristic of the \(\text{O-H}\) stretching vibration in an alcohol, and the band at 1050 cm\(^{-1}\) corresponds to the \(\text{C-O}\) bond stretch. The absence of a carbonyl peak (\(\text{C=O}\)) in the 1680–1750 cm\(^{-1}\) range confirms the functional group is an **alcohol / hydroxyl group**.

A primary alcohol isomer of \(\text{C}_4\text{H}_{10}\text{O}\) can be either **butan-1-ol** or **2-methylpropan-1-ol**.

Displayed structure of butan-1-ol:
```
H H H H
| | | |
H-C - C - C - C-O-H
| | | |
H H H H
```
Skeletal structure of butan-1-ol can be represented as a chain of 4 carbons ending with an -OH group:
`\/\/OH`

**Part (a)** [3 marks]
* **M1**: Calculates moles of C (0.0800 mol) or mass of C (0.960 g). [1 mark]
* **M2**: Calculates moles of H (0.200 mol) or mass of H (0.200 g). [1 mark]
* **M3**: Subtracts to find mass of O (0.320 g), moles of O (0.0200 mol) and gives empirical formula \(\text{C}_4\text{H}_{10}\text{O}\). [1 mark]

**Part (b)** [3 marks]
* **(i) M1**: Deduces molecular formula is \(\text{C}_4\text{H}_{10}\text{O}\) with justification (empirical mass matches \(m/z\) value). [1 mark]
* **(ii) M2**: Identifies the peak at \(m/z = 75\) as the \(M+1\) peak. [1 mark]
* **(ii) M3**: Attributes this peak to the presence of the carbon-13 (\(^{13}\text{C}\)) isotope. [1 mark]

**Part (c)** [2 marks]
* **M1**: Identifies the functional group as an alcohol / hydroxyl / \(\text{-OH}\) group. [1 mark]
* **M2**: Correctly draws the skeletal or displayed structure of a primary alcohol isomer of \(\text{C}_4\text{H}_{10}\text{O}\) (either butan-1-ol or 2-methylpropan-1-ol). [1 mark]
(Reject secondary/tertiary isomers, e.g. butan-2-ol or 2-methylpropan-2-ol).