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A large increase (jump) occurs between the 3rd and 4th ionization energies (from 2745 to 11577 kJ mol\(^{-1}\)). This indicates that the 4th electron is removed from an inner quantum shell, closer to the nucleus. Therefore, the element has three valence electrons in its outer shell. In Period 3, the element with three outer shell electrons is Aluminium.

C is correct. Award 1 mark for identifying the large increase between the 3rd and 4th ionization energies, which corresponds to three outer shell electrons (Group 13/3, Aluminium).

Let the fractional abundance of \(^{107}\text{Ag}\) be \(x\). The fractional abundance of \(^{109}\text{Ag}\) is then \(1 - x\).

Using the formula for relative atomic mass:
\(107x + 109(1 - x) = 107.87\)
\(107x + 109 - 109x = 107.87\)
\(-2x = -1.13\)
\(x = 0.565\)

Multiplying by 100 to get percentage gives 56.5%.

C is correct. Award 1 mark for the correct calculation showing 56.5%.

\(\text{BF}_3\) (boron trifluoride) has 3 bonding pairs and 0 lone pairs around the central boron atom, giving it a trigonal planar geometry with a bond angle of exactly 120\(^\circ\). \(\text{PF}_3\) is trigonal pyramidal (bond angle < 109.5\(^\circ\)), \(\text{SF}_6\) is octahedral (90\(^\circ\)), and \(\text{ClF}_3\) is T-shaped (angles < 90\(^\circ\)).

A is correct. Award 1 mark for identifying that \(\text{BF}_3\) has trigonal planar geometry with bond angles of 120\(^\circ\).

HF has the highest boiling temperature because fluorine is highly electronegative, enabling HF molecules to form strong intermolecular hydrogen bonds. HCl, HBr, and HI cannot form hydrogen bonds and rely on weaker permanent dipole-dipole interactions and London forces.

A is correct. Award 1 mark for identifying that hydrogen bonding in HF is much stronger than the intermolecular forces in other hydrogen halides.

1. Find moles of \(\text{CaCO}_3\):
\(n(\text{CaCO}_3) = \frac{10.0\text{ g}}{100.1\text{ g mol}^{-1}} = 0.0999\text{ mol}\)

2. Since the molar ratio of reactant to product is 1:1, the amount of \(\text{CO}_2\) produced is 0.0999 mol.

3. Calculate the volume of \(\text{CO}_2\):
\(V = 0.0999\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 2.40\text{ dm}^3\).

C is correct. Award 1 mark for the correct calculation of gas volume using molar ratio and molar volume of gas at r.t.p.

The equation for % atom economy is:
\(\text{Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Total molar mass of all reactants (or products)}} \times 100\%\)

Total mass of reactants = \(2 \times 17.0\text{ (for }2\text{NH}_3) + 74.5\text{ (for NaClO)} = 34.0 + 74.5 = 108.5\text{ g mol}^{-1}\).

Mass of desired product (\(\text{N}_2\text{H}_4\)) = 32.0 \(\text{g mol}^{-1}\).

\(\text{Atom Economy} = \frac{32.0}{108.5} \times 100\% = 29.49\% \approx 29.5\%\).

A is correct. Award 1 mark for finding the total mass of reactants and correctly dividing the desired mass of hydrazine by this total.

Down Group 2, the ionic radius of the metal cation increases while maintaining the same 2+ charge, so the charge density decreases. A cation with lower charge density has less polarising power and thus distorts the carbonate ion less, making the carbonate more thermally stable. Therefore, thermal stability increases down the group.

B is correct. Award 1 mark for explaining that thermal stability increases down the group because a larger cation radius results in less polarisation of the carbonate ion.

In elemental chlorine, \(\text{Cl}_2\), the oxidation state of Cl is 0.

In sodium chloride, \(\text{NaCl}\), Na is +1, so Cl is -1.

In sodium chlorate(I), \(\text{NaClO}\), Na is +1 and O is -2, so Cl must be +1 (since \(1 + \text{Cl} - 2 = 0\)).

Therefore, the oxidation states are 0, -1, and +1. This is a disproportionation reaction.

A is correct. Award 1 mark for identifying the correct oxidation states for chlorine in each species.

(a) First ionization energy is the energy required to remove one mole of electrons (1 mark) from one mole of gaseous atoms (1 mark) to form one mole of gaseous 1+ ions (1 mark).

(b) Group 2 (1 mark). There is a very large increase/jump between the second and third ionization energies (1 mark). This indicates that the third electron is being removed from an inner shell closer to the nucleus, experiencing significantly more attraction, hence there are 2 valence electrons (1 mark).

(c) \(\text{RAM} = \frac{(24 \times 79.0) + (25 \times 10.0) + (26 \times 11.0)}{100}\) (1 mark)
\(\text{RAM} = \frac{1896 + 250 + 286}{100} = \frac{2432}{100}\) (1 mark)
\(\text{RAM} = 24.32\) (1 mark)

(d) The neutral M (magnesium) has 12 electrons. The \(\text{M}^{2+}\) ion has 10 electrons: \(1s^2 2s^2 2p^6\) (1 mark).

(e) Phosphorus has a outer electronic configuration of \(3s^2 3p^3\) with singly occupied \(3p\) orbitals, whereas sulfur has \(3s^2 3p^4\) (1 mark). In sulfur, one of the \(3p\) orbitals contains a pair of electrons. The mutual repulsion between these paired electrons makes it easier to remove one of them, resulting in a lower first ionization energy (1 mark).

(a)
- M1: Energy/enthalpy change required to remove one mole of electrons
- M2: From one mole of gaseous atoms
- M3: To form one mole of gaseous 1+ ions (equation \(\text{X}(g) \rightarrow \text{X}^+(g) + e^-\) with state symbols can award M2 and M3 if complete).

(b)
- M1: Group 2
- M2: Large jump/gap between 2nd and 3rd ionization energies
- M3: Indicates removal of an electron from a shell closer to the nucleus / shell with lower principal quantum number / 2 electrons in outer shell.

(c)
- M1: Correct expression \(\frac{(24 \times 79.0) + (25 \times 10.0) + (26 \times 11.0)}{100}\)
- M2: Correct calculation of intermediate sum (2432)
- M3: Final evaluation to 2 decimal places: 24.32 (no units required, reject other decimal places).

(d)
- M1: \(1s^2 2s^2 2p^6\) (allow upper case, ignore orbital subscript/superscript formatting if meaning is clear).

(e)
- M1: Phosphorus has 3 singly occupied p-orbitals, whereas Sulfur has a pair of electrons in a p-orbital / Sulfur has paired electrons in a 3p orbital.
- M2: Spin-pair repulsion between the paired electrons in the 3p orbital of sulfur makes the electron easier to remove.

(a) Mass of water lost = \(4.88 - 4.16 = 0.72\text{ g}\) (1 mark)
Molar mass of \(\text{BaCl}_2 = 137.3 + (2 \times 35.5) = 208.3\text{ g mol}^{-1}\)
Moles of \(\text{BaCl}_2 = \frac{4.16}{208.3} = 0.01997\text{ mol}\) (1 mark)
Moles of \(\text{H}_2\text{O} = \frac{0.72}{18.0} = 0.0400\text{ mol}\) (1 mark)
Ratio of \(\text{H}_2\text{O} : \text{BaCl}_2 = \frac{0.0400}{0.01997} = 2.003\text{, so } x = 2\) (1 mark).

(b) \(\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)\) (1 mark for species, 1 mark for state symbols).

(c) Use \(PV = nRT\)
Convert units:
\(P = 101 \times 10^3\text{ Pa}\) or \(101000\text{ Pa}\)
\(T = 97 + 273 = 370\text{ K}\) (1 mark for both conversions)
\(V = 37.0 \times 10^{-6}\text{ m}^3\) (1 mark)
\(n = \frac{PV}{RT} = \frac{101000 \times (37.0 \times 10^{-6})}{8.31 \times 370}\) (1 mark)
\(n = \frac{3.737}{3074.7} = 0.001215\text{ mol}\) (1 mark)
\(M = \frac{m}{n} = \frac{0.105}{0.001215} = 86.42\text{ g mol}^{-1} \approx 86.4\text{ g mol}^{-1}\) (1 mark).

(d) Any one of:
- The intermolecular forces between the gas molecules are negligible.
- The volume occupied by the gas molecules themselves is negligible compared to the volume of the container.
- Collisions between molecules are perfectly elastic.

(a)
- M1: Calculate mass of water lost = 0.72 g
- M2: Calculate moles of anhydrous \(\text{BaCl}_2\) = 0.01997 mol (accept 0.020 mol)
- M3: Calculate moles of water = 0.0400 mol
- M4: Ratio to give whole number \(x = 2\) (must be an integer).

(b)
- M1: Correct ionic equation: \(\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}\)
- M2: Correct state symbols: \((aq)\) on reactants, \((s)\) on product (dependent on M1 or nearly correct equation).

(c)
- M1: Correctly convert temperature to 370 K AND pressure to 101000 Pa
- M2: Correctly convert volume to \(3.70 \times 10^{-5}\text{ m}^3\) (or \(37.0 \times 10^{-6}\text{ m}^3\))
- M3: Rearrange ideal gas equation correctly: \(n = \frac{PV}{RT}\)
- M4: Calculate \(n = 0.001215\text{ mol}\) (accept 0.00121 to 0.00122)
- M5: Calculate \(M = 86.4\text{ g mol}^{-1}\) (must be 3 sig figs, accept 86.1 to 86.7 depending on intermediate rounding. Correct answer with no working scores 5).

(d)
- M1: Accept: gas molecules have negligible volume / zero volume OR no attractive/intermolecular forces between molecules OR elastic collisions.

(a) (i) Sulfur is the central atom, sharing 4 single covalent bonds with 4 fluorine atoms, leaving one lone pair on the sulfur atom. S will have 10 electrons in its outer shell (expanded octet). (1 mark for 4 shared pairs with F, 1 mark for lone pair on S).
(ii) Shape: Seesaw (or distorted tetrahedral) (1 mark).
Bond angle: Accept any value between \(100^\circ\) and \(104^\circ\) (for equatorial-equatorial) and \(170^\circ\) to \(178^\circ\) (axial-axial) OR simply '< 120' and '< 90' (1 mark).
Explanation: S has 5 pairs of electrons (4 bonding, 1 lone pair) in its outer shell. Electron pairs repel to be as far apart as possible. Lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, distorting the bond angles (1 mark).

(b) HF has a much higher boiling temperature than expected because HF molecules can form hydrogen bonds with each other (1 mark). Hydrogen bonds are the strongest type of intermolecular force and require significant energy to break (1 mark).
From HCl to HI, the boiling temperatures increase (1 mark). As we go down the group, the molecules have more electrons (1 mark). This increases the strength of London forces (instantaneous dipole-induced dipole forces) between molecules, requiring more energy to separate them (1 mark).

(c) In graphite, each carbon atom forms three covalent bonds, leaving one delocalised electron per carbon atom free to move and carry a charge along the layers (1 mark). In diamond, each carbon atom is tetrahedrally bonded to four other carbon atoms, so all valence electrons are fixed in localized covalent bonds and cannot move (1 mark).

(a) (i)
- M1: 4 S-F single covalent bonds shown correctly (one electron from S, one from F).
- M2: One lone pair shown on S, and all F atoms have 3 lone pairs (6 non-bonding electrons) completing their octets.
(ii)
- M1: Seesaw shape
- M2: Angle: accept any value < 120° (e.g. 102°) and/or < 90° (e.g. 87° / 173°)
- M3: Explanation: 5 regions of electron density (4 bp, 1 lp) / electron pairs repel to positions of minimum repulsion / lone pairs repel more than bonding pairs.

(b)
- M1: HF has hydrogen bonding (between molecules)
- M2: Hydrogen bonding is significantly stronger than other intermolecular forces, requiring more energy to break.
- M3: From HCl to HI, boiling points increase.
- M4: Due to an increase in the number of electrons (in the molecules)
- M5: Leading to stronger London forces (instantaneous dipole-induced dipole forces) requiring more energy to overcome.

(c)
- M1: Graphite: each carbon bonded to 3 others, leaving 1 delocalised electron per carbon atom free to move.
- M2: Diamond: each carbon bonded to 4 others, no delocalised/mobile electrons.

(a) Thermal stability of Group 2 carbonates increases down the group (1 mark).
Down the group, the ionic radius of the Group 2 cation increases (1 mark) while the 2+ charge remains constant, so the charge density of the cation decreases (1 mark).
This means the cation has a smaller polarising effect on the large carbonate anion, weakening the C-O bonds less (1 mark). Consequently, more energy is required to decompose the carbonate down the group.

(b) \(\text{MgCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)\)
(1 mark for correct species and balancing; 1 mark for correct state symbols).

(c) (i) \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) (2 marks. 1 mark for correct reactants and products; 1 mark for correct balancing).
(ii) This is a disproportionation reaction (1 mark).
The oxidation number of chlorine in \(\text{Cl}_2\) is 0 (1 mark).
In \(\text{NaCl}\), the oxidation number of chlorine is -1, which is a reduction (1 mark).
In \(\text{NaClO}\), the oxidation number of chlorine is +1, which is an oxidation (1 mark).
Since the same chlorine species is simultaneously oxidized and reduced, it is a disproportionation reaction.

(a)
- M1: Thermal stability increases down the group.
- M2: Cation size/ionic radius increases down the group.
- M3: Cation charge density decreases (or polarising power of cation decreases).
- M4: Carbonate ion is polarised less / C-O bond is weakened less, requiring more thermal energy to break.

(b)
- M1: Correct species and balancing: \(\text{MgCO}_3 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\)
- M2: State symbols: \((s)\), \((aq)\) on left, \((aq)\), \((g)\), \((l)\) on right (dependent on M1 or minor balancing slip).

(c) (i)
- M1: Reactants and products: \(\text{Cl}_2\) and \(\text{NaOH}\) yielding \(\text{NaCl}\), \(\text{NaClO}\) (or \(\text{NaOCl}\)), and \(\text{H}_2\text{O}\).
- M2: Correctly balanced: \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\)
(ii)
- M1: Disproportionation
- M2: Oxidation number of Cl in \(\text{Cl}_2\) is 0
- M3: Cl is reduced to -1 in \(\text{NaCl}\)
- M4: Cl is oxidised to +1 in \(\text{NaClO}\) (or sodium chlorate(I)).

(a) Moles of \(\text{S}_2\text{O}_3^{2-}\) reacted = \(\text{concentration} \times \text{volume} = 0.120 \times \frac{21.40}{1000} = 2.568 \times 10^{-3}\text{ mol}\) (1 mark)
From the titration equation, 1 mole of \(\text{I}_2\) reacts with 2 moles of \(\text{S}_2\text{O}_3^{2-}\).
Moles of \(\text{I}_2\) produced = \(\frac{2.568 \times 10^{-3}}{2} = 1.284 \times 10^{-3}\text{ mol}\) (1 mark)
From the first equation, 1 mole of \(\text{ClO}^-\)(aq) produces 1 mole of \(\text{I}_2\)(aq).
Moles of \(\text{ClO}^-\)(aq) in \(10.0\text{ cm}^3\) bleach = \(1.284 \times 10^{-3}\text{ mol}\) (1 mark)
Concentration of \(\text{ClO}^-\)(aq) = \(\frac{\text{moles}}{\text{volume in dm}^3} = \frac{1.284 \times 10^{-3}}{0.0100} = 0.1284\text{ mol dm}^{-3}\) (1 mark)
Rounding to 3 sig figs: \(0.128\text{ mol dm}^{-3}\) (1 mark for correct rounding/final value).

(b) Indicator: Starch indicator (1 mark).
Colour change: Blue-black to colourless (1 mark) (reject: clear / starch added too early / yellow to clear).

(c) (i) Let oxidation state of S be \(x\): \(2x + 3(-2) = -2 \Rightarrow 2x = +4 \Rightarrow x = +2\) (1 mark).
(ii) Let oxidation state of S be \(y\): \(4y + 6(-2) = -2 \Rightarrow 4y = +10 \Rightarrow y = +2.5\) or \(+2 \frac{1}{2}\) (1 mark).

(d) Oxidation half-equation: \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\) (1 mark)
Reduction half-equation: \(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\) (1 mark)
Multiply oxidation half-equation by 5 and combine:
\(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\) (1 mark).

(a)
- M1: Calculate moles of thiosulfate: \(2.568 \times 10^{-3}\text{ mol}\)
- M2: Calculate moles of iodine: \(1.284 \times 10^{-3}\text{ mol}\)
- M3: State moles of hypochlorite = moles of iodine (\(1.284 \times 10^{-3}\text{ mol}\))
- M4: Divide moles by volume (\(0.0100\text{ dm}^3\)) to find concentration
- M5: Final answer: \(0.128\text{ mol dm}^{-3}\) (allow \(0.1284\text{ mol dm}^{-3}\))

(b)
- M1: Starch (added near the end-point)
- M2: Blue-black to colourless

(c)
- M1: +2 (or II)
- M2: +2.5 (or +5/2)

(d)
- M1: Correctly states or uses \(\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-\)
- M2: Correctly states or uses \(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}\)
- M3: Correct combined overall ionic equation: \(\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}\) (ignore state symbols)

(a) (i) Moles of \(\text{CaCO}_3 = \frac{5.00}{100.1} = 0.04995\text{ mol}\) (1 mark)
Theoretical moles of \(\text{Ca(NO}_3)_2 = 0.04995\text{ mol}\) (from 1:1 stoichiometry) (1 mark)
Theoretical mass of \(\text{Ca(NO}_3)_2 = 0.04995 \times 164.1 = 8.197\text{ g}\) (1 mark)
Percentage yield = \(\frac{\text{Actual Mass}}{\text{Theoretical Mass}} \times 100 = \frac{6.85}{8.197} \times 100 = 83.56\% \approx 83.6\%\) (1 mark).

(ii) \(\text{Atom economy} = \frac{M_r(\text{desired product})}{\text{Total } M_r \text{ of reactants}} \times 100\)
Desired product: \(\text{Ca(NO}_3)_2\) (\(M_r = 164.1\)) (1 mark)
Reactants: \(\text{CaCO}_3 + 2\text{HNO}_3 = 100.1 + (2 \times 63.0) = 226.1\) (1 mark)
\(\text{Atom economy} = \frac{164.1}{226.1} \times 100 = 72.58\% \approx 72.6\%\) (1 mark).

(b) Moles of sulfamic acid in original \(250.0\text{ cm}^3\) solution:
\(n = \frac{\text{mass}}{M_r} = \frac{1.45}{97.1} = 0.014933\text{ mol}\) (1 mark)
Moles of sulfamic acid in \(25.00\text{ cm}^3\) portion used in titration:
\(n_{titre} = 0.014933 \times \frac{25.00}{250.0} = 1.4933 \times 10^{-3}\text{ mol}\) (1 mark)
Since sulfamic acid is monoprotic, the reaction with \(\text{NaOH}\) is 1:1:
\(\text{HSO}_3\text{NH}_2 + \text{NaOH} \rightarrow \text{NaSO}_3\text{NH}_2 + \text{H}_2\text{O}\)
Moles of \(\text{NaOH}\) reacted = \(1.4933 \times 10^{-3}\text{ mol}\) (1 mark)
Volume of \(\text{NaOH} = 22.50\text{ cm}^3 = 0.02250\text{ dm}^3\)
Concentration of \(\text{NaOH} = \frac{1.4933 \times 10^{-3}}{0.02250} = 0.06637\text{ mol dm}^{-3}\) (1 mark)
Rounding to 3 significant figures: \(0.0664\text{ mol dm}^{-3}\) (1 mark).

(a) (i)
- M1: Calculate moles of \(\text{CaCO}_3\) = 0.04995 mol
- M2: State theoretical moles of \(\text{Ca(NO}_3)_2\) is also 0.04995 mol
- M3: Calculate theoretical mass of \(\text{Ca(NO}_3)_2\) = 8.197 g (accept 8.20 g)
- M4: Calculate percentage yield = 83.6% (accept 83.5% to 84.0% depending on rounding, must be 3 sig figs).
(ii)
- M1: Correctly identifies desired product \(M_r = 164.1\)
- M2: Correctly calculates sum of reactants \(M_r = 226.1\)
- M3: Calculates atom economy = 72.6% (accept 72.58%, 73%)

(b)
- M1: Calculate moles of sulfamic acid in flask = \(0.014933\text{ mol}\)
- M2: Calculate moles in pipette (25.00 cm³) = \(1.4933 \times 10^{-3}\text{ mol}\)
- M3: State 1:1 ratio so moles of \(\text{NaOH}\) = \(1.4933 \times 10^{-3}\text{ mol}\)
- M4: Divide moles of \(\text{NaOH}\) by volume in dm³ (0.02250 dm³)
- M5: Final answer: \(0.0664\text{ mol dm}^{-3}\) (accept 0.0663 to 0.0665, 3 sig figs).

The bonds broken are 1 \(C=C\) (614 kJ mol\(^{-1}\)) and 1 \(H-H\) (436 kJ mol\(^{-1}\)), giving a total energy input of 1050 kJ mol\(^{-1}\). The bonds formed are 1 \(C-C\) (348 kJ mol\(^{-1}\)) and 2 \(C-H\) (2 \(\times\) 413 = 826 kJ mol\(^{-1}\)), giving a total energy output of 1174 kJ mol\(^{-1}\). Enthalpy change \(\Delta H\) = energy of bonds broken - energy of bonds formed = 1050 - 1174 = -124 kJ mol\(^{-1}\).

1 mark: correct calculation of enthalpy change showing the negative sign.

The structural isomers of \(\text{C}_4\text{H}_9\text{Br}\) are 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane, and 2-bromo-2-methylpropane. Only 2-bromobutane has a chiral carbon atom (carbon 2 is bonded to -H, -CH3, -CH2CH3, and -Br). Thus, only 1 structural isomer contains a chiral carbon atom.

1 mark: correctly identifies that only 1 structural isomer (2-bromobutane) is chiral.

A catalyst provides an alternative reaction pathway with a lower activation energy, which increases the rate of reaction. It does not alter the average kinetic energy of the molecules (which depends on temperature) or change the enthalpy change or equilibrium constant of the reaction.

1 mark: correct identification of the function of a catalyst.

The value of the equilibrium constant \(K_c\) is only affected by temperature. Since the forward reaction is exothermic (\(\Delta H < 0\)), decreasing the temperature will shift the equilibrium position to the right to oppose the change, increasing both the yield of \(\text{SO}_3\) and the value of \(K_c\).

1 mark: correct identification of temperature decrease as the only factor increasing both Kc and yield for an exothermic reaction.

The balanced equation is \(\text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}\). Moles of ethanol = 1.20 / 46.0 = 0.0261 mol. Moles of \(\text{CO}_2\) produced = 2 \(\times\) 0.0261 = 0.0522 mol. Mass of \(\text{CO}_2\) = 0.0522 mol \(\times\) 44.0 g mol\(^{-1}\) = 2.30 g.

1 mark: correct stoichiometry and molar mass calculations leading to 2.30 g.

During electrophilic addition to propene, the hydrogen ion adds to the terminal carbon to form the more stable secondary carbocation intermediate, \(\text{CH}_3\text{CH}^+\text{CH}_3\), rather than the less stable primary carbocation.

1 mark: correct secondary carbocation intermediate identified.

Total mass of solution = 50.0 + 50.0 = 100.0 g. Heat released \(q = m c \Delta T = 100.0 \times 4.18 \times 6.5 = 2717\text{ J} = 2.717\text{ kJ}\). Moles of \(\text{H}_2\text{O}\) formed = \(c \times V = 1.00 \times 0.0500 = 0.0500\text{ mol}\). Enthalpy change \(\Delta H_{neut} = -q / n = -2.717 / 0.0500 = -54.34\text{ kJ mol}^{-1}\), which rounds to -54.3 kJ mol\(^{-1}\).

1 mark: correct calculation of enthalpy of neutralisation with negative sign and appropriate rounding.

Ketones are formed by the oxidation of secondary alcohols. Propan-2-ol is a secondary alcohol, whereas propan-1-ol and 2-methylpropan-1-ol are primary alcohols (which oxidize to aldehydes/carboxylic acids), and 2-methylpropan-2-ol is a tertiary alcohol (which resists oxidation).

1 mark: identifies propan-2-ol as the secondary alcohol that oxidises to a ketone.

Part (a): The complete combustion of propan-1-ol produces carbon dioxide and water. The balanced equation with state symbols is:
\(\text{C}_3\text{H}_7\text{OH(l)} + 4.5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)}\)

Part (b)(i): \(q = m c \Delta T\)
Mass of water \(m = 150\text{ g}\)
\(\Delta T = 55.2 - 19.5 = 35.7\ ^\circ\text{C}\)
\(q = 150 \times 4.18 \times 35.7 = 22383\text{ J} = 22.383\text{ kJ}\)
Rounding to 3 sig figs gives \(22.4\text{ kJ}\).

Part (b)(ii): Mass of propan-1-ol burned = \(122.45 - 121.73 = 0.72\text{ g}\).
Molar mass (\(M_r\)) of propan-1-ol (\(\text{C}_3\text{H}_8\text{O}\)) = \((3 \times 12.0) + (8 \times 1.0) + 16.0 = 60.0\text{ g mol}^{-1}\).
Moles of propan-1-ol = \(0.72 / 60.0 = 0.0120\text{ mol}\).

Part (b)(iii): \(\Delta_c H = -\frac{q}{n} = -\frac{22.383\text{ kJ}}{0.0120\text{ mol}} = -1865.25\text{ kJ mol}^{-1}\).
Rounding to 3 sig figs gives \(-1870\text{ kJ mol}^{-1}\).

Part (c): 1. Incomplete combustion of propan-1-ol, which produces carbon monoxide or soot instead of carbon dioxide, releasing less energy.
2. Evaporation of propan-1-ol from the wick of the burner after weighing but before lighting or after extinguishing, which leads to an overestimation of the fuel actually burned.

Part (d): The student can minimize heat loss by surrounding the apparatus with wind shields (or draft screens) to shield the flame and by placing a lid on top of the copper calorimeter to prevent evaporation and convection losses.

Part (a) [2 marks]:
- Correctly balanced equation with correct molecular formulae [1]
- State symbols correct for all reactants and products (l, g, g, l) [1]

Part (b)(i) [2 marks]:
- Calculation of temperature change (\(35.7\ ^\circ\text{C}\)) and correct substitution into the formula \(q = 150 \times 4.18 \times 35.7\) [1]
- Correct value of \(22.4\text{ kJ}\) (or \(22383\text{ J}\)) [1]

Part (b)(ii) [2 marks]:
- Correctly calculates mass of propan-1-ol burned (\(0.72\text{ g}\)) and states molar mass is \(60.0\text{ g mol}^{-1}\) [1]
- Correct number of moles: \(0.0120\text{ mol}\) [1]

Part (b)(iii) [2 marks]:
- Divides heat energy by moles: \(\frac{22.383}{0.0120}\) [1]
- Correct final value with negative sign and to 3 s.f.: \(-1870\text{ kJ mol}^{-1}\) [1]

Part (c) [2 marks]:
- Incomplete combustion (which produces carbon monoxide or carbon instead of carbon dioxide, releasing less energy) [1]
- Evaporation of fuel from the wick of the burner (which exaggerates the mass of fuel burned) [1]

Part (d) [2 marks]:
- Use of draft screens or wind shield (to prevent flame wavering) [1]
- Placing a lid on the copper calorimeter (to reduce heat loss from the top) [1]

Part (a): The mechanism is nucleophilic substitution (specifically \(\text{S}_\text{N}1\)). The hydroxide ion (\(\text{OH}^-\)) acts as a nucleophile by donating a lone pair of electrons to the electron-deficient carbon atom.

Part (b): The mechanism proceeds in two steps:
1. Heterolytic fission of the carbon-bromine bond: A curly arrow starts from the C-Br bond and points to the Br atom. The carbon is electron-deficient (\(\delta+\)) and the bromine is electron-rich (\(\delta-\)). This step forms a tertiary carbocation intermediate, \(\text{(CH}_3)_3\text{C}^+\), and a bromide ion, \(\text{Br}^-\).
2. Nucleophilic attack: A curly arrow starts from a lone pair on the oxygen of the hydroxide ion, \(\text{OH}^-\), and points directly to the positively charged carbon atom of the carbocation. This forms 2-methylpropan-2-ol.

Part (c): Under hot ethanolic conditions, an elimination reaction takes place. The hydroxide ion behaves as a base by accepting a proton (hydrogen ion) from one of the adjacent methyl carbon atoms, leading to the formation of a carbon-carbon double bond and expulsion of the bromide ion.

Part (d): 2-bromo-2-methylpropane is a tertiary halogenoalkane. It undergoes nucleophilic substitution via the \(\text{S}_\text{N}1\) pathway because it can form a highly stable tertiary carbocation intermediate, stabilized by the electron-releasing inductive effect of the three methyl groups. 1-bromobutane is a primary halogenoalkane; primary carbocations are extremely unstable, preventing the reaction from taking place via \(\text{S}_\text{N}1\). Instead, 1-bromobutane undergoes \(\text{S}_\text{N}2\) substitution because the primary carbon is accessible (minimal steric hindrance) for a direct backside attack by the nucleophile while the C-Br bond breaks simultaneously.

Part (a) [2 marks]:
- Identifies the mechanism as nucleophilic substitution (accept \(\text{S}_\text{N}1\)) [1]
- Identifies the role of the hydroxide ion as a nucleophile [1]

Part (b) [4 marks]:
- Curly arrow from C-Br bond to the Br atom [1]
- Dipoles shown (\(\delta+\) on C, \(\delta-\) on Br) AND correct structure of tertiary carbocation intermediate shown [1]
- Lone pair on oxygen of \(\text{OH}^-\) shown, and curly arrow from the lone pair to the positive carbon atom of the carbocation [1]
- Correct products shown (2-methylpropan-2-ol and \(\text{Br}^-\)) [1]

Part (c) [2 marks]:
- Identifies the type of reaction as elimination [1]
- Identifies the role of the hydroxide ion as a base [1]

Part (d) [4 marks]:
- Mentions that 2-bromo-2-methylpropane is tertiary and 1-bromobutane is primary [1]
- Mentions that tertiary carbocations are highly stable [1] due to the electron-releasing inductive effect of the three methyl/alkyl groups [1]
- Mentions that primary carbocations are highly unstable, but primary halogenoalkanes have less steric hindrance which allows for direct attack in the \(\text{S}_\text{N}2\) mechanism [1]

Part (a): The equilibrium constant expression is:
\(K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}\)
To find the units: \(\frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})} = \text{mol}^{-1}\text{ dm}^3 = \text{dm}^3\text{ mol}^{-1}\).

Part (b): Using the stoichiometry of the reaction:
\(2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}\)
Let the change in moles of \(\text{SO}_3\) be \(+2x\).
Since equilibrium moles of \(\text{SO}_3 = 1.60\text{ mol}\), then \(2x = 1.60 \implies x = 0.80\text{ mol}\).
- Equilibrium moles of \(\text{SO}_2 = 2.00 - 2x = 2.00 - 1.60 = 0.40\text{ mol}\)
- Equilibrium moles of \(\text{O}_2 = 1.50 - x = 1.50 - 0.80 = 0.70\text{ mol}\)

Part (c): To calculate \(K_c\), convert equilibrium moles to concentration by dividing by the volume of \(5.00\text{ dm}^3\):
- \([\text{SO}_2] = \frac{0.40}{5.00} = 0.080\text{ mol dm}^{-3}\)
- \([\text{O}_2] = \frac{0.70}{5.00} = 0.140\text{ mol dm}^{-3}\)
- \([\text{SO}_3] = \frac{1.60}{5.00} = 0.320\text{ mol dm}^{-3}\)

Substitute into \(K_c\):
\(K_c = \frac{(0.320)^2}{(0.080)^2 \times 0.140} = \frac{0.1024}{0.0064 \times 0.140} = \frac{0.1024}{0.000896} = 114.2857\)
Rounding to 3 sig figs gives \(114\) (units: \(\text{dm}^3\text{ mol}^{-1}\)).

Part (d)(i): Since the forward reaction is exothermic (\(\Delta H = -197\text{ kJ mol}^{-1}\)), increasing the temperature shifts the equilibrium in the endothermic (reverse) direction to absorb the added heat.
- Therefore, the equilibrium yield of \(\text{SO}_3\) decreases.
- The value of \(K_c\) decreases because the ratio of product concentration to reactant concentrations decreases.

Part (d)(ii): A catalyst increases the rate of both the forward and reverse reactions by the same factor.
- It has no effect on the position of equilibrium, so the yield of \(\text{SO}_3\) remains unchanged.
- The value of \(K_c\) remains unchanged.

Part (a) [2 marks]:
- Correct expression for \(K_c\) [1]
- Correct units of \(\text{dm}^3\text{ mol}^{-1}\) (or \(\text{mol}^{-1}\text{ dm}^3\)) [1]

Part (b) [2 marks]:
- Equilibrium moles of \(\text{SO}_2 = 0.40\text{ mol}\) [1]
- Equilibrium moles of \(\text{O}_2 = 0.70\text{ mol}\) [1]

Part (c) [3 marks]:
- Divides equilibrium moles of all three species by the volume of \(5.00\text{ dm}^3\) to obtain concentrations [1]
- Correct substitution of concentrations into \(K_c\) expression: \(\frac{(0.320)^2}{(0.080)^2 \times 0.140}\) [1]
- Correct calculation of \(K_c = 114\) (accept 114.3) [1] (Allow consequential error from b)

Part (d)(i) [3 marks]:
- Yield of \(\text{SO}_3\) decreases [1]
- Equilibrium shifts to the left because the forward reaction is exothermic [1]
- \(K_c\) decreases [1]

Part (d)(ii) [2 marks]:
- Yield of \(\text{SO}_3\) is unchanged [1]
- \(K_c\) is unchanged (as a catalyst only increases reaction rate) [1]

Part (a): Activation energy is the minimum energy that colliding particles must possess in order for a reaction to occur (by successfully breaking the chemical bonds in reactants).

Part (b): According to collision theory, increasing the temperature increases the average kinetic energy of the reacting particles. This has two effects:
1. The particles move faster, resulting in more frequent collisions per unit time.
2. Much more significantly, a much larger fraction of the colliding particles possess energy greater than or equal to the activation energy (\(E_a\)). Therefore, a much higher proportion of collisions are successful, which increases the rate of reaction.

Part (c):
- At temperature \(T_1\), the Maxwell-Boltzmann distribution curve starts at the origin, rises to a single peak (representing the most probable energy), and then curves downwards towards the energy axis, asymptotically approaching but never touching it.
- When the temperature is increased to \(T_2\) (where \(T_2 > T_1\)):
1. The peak of the curve shifts to the right (representing a higher most probable energy).
2. The height of the peak decreases (flattens), so the curve is broader.
3. The area under the curve to the right of the activation energy (\(E_a\)) increases significantly, representing a much larger fraction of molecules with energy \(\ge E_a\).

Part (d): A catalyst increases the rate of reaction by providing an alternative reaction pathway that has a lower activation energy. Consequently, at the same temperature, a larger proportion of particles have energy greater than or equal to this lower activation energy, resulting in a higher frequency of successful collisions.

Part (a) [2 marks]:
- Mentions 'minimum energy' [1]
- Required for colliding particles to react / for a collision to be successful [1]

Part (b) [4 marks]:
- Particles have higher kinetic energy / move faster [1]
- Increased frequency of collisions (more collisions per second) [1]
- A much larger fraction/proportion of particles have energy \(\ge E_a\) [1]
- A higher proportion/fraction of collisions are successful (lead to reaction) [1]

Part (c) [4 marks]:
- Curve starts at the origin and asymptotically approaches the x-axis [1]
- At higher temperature \(T_2\), the peak shifts to the right [1]
- At higher temperature \(T_2\), the peak height is lower [1]
- The fraction of molecules with energy \(\ge E_a\) (area under the curve to the right of \(E_a\)) is larger [1]

Part (d) [2 marks]:
- Provides an alternative pathway/mechanism [1]
- With a lower activation energy [1]

Part (a)(i):
- Mass of carbon: \(2.64\text{ g of }\text{CO}_2 \times \frac{12.0\text{ g mol}^{-1}}{44.0\text{ g mol}^{-1}} = 0.72\text{ g}\) of carbon.
- Mass of hydrogen: \(1.62\text{ g of }\text{H}_2\text{O} \times \frac{2.0\text{ g mol}^{-1}}{18.0\text{ g mol}^{-1}} = 0.18\text{ g}\) of hydrogen.
- Mass of oxygen: \(1.38\text{ g} - (0.72\text{ g} + 0.18\text{ g}) = 0.48\text{ g}\) of oxygen.

Part (a)(ii):
Convert masses to moles:
- Moles of C = \(\frac{0.72}{12.0} = 0.060\text{ mol}\)
- Moles of H = \(\frac{0.18}{1.0} = 0.180\text{ mol}\)
- Moles of O = \(\frac{0.48}{16.0} = 0.030\text{ mol}\)

Divide by the smallest value (0.030):
- Ratio C : H : O = \(\frac{0.060}{0.030} : \frac{0.180}{0.030} : \frac{0.030}{0.030} = 2 : 6 : 1\).
The empirical formula of compound **X** is \(\text{C}_2\text{H}_6\text{O}\).

Part (b)(i):
- Ideal gas equation: \(pV = nRT\)
- Convert values to standard SI units:
- \(p = 101\text{ kPa} = 101000\text{ Pa}\)
- \(V = 138\text{ cm}^3 = 138 \times 10^{-6}\text{ m}^3\)
- \(T = 373\text{ K}\)
- Rearrange for \(n\):
\(n = \frac{pV}{RT} = \frac{101000 \times 138 \times 10^{-6}}{8.31 \times 373} = \frac{13.938}{3099.63} = 4.4967 \times 10^{-3}\text{ mol} \approx 4.50 \times 10^{-3}\text{ mol}\).

Part (b)(ii):
- Molar mass (\(M\)) = \(\frac{\text{mass}}{n} = \frac{0.207\text{ g}}{4.4967 \times 10^{-3}\text{ mol}} = 46.03\text{ g mol}^{-1} \approx 46.0\text{ g mol}^{-1}\).
- The empirical formula mass of \(\text{C}_2\text{H}_6\text{O}\) is \((2 \times 12.0) + (6 \times 1.0) + 16.0 = 46.0\text{ g mol}^{-1}\).
- Since the calculated molar mass matches the empirical formula mass, the molecular formula of **X** is also \(\text{C}_2\text{H}_6\text{O}\).

Part (a)(i) [3 marks]:
- Correct mass of C (\(0.72\text{ g}\)) [1]
- Correct mass of H (\(0.18\text{ g}\)) [1]
- Correct mass of O (\(0.48\text{ g}\)) [1]

Part (a)(ii) [3 marks]:
- Division of masses by atomic masses to find moles (C: 0.060, H: 0.180, O: 0.030) [1]
- Simplification to find mole ratio (2 : 6 : 1) [1]
- Correct empirical formula: \(\text{C}_2\text{H}_6\text{O}\) [1] (allow consequential error from a(i))

Part (b)(i) [4 marks]:
- Ideal gas equation stated: \(pV = nRT\) (or rearranged) [1]
- Correct conversion of pressure to Pa AND volume to \(\text{m}^3\) [1]
- Substitution into the formula: \(n = \frac{101000 \times 138 \times 10^{-6}}{8.31 \times 373}\) [1]
- Correct calculated value of \(n = 4.50 \times 10^{-3}\text{ mol}\) (accept range \(4.49 \times 10^{-3}\) to \(4.50 \times 10^{-3}\)) [1]

Part (b)(ii) [2 marks]:
- Calculation of molar mass: \(\frac{0.207}{4.4967 \times 10^{-3}} = 46.0\text{ g mol}^{-1}\) [1]
- Deducing that molecular formula is \(\text{C}_2\text{H}_6\text{O}\) because molar mass matches empirical formula mass [1]

Part (a): The three structural isomers of \(\text{C}_4\text{H}_8\) that are alkenes are:
1. But-1-ene: \(\text{CH}_2=\text{CH-CH}_2\text{-CH}_3\) (skeletal: a zig-zag of four carbons with a double bond at the end).
2. But-2-ene: \(\text{CH}_3\text{-CH}=\text{CH-CH}_3\) (skeletal: a zig-zag of four carbons with a double bond in the middle).
3. 2-Methylpropene: \(\text{CH}_2=\text{C(CH}_3)_2\) (skeletal: a three-carbon chain with a double bond and a methyl group on the middle carbon).

Part (b)(i): The structural isomer that exhibits stereoisomerism is but-2-ene.
- *E*-but-2-ene has the methyl groups on opposite sides of the double bond.
- *Z*-but-2-ene has the methyl groups on the same side of the double bond.

Part (b)(ii): Stereoisomerism occurs because:
1. There is restricted rotation about the carbon-carbon double bond (\(\text{C}=\text{C}\)) due to the presence of the \(\pi\) bond.
2. Each of the carbon atoms in the double bond is attached to two different groups (a methyl group, \(-\text{CH}_3\), and a hydrogen atom, \(-\text{H}\)).

Part (c)(i): Compound **Y** is but-1-ene (or 2-methylpropene). Both are unsymmetrical alkenes that give two structural isomers upon reaction with \(\text{HCl}\).

Part (c)(ii): If **Y** is but-1-ene:
- **Z1** is 2-chlorobutane, structural formula: \(\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3\).
- Explanation: The reaction is an electrophilic addition that goes via a carbocation intermediate. The major product **Z1** is formed via the secondary carbocation intermediate (\(\text{CH}_3\text{CH}_2\text{CH}^+\text{CH}_3\)), which is more stable than the primary carbocation intermediate (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^+\)) due to the electron-releasing inductive effect of the two surrounding alkyl groups.

Part (a) [3 marks]:
- Correct skeletal formula for but-1-ene [1]
- Correct skeletal formula for but-2-ene [1]
- Correct skeletal formula for 2-methylpropene [1]

Part (b)(i) [3 marks]:
- Identifies but-2-ene [1]
- Correct skeletal drawing of *E*-but-2-ene with label [1]
- Correct skeletal drawing of *Z*-but-2-ene with label [1]

Part (b)(ii) [2 marks]:
- Mentions restricted rotation around the \(\text{C}=\text{C}\) double bond [1]
- Mentions that each carbon of the double bond is attached to two different groups (or hydrogens and methyls) [1]

Part (c)(i) [1 mark]:
- But-1-ene (accept 2-methylpropene) [1]

Part (c)(ii) [3 marks]:
- Correct structural formula of 2-chlorobutane: \(\text{CH}_3\text{CH}_2\text{CH(Cl)CH}_3\) (or 2-chloro-2-methylpropane if 2-methylpropene is chosen: \(\text{(CH}_3)_3\text{CCl}\)) [1]
- Explanation refers to carbocation stability: major product goes via secondary carbocation (or tertiary carbocation if methylpropene is chosen) [1]
- Mentions that the secondary (or tertiary) carbocation is more stable than the primary carbocation due to the electron-releasing inductive effect of alkyl groups [1]