2025 OCR AS Level Chemistry B (Salters) - H033 模拟试题及答案详解

The titre is a difference of two readings, so the uncertainties add: \(0.05 + 0.05 = 0.10\ \mathrm{cm^3}\). Percentage uncertainty \(= 0.10 / 20.00 \times 100 = 0.50\%\). Answer B.

B (1 mark). Accuracy mark only. Common wrong answer A arises from using only one reading uncertainty (0.05/20.00); D from 0.10/100; C from doubling again.

\(n(\mathrm{C_3H_8}) = 2.20/44.0 = 0.0500\) mol. \(\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}\), so \(n(\mathrm{CO_2}) = 0.150\) mol. Volume \(= 0.150 \times 24.0 = 3.60\ \mathrm{dm^3}\). Answer C.

C (1 mark). Method: moles of propane, 1:3 ratio, multiply by 24.0. Distractor A comes from forgetting the 1:3 ratio (0.05 x 24.0); D from multiplying by 9 instead of 3.

In \(\mathrm{Cl_2}\) the oxidation state is 0. In NaCl it is \(-1\) (reduction) and in NaClO it is \(+1\) (oxidation). The same element is both oxidised and reduced (disproportionation). Sodium stays at +1. Answer A.

A (1 mark). Accuracy mark only. Distractor C uses +3 (that would be NaClO2); D is wrong because sodium is +1 throughout.

\(n(\mathrm{MgSO_4}) = 1.20/120 = 0.0100\) mol. Mass of water \(= 2.46 - 1.20 = 1.26\) g, so \(n(\mathrm{H_2O}) = 1.26/18.0 = 0.0700\) mol. Ratio \(= 7\). Answer D.

D (1 mark). Method: moles of anhydrous salt, moles of water from mass lost, ratio. Distractor A or B arise from arithmetic slips; C from using 1.26/0.126.

\(K_c\) depends only on temperature, so it is unchanged. Higher pressure favours the side with fewer gas moles (1 mol \(\mathrm{N_2O_4}\) versus 2 mol \(\mathrm{NO_2}\)), so the position shifts left. Answer B.

B (1 mark). Both parts of the statement must be correct. Distractors A and C wrongly make Kc depend on pressure; D has the wrong direction.

Bonds broken: \(436 + 243 = 679\) kJ. Bonds made: \(2 \times 432 = 864\) kJ. \(\Delta H = 679 - 864 = -185\ \mathrm{kJ\,mol^{-1}}\). Answer A.

A (1 mark). Method: broken minus made. Distractor B has the sign reversed; C and D come from using only one 432 value (679 - 432 = 247).

E/Z isomerism needs two different groups on each carbon of the C=C. Pent-2-ene, \(\mathrm{CH_3CH=CHCH_2CH_3}\), has H and \(\mathrm{CH_3}\) on one carbon and H and \(\mathrm{C_2H_5}\) on the other. The others have two identical groups (two H or two \(\mathrm{CH_3}\)) on one carbon. Answer C.

C (1 mark). Accuracy mark only.

The configuration has 18 electrons. \(\mathrm{S^{2-}}\) (16 + 2 = 18) matches. \(\mathrm{Mg^{2+}}\) and \(\mathrm{Na^+}\) have 10 electrons and \(\mathrm{Br^-}\) has 36. Answer D.

D (1 mark). Accuracy mark only. Mg2+ and Na+ are common wrong choices from confusing the neon and argon configurations.

The titre is the difference of two readings, so the absolute uncertainty is \(2\times0.05=0.10\ \text{cm}^3\). Percentage uncertainty = \(0.10/25.00\times100=0.40\%\).

B (1 mark). Accuracy mark only. Common errors: A uses a single reading uncertainty (0.05); C and D are arithmetic slips. ALLOW 0.4%.

\(V=5.00\times10^{-4}\ \text{m}^3\), \(p=1.00\times10^{5}\ \text{Pa}\). \(n=pV/RT=50.0/(8.314\times300)=0.0200\ \text{mol}\). \(M_r=0.88/0.0200=44\).

C (1 mark). Accuracy mark only. Distractors arise from unit-conversion slips or arithmetic errors.

In NaCl chlorine is \(-1\) (reduced); in NaClO chlorine is \(+1\) (oxidised). Chlorine is both oxidised and reduced (disproportionation).

C (1 mark). D is the reaction with hot concentrated alkali, forming chlorate(V).

Mass of water \(=2.46-1.20=1.26\ \text{g}\), \(n=1.26/18=0.0700\ \text{mol}\). \(n(\text{MgSO}_4)=1.20/120=0.0100\ \text{mol}\). Ratio \(=7\).

C (1 mark). Accuracy mark only.

\(K_c\) depends only on temperature. Raising the temperature favours the endothermic direction, so the equilibrium shifts right and \(K_c\) increases. Pressure, catalyst and concentration changes do not alter \(K_c\).

A (1 mark). Accuracy mark only.

\(q=mc\Delta T=100\times4.18\times30.0=12\,540\ \text{J}=12.54\ \text{kJ}\). \(n=0.46/46.0=0.0100\ \text{mol}\). \(\Delta_cH=-12.54/0.0100\approx-1250\ \text{kJ mol}^{-1}\) (exothermic, so negative).

C (1 mark). A omits the division by moles; D has the wrong sign.

On the Boltzmann distribution, a small rise in temperature greatly increases the fraction of molecules with \(E\geq E_a\), so many more collisions are successful. Activation enthalpy and \(\Delta H\) are unchanged by temperature.

B (1 mark). A describes the effect of a catalyst; C is insufficient as collision frequency changes only slightly.

E/Z isomerism needs two different groups on each carbon of C=C. Pent-2-ene, \(\text{CH}_3\text{CH}=\text{CHCH}_2\text{CH}_3\), has H and an alkyl group on each. The others each have a C=C carbon with two identical groups (two H or two CH\(_3\)).

B (1 mark). Accuracy mark only.

\(K_c = \dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \dfrac{(0.020)^2}{0.040} = 0.010\) mol dm\(^{-3}\). B is just [NO\(_2\)], not a Kc value; C uses \([\text{NO}_2]/[\text{N}_2\text{O}_4]\) without squaring; D is the inverse expression \([\text{N}_2\text{O}_4]/[\text{NO}_2]^2\).

A (1 mark). Accuracy mark only: correct Kc expression applied with the square of [NO2]. DO NOT ALLOW B, C or D.

In Cl\(^-\) chlorine is \(-1\) (gain of electrons, reduction). In ClO\(^-\), O is \(-2\) so Cl is \(+1\) (loss of electrons, oxidation). Chlorine is both oxidised and reduced (disproportionation).

A (1 mark). Correct oxidation states and correct assignment of oxidation and reduction. DO NOT ALLOW C (terms reversed).

E/Z isomerism needs restricted rotation about C=C with two different groups on each carbon. Pent-2-ene, CH\(_3\)CH=CHCH\(_2\)CH\(_3\), has H and CH\(_3\) on one carbon and H and C\(_2\)H\(_5\) on the other. In A, C2 carries two methyl groups; in C, C1 carries two hydrogens; in D, C1 carries two hydrogens (and C2 two methyls).

B (1 mark). DO NOT ALLOW A, C or D: each has two identical groups on one C=C carbon.

The titre is the difference of two readings, so the uncertainties add: \(0.05 + 0.05 = \pm 0.10\) cm\(^3\). Percentage uncertainty \(= \dfrac{0.10}{24.50} \times 100 = 0.41\%\). A uses only one reading uncertainty; C is the absolute uncertainty; D doubles the answer again.

B (1 mark). Accuracy mark only: two readings' uncertainties combined, then divided by titre. DO NOT ALLOW A (single reading counted).

(a) In NaClO, Na is +1 and O is -2, so Cl is +1. In Cl- it is -1.
(b) \(\mathrm{ClO^-(aq) + 2I^-(aq) + 2H^+(aq) \rightarrow Cl^-(aq) + I_2(aq) + H_2O(l)}\)
(c) Starch indicator, added when the iodine colour has faded to pale yellow; blue-black to colourless.
(d) n(S2O3 2-) = 0.100 x 22.40/1000 = 2.240 x 10^-3 mol. n(I2) = 1.120 x 10^-3 mol (ratio 1:2). n(ClO-) in 25.0 cm3 = 1.120 x 10^-3 mol (1:1 with I2). In 250.0 cm3: x10 = 1.120 x 10^-2 mol, all from the 10.0 cm3 of bleach. Concentration = 1.120 x 10^-2 / 0.0100 = 1.12 mol dm-3.
(e) Titre is the difference of two readings, so the uncertainty is 2 x 0.05 = 0.10 cm3. Percentage = 0.10/22.40 x 100 = 0.45%.

(a) +1 (1); -1 (1). ALLOW +I and -I. DO NOT ALLOW 1+ or 1-.
(b) Correct species on both sides (1); balanced with correct state symbols (1). ALLOW ECF on state symbols only if species correct.
(c) Starch (1); blue-black to colourless (1). DO NOT ALLOW blue to clear unless 'blue-black' or 'blue' is accepted as AW; ALLOW 'dark blue'. IGNORE 'clear' for colourless.
(d) FIRST CHECK ANSWER ON ANSWER LINE: 1.12 mol dm-3 (3 s.f.) earns all 5 marks.
n(S2O3 2-) = 2.24 x 10^-3 mol (1); n(I2) = half = 1.12 x 10^-3 mol (1); n(ClO-) in 25.0 cm3 = 1.12 x 10^-3 mol (1); x10 for 250 cm3 = 1.12 x 10^-2 mol (1); divide by 0.0100 dm3 = 1.12 mol dm-3 (1). ALLOW 2 or more s.f. (answer 1.1 acceptable). ECF from an earlier step, but not for a wrong final dilution factor already penalised. Standard rounding slip: 1.1 or 1.12 both fine.
(e) Uncertainty of 0.10 cm3 (two readings) (1); 0.10/22.40 x 100 = 0.45% (1). ALLOW 0.446%. If only 0.05 used, 0.22%: ALLOW 1 mark.

(a) The enthalpy change when one mole of a substance burns completely in oxygen under standard conditions (100 kPa, stated temperature, usually 298 K), with all reactants and products in their standard states.
(b) Temperature rise = 26.0 K. q = mc\(\Delta\)T = 150 x 4.18 x 26.0 = 16 300 J = 16.3 kJ. n = 0.85/60.0 = 0.01417 mol. \(\Delta_cH\) = -16.3/0.01417 = -1150 kJ mol-1 = -1.15 x 10^3 kJ mol-1.
(c) \(\mathrm{C_3H_7OH(l) + 4\tfrac{1}{2}O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)}\). \(\Delta_cH\) = [3(-394) + 4(-286)] - (-303) = -1182 - 1144 + 303 = -2023 kJ mol-1.
(d) Heat loss to surroundings / incomplete combustion (soot, CO) / evaporation of fuel / heat absorbed by can. Improvement: add a draught shield/insulation around the flame and can, or place the can closer to the flame.

(a) One mole of substance burns completely in oxygen (1); under standard conditions / all substances in standard states (1). ALLOW 'enthalpy change when one mole is burnt in excess oxygen'. IGNORE 'heat given out' without enthalpy change reference.
(b) FIRST CHECK ANSWER ON ANSWER LINE: -1.15 x 10^3 (kJ mol-1) = 4 marks. q = 150 x 4.18 x 26.0 = 16 300 J/16.3 kJ (1); n = 0.85/60.0 = 0.0142 mol (1); q/n = 1150 kJ mol-1 with correct J to kJ conversion (1); negative sign and 3 s.f. (1). ALLOW -1151 or -1.15 x 10^3. ECF from wrong q or n. Missing negative sign loses the final mark only.
(c) Balanced equation (1) (ALLOW doubled equation if value scaled; state symbols not required for the mark); correct cycle/expression: sum of products minus reactant (1); -2023 (1). ALLOW -2020 and -2.02 x 10^3. ECF on arithmetic slip, sign error loses 1.
(d) Any two from: heat loss to surroundings/air (1); incomplete combustion (1); heat absorbed by the can/thermometer/not included (1); evaporation of propan-1-ol/mass loss from wick before weighing (1). Improvement: draught shield/lid/insulation/copper can/stirring (1). Max 3. DO NOT ALLOW human error or 'not at standard conditions' alone.

(a) But-1-ene, E-but-2-ene, Z-but-2-ene (cis/trans), 2-methylpropene.
(b) The C=C double bond cannot rotate (restricted rotation, pi bond). In but-2-ene each carbon of the C=C carries two different groups (H and CH3), so two arrangements exist. In but-1-ene one carbon of the C=C has two H atoms, so there is only one arrangement.
(c) The pi electrons of C=C attack the delta+ H of H-delta+Br-delta-; the H-Br bond breaks, both electrons going to Br, giving a secondary carbocation CH3CH(+)CH2CH3 and Br-. A lone pair on Br- attacks the C+ to form CH3CHBrCH2CH3.
(d) Bromine is polarised by the alkene and the first step gives Br- (or Br+ added to the carbon) and a positively charged intermediate (carbocation/bromonium). This intermediate is then attacked by any available anion. Cl- is present in the solution and competes with Br-, giving the bromochloro product. If the two Br atoms added simultaneously in one step, no chlorine would be incorporated.
(e) Orange/brown to colourless.
(f) -CH2-C(CH3)2- with continuation bonds.

(a) Any three correct names (1); all four (2). ALLOW cis/trans for Z/E, 2-methylpropene, methylpropene. IGNORE but-2-ene without E/Z.
(b) Restricted/no rotation about C=C (1); each carbon of C=C in but-2-ene has two different groups attached (AW) / but-1-ene has two identical H on one C=C carbon (1). DO NOT ALLOW 'double bond cannot rotate' linked to the wrong reason, e.g. only mentioning a chiral centre.
(c) Curly arrow from C=C to H of HBr (1); dipole delta+ H, delta- Br shown and curly arrow from H-Br bond to Br (1); carbocation intermediate with + on the correct carbon (1); curly arrow from lone pair on Br- to C+ and correct product (1). Arrow must start from the double bond or lone pair, not an atom. DO NOT ALLOW arrows to the wrong atom or a radical mechanism.
(d) Mechanism proceeds via an intermediate/carbocation or cation (1); chloride ion is a nucleophile/anion that competes with bromide for the intermediate (1); if Br2 added in one step/without an ionic intermediate no Cl would appear in the product, so evidence for a two-step mechanism (1). ALLOW bromonium ion. ALLOW AW.
(e) Orange/brown/yellow to colourless (1). DO NOT ALLOW clear alone or 'decolourised' without start colour; IGNORE 'discolours'.
(f) Repeat unit -CH2-C(CH3)2- with open end bonds drawn (1). DO NOT ALLOW a C=C remaining or missing end bonds.

(a) \(K_c = \dfrac{[\mathrm{HI}]^2}{[\mathrm{H_2}][\mathrm{I_2}]}\)
(b) \(K_c = (0.0300)^2 / (0.0050 \times 0.0080) = 9.0\times10^{-4} / 4.0\times10^{-5} = 22.5\)
(c) The forward reaction is exothermic, so raising the temperature favours the endothermic (reverse) direction to oppose the change. The position moves left (less HI), so \(K_c\) decreases.
(d) There are 2 mol of gas on each side, so pressure change does not favour either direction; the position does not change. \(K_c\) depends only on temperature, so it is unchanged.
(e) Rate of forward reaction equals rate of reverse reaction; concentrations of reactants and products remain constant; occurs in a closed system.
(f) Adding H2 raises the denominator, so the ratio [HI]^2/([H2][I2]) falls below \(K_c\). To restore \(K_c\), [HI] must increase (H2 and I2 react), so the amount of HI increases.

(a) Kc = [HI]^2 / ([H2][I2]) (1). Square brackets required; DO NOT ALLOW round brackets for concentration. ALLOW HI2 written as [HI]x[HI].
(b) FIRST CHECK ANSWER ON ANSWER LINE: 22.5 = 2 marks. Correct substitution (1); 22.5 (1). ALLOW 23 or 22.5 to 2+ s.f. No units required; IGNORE units given. ECF from an incorrect expression in (a) for the substitution mark only.
(c) Kc decreases (1); reaction is exothermic in forward direction / reverse reaction is endothermic (1); equilibrium moves left to oppose the temperature rise / absorb heat (1). ALLOW position shifts to reactants. DO NOT ALLOW Kc increases.
(d) No change in position because equal number of moles of gas on both sides (1); Kc unchanged as it only depends on temperature (1). ALLOW 'both sides have 2 mol gas'.
(e) Any two: forward and reverse reaction rates equal (1); concentrations of reactants and products constant (1); closed system (1); both reactions continue / dynamic. Max 2. DO NOT ALLOW 'concentrations equal'.
(f) Ratio/quotient [HI]^2/([H2][I2]) becomes less than Kc (1); so equilibrium shifts right / more HI forms until ratio again equals Kc (1). ALLOW AW. DO NOT ALLOW answers by Le Chatelier alone without reference to the Kc expression.

Convert units: \(V = 180\times10^{-6} = 1.80\times10^{-4}\) m\(^3\); \(p = 101000\) Pa.
\(n = pV/RT = (101000 \times 1.80\times10^{-4})/(8.314\times300) = 18.18/2494.2 = 7.29\times10^{-3}\) mol.
\(M_r = 0.320 / 7.29\times10^{-3} = 43.9\).
C\(_n\)H\(_{2n+2}\): \(14n+2 = 44\), \(n=3\), so propane, C\(_3\)H\(_8\).

FIRST CHECK ANSWER ON ANSWER LINE: \(M_r\) = 43.9 (allow 44 / 43.8 to 44.0) and propane = 3 marks.
✓ Converts units correctly (cm\(^3\) to m\(^3\) and kPa to Pa) and uses \(n = pV/RT\) to get \(n = 7.29\times10^{-3}\) mol (allow 7.3 x 10^-3). DO NOT ALLOW if units are not converted.
✓ \(M_r\) = mass / n = 43.9 (ECF from n; allow two or more significant figures).
✓ Identifies C\(_3\)H\(_8\) / propane from \(M_r\) (ECF from a sensible \(M_r\) value for an alkane, e.g. by using \(14n+2\)). ALLOW working shown as C\(_n\)H\(_{2n+2}\).

\(q = mc\Delta T = 150\times4.18\times18.5 = 11599.5\) J = 11.6 kJ.
\(n(\text{ethanol}) = 1.15/46.0 = 0.0250\) mol.
\(\Delta_c H = -11.5995/0.0250 = -464\) kJ mol\(^{-1}\) (exothermic, so negative).

FIRST CHECK ANSWER ON ANSWER LINE: -464 kJ mol\(^{-1}\) = 3 marks.
✓ \(q = mc\Delta T = 11600\) J / 11.6 kJ (allow 11599.5).
✓ \(n\) = 0.0250 mol.
✓ \(q/n\) = 464 with negative sign and units (kJ mol\(^{-1}\)). If the sign is missing or positive, award 2 marks maximum. ALLOW -463 to -464. ECF from earlier steps. ALLOW -464000 J mol\(^{-1}\). IGNORE significant figures if at least 2 s.f. given, but 3 s.f. expected for the final mark.

\(n(\text{S}_2\text{O}_3^{2-}) = 0.02240\times0.100 = 2.24\times10^{-3}\) mol.
\(n(\text{I}_2) = 2.24\times10^{-3}/2 = 1.12\times10^{-3}\) mol in 25.0 cm\(^3\).
Concentration = \(1.12\times10^{-3}/0.0250 = 0.0448\) mol dm\(^{-3}\).
Mass concentration = \(0.0448\times253.8 = 11.4\) g dm\(^{-3}\).

FIRST CHECK ANSWER ON ANSWER LINE: 11.4 g dm\(^{-3}\) (allow 11.37 / 11.3 to 11.4) = 3 marks.
✓ \(n(\text{thiosulfate})\) = 2.24 x 10^-3 mol.
✓ Uses 1:2 ratio to give \(n(\text{I}_2)\) = 1.12 x 10^-3 mol, then concentration = 0.0448 mol dm^-3. DO NOT ALLOW if ratio is inverted.
✓ Multiplies by 253.8 to give 11.4 g dm^-3. ECF from earlier steps. Answer of 0.0448 mol dm^-3 only (no conversion to g dm^-3) scores 2 marks.

\(K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \dfrac{(0.0480)^2}{0.0120\times0.0080} = \dfrac{2.304\times10^{-3}}{9.6\times10^{-5}} = 24.0\).
Units: mol dm\(^{-3}\) cancel (concentration terms on top and bottom have the same powers), so no units.

✓ Correct \(K_c\) expression: [HI]^2 / ([H2][I2]). DO NOT ALLOW round brackets used as anything other than concentrations; ALLOW square brackets only.
✓ Correct substitution and value 24.0 (allow 24). FIRST CHECK ANSWER: 24.0 earns this mark and the expression mark if the expression is seen. ECF from an incorrect expression.
✓ No units because the units cancel / same number of moles on each side. ALLOW powers of concentration are equal top and bottom.

Mass of water = 2.86 - 1.06 = 1.80 g.
\(M_r(\text{Na}_2\text{CO}_3) = 106.0\), so \(n = 1.06/106.0 = 0.0100\) mol.
\(n(\text{H}_2\text{O}) = 1.80/18.0 = 0.100\) mol.
Ratio = 0.100 : 0.0100 = 10 : 1, so \(x = 10\).

FIRST CHECK ANSWER ON ANSWER LINE: \(x\) = 10 = 3 marks.
✓ Mass of water = 1.80 g and \(n(\text{H}_2\text{O})\) = 0.100 mol.
✓ \(n(\text{Na}_2\text{CO}_3)\) = 0.0100 mol (\(M_r\) = 106.0).
✓ Ratio gives \(x\) = 10. ECF from earlier errors, but \(x\) must be a whole number to be awarded if the working is otherwise correct. ALLOW working via mass ratio of \(M_r\) values.

Equation: \( \text{MgCO}_3(s) \rightarrow \text{MgO}(s) + \text{CO}_2(g) \). Both ions have a 2+ charge, but the ionic radius increases down the group, so Mg2+ has a higher charge density than Ba2+. A more highly charged-dense cation polarises (distorts) the electron cloud of the carbonate ion more, weakening a C-O bond, so less energy is needed to decompose the carbonate. MgCO3 therefore decomposes at a lower temperature and releases CO2 sooner, turning limewater cloudy faster; BaCO3 is more thermally stable and needs stronger heating or longer.

Total 4.5 marks: four points worth 1 mark each plus one point worth 0.5 mark.
1. Correct equation MgCO3(s) -> MgO(s) + CO2(g) (1) ALLOW correct equation for another carbonate only if Mg is not named; state symbols not required. DO NOT ALLOW unbalanced equation.
2. Mg2+ is smaller (radius) than Ba2+ / ionic radius increases down the group, so Mg2+ has the higher charge density (1) ALLOW same charge but smaller size. IGNORE 'atomic' radius alone.
3. Mg2+ polarises / distorts the carbonate ion (electron cloud) more than Ba2+ (1) ORA for Ba2+. DO NOT ALLOW 'attracts the carbonate ions more strongly' without polarisation.
4. C-O bond in the carbonate ion is weakened more / less energy needed to decompose, so MgCO3 decomposes at lower temperature / is less thermally stable (1) ORA.
5. Link to observation: limewater turns cloudy sooner with MgCO3 because CO2 is released sooner / at a lower temperature (0.5) AW.
IGNORE references to electronegativity or shielding unless linked to the above.

Kc = \( \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \). Kc depends only on temperature. The forward reaction is endothermic, so raising the temperature shifts the position to the right (to oppose the change by absorbing heat); [NO2] rises and [N2O4] falls, so Kc increases and the mixture becomes darker brown. Raising pressure (e.g. compressing the syringe) increases both concentrations, but [NO2] is squared, so the quotient becomes larger than Kc. The system must shift left (towards the side with fewer moles of gas, 1 rather than 2) until the quotient again equals Kc. Kc is therefore unchanged. The student is correct for temperature only.

Total 4.5 marks: four points worth 1 mark each plus one point worth 0.5 mark.
1. Verdict: statement is only partly correct / Kc changes with temperature but not with pressure (1) AW.
2. Kc = [NO2]^2 / [N2O4] stated or used correctly (1) ALLOW square brackets omitted only if concentration is clearly indicated. DO NOT ALLOW expression upside down or with 2[NO2].
3. Higher temperature: forward reaction endothermic so position shifts right / more NO2, and Kc increases because the ratio [NO2]^2/[N2O4] is larger at equilibrium (1) AW.
4. Higher pressure: position shifts left / towards N2O4 because fewer moles of gas (1 vs 2) on the left (1) AW. ALLOW explanation in terms of opposing the pressure increase.
5. Kc unchanged by pressure because on compression the quotient exceeds Kc so the system moves left until the ratio returns to the same value of Kc (0.5) AW.
Mark awarded for only one of points 3 and 4 explained without reference to Kc cannot exceed the marks listed. IGNORE rate or collision-theory arguments. ALLOW colour reference (darker brown at higher temperature) as support but no mark.

(c) n(HCl) = 0.100 x 20.45/1000 = 2.045 x 10^-3 mol. n(Na2CO3) in 25.0 cm3 = 1.0225 x 10^-3 mol (1:2 ratio). In 250 cm3: 1.0225 x 10^-2 mol. Molar mass of crystals = 2.50 / 1.0225 x 10^-2 = 244.5 g/mol. Mr(Na2CO3) = 106.0, so water = 138.5 g/mol, x = 138.5/18.0 = 7.69, so x = 8.
(e)(i) Two readings: total uncertainty = 0.10 cm3; 0.10/20.45 x 100 = 0.49%.
(f) A 0.200 mol dm^-3 acid would give a titre of about 10.2 cm3, so 0.10/10.2 = about 0.98%, i.e. the percentage uncertainty doubles; the claim is wrong.

(a) Max 3: weigh the solid accurately (by difference) and dissolve in distilled water in a beaker/with stirring ✓; transfer to a 250 cm3 volumetric flask using a funnel, rinsing the beaker and funnel into the flask ✓; make up to the mark with water so the bottom of the meniscus is on the line (add last drops with a dropping pipette) ✓; stopper and invert several times to mix ✓. DO NOT ALLOW a measuring cylinder/conical flask.
(b) Rough titre (1) is excluded/not concordant ✓. Titres 2-4 are within 0.10 cm3 of each other (concordant); mean = 20.45 cm3 ✓. ALLOW 20.5 if working shown. IGNORE units errors.
(c) FIRST CHECK ANSWER: x = 7.69 (or 8) gets full marks. n(HCl) = 2.045 x 10^-3 ✓; n(Na2CO3) in 25.0 cm3 = half of this = 1.0225 x 10^-3 ✓; x10 = 1.0225 x 10^-2 mol in whole sample ✓; M = 2.50/1.0225 x 10^-2 = 244.5 (ALLOW 244-245) ✓; x = (244.5 - 106.0)/18.0 = 7.69 and 8 ✓ (final mark needs 3 s.f. value and integer). ECF from an earlier step but not for the ratio omitted. A missing x10 scaling gives M = 24.45 and a negative x: no final mark.
(d) Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g): correct formulae and balancing ✓; all state symbols correct ✓. ALLOW ionic equation CO3^2-(aq) + 2H+(aq) -> H2O(l) + CO2(g) for both marks.
(e)(i) Two readings used, so 2 x 0.05 = 0.10 cm3 ✓; 0.10/20.45 x 100 = 0.49% (ALLOW 0.5%, 0.489) ✓. Using 0.05 only gives 0.24%: 1 mark only (ECF).
(ii) Use more dilute acid (e.g. 0.0500 mol dm^-3) / larger titre / use a larger mass of crystals in same volume ✓ (any one). DO NOT ALLOW 'use a more precise burette' unless a smaller scale division is stated. IGNORE repeat titrations.
(f) Claim is not correct ✓ (verdict must be stated). Titre would halve to about 10 cm3 so the fixed 0.10 cm3 uncertainty is a larger percentage (about 1%, doubled) ✓. AW.

(c) n(S2O3^2-) = 0.0500 x 0.01840 = 9.20 x 10^-4 mol. n(I2) = 4.60 x 10^-4 mol. n(ClO-) in 25.0 cm3 = 4.60 x 10^-4 mol (1:1). In 250 cm3 = 4.60 x 10^-3 mol, all from 10.0 cm3 bleach. Concentration = 4.60 x 10^-3 / 0.0100 = 0.460 mol dm^-3. Mass concentration = 0.460 x 74.5 = 34.3 g dm^-3.
(d)(ii) NaOH removes H+, lowering [H+]. The quotient [HClO][H+][Cl-]/[Cl2] becomes smaller than Kc. Kc is constant at constant temperature, so equilibrium moves to the right to restore the ratio.

(a)(i) Chlorine has oxidation state +1 ✓. (ii) +1 and -1 (both needed) ✓. ALLOW +I / I+.
(b) ClO- + 2H+ + 2e- -> Cl- + H2O: species correct ✓; balanced for atoms and charge (electrons on the left) ✓. ALLOW 2e with no sign error penalty only if charge balances. DO NOT ALLOW electrons on right.
(c) FIRST CHECK ANSWER: 0.460 mol dm^-3 and 34.3 g dm^-3 gets full marks. n(S2O3^2-) = 9.20 x 10^-4 ✓; n(I2) = half = 4.60 x 10^-4 ✓; n(ClO-) = n(I2) (1:1) = 4.60 x 10^-4 in the 25.0 cm3 ✓; x10 for dilution and divide by 0.0100 dm3 gives 0.460 mol dm^-3 ✓; x74.5 = 34.3 g dm^-3 (ALLOW 34.2-34.3; ECF from the mol dm^-3 value) ✓. If x10 omitted, concentration = 0.0460: max 4 with ECF.
(d)(i) Kc = [HClO(aq)][H+(aq)][Cl-(aq)] / [Cl2(aq)] ✓. DO NOT ALLOW round brackets or inclusion of H2O. ALLOW omission of state symbols.
(ii) OH- reacts with/removes H+ so [H+] falls ✓; the ratio of concentrations (the reaction quotient) is now less than Kc ✓; Kc is unchanged at constant temperature, so the equilibrium shifts right (more products formed) to restore the value of Kc ✓.
(iii) Chlorine is both oxidised and reduced / same element in the same reaction ✓; oxidation state 0 to +1 (in HClO) and 0 to -1 (in Cl-) ✓.
(e) Risk: toxic/poisonous/corrosive gas, hazardous if leaked ✓. Benefit: kills bacteria/microorganisms/sterilises water, preventing disease ✓. IGNORE 'harmful to the environment' alone.

(b) q = mcΔT = 150 x 4.18 x 24.5 = 15361.5 J = 15.36 kJ. n = 0.74/74.0 = 0.0100 mol. ΔcH = -15.36/0.0100 = -1536, so -1540 kJ/mol (3 s.f.).
(d) Formation: 4C(s) + 5H2(g) + 1/2 O2(g) -> C4H9OH(l). Cycle: elements -> 4CO2 + 5H2O (4 x -394 + 5 x -286 = -1576 - 1430 = -3006) equals ΔfH + (-2676). ΔfH = -3006 + 2676 = -330 kJ/mol.
(e) C4H9OH + 6O2 -> 4CO2 + 5H2O. n(CO2) = 4 x 0.0100 = 0.0400 mol; V = 0.0400 x 24.0 = 0.960 dm3.

(a) Enthalpy change when one mole ✓ of a substance is completely burned/reacts fully in oxygen with all reactants and products in their standard states (at 100 kPa, stated temperature, usually 298 K) ✓. ALLOW heat energy change for enthalpy change. IGNORE 'under standard conditions' if no detail.
(b) FIRST CHECK ANSWER: -1540 (kJ mol^-1) gets full marks. q = 150 x 4.18 x 24.5 = 15361.5 J (15.4 kJ) ✓; n = 0.74/74.0 = 0.0100 mol ✓; divide q by n: 1536 kJ mol^-1 ✓; negative sign (exothermic) and 3 s.f. = -1540 (ALLOW -1.54 x 10^3) ✓. If 24.5 is replaced by 297.5 K or mass 0.74 used as water, no method mark. ECF from n.
(c) Any two: heat lost to surroundings/air/the can/not all heat transferred to water ✓; incomplete combustion (soot/CO) ✓; evaporation of alcohol / not standard conditions / water not stirred ✓. DO NOT ALLOW 'human error' or 'impurities' alone.
(d) Cycle drawn correctly (elements at top, products CO2 and H2O at bottom, with arrows in correct directions and values) ✓; ΔfH = 4(-394) + 5(-286) - (-2676) ✓; = -330 kJ mol^-1 (sign needed) ✓. ECF for arithmetic slips; ALLOW 1 mark for +330.
(e) C4H9OH + 6O2 -> 4CO2 + 5H2O ✓; n(CO2) = 0.0400 mol ✓; volume = 0.960 dm3 (ALLOW 960 cm3, 0.96) ✓. ECF using the moles in (b).
(f) Max 4: at higher temperature the curve peak is lower and shifts to the right/ greater spread ✓; total area under both curves is the same (same number of molecules) ✓; the Ea line is fixed, and a larger proportion/number of molecules has energy greater than or equal to Ea ✓; more frequent successful/effective collisions, so rate increases ✓. DO NOT ALLOW 'molecules collide more often' as the only explanation.

(a) The C=C bond cannot rotate, and each carbon carries two different groups (H and CH3), so two arrangements exist: E (CH3 groups on opposite sides) and Z (same side).
(b)(i) The C=C pi electrons attack the H(delta+) of H-Br (curly arrow from double bond to H, and from H-Br bond to Br); Br- leaves, giving CH3CH+CH2CH3 and Br-. Br- then attacks C+ with a curly arrow from its lone pair to give CH3CHBrCH2CH3.
(ii) Cl- ions can only be attached if a positively charged intermediate forms in step one; Cl- competes with Br- to attack it, so mixed product forms.
(e) Butan-2-ol is a secondary alcohol, and oxidation gives butanone. Orange Cr2O7^2- is reduced to green Cr3+. Reflux lets the mixture boil without loss of volatile alcohol/ketone, giving complete oxidation. IR: broad O-H at 3200-3600 cm^-1 disappears, strong C=O at 1640-1750 cm^-1 appears.

(a)(i) Restricted/no rotation about the C=C double bond ✓; each carbon of C=C is bonded to two different groups/atoms (H and CH3) ✓.
(ii) Two skeletal formulae each with a correct zig-zag chain and a C=C ✓; E labelled for CH3 groups on opposite sides and Z for the same side ✓ (both needed). DO NOT ALLOW both labelled the same way.
(b)(i) Dipole shown on H-Br (H delta+, Br delta-) ✓; curly arrow from C=C double bond to H atom ✓; curly arrow from the H-Br bond to Br ✓; correct carbocation CH3CH+CH2CH3 and Br- ✓. Final step (arrow from Br- lone pair to C+) needed for product with 2-bromobutane - award this in place of the dipole mark if all other marks are scored (max 4). DO NOT ALLOW arrows starting from a carbon atom rather than the bond.
(ii) Cl- as well as Br- is present in the solution and attacks the intermediate (carbocation/positive ion) ✓; if Br2 added in a single step to both carbons simultaneously, only BrCH2CH2Br would form, so the Cl-containing product shows the intermediate is formed first ✓. AW.
(c) Orange/brown/yellow to colourless ✓. DO NOT ALLOW 'clear' for the final colour. IGNORE decolourises alone.
(d) -CH(CH3)-CH(CH3)- with continuation bonds shown ✓ and both methyl groups on separate carbons of the 2-carbon backbone ✓. ALLOW brackets with n. DO NOT ALLOW a double bond in the repeat unit.
(e)(i) Butanone ✓; orange to green ✓; reflux allows heating for a long time/ without losing volatile reactants/ products (condensed back into flask) so the alcohol is fully oxidised ✓. ALLOW butan-2-one. DO NOT ALLOW butanal or butanoic acid.
(ii) Broad O-H absorption at about 3200-3600 cm^-1 (alcohol) is absent in the product ✓; strong C=O absorption present at about 1640-1750 cm^-1 ✓. ALLOW 1700 cm^-1 and 3300 cm^-1 as a stated wavenumber. IGNORE C-H absorptions.

\( K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \dfrac{(0.0940)^2}{(0.0120)(0.0120)} = \dfrac{8.836\times10^{-3}}{1.44\times10^{-4}} = 61.4 \)

(i) Adding H\(_2\) increases the denominator, so the ratio \( [\text{HI}]^2/([\text{H}_2][\text{I}_2]) \) falls below \( K_c \). To restore the ratio to \( K_c \) (constant at constant temperature), HI must increase and H\(_2\)/I\(_2\) decrease, so the position shifts to the right (net forward reaction).

(ii) \( K_c \) is lower (61.4) at the higher temperature than at 500 K (160), so the position shifts toward reactants as temperature rises. Raising temperature favours the endothermic direction, so the backward reaction is endothermic and the forward reaction is exothermic (\( \Delta H \) negative).

(iii) There are 2 mol gas on the left and 2 mol on the right, so changing the pressure changes all concentrations by the same factor and the ratio in the \( K_c \) expression is unchanged. No change in position.

LEVEL OF RESPONSE (6 marks)

FIRST CHECK ANSWER: \( K_c = 61.4 \) (ALLOW 61 or 61.36 to 2+ s.f.; units not required, DO NOT ALLOW a wrong unit if one is given).

Level 3 (5-6 marks): Correct Kc calculation AND all three explanations (i), (ii), (iii) given, with (i) clearly linked to the Kc expression/ratio. The reasoning is logical, complete and clearly communicated. Minor slips only.

Level 2 (3-4 marks): Correct Kc (or ECF method) and two of the three explanations OR all three with weak/incomplete reasoning (e.g. (i) stated without reference to Kc). Some logical structure.

Level 1 (1-2 marks): Kc expression or calculation attempted, and/or one explanation with correct conclusion. Reasoning limited or muddled.

Level 0 (0 marks): No relevant response.

Indicative scientific points:
1. \( K_c = [\text{HI}]^2/([\text{H}_2][\text{I}_2]) \) correct expression (ALLOW square brackets omitted only if concentrations clearly meant) ✓
2. Substitution and answer 61.4 ✓ (ECF from a wrong expression)
3. (i) Adding H2 makes the ratio/quotient smaller than Kc ✓
4. Net forward reaction / equilibrium shifts right until ratio = Kc again (Kc constant at constant T) ✓
5. (ii) Kc decreases as T increases, so position moves left/toward reactants at higher T ✓
6. Forward reaction is exothermic (\( \Delta H \) negative) because the endothermic (backward) direction is favoured by higher T ✓
7. (iii) No change in position as moles of gas are equal (2 and 2) ✓, concentration terms cancel in Kc so ratio unchanged ✓

IGNORE: Le Chatelier statements alone with no reference to Kc for (i) (limits to Level 2). ALLOW AW throughout. DO NOT ALLOW 'Kc changes when H2 is added'. Science decides the level; quality of communication decides the mark within the level.

Product A (aldehyde): Place acidified dichromate(VI) in a flask with a dropping funnel / add it gradually to warm propan-1-ol, with the flask set up for distillation (still head, thermometer, condenser, receiver). Propanal has no hydrogen bonding between molecules so it has a much lower boiling point (about 49 °C) than propan-1-ol (about 97 °C), and is distilled off as it forms. Removing it from the oxidising mixture prevents further oxidation. Alcohol is in excess.

Product B (carboxylic acid): Use an excess of acidified dichromate(VI) and heat the mixture under reflux (flask, vertical condenser with water in at the bottom, heating mantle or water bath, anti-bumping granules). Reflux keeps the volatile alcohol and aldehyde in the flask so they are oxidised completely to the acid. The acid is then distilled off after reflux.

Colour change: orange (Cr\(_2\)O\(_7^{2-}\)) to green (Cr\(^{3+}\)).

Equations: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{CHO} + \text{H}_2\text{O} \); \( \text{CH}_3\text{CH}_2\text{CHO} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{COOH} \).

Purity: the IR spectrum of Product B should show a very broad O–H absorption (about 2500–3300 cm\(^{-1}\)) and C=O (about 1680–1750 cm\(^{-1}\)); a separate sharp alcohol O–H band near 3200–3600 cm\(^{-1}\) would indicate unreacted alcohol. Alternatively, a boiling-point check: pure propanoic acid boils at about 141 °C, so a constant temperature at this value on redistillation shows no lower-boiling aldehyde or alcohol.

LEVEL OF RESPONSE (6 marks)

Level 3 (5-6 marks): Both preparations described with correct apparatus/conditions (distillation for A; reflux with excess oxidant for B), a clear explanation linking method to product (volatility/boiling points, further oxidation), correct colour change, both equations, and a valid purity check. Logical, well-structured.

Level 2 (3-4 marks): Both methods identified with some correct conditions and an explanation of at least one difference; some of the colour change, equations or purity test present. May have gaps or one error.

Level 1 (1-2 marks): Some correct points (e.g. orange to green, one equation, or reflux/distillation named) but little explanation linking method to product.

Level 0 (0 marks): No relevant response.

Indicative scientific points:
1. A: distillation set-up/distil off product as it forms ✓
2. A: aldehyde has lower boiling point than alcohol (no hydrogen bonding) so it distils off and avoids further oxidation ✓ (ORA for alcohol; comparison needed)
3. A: excess alcohol / controlled (dropwise) addition of oxidant and gentle warming ✓
4. B: heat under reflux with excess oxidising agent ✓
5. B: reflux condenses vapours back so aldehyde is retained and fully oxidised to acid ✓
6. Colour: orange to green (Cr2O7 2- to Cr3+) ✓ (DO NOT ALLOW 'blue' or 'purple')
7. CH3CH2CH2OH + [O] -> CH3CH2CHO + H2O ✓
8. CH3CH2CHO + [O] -> CH3CH2COOH (ALLOW CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O) ✓
9. Purity: IR (no sharp alcohol O-H; broad acid O-H present; no aldehyde C=O alone) OR boiling-point/redistillation check at about 141 °C ✓ (ALLOW 2,4-DNP or Tollens' test negative; DO NOT ALLOW test that is positive for acids as well)

ALLOW AW. ECF for names/formulae. IGNORE safety precautions unless linked to method. Science decides the level; quality of communication decides the mark within the level.

Axes: concentration / mol dm\(^{-3}\) on the x-axis (0 to 1.00, e.g. 2 cm per 0.10), initial rate / cm\(^3\) s\(^{-1}\) on the y-axis (0 to 0.70). Plot all five points. The points at 0.20, 0.40, 0.80 and 1.00 mol dm\(^{-3}\) lie close to a straight line through the origin, \(\text{rate} \approx 0.61 \times [\text{H}_2\text{O}_2]\), consistent with rate being proportional to concentration. At 0.60 mol dm\(^{-3}\) the line gives about 0.37 cm\(^3\) s\(^{-1}\), but the measured value is 0.46 cm\(^3\) s\(^{-1}\), about 0.09 above the line and far larger than the scatter of the other points. So (0.60, 0.46) is the anomaly. The line of best fit ignores this point.

Mark points (3 marks):

1. Axes labelled with quantity and units (concentration / mol dm\(^{-3}\) on x; initial rate / cm\(^3\) s\(^{-1}\) on y), scale uses more than half of each axis, and all five points plotted accurately to within half a small square. ALLOW the axes the other way round only if the student is following the question wording; the question specifies rate on the vertical axis. DO NOT ALLOW awkward scales (e.g. 3 or 7 squares per unit) or points not plotted within tolerance.

2. A single straight line of best fit (ruled, with a balance of points either side), ignoring the anomalous point. ALLOW a line through the origin or one close to it. DO NOT ALLOW a line that joins dot to dot, a thick or feathered line, or a curve. ECF on plotted points.

3. Point at 0.60 mol dm\(^{-3}\) (0.46 cm\(^3\) s\(^{-1}\)) circled AND justified: it lies well off / far above the line of best fit (AW), while the other points lie close to the line. ALLOW 'rate expected to be about 0.37'. IGNORE vague statements such as 'it is wrong' with no reference to the line. No mark for the circle without justification.

\(n(H_2SO_4) = 0.0500 \times 18.40/1000 = 9.20 \times 10^{-4}\) mol.
\(n(NaOH) = 2 \times 9.20 \times 10^{-4} = 1.84 \times 10^{-3}\) mol.
Concentration = \(1.84 \times 10^{-3} / 0.0250 = 0.0736\) mol dm\(^{-3}\).
\(M_r(NaOH) = 40.0\), so concentration = \(0.0736 \times 40.0 = 2.94\) g dm\(^{-3}\).

FIRST CHECK ANSWER ON ANSWER LINE: 2.94 g dm\(^{-3}\) = 3 marks.
1 mark: \(n(H_2SO_4) = 9.20 \times 10^{-4}\) mol and doubled to give \(n(NaOH) = 1.84 \times 10^{-3}\) mol (method; ratio 2:1).
1 mark: concentration in mol dm\(^{-3}\) = 0.0736 (ALLOW 0.074). ECF from the first step.
1 mark: multiplied by 40.0 to give 2.94 g dm\(^{-3}\) (ALLOW 2.9 or more sig figs, e.g. 2.944).
DO NOT ALLOW answers with 1 sig fig. ECF from earlier steps but not later ones. Missing 2:1 ratio gives 1.47 g dm\(^{-3}\): max 2 marks.

\(V = 285 \times 10^{-6}\) m\(^3\), \(p = 101000\) Pa.
\(n = pV/RT = (101000 \times 2.85 \times 10^{-4})/(8.314 \times 300) = 0.01154\) mol.
\(M_r = 0.480 / 0.01154 = 41.6\).

FIRST CHECK ANSWER ON ANSWER LINE: 41.6 = 2 marks.
1 mark: \(n = pV/RT\) with correct unit conversions (Pa and m\(^3\)) giving \(n = 0.0115\) mol (ALLOW 0.01154).
1 mark: \(M_r = 0.480/n = 41.6\). ECF from the first step.
ALLOW 41.5 to 41.7 from rounding. A power-of-ten error in unit conversion loses the first mark; ECF for the second.

\(n(S_2O_3^{2-}) = 0.100 \times 24.80/1000 = 2.48 \times 10^{-3}\) mol.
\(n(I_2) = 1.24 \times 10^{-3}\) mol.
\(n(ClO^-) = n(I_2) = 1.24 \times 10^{-3}\) mol in 10.0 cm\(^3\).
Concentration = \(1.24 \times 10^{-3}/0.0100 = 0.124\) mol dm\(^{-3}\).

FIRST CHECK ANSWER ON ANSWER LINE: 0.124 mol dm\(^{-3}\) = 3 marks.
1 mark: \(n(S_2O_3^{2-}) = 2.48 \times 10^{-3}\) mol.
1 mark: \(n(I_2) = 1.24 \times 10^{-3}\) mol (ratio 1:2) and \(n(ClO^-) = n(I_2)\) (ratio 1:1). ECF.
1 mark: concentration = 0.124 mol dm\(^{-3}\) (ALLOW 0.12). ECF.
If the 1:2 ratio is omitted the answer is 0.248: max 2 marks.

\(M_r\) of reactants = 16.0 + 18.0 = 34.0.
\(M_r\) of desired product = \(3 \times 2.0 = 6.0\).
Atom economy = \(6.0/34.0 \times 100 = 17.6\%\).

FIRST CHECK ANSWER ON ANSWER LINE: 17.6% = 2 marks.
1 mark: total \(M_r\) of reactants 34.0 and \(M_r\) of \(3H_2\) = 6.0 (must include the factor of 3).
1 mark: \(6.0/34.0 \times 100 = 17.6\%\). ALLOW 17.7 or 17.65.
Using only one \(H_2\) (2.0/34.0 = 5.88%): max 1 mark. ECF for the second mark.

Mass of mixture = 100 g.
\(q = mc\Delta T = 100 \times 4.18 \times 6.8 = 2842\) J = 2.84 kJ.
\(n(H_2O) = 1.00 \times 0.0500 = 0.0500\) mol.
\(\Delta_{neut}H = -2.842/0.0500 = -56.8\) kJ mol\(^{-1}\).

FIRST CHECK ANSWER ON ANSWER LINE: \(-56.8\) kJ mol\(^{-1}\) = 3 marks.
1 mark: \(q = 100 \times 4.18 \times 6.8 = 2842\) J (mass 100 g, not 50 g).
1 mark: \(n = 0.0500\) mol.
1 mark: \(-56.8\) kJ mol\(^{-1}\) with negative sign and conversion to kJ. ALLOW \(-56.9\) or \(-57\). ECF.
DO NOT ALLOW a positive value for the final mark. Answer of \(-56800\) with J units loses the conversion mark.

\(K_c = [HI]^2/([H_2][I_2]) = (0.0960)^2/(0.0120 \times 0.0150) = 9.216 \times 10^{-3}/1.80 \times 10^{-4} = 51.2\). There are equal numbers of moles on each side so \(K_c\) has no units.

FIRST CHECK ANSWER ON ANSWER LINE: 51.2 = 2 marks.
1 mark: \(K_c = [HI]^2/([H_2][I_2])\) (square brackets needed; DO NOT ALLOW round brackets).
1 mark: substitution and value 51.2 (ALLOW 51). ECF from an incorrect expression, e.g. omitting the square gives 6.4 (1 mark max overall if the expression is wrong).
IGNORE units; if given they must be 'none'.

HBr is polar, H\(^{\delta+}\)-Br\(^{\delta-}\), so the hydrogen is electrophilic. The electron-rich C=C attacks H. The electron pair in the pi bond forms a new C-H bond to the terminal carbon (CH\(_2\)), leaving a positive charge on the second carbon, which gives a secondary carbocation. Simultaneously the H-Br bond breaks heterolytically and both electrons go to Br, forming Br\(^-\). A lone pair on Br\(^-\) then attacks the carbocation, forming the C-Br bond in 2-bromobutane.

1 mark: curly arrow starting from the C=C double bond (the bond, not an atom) and pointing to the H of HBr; partial charges H\(^{\delta+}\) and Br\(^{\delta-}\) shown on HBr. ALLOW the arrow to point towards H with its head on or near the H atom. DO NOT ALLOW an arrow starting from a carbon atom.
1 mark: curly arrow from the H-Br bond to the Br atom (head at or near Br).
1 mark: correct carbocation \(\mathrm{CH_3CH_2C^+HCH_3}\) with a + charge on C2 (secondary), and Br\(^-\) shown with a negative charge and lone pair. ALLOW a full + on the correct carbon in a displayed or skeletal formula. DO NOT ALLOW the primary carbocation. ECF not available for wrong carbocation in the final mark unless the arrow and product are consistent.
1 mark: curly arrow from a lone pair on Br\(^-\) (or from the negative charge) to C\(^+\), and the product 2-bromobutane shown correctly. DO NOT ALLOW an arrow starting from the Br atom without a lone pair or charge.
IGNORE the use of a solvent; IGNORE the minor product 1-bromobutane.