Reaction kinetics

3(a) 3 bonding-pair centres and one lone pair (on iodine) 1 3(b) 3I + 6NaOH → NaIO + 5NaI + 3H O 1 2 3 2 © UCLES 2021 Page 6 of 13 3(c)(i) M1: E⦵ for IO – / H O = –0.68 + 1.19 = +0.51 (∴ feasible) 3 cell 3 2 2 M2: E⦵ for H O / I = +1.77 – 1.19 = +0.58 (∴ feasible) cell 2 2 2 M3: 5H O + I → 4H O + 2IO – + 2H+ 2 2 2 2 3 3(c)(ii) 2H O → 2H O + O 1 2 2 2 2 3(d)(i) M1: first order w.r.t. H O AND change in conc. × 1.5 gives increase rate × 1.5 (expts 3 / 4) 3 2 2 M2: first order w.r.t. IO – AND change in conc. × 2 gives increase rate × 2 (as reaction first order w.r.t. H O ) (expts 1 / 3) 3 2 2 M3: zeroth order w.r.t. H+ AND change in conc. has no effect on rate (expts 1 / 3 / 4 and 2) 3(d)(ii) rate = k[H O ][IO –] ecf 1 2 2 3 3(d)(iii) M1: k = 8.82 × 10–5 ÷ (0.150 × 0.140) = 4.20 × 10–3 min 2sf ecf 2 M2: mol–1 dm3 s–1 ecf 3(e)(i) K = [Pb2+][IO –]2 1 sp 3 3(e)(ii) M1: 3.69 × 10–13 = x(2x)2 OR x = ∛(3.69 × 10–13 ÷ 4) 2 M2: = 4.5(2) × 10–5 (mol dm–3) min 2sf ecf 3(f)(i) M1: ΔS = ½(192) + ½(205) + ½(261) + 2(70) – 42 2 M2: (+)427 (J K–1 mol–1) ecf 3(f)(ii) ΔG (always) negative because 1 • ΔH < 0 / negative OR exothermic AND • ΔS > 0 / positive OR –TΔS < 0 for all T Question Answer Marks

5(a) measure volume / amount of oxygen formed / mass lost / and time / against time / per unit time 1 OR measure absorbance / transmission against time / per unit time 5(b)(i) time taken for the concentration / mass / amount 1 of a reactant to fall to half (its original value) / to halve 5(b)(ii) t = 150 s 1 1 / 2 AND evidence on graph / paper of one half-life 5(b)(iii) no change 1 5(c)(i) M1: evidence on graph of tangent AND 4 to 5 × 10–4 2 M2: mol dm–3 s–1 © UCLES 2021 Page 11 of 17 5(c)(ii) (c)(i) / 0.10 AND s–1 1 5(d) M1: NO + O → NO + O 2 2 3 3 2 M2: NO + NO → N O 2 3 2 5 Question Answer Marks

7(a)(i) the power to which a concentration of a reactant is raised in the rate equation / law 1 7(a)(ii) M1: (using expt 1 and 3) as [ClO ] × 2.5 rate × 6.25 so 2nd order 2 2 M2: (using expt 1 and 2) as [OH–] × 4 rate × 4 so 1st order 7(a)(iii) rate = k[ClO ]2[OH–] 1 2 7(a)(iv) M1: 2 k = rate / [ClO ]2[OH–] 2 k = 7.20 × 10-4 / (0.02)2(0.03) k = 60 M2: mol–2 dm6 min–1 © UCLES 2021 Page 13 of 16 7(b)(i) 1 structure of phenol: C H OH OR 6 5 7(b)(ii) tangent drawn correctly AND rate = 0.015 / 260 = 5.8 × 10–5 1 ALLOW values consistent with tangent drawn at 100 sec 7(c) 1 AND half-life decreases (1st box) © UCLES 2021 Page 14 of 16

5(a) measure volume / amount of oxygen formed / mass lost / and time / against time / per unit time 1 OR measure absorbance / transmission against time / per unit time 5(b)(i) time taken for the concentration / mass / amount 1 of a reactant to fall to half (its original value) / to halve 5(b)(ii) t = 150 s 1 1 / 2 AND evidence on graph / paper of one half-life 5(b)(iii) no change 1 5(c)(i) M1: evidence on graph of tangent AND 4 to 5 × 10–4 2 M2: mol dm–3 s–1 © UCLES 2021 Page 11 of 17 5(c)(ii) (c)(i) / 0.10 AND s–1 1 5(d) M1: NO + O → NO + O 2 2 3 3 2 M2: NO + NO → N O 2 3 2 5 Question Answer Marks

5(a)(i) CH CHClCOOH + H O ⇌ CH CHClCOO– + H O+ 1 3 2 3 3 OR CH CHClCOOH ⇌ CH CHClCOO– + H+ 3 3 5(a)(ii) M1: [H+] = 10–1.51 = 0.0309 (mol dm– 3) 2 M2: K = 0.03092/0.60 = 1.592 × 10–3 ecf a pK = –log 1.592 × 10–3 = 2.80 ecf a 5(a)(iii) CH CHCl COOH is a stronger acid (than propanoic acid, owing to higher [H+] [1] 2 3 because electron-withdrawing effect of Cl (substituent) AND weakens O—H / carboxylate anion stabilised [1] 5(b)(i) pH at which a molecule has no overall charge / is neutral OR pH at which it exists as a zwitterion / dipolar ion 1 5(b)(ii) CH CH(N+H )COOH 1 3 3 5(b)(iii) 1 5(b)(iv) 2 middle amide group shown displayed [1] rest of structure correct [1] © UCLES 2022 Page 10 of 12 5(b)(v) condensation 1 5(b)(vi) C is biodegradable / easily hydrolysed 1 5(c)(i) pair I = CH CHCl COOH (acid) and CH CHCl COO– (c.base) 1 3 3 pair II = NH (base) and NH + (c.acid) 3 4 5(c)(ii) evidence on graph / paper of one half-life (use of data) / t = 500 s [1] 2 1/2 constant half-life (= first order) [1] 5(c)(iii) so that [NH ] is (effectively) constant AND doesn’t affect the rate / zero order 1 3 5(c)(iv) M1: when [NH ] increases ×1.5, rate increase ×1.5 AND first order (w.r.t. [NH ]) / rate is proportional to [NH ] 2 3 3 3 M2: S 2 DEP on M1 N (because rate is first order w.r.t. to [2-chloropropanoate] and [NH ]) 3 5(c)(v) greater proportion of particles have E ⩾ E [1] 2 A frequency of (effective) collisions increases AND rate increases OR rate of collisions increases AND rate increases [1] 5(c)(vi) 1 Question Answer Marks

2(a)(i) (homogeneous is in the) same phase / state as reactants 1 AND (heterogeneous is in a) different phase / state to reactants 2(a)(ii) 1 S O 2– + 2Fe2+ → 2Fe3+ + 2SO 2– [1] 2 2 8 4 2 2I– + 2Fe3+ → 2Fe2+ + I [1] 2 2(a)(iii) reactants are both anions / negatively charged 1 AND so they repel each other OWTTE 2(b)(i) rate = k[NO]2[O ] OR rate = 8.6  106 [NO]2[O ] 1 2 2 2(b)(ii) rate = 8.6  106  (7.2  10–4)2  1.9  10–3 1 rate = 8.47  10–3 (mol dm–3 s–1) min 2sf 2(c)(i) (reaction is) first order wrt cisplatin / overall 1 OR rate is directly proportional to concentration of cisplatin 2(c)(ii) 0.693 / 2.50  10–5 = (2.77  104 s) 1 OR In 2 / 2.50  10–5 = (2.77  104 s) © UCLES 2022 Page 6 of 16 2(c)(iii) initial concentration is 8.0  10–5 mol dm–3 2 – five half-life periods have elapsed [1] time = 5  27720 = 1.39  105 s [1] min 2 sf Question Answer Marks

2(a)(i) rate = k[NO][O ] 1 3 2(a)(ii) 1.66  10–8 [1] 2 mol dm–3 s–1 [1] 2(a)(iii) not constant 1 AND overall second order / not overall first order 2(b)(i) graph is straight line clearly parallel to x-axis 1 2(b)(ii) graph is straight line with negative gradient 1 © UCLES 2022 Page 5 of 11 2(b)(iii) adsorption of reactants onto catalyst surface [1] 3 bonds in reactants weaken [1] reaction occurs followed by desorption of products [1] 2(b)(iv) all active sites on catalyst surface are occupied 1 Question Answer Marks

2(a)(i) (homogeneous is in the) same phase / state as reactants 1 AND (heterogeneous is in a) different phase / state to reactants 2(a)(ii) 1 S O 2– + 2Fe2+ → 2Fe3+ + 2SO 2– [1] 2 2 8 4 2 2I– + 2Fe3+ → 2Fe2+ + I [1] 2 2(a)(iii) reactants are both anions / negatively charged 1 AND so they repel each other OWTTE 2(b)(i) rate = k[NO]2[O ] OR rate = 8.6  106 [NO]2[O ] 1 2 2 2(b)(ii) rate = 8.6  106  (7.2  10–4)2  1.9  10–3 1 rate = 8.47  10–3 (mol dm–3 s–1) min 2sf 2(c)(i) (reaction is) first order wrt cisplatin / overall 1 OR rate is directly proportional to concentration of cisplatin 2(c)(ii) 0.693 / 2.50  10–5 = (2.77  104 s) 1 OR In 2 / 2.50  10–5 = (2.77  104 s) © UCLES 2022 Page 6 of 16 2(c)(iii) initial concentration is 8.0  10–5 mol dm–3 2 – five half-life periods have elapsed [1] time = 5  27720 = 1.39  105 s [1] min 2 sf Question Answer Marks

3(a)(i) the power to which the concentration of a reactant is raised in the rate equation 1 © UCLES 2023 Page 9 of 17 3(a)(ii) 1 the order of reaction with respect to [IO –] 1 3 the order of reaction with respect to [H+] 2 the order of reaction with respect to [I–] 2 the overall order of reaction 5 All correct for one mark 3(a)(iii) 1 3(a)(iv) M1 k = rate / [I–]2[IO –][H+]2 = (4.20  10-2)/(0.0252  0.04x0.0152) = 7.47  106 2 3 M2 units = mol–4 dm12 min–1 3(a)(v) 0.0709 = k  0.12  [H+]2  0.01252 1 [H+] = 2.25  10-2 3(a)(vi) x = 102 / 1 = 100 1 3(b) M1 step 1 Fe3+ + I–  FeI2+ 3 M2 step 2 FeI2+ + I–  Fe2+ + I – OR FeI2+ + I–  FeI + AND slowest step = step 2 2 2 M3 step 3 Fe3+ + I –  Fe2+ + I OR FeI + + Fe3+  2Fe2+ + I 2 2 2 2 © UCLES 2023 Page 10 of 17

4(a)(i) the sum/total of the power / exponent to which a concentration of a reactant is raised in the rate equation / law 1 4(a)(ii) M1 (expt 1 and 2) when [Fe3+]  3, rate  3 so first order w.r.t. Fe3+ 2 M2 (expt 1 and 3) when [Fe3+]  2, [I–]  2, rate  8 so second order w.r.t. I– OR (expt 2 and 3) when [Fe3+] ÷ 1.5, [I–]  2, rate  2.7 so second order w.r.t. I– 4(a)(iii) (rate =) k[Fe3+][I–]2 [1] ecf (a)(ii) 1 4(a)(iv) k = rate / [Fe3+][I–]2 = (2.64  10–4) / (0.04  0.022) 2 k = 16.5 min 2sf ecf units = mol-2 dm6 s–1 ecf 4(a)(v) (k and rate of reaction) both increase 1 © UCLES 2023 Page 11 of 19 4(b)(i) step 1 as this has one mole of H O and one mole of I– 1 2 2 OR step 1 as correct stoichiometry / coefficients of 1 H O and 1 I– 2 2 OR step 1 as number of moles of each reactant are consistent with rate equation / their orders 4(b)(ii) step 1 I oxid no –1  +1 AND O oxid no –1  –2 1 OR step 3 I oxid no –1  0 AND I oxid no +1  0 4(b)(iii) intermediate AND formed (in step 2) and used up (in step 3) 1 ALLOW oxidising agent (in step 3) AND oxidises I– (to I ) 2 Question Answer Marks

3(a)(i) the power to which the concentration of a reactant is raised in the rate equation 1 © UCLES 2023 Page 9 of 17 3(a)(ii) 1 the order of reaction with respect to [IO –] 1 3 the order of reaction with respect to [H+] 2 the order of reaction with respect to [I–] 2 the overall order of reaction 5 All correct for one mark 3(a)(iii) 1 3(a)(iv) M1 k = rate / [I–]2[IO –][H+]2 = (4.20  10-2)/(0.0252  0.04x0.0152) = 7.47  106 2 3 M2 units = mol–4 dm12 min–1 3(a)(v) 0.0709 = k  0.12  [H+]2  0.01252 1 [H+] = 2.25  10-2 3(a)(vi) x = 102 / 1 = 100 1 3(b) M1 step 1 Fe3+ + I–  FeI2+ 3 M2 step 2 FeI2+ + I–  Fe2+ + I – OR FeI2+ + I–  FeI + AND slowest step = step 2 2 2 M3 step 3 Fe3+ + I –  Fe2+ + I OR FeI + + Fe3+  2Fe2+ + I 2 2 2 2 © UCLES 2023 Page 10 of 17

1(a)(i) M1:F + ClO → FClO + F [1] 2 2 2 2 M2:ClO + F → FClO [1] 2 2 OR M1: F + ClO → F ClO 2 2 2 2 M2: F ClO + ClO → 2FClO 2 2 2 2 two balanced equations MUST add to give the overall equation 1(a)(ii) first step AND 1 has one mole / molecule of F and ClO [1] 2 2 OR same moles of reactants as orders in rate equation 1(b)(i) rate = k [F ][Cl O ] / rate = 1.22 [F ][Cl O ] 1 2 2 2 2 AND second / 2 BOTH[1] 1(b)(ii) rate = 1.22  2  10-3  2  10-3 = 4.88 x 10-6 [1] 1 min 2sfecf 1(b)(i) 1(c)(i) k = (0.693 / 4) = 0.173 OR 1.73  10–1[1] 2 1 MUST BE 3SF s–1 [1] 1(c)(ii) at 4 s = 0.001 1 at 8 s = 0.0005 at 12 s = 0.00025 smooth curve ALL correct [1] 1(c)(iii) tangent drawn at 0.00100 mol dm–3 1 AND gradient dependent on their rate = Y / X in the range = 1.5–2.0 10–4 [1] IGNORE sign © UCLES 2023 Page 5 of 18

1(a) 1 2 0 [1] 1 2 [1] 1(b)(i) The line should start at 1  10–5 and decrease to 0.773  10–5 / 7.73  10–6 and be straight for 10 s [1] 1 1(b)(ii) its not possible AND the concentrations in the rate law are not known [1] 1 1(c) 9.24  10–7 [1] 1 1(d)(i) 630 [1] 1 1(d)(ii) Three halvings taking equal times, these times to agree with their answer to di.[1] 1 1(e)(i) CH COH=CH + I → CH C+(OH)CH I + I– [1] 1 3 2 2 3 2 1(e)(ii) step 1 AND has (both) substances in rate law.[1] 1 1(e)(iii) CH C+OHCH and CH COCH OR 1 3 3 3 3 CH C+(OH)CH and CH C(OH)=CH OR 3 3 3 2 CH C+(OH)CH I and CH COCH I [1] 3 2 3 2 Question Answer Marks

1(a)(i) M1:F + ClO → FClO + F [1] 2 2 2 2 M2:ClO + F → FClO [1] 2 2 OR M1: F + ClO → F ClO 2 2 2 2 M2: F ClO + ClO → 2FClO 2 2 2 2 two balanced equations MUST add to give the overall equation 1(a)(ii) first step AND 1 has one mole / molecule of F and ClO [1] 2 2 OR same moles of reactants as orders in rate equation 1(b)(i) rate = k [F ][Cl O ] / rate = 1.22 [F ][Cl O ] 1 2 2 2 2 AND second / 2 BOTH[1] 1(b)(ii) rate = 1.22  2  10-3  2  10-3 = 4.88 x 10-6 [1] 1 min 2sfecf 1(b)(i) 1(c)(i) k = (0.693 / 4) = 0.173 OR 1.73  10–1[1] 2 1 MUST BE 3SF s–1 [1] 1(c)(ii) at 4 s = 0.001 1 at 8 s = 0.0005 at 12 s = 0.00025 smooth curve ALL correct [1] 1(c)(iii) tangent drawn at 0.00100 mol dm–3 1 AND gradient dependent on their rate = Y / X in the range = 1.5–2.0 10–4 [1] IGNORE sign © UCLES 2023 Page 5 of 18

5(a)(i) tangent drawn at t = 0 AND gradient of tangent calculated at t = 0 1 answer between 0.016 to 0.040 5(a)(ii) [I–] stays constant / [I–] does not change 1 OR the overall order is one under these conditions 5(b) M1 equation 1: 2Fe2+ + S O 2– → 2Fe3+ + 2SO 2– 2 2 8 4 M2 equation 2: 2Fe3+ + 2I– → 2Fe2+ + I 2 5(c) the rate constant and rate of reaction 1 will both increase 1 5(d) k = 0.693 ÷ t 1 ½ t = 0.693 ÷ 0.0158 = 43.9 s ½ 5(e) NO + Br → NOBr OR NO + Br → NOBr + Br 2 2 2 2 NOBr + NO → 2NOBr OR Br + NO → NOBr 2 M1 two equations adding up to the overall equation M2 step 1 has one NO and one Br only and is identified as slow step 2 © Cambridge University Press & Assessment 2024 Page 9 of 18

5(a)(i) M1 (using exp 1 and 2) [NO]  3, rate  9 so 2nd order to NO 3 M2 (using exp 1 and 3) [NO]  2, [Cl ]  4, rate  16 so 1st order to Cl 2 2 OR (using exp 2 and 3) [NO]  2 / 3, [Cl ]  4, rate  1.8 so 1st order to Cl 2 2 M3 (rate =) k [NO]2 [Cl ] ecf 2 5(a)(ii) M1 rate = k [NO]2[Cl ] 2 2 k = (3.68  10–2) ÷ (0.0252  0.015) = 3925.33 ecf (a)(i) min 2sf M2 dm6 mol–2 min–1 ecf (a)(i) 5(b) 2 Any two [1] all three [2] 5(c) M1 O + NO → NO + O ALLOW O + NO → NO 2 3 2 3 2 3 2 5 M2 NO + NO → N O ALLOW NO + NO → N O + O 3 2 2 5 5 2 2 5 2 Question Answer Marks

5(a)(i) tangent drawn at t = 0 AND gradient of tangent calculated at t = 0 1 answer between 0.016 to 0.040 5(a)(ii) [I–] stays constant / [I–] does not change 1 OR the overall order is one under these conditions 5(b) M1 equation 1: 2Fe2+ + S O 2– → 2Fe3+ + 2SO 2– 2 2 8 4 M2 equation 2: 2Fe3+ + 2I– → 2Fe2+ + I 2 5(c) the rate constant and rate of reaction 1 will both increase 1 5(d) k = 0.693 ÷ t 1 ½ t = 0.693 ÷ 0.0158 = 43.9 s ½ 5(e) NO + Br → NOBr OR NO + Br → NOBr + Br 2 2 2 2 NOBr + NO → 2NOBr OR Br + NO → NOBr 2 M1 two equations adding up to the overall equation M2 step 1 has one NO and one Br only and is identified as slow step 2 © Cambridge University Press & Assessment 2024 Page 9 of 18

3(a)(i) rate = k[A][B] [1] 2 rate = k[A]2 / k[A]2[B]0 AND rate = k[B]2 / k[A]0[B]2 [1] BOTH 3(a)(ii) k = 7.62 / 7.625 / 7.63 [1] ecf from 3a(i) min 3sf 2 mol–1 dm3 s–1 [1] ecf from 3a(i) 3(a)(iii) reaction is 2nd order 1 OR reaction is not 1st order OR reaction does not have constant half-life [1] 3(b)(i) Pt, Pd, Rh any two [1] 1 3(b)(ii) • adsorption (of reactants) 2 / bond forms between the surface/catalyst and reactants • bond weakening (in reactants) • desorption (of reactants / products) / bond breaks between the surface/catalyst and products any two [1] all three [2] 3(b)(iii) S O 2– + 2Fe2+ → 2SO 2– + 2Fe3+ [1] any multiple 2 2 8 4 2Fe3+ + 2I– → 2Fe2+ + I [1] any multiple 2 3(b)(iv) (heterogeneous) catalyst and reactants in different states / phases 1 AND (homogeneous) catalyst and reactants in same states / phases [1] BOTH 3(c)(i) similar energy of the (3)d and (4)s subshells / orbitals [1] 1 © Cambridge University Press & Assessment 2024 Page 8 of 14 3(c)(ii) M1M2: 4 • Cl O– chosen • O chosen 2 • E⦵ values 1.45 (with Cl O–) cell • E⦵ values 0.96 (with O ) cell 2 any two [1] all four [2] M3: 2Fe(OH) + Cl O– + H O → Cl– + 2Fe(OH) [1] 2 2 3 M4: 4Fe(OH) + O + 2H O → 4Fe(OH) [1] 2 2 2 3 Question Answer Marks

1(a)(i) Change in concentration = 0.135 [1] 2 0.135 / 1800 = 7.5  10–5 [1] 1(a)(ii) Rate = k[X] [1] 1 1(a)(iii) k = 0.693 / t / k = 0.693 / 900 / k = ln2 / t / k = ln2 / 900 / t = ln2 / k [1] 1 ½ ½ ½ 1(a)(iv) 1.16  10–4 [1] 2 mol dm–3 s–1 [1] 1(b)(i) Any two from: 1 • variable oxidation state • vacant / empty / unfilled d orbitals [1] • can form dative bonds / can accept electrons 1(b)(ii) Iron AND 1 iron is solid, reactants are gases OR catalyst and reactants are in different phases / states [1] 1(b)(iii) Two for one mark, three for two marks: 2 • adsorption of reactants by Pt • bonds of reactants weaken [1] • desorption of products [1] 1(b)(iv) NO + SO → NO + SO AND 1 2 2 3 2NO + O → 2NO [1] 2 2 1(c) BiO+ [1] 3 3H SO + 2BiO+ + H O → 3SO 2- + 8H+ + 2Bi [1] 2 3 2 4 0.11 V [1] © Cambridge University Press & Assessment 2024 Page 5 of 12

3(a)(i) rate = k[A][B] [1] 2 rate = k[A]2 / k[A]2[B]0 AND rate = k[B]2 / k[A]0[B]2 [1] BOTH 3(a)(ii) k = 7.62 / 7.625 / 7.63 [1] ecf from 3a(i) min 3sf 2 mol–1 dm3 s–1 [1] ecf from 3a(i) 3(a)(iii) reaction is 2nd order 1 OR reaction is not 1st order OR reaction does not have constant half-life [1] 3(b)(i) Pt, Pd, Rh any two [1] 1 3(b)(ii) • adsorption (of reactants) 2 / bond forms between the surface/catalyst and reactants • bond weakening (in reactants) • desorption (of reactants / products) / bond breaks between the surface/catalyst and products any two [1] all three [2] 3(b)(iii) S O 2– + 2Fe2+ → 2SO 2– + 2Fe3+ [1] any multiple 2 2 8 4 2Fe3+ + 2I– → 2Fe2+ + I [1] any multiple 2 3(b)(iv) (heterogeneous) catalyst and reactants in different states / phases 1 AND (homogeneous) catalyst and reactants in same states / phases [1] BOTH 3(c)(i) similar energy of the (3)d and (4)s subshells / orbitals [1] 1 © Cambridge University Press & Assessment 2024 Page 8 of 14 3(c)(ii) M1M2: 4 • Cl O– chosen • O chosen 2 • E⦵ values 1.45 (with Cl O–) cell • E⦵ values 0.96 (with O ) cell 2 any two [1] all four [2] M3: 2Fe(OH) + Cl O– + H O → Cl– + 2Fe(OH) [1] 2 2 3 M4: 4Fe(OH) + O + 2H O → 4Fe(OH) [1] 2 2 2 3 Question Answer Marks

2(a)(i) mol–2 dm6 s–1 1 2(a)(ii) M1: 2.57  10–6 = 26.4  [concentration]3 2 M2: [concentration] = 4.60  10–3 2(a)(iii) 2.57  10–3 1 2(b)(i) overall 1st order / pseudo 1st order 1 OR 1st order with respect to Cl AND concentration of NO doesn’t change 2 2(b)(ii) ln2 / 105.6 or 0.693 / 105.6 1 2(b)(iii) 0.0262 / 2.62  10–2 1 © Cambridge University Press & Assessment 2025 Page 7 of 16 2(c) M1: homogeneous catalyst OR catalyst because it is regenerated 3 M2: 2NO + O → 2NO 2 2 M3: NO + SO → NO + SO 2 2 3 Question Answer Marks

4(a)(i) the power/exponent to which a concentration AND of a reactant is raised in the rate equation 1 4(a)(ii) 2 Any two [1] all three [2] © Cambridge University Press & Assessment 2025 Page 9 of 17 4(a)(iii) 1 negative gradient curve 4(b)(i) rate = 4  4 = 16 times 1 4(b)(ii) slowest step / slow step in the mechanism / multi-step reaction 1 4(b)(iii) M1: step 1 O + NO → NO + O 2 3 2 3 2 M2: step 2 NO + NO → N O AND sum of step 1 and 2 is consistent with equation for reaction 2 3 2 2 5 4(c)(i) Use of graph k = 12  10–4 / 0.04 = 0.03 1 4(c)(ii) k = 0.693 / t 1 1/2 t = 0.693 / 0.030 = 23.1 (s) min 2sf 1/2 Question Answer Marks

2(a)(i) mol–2 dm6 s–1 1 2(a)(ii) M1: 2.57  10–6 = 26.4  [concentration]3 2 M2: [concentration] = 4.60  10–3 2(a)(iii) 2.57  10–3 1 2(b)(i) overall 1st order / pseudo 1st order 1 OR 1st order with respect to Cl AND concentration of NO doesn’t change 2 2(b)(ii) ln2 / 105.6 or 0.693 / 105.6 1 2(b)(iii) 0.0262 / 2.62  10–2 1 © Cambridge University Press & Assessment 2025 Page 7 of 16 2(c) M1: homogeneous catalyst OR catalyst because it is regenerated 3 M2: 2NO + O → 2NO 2 2 M3: NO + SO → NO + SO 2 2 3 Question Answer Marks

3(a) heterogeneous AND MnO is in a different state / phase to the reactants / H O 1 2 2 2 3(b)(i) M1: (expt 2 and 3) [I–]  2 initial rate  2 so 1st order wrt [I–] 4 M2: (expt 1 and 2) [I–]  2, [H O ]  0.5 initial rate  1 so 1st order wrt [H O ] 2 2 2 2 OR (expt 1 and 4) [I–]  4 initial rate  4 AND [H+]  4 so zero order wrt [H+] OR (expt 1 and 3) [I–]  4, [H O ]  0.5 initial rate  2 so 1st order wrt [H O ] 2 2 2 2 M3: (expt 1 and 4) [I–] x 4 , [H+]  4 initial rate  4 so zero order wrt [H+] OR (expt 3 and 4) [H O ]  2 initial rate  2AND [H+] x 4 so zero order wrt [H+] 2 2 OR (expt 3 and 4) [H O ]  2, [H+]  4 initial rate  2 so 1st order wrt [H O ] 2 2 2 2 M4: rate = k [H O ] [I–] OR rate = k [H O ] [I–] [H+]0 2 2 2 2 3(b)(ii) M1: k = 2.42  10–3 ÷ (0.045  0.030) = 1.79 min 2sf 2 M2: units = dm3 mol-1 s-1 3(c)(i) M1: calculation of one t = 60 s ± 5 s 2 1/2 M2: two t calculated that are constant 1/2 3(c)(ii) k = In2 ÷ 60 = 0.0116 OR k = 0.693 ÷ 60 = 0.0116 min 2sf 1 3(d) increases k and increases the rate of reaction 1

2(a)(i) homolytic fission 1 2(a)(ii) 1 2(a)(iii) 115–120o 1 2(b) M1: 3.825  10–3 / 0.003825 / 0.00383 / 0.0038 / 3.8  10–3 2 M2: units = mol–1 dm3 s–1 2(c) not constant half-life because overall second order 1 © Cambridge University Press & Assessment 2025 Page 7 of 15 2(d) 8.17 / 8.2 1 2(e) • reactants are adsorbed onto the catalyst surface 2 • bonds within reactant molecules are weakened • products are desorbed from catalyst surface Any two [1], all three [2] Question Answer Marks

3(a) heterogeneous AND MnO is in a different state / phase to the reactants / H O 1 2 2 2 3(b)(i) M1: (expt 2 and 3) [I–]  2 initial rate  2 so 1st order wrt [I–] 4 M2: (expt 1 and 2) [I–]  2, [H O ]  0.5 initial rate  1 so 1st order wrt [H O ] 2 2 2 2 OR (expt 1 and 4) [I–]  4 initial rate  4 AND [H+]  4 so zero order wrt [H+] OR (expt 1 and 3) [I–]  4, [H O ]  0.5 initial rate  2 so 1st order wrt [H O ] 2 2 2 2 M3: (expt 1 and 4) [I–] x 4 , [H+]  4 initial rate  4 so zero order wrt [H+] OR (expt 3 and 4) [H O ]  2 initial rate  2AND [H+] x 4 so zero order wrt [H+] 2 2 OR (expt 3 and 4) [H O ]  2, [H+]  4 initial rate  2 so 1st order wrt [H O ] 2 2 2 2 M4: rate = k [H O ] [I–] OR rate = k [H O ] [I–] [H+]0 2 2 2 2 3(b)(ii) M1: k = 2.42  10–3 ÷ (0.045  0.030) = 1.79 min 2sf 2 M2: units = dm3 mol-1 s-1 3(c)(i) M1: calculation of one t = 60 s ± 5 s 2 1/2 M2: two t calculated that are constant 1/2 3(c)(ii) k = In2 ÷ 60 = 0.0116 OR k = 0.693 ÷ 60 = 0.0116 min 2sf 1 3(d) increases k and increases the rate of reaction 1

4(a)(i) 1 the order of reaction with respect to [H ] 1 2 the order of reaction with respect to [NO] 2 overall order of the reaction 3 4(a)(ii) 1 1 rate  4 4(a)(iii) rate  27 1 4(a)(iv) a four-particle collision is unlikely 1 4(a)(v) step 1 2NO → N O 2 2 2 step 2 N O + H → N O + H O 2 2 2 2 2 step 3 N O + H → N + H O 2 2 2 2 4(a)(vi) intermediate formed (step 2) then used up later (step 3) 1 4(b) k = 0.693 / 720 = 9.625  10–4 min 2sf 1 AND units: s–1 4(c)(i) homogeneous 1 AND Co3+ / catalyst / it is in the same phase / same state as the reactants 4(c)(ii) M1: (equation 1) 2Co3+ + 2I– → 2Co2+ + I 2 2 M2: (equation 2) 2Co2+ + S O 2– → 2Co3+ + 2SO 2– 2 8 4 4(c)(iii) repulsion of two negative / same charge ions slows the reaction / raises E 1 A © Cambridge University Press & Assessment 2025 Page 11 of 16

3(a) 2 1 1 3 3(b) M1 1.4x10–6 2 M2 mol–2dm6s–1 I mol–2dm6s– . 3(c)(i) (overall) first order and 1 I ‘[H ] first order’. 2 [NO] is (almost) constant I [NO] zero order. I [NO] not a limiting factor. I [H ] is a limiting factor. 2 A [NO] pseudo zero order. 3(c)(ii) M1 33 000 / 32 500 / 32 542 / 32 540 / 32 535 / 3.25  104, etc. M2 130 000 / 130 141 / 1.3  105 / 1.32  105, etc. ecf © Cambridge University Press & Assessment 2026 Page 10 of 21 t 12 2 32 542 if ln2 used. answer = t 12  4 3(d)(i) M1 one N O and one H on L or one N O and one H O on R 2 (M1 is for either correct LHS or RHS). 2 2 2 2 2 M2 N O + H → N O + H O 2 2 2 2 2 Trace ID: PC-F9EA0E463D Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/41 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 3(d)(ii) no H involved in step one (and H in rate equation) or 1 I ‘does not agree with rate equation’. 2 2 only one species from rate equation or I reference to NO without reference to does not contain species in the same molar proportion as rate equation or order of reaction. if step 1 is RDS then order wrt [H ] = 0 or I N O not in the rate equation. 2 2 2 step 1 does not include all species in rate equation or rate would be second order wrt NO only or reaction would be overall second order Question Answer Marks Guidance

3(a)(i) M1 Exp 1 & 2 [CH OH]  1.5 rate  1.5 or [CH OH]  2 / 3 rate  2 / 3 4 Reasoning must be linked to correct expts – 3 3 and so first order to [CH OH] [1] u / c CON once only for M1-M3. 3 M2 Exp 3 & 4 [H+]  1.33 rate  1.33 or [H+]  3 / 4 rate  3 / 4 A any viable mathematical approach. and so first order to [H+] [1] u / c M3 Exp 2 & 3 [CH CHO]  0.5, [CH OH]  2 and [H+]  3 rate  3 A any correct alternates to M3. 3 3 and so first order to [CH CHO] [1] ecf –all conditional of orders in M1M2 e.g. 3 Exp 2 & 4 M3 [CH 3 CHO]  2, [CH 3 OH]  0.5 A second part of M1 M2 M3 and [H+]  1 / 4 rate  1 / 4 ‘and first order’ can be read in from their rate equation and so first order to [CH CHO]. 3 M4 Rate = k[CH OH][H+][CH CHO] [1] u / c ecf Exp 1 & 3 M3 [CH CHO]  0.5, [CH OH]  3 3 3 3 3 or rate = 1.25 [CH OH][H+][CH CHO] and [H+]  3 rate  9 / 2 3 3 and so first order to [CH CHO]. 3 I any annotations on the table unless directed by the candidates Exp 1 & 4 M3 [CH CHO]  0.5, [CH OH]  3 3 3 and [H+]  4 rate  6 and so first order to [CH CHO]. 3 3(a)(ii) mol–2dm6s–1 u / c ecf from 3(a)(i) 1 3(a)(iii) initial rate = 0.0195 u / c ecf from 3(a)(i) min 2sf 1 A methods with calculations of k first then rate. 3(b) 2 I any curly arrows even if incorrect. A H+ + Br- for HBr. each step 2 x [1] © Cambridge University Press & Assessment 2026 Page 12 of 25 Trace ID: PC-9D76E91B48 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 3(c) • (reactants) adsorb (to surface of catalyst) 2 I absorbed. • bonds (in reactant) weaken A bullet 1 ‘binds to the surface of the catalyst’ for ‘adsorbed’. • (reaction occurs and the) product(s) are desorbed or reaction occurs and substances are desorbed I lowers E a . A product(s) deadsorbed ● ✓ ✓ I bonds between reactants. I ‘bonds are broken’ (happens in all reactions). © Cambridge University Press & Assessment 2026 Page 13 of 25 Trace ID: PC-9D76E91B48 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/42 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

3(a) 2 1 1 3 3(b) M1 1.4x10–6 2 M2 mol–2dm6s–1 I mol–2dm6s– . 3(c)(i) (overall) first order and 1 I ‘[H ] first order’. 2 [NO] is (almost) constant I [NO] zero order. I [NO] not a limiting factor. I [H ] is a limiting factor. 2 A [NO] pseudo zero order. 3(c)(ii) M1 33 000 / 32 500 / 32 542 / 32 540 / 32 535 / 3.25  104, etc. M2 130 000 / 130 141 / 1.3  105 / 1.32  105, etc. ecf © Cambridge University Press & Assessment 2026 Page 10 of 21 t 12 2 32 542 if ln2 used. answer = t 12  4 3(d)(i) M1 one N O and one H on L or one N O and one H O on R 2 (M1 is for either correct LHS or RHS). 2 2 2 2 2 M2 N O + H → N O + H O 2 2 2 2 2 Trace ID: PC-F2CDA313A1 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/43 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 3(d)(ii) no H involved in step one (and H in rate equation) or 1 I ‘does not agree with rate equation’. 2 2 only one species from rate equation or I reference to NO without reference to does not contain species in the same molar proportion as rate equation or order of reaction. if step 1 is RDS then order wrt [H ] = 0 or I N O not in the rate equation. 2 2 2 step 1 does not include all species in rate equation or rate would be second order wrt NO only or reaction would be overall second order Question Answer Marks Guidance

3(a) 1 / first 1 Only answer. 1 / first 2 / second 3(b)(i) M1 … 0.120 … unc 2 No efc from 3(a). M2 … … 6.90  10–6 unc min 2sf 3(b)(ii) M1 0.0046 / 4.6  10–3 min 2sf unc 2 No ecf. M2 mol–1 dm3 s–1 unc Any order, but no work left for examiner to do so I s–1 / mol1 dm–3 or mol–1 dm3 / s CON mol– or s– 3(c) 69 / 68.6 / 68.61 / 68.63 min 2sf 1 68.63 if ln2 used rather than 0.693. A any answer between 68.61–68.63. 3(d) M1 e.g. NO + F → NO F or NO + F → NO F + F 2 Equations must be balanced for charge as 2 2 2 2 2 2 2 NO + NO F → 2NO F F + NO → NO F well as species for M2. 2 2 2 2 2 2 step 1 step 1 I radical dots. I equations with cancellable species. first step has one NO and one F only as reactants (ignore products) and step 2 2 1 / their equation for step 1 is slowest M2 two steps add up to equation © Cambridge University Press & Assessment 2026 Page 11 of 21 Trace ID: PC-0F078E40FE Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P97a01p/4a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9701/44 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/