A-LevelChemistryReaction kinetics9701/42 · Feb/Mar 2025 · Q215 Marks

Propanone, CH3COCH3, is a common organic solvent and reagent.

(a)(i)Propanone reacts with methanol, CH3OH, under acidic conditions to form compound A, as shown by reaction 1. [Figure 2.1] The overall order of reaction 1 can be found by studying experimental data. Table 2.1 shows how the initial rate of reaction changes as [CH3OH] and [H+] are varied. In each experiment, a large excess of CH3COCH3 is used. Table 2.1 experiment [CH3OH] /moldm-3 [H+] /mol dm-3 relative initial rate of reaction 1 0.010 0.010 1.00 2 0.015 0.015 2.25 3 0.015 0.020 3.00 Explain why a large excess of CH3COCH3 is used in each experiment.[1]
(a)(ii)Use the data in Table 2.1 to determine the order of reaction 1 with respect to CH3OH and to H+ ions. Explain your answers.[2]
(a)(iii)In a separate experiment, a large excess of CH3OH and H+ ions are added to a solution containing a known concentration of CH3COCH3. [Figure 2.2] Use Fig. 2.2 to show how, under these conditions, reaction 1 is first order with respect to CH3COCH3.[1]
(b)Propanone also reacts with acidified cyanide ions to form the hydroxynitrile compound B, as shown by reaction 2. [Figure 2.3] The following rate equation is determined for reaction 2. rate = k [CH3COCH3] [H+] Four possible mechanisms for reaction 2 are shown in Table 2.2. [Table 2.2] Suggest which of these mechanisms is consistent with the rate equation for reaction 2. Explain your answer. proposed reaction mechanism ............................................................................................................................................ explanation .......................................................................................................................................................................[3]
(c)(i)Carboxylic acid C, C4H8O3, forms when B is hydrolysed under hot acidic conditions. Draw the structure of C.[1]
(c)(ii)The pKa of C is 3.95. Calculate the pH of a 0.500 mol dm⁻³ solution of C. Show your working. pH = ..............................................[2]
(d)(i)C can be used to form buffer solution D. Define a buffer solution.[2]
(d)(ii)Buffer solution D is made when 20.0 cm³ of 1.00 moldm⁻³ NaOH(aq) is added to 100 cm³ of a 0.500 mol dm⁻³ solution of C. The pKa of C is 3.95. Calculate the pH of buffer solution D. Show your working. pH = ..............................................[3]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

[1] [CH3COCH3] concentration stays (almost) constant / does not change OR [CH3COCH3] concentration does not affect rate
[2] M1: first order w.r.t. H+ AND rate × 4/3 and [H+] × 4/3 (in expts 2 and 3)
M2: first order w.r.t. CH3OH AND [H+] × 1.5, [CH3OH] × 1.5 and rate × 2.25 / 1.52 (expts 1 and 2) / [H+] × 2, [CH3OH] × 1.5 and rate × 3 / (2 × 1.5) (expts 1 and 3)
[1] two quoted half-lives within range / 155–170 s AND (roughly) constant
[3] M1: proposed mechanism = 4
M2: slow step contains only species from the rate equation/law OR slow step contains CH3COCH3 and H+ from the rate equation/law
M3: stoichiometric / (mole) ratio / amounts AND corresponds to order of reaction / to that in the rate equation
[1] HO COOH / H3C C CH3 [Structure of 2-hydroxy-2-methylpropanoic acid]
[2] Ka= 10-3.95 = 1.122 × 10-4 [H+]² = (1.122 × 10-4 × 0.5)
[H+] = √(1.122 × 10–4 × 0.5) = 7.49 × 10–3
pH = -log (7.49 × 10-3) pH = 2.13 min 2sf
[2] a solution that resists / minimises / opposes changes in pH when small amounts of (strong) acid or base are added
[3] M1: moles of conjugate base = 20 / 1000 × 1.00 = 0.0200 mol AND moles of acid C left = 100 / 1000 × 0.500 – 0.0200 = 0.0300 mol
M2: Ka = [H+] × 0.0200 / 0.0300 [H+] = 3/2 × 10–3.95 = 1.68 × 10-4 ecf from M1
M3: pH = -log (1.68 × 10-4) pH = 3.77 min 2sf ecf from a calculated [H+] M2