Zadanie 3e, Paper 2, listopad 2025, TZ1, Chemia IB HL
Wzorowane na: Chemistry HL Paper 2, November 2025, TZ1, pytanie 3(e). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
Many esters are still known by trivial names. Propyl ethanoate, CH3COOCH2CH2CH3, which has the odour of pears, is manufactured from propan-1-ol and ethanoic acid in the presence of an acid catalyst.
CH3CH2CH2OH(l) + CH3COOH(l) ⇌ CH3COOCH2CH2CH3(l) + H2O(l)
Both reactants can be synthesized from alkanes: propan-1-ol from propane via 1-bromopropane, and ethanoic acid from ethane via bromoethane and ethanol. The overall equation for the synthesis of the ester from the two alkanes is
Polecenie
(i) To check the value of K found in (d)(viii), a student set up an experiment: 0.500 mol each of propan-1-ol and ethanoic acid were mixed with a small amount of concentrated sulfuric acid at 25 °C, and the mixture's volume was kept constant throughout. Thirty minutes later, 0.300 mol of ethanoic acid was found to remain in the mixture.
From this data, determine the value of K that the experiment gives.
(ii) The value of K measured experimentally in (e)(i) differs from the value calculated in (d)(viii). Suggest why. (If you have no values for these, take (d)(viii) as 5.5 and (e)(i) as 0.60 for the purpose of this question, though neither is the actual value.)
(iii) State the correct term for the quantity calculated in (e)(i).
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Odpowiedź z klucza
(i) «amount of» propan-1-ol «after 30 minutes» = 0.300 «mol», «amount of» propyl ethanoate = 0.200 «mol», «amount of» water = 0.200 «mol» ✔
K = « = » ✔
0.44 ✔
(ii) «the experimental value is lower than the calculated K because» insufficient time was allowed for the system to reach equilibrium / the reaction was still proceeding in the forward direction when the sample was analysed ✔
(iii) reaction quotient / Q ✔
Schemat punktowania
(i) M1: amounts of the three other species at the time of sampling; M2: correct expression with the amounts substituted (the volume cancels); M3: 0.44 (accept 0.444). Award [3] for the correct final answer; M1 can be awarded for a correct expression in M2. (ii) [1] for insufficient time / equilibrium not yet reached. (iii) [1] for reaction quotient or Q.
Komentarz. Because the equation has two moles on each side the volume cancels in the expression, so amounts can be used directly; the value is lower than K, which is why the mixture had not reached equilibrium.