Zadanie 3d, Paper 2, listopad 2025, TZ1, Chemia IB HL
Wzorowane na: Chemistry HL Paper 2, November 2025, TZ1, pytanie 3(d). 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) Calculate the enthalpy change, ΔH, for the overall synthesis of propyl ethanoate from propane and ethane shown above. Use section 12 of the data booklet.
(ii) Two students each calculated the enthalpy change for the reaction between propan-1-ol and ethanoic acid by a different route, and both reached a correct result: one used bond enthalpy data and found 0 kJ mol−1, while the other used enthalpy of formation data and found −3 kJ mol−1.
C3H8O(l) + C2H4O2(l) ⇌ C5H10O2(l) + H2O(l)
Explain how both calculations can be correct even though the values differ.
(iii) Propyl ethanoate, C5H10O2, has a standard enthalpy change of formation, ΔHf⊖, that can be found from this reaction's standard enthalpy change, ΔHr⊖ = −3 kJ mol−1. Calculate this value, in kJ mol−1. Use sections 1 and 13 of the data booklet and ΔHf⊖ of propan-1-ol = −303 kJ mol−1.
(iv) For this reaction, calculate the standard entropy change, ΔSr⊖, in J K−1 mol−1. Use section 13 of the data booklet together with S⊖ of propan-1-ol = 193 J K−1 mol−1 and S⊖ of propyl ethanoate = 288 J K−1 mol−1.
(v) Using ΔHr⊖ = −3 kJ mol−1 together with your answer to (d)(iv), calculate the standard Gibbs energy change, ΔG⊖, in kJ mol−1, for this reaction. (If (d)(iv) was not answered, take ΔSr⊖ as −8 J K−1 mol−1 instead, noting that this is not the actual value.) Use sections 1 and 4 of the data booklet.
(vi) If the temperature of this reaction were increased, explain what effect this would have, if any, on its spontaneity. Use section 1 of the data booklet.
(vii) Chemists monitored this reaction continuously, starting from the moment the reactants were mixed and continuing until the system reached equilibrium. Sketch a graph to show how the forward and reverse reaction rates each change over this period, from the start of the reaction to the point where equilibrium is established.
Powiększ(viii) Using your answer to (d)(v), determine the value of the equilibrium constant, K, for this reaction at 25 °C. (If (d)(v) was not answered, take ΔG⊖ as −7.0 kJ mol−1 instead, noting that this is not the actual value.) Use sections 1 and 2 of the data booklet.
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Pokaż rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) «bonds broken» 1½ O=O + 4 C–H + Br–Br / 1.5 × 498 + 4 × 414 + 193 / 2596 «kJ mol−1» ✔
«bonds formed» C=O + 2 C–O + 2 O–H + 2 H–Br / 804 + 2 × 358 + 2 × 463 + 2 × 366 / 3178 «kJ mol−1» ✔
ΔH = «2596 − 3178 =» −582 «kJ mol−1» ✔
OR
«all bonds broken» 3 C–C + 14 C–H + Br–Br + 1½ O=O / 3 × 346 + 14 × 414 + 193 + 1.5 × 498 / 7774 «kJ mol−1» ✔
«all bonds formed» 3 C–C + 10 C–H + C=O + 2 C–O + 2 H–Br + 2 O–H / 3 × 346 + 10 × 414 + 804 + 2 × 358 + 2 × 366 + 2 × 463 / 8356 «kJ mol−1» ✔
ΔH = «7774 − 8356 =» −582 «kJ mol−1» ✔
(ii) the bond enthalpies used are typical values averaged over many compounds, not measurements taken for the actual bonds in propan-1-ol, ethanoic acid or the ester
OR
bond enthalpy data refers to substances in the gas phase, while this reaction happens between liquids
OR
enthalpies of formation are measured separately for each specific substance taking part in the reaction ✔
(iii) −3 = ΔHf⊖(ester) + (−286) − (−303 + (−484)) «kJ mol−1» ✔
ΔHf⊖(ester) = «−3 + 286 − 303 − 484 =» −504 «kJ mol−1» ✔
(iv) ΔSr⊖ = «(288 + 70) − (193 + 160) =» +5 «J K−1 mol−1» ✔
(v) conversion to common units «ΔS = 0.005 kJ K−1 mol−1» ✔
ΔG⊖ = «−3 kJ mol−1 − 298 K × 0.005 kJ K−1 mol−1 =» −4.5 «kJ mol−1» ✔
(vi) since ΔS is positive here, raising T makes the −TΔS term more negative, which pulls ΔG down further ✔
so the reaction gains spontaneity «as it gets hotter» ✔
(vii) forward rate starts high and decreases, reverse rate starts at zero and increases, both curves levelling off at the same rate ✔
upper curve labelled forward reaction AND lower curve labelled reverse reaction ✔
(viii) ln K = «−ΔG⊖/(RT) = 4490 J mol−1 ÷ (8.31 J K−1 mol−1 × 298 K) =» 1.81 ✔
K = «e1.81 =» 6.1 ✔
Schemat punktowania
(i) M1: bonds broken 2596 kJ; M2: bonds formed 3178 kJ; M3: −582 kJ mol⁻¹ (reactants minus products). Award [3] for the correct final answer, [2 max] for +582. Accept breaking and re-forming all bonds (7774 − 8356). (ii) [1] for any one valid reason. (iii) M1: correct expression or rearrangement; M2: −504 kJ mol⁻¹; award [1] for +504. Award [2] for the correct final answer. (iv) [1] for +5 J K⁻¹ mol⁻¹. (v) M1: entropy change converted to kJ K⁻¹ mol⁻¹; M2: −4.5 kJ mol⁻¹ (accept −4.49); with the fallback −8 J K⁻¹ mol⁻¹ the answer is −0.6 kJ mol⁻¹. Award [2] for the correct final answer. (vi) M1: −TΔS more negative so ΔG more negative; M2: increase in spontaneity (accept feasibility); if the fallback value is used, the reversed conclusion earns the marks; do not accept 'spontaneous' with no reference to the change. (vii) M1: shapes correct and both curves finishing at the same rate (accept straight lines to the equilibrium rate); M2: upper curve labelled forward, lower curve labelled reverse (labels such as 'reactants/products' or formulas are not accepted). (viii) M1: ln K = 1.81; M2: K = 6.1 (accept 6.1 to 6.2). With the fallback −7.0 kJ mol⁻¹ the answer is 17 (16.9). Award [1 max] for K = 1.00. Award [2] for the correct final answer.
Komentarz. In (v) the entropy change must be in kJ K⁻¹ mol⁻¹ before it is multiplied by 298 K; in (viii) ΔG must be in J mol⁻¹ to match R = 8.31 J K⁻¹ mol⁻¹.