(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$^{-1}$» ✔

$\Delta H$ = «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$^{-1}$» ✔

$\Delta H$ = «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 = $\Delta H_\mathrm{f}^\ominus$(ester) + (−286) − (−303 + (−484)) «kJ mol$^{-1}$» ✔

$\Delta H_\mathrm{f}^\ominus$(ester) = «−3 + 286 − 303 − 484 =» −504 «kJ mol$^{-1}$» ✔

(iv) $\Delta S_\mathrm{r}^\ominus$ = «(288 + 70) − (193 + 160) =» +5 «J K$^{-1}$ mol$^{-1}$

(v) conversion to common units «$\Delta S$ = 0.005 kJ K$^{-1}$ mol$^{-1}$» ✔

$\Delta G^\ominus$ = «−3 kJ mol$^{-1}$ − 298 K × 0.005 kJ K$^{-1}$ mol$^{-1}$

(vi) since $\Delta S$ is positive here, raising $T$ makes the −T$\Delta S$ term more negative, which pulls $\Delta G$ 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 $K$ = «−$\Delta G^\ominus$/(RT) = 4490 J mol$^{-1}$ ÷ (8.31 J K$^{-1}$

$K$ = «e$^{1.81}$ =» 6.1 ✔

(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.