(i) $\ce{Br2(g) -> 2Br^{.}(g)}$ ✔

(ii) homolytic «fission» ✔

(iii) bonds broken: $\ce{C#O}$ + $\ce{Br-Br}$ = 1077 + 193 = 1270 «$\mathrm{kJ\,mol^{-1}}$» ✔

bonds formed: $\ce{C=O}$ + 2 $\ce{C-Br}$ = 804 + 2 × 285 = 1374 «$\mathrm{kJ\,mol^{-1}}$» ✔

$\Delta H$ «= 1270 − 1374» = −104 «$\mathrm{kJ\,mol^{-1}}$» ✔

(iv) exothermic profile: products drawn below the reactants, with an activation-energy barrier between them ✔

$\Delta H$ shown as the vertical gap between the reactant and product levels AND "Reactants" and "Products" labelled ✔

(v) $\Delta S^\ominus$ «= 309 − (198 + 245)» = −134 «$\mathrm{J\,K^{-1}\,mol^{-1}}$» ✔

(vi) $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus = -104\,000 - 298 \times (-134) = -104\,000 + 39\,900 = -64\,100$

$\ln K = -\dfrac{\Delta G^\ominus}{RT} = \dfrac{64\,100}{8.31 \times 298} = 25.9$

$K = e^{25.9} \approx 1.8 \times 10^{11}$ «$1.7 \times 10^{11}$

(i) [1] the radical dot must be shown; accept ½Br2 -> Br•. (ii) [1]. (iii) [3] M1 sum of bonds broken; M2 sum of bonds formed; M3 ΔH with sign; award [3] for a correct final answer. (iv) [2] M1 exothermic shape with a barrier; M2 ΔH and both labels. (v) [1] the entropy of CO (198) must be taken from section 13. (vi) [3] M1 ΔG in consistent units; M2 ln K; M3 K with the conclusion 'completion'. Accept K in the range 1.5 to 1.9 × 10^11. With the substitute values: ΔG = −57 200 J mol-1, ln K = 23.1, K = 1 × 10^10, completion.