Zadanie 3a, Paper 2, maj 2025, TZ2, Chemia IB HL
Wzorowane na: Chemistry HL Paper 2, May 2025, TZ2, pytanie 3(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
Carbonyl dibromide, Br2CO, is a reactive compound used as a brominating agent in organic synthesis.
Polecenie
Suppose carbonyl dibromide were made by a free-radical chain reaction of carbon monoxide, CO, with bromine vapour, Br2, started by ultraviolet light. The questions below treat this hypothetical route with the bond-enthalpy model and the entropy data given.
(i) Write an equation for the initiation step.
(ii) State the type of bond fission that occurs in this step.
(iii) Use bond enthalpies from section 12 of the data booklet to determine the enthalpy change for the overall reaction CO(g) + Br2(g) → Br2CO(g).
(iv) On the axes below, sketch the energy profile of this reaction. Mark the positions of the reactants and the products and show ΔH on your sketch.
Powiększ(v) The standard entropies of bromine vapour and of carbonyl dibromide are given below; the value for carbon monoxide is in section 13 of the data booklet. Calculate ΔS⊖ for the reaction in (a)(iii).
| Substance | S⊖ / J K−1 mol−1 |
|---|---|
| Br2(g) | 245 |
| Br2CO(g) | 309 |
(vi) Using your answers to (a)(iii) and (a)(v) and sections 1 and 2 of the data booklet, deduce what this model predicts for 298 K: that the formation of carbonyl dibromide goes essentially to completion, or that it stops at an equilibrium mixture.
Should you have no answer to (a)(iii) or (a)(v), take (a)(iii) = −90 kJ mol−1 and (a)(v) = −110 J K−1 mol−1; these are not the correct values.
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Pokaż rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) Br2(g) → 2 Br ∙ (g) ✔
(ii) homolytic «fission» ✔
(iii) bonds broken: C≡O + Br−Br = 1077 + 193 = 1270 «kJ mol−1» ✔
bonds formed: C=O + 2 C−Br = 804 + 2 × 285 = 1374 «kJ mol−1» ✔
ΔH «= 1270 − 1374» = −104 «kJ mol−1» ✔
(iv) exothermic profile: products drawn below the reactants, with an activation-energy barrier between them ✔
ΔH shown as the vertical gap between the reactant and product levels AND "Reactants" and "Products" labelled ✔
(v) ΔS⊖ «= 309 − (198 + 245)» = −134 «J K−1 mol−1» ✔
(vi) ΔG⊖ = ΔH⊖ − TΔS⊖ = −104 000 − 298 × (−134) = −104 000 + 39 900 = −64 100 «J mol−1» ✔
✔
K = e25.9 ≈ 1.8 × 1011 «1.7 × 1011 with unrounded values», which is so large that the model predicts the reaction effectively goes to completion ✔
Schemat punktowania
(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.
Komentarz. The route is presented as hypothetical on purpose: the bond-enthalpy estimate overstates the stability of carbonyl dibromide, which in reality decomposes to CO and Br2 at room temperature. The chemistry the part tests (radical initiation, bond enthalpies, entropy, Gibbs energy and K) is unaffected.