Zadanie 4b
Wzorowane na: Chemistry HL Paper 2, May 2026, TZ3, pytanie 4(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 4a-4f.
Organic compounds are made from many different starting materials and by a variety of reactions, and some of them are used as fuels.
Polecenie
Propane undergoes complete combustion.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
(i) Calculate the standard enthalpy change of reaction, ΔH⊖, in kJ mol−1, for the combustion of propane. Use section 13 of the data booklet.
(ii) The standard entropy change, ΔS⊖, for the combustion of propane is −374 J K−1 mol−1. Determine the standard Gibbs energy change, ΔG⊖, in kJ mol−1, for the combustion of propane at 50 °C using your answer from part (i). Use sections 1 and 4 of the data booklet. If you did not obtain an answer in part (i), use −2000 kJ mol−1, although this is not the correct answer.
(iii) State the oxidation state of carbon-2 in propane and of carbon in carbon dioxide.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz CKE i punktacja
Odpowiedź z klucza
(i) «ΣΔHf⊖ products = 3 × (−394) + 4 × (−286) =» −2326 kJ mol−1 AND «ΣΔHf⊖ reactants =» −105 kJ mol−1 ✔
«−2326 − (−105) =» −2221 kJ mol−1 ✔
(ii) «T = 323 K AND ΔS⊖ = −0.374 kJ K−1 mol−1» ✔
«ΔG⊖ = ΔH⊖ − TΔS⊖ = −2221 − (323 × −0.374) =» −2100 kJ mol−1 ✔
(iii) propane (C-2): −2 ✔
carbon dioxide: +4 ✔
Schemat punktowania
(i) Award [2] for the correct final answer. M1: both sums; M2: the difference, with the sign. (ii) Award [2] for the correct final answer; if −2000 is used the answer is −1880 kJ mol−1 (ECF). M1: temperature in K and ΔS in kJ; M2: result. (iii) [1] each; do not accept the sign written after the number (2-, 4+); accept IV for +4.
Komentarz
The commonest error is leaving the entropy in J while the enthalpy is in kJ, which makes the TΔS term about a thousand times too large.