Zadanie 4f
Wzorowane na: Chemistry HL Paper 2, May 2026, TZ3, pytanie 4(f). 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
Ethanol can be used as a fuel.
(i) Calculate the enthalpy change for the combustion of ethanol using section 12 of the data booklet.
C2H5OH(g) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)
(ii) The enthalpy of formation of ethanol is defined from its elements. State, with a reason, the value of the standard enthalpy of formation, ΔHf⊖, of carbon in the form of graphite.
(iii) This is a reaction for the incomplete combustion of ethanol:
C2H5OH(g) + 2 O2(g) → 2 CO(g) + 3 H2O(g) ΔH = −713 kJ mol−1
Suggest why the incomplete combustion of ethanol releases less energy than its complete combustion.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz CKE i punktacja
Odpowiedź z klucza
(i) «bonds broken:» 5 C–H + C–C + C–O + O–H + 3 O=O = 5 × 414 + 346 + 358 + 463 + 3 × 498 = 4731 «kJ mol−1»
AND
«bonds formed:» 4 C=O + 6 O–H = 4 × 804 + 6 × 463 = 5994 «kJ mol−1» ✔
«ΔH = 4731 − 5994 =» −1263 kJ mol−1 ✔
(ii) zero AND graphite is an element in its standard state ✔
(iii) Any two of:
carbon in CO is not fully oxidized «oxidation state +2 instead of +4» ✔
the triple bond in CO is weaker than the two double bonds in CO2 / less energy is released in forming the bonds ✔
CO can burn further, releasing more energy / the combustion of CO is exothermic ✔
Schemat punktowania
(i) Award [2] for the correct final answer. M1: both totals, from the correct numbers of each bond; M2: the difference, broken minus formed. (ii) [1]: the value and the reason are both required; accept "graphite is the most stable form of carbon". (iii) [2 max]; accept "fewer / weaker bonds formed in CO than in CO2".