Zadanie 5d, Paper 2, specimen 2025, Chemia IB HL
Wzorowane na: Chemistry HL Paper 2, specimen 2025, pytanie 5(d). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
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Pentadecane, C15H32, is a component of diesel fuel and is also produced naturally by some cyanobacteria.
Polecenie
Pentadecane can be cracked into smaller molecules. Consider the reaction:
C15H32(g) → 3 C4H8(g) + C3H8(g)
where C4H8 is but-1-ene.
(i) Use the bond enthalpies in section 12 of the data booklet to determine ΔH⊖ for this reaction.
(ii) Determine ΔH for the same reaction from enthalpies of formation, using the value below together with section 13 of the data booklet.
ΔHf⊖ (C15H32(g)) = −352.7 kJ mol−1
(iii) Comment on why the values obtained in (d)(i) and (d)(ii) differ.
(iv) Predict the sign of ΔS for this reaction and justify your prediction.
(v) Using your answers to (d)(ii) and (d)(iv), discuss whether raising the temperature makes this reaction more or less likely to be spontaneous.
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Odpowiedź z klucza
(i) bonds broken: 6 (C–C) / 6 × 346 «kJ» ✔
bonds formed: 3 (C=C) / 3 × 614 «kJ» ✔
ΔH⊖ = «6 × 346 kJ − 3 × 614 kJ = 2076 kJ − 1842 kJ =» «+»234 «kJ» ✔
(ii) «ΔH = ΣΔHf(products) − ΣΔHf(reactants)»
ΔH = [3(+0.1) + (−105)] − (−352.7) «kJ mol−1» ✔
ΔH = «+»248.0 «kJ mol−1» ✔
(iii) the value from enthalpies of formation, (d)(ii), is the more reliable one because the bond enthalpies in section 12 are averages taken over many compounds
OR
because the bond enthalpies do not apply exactly to the bonds in these particular molecules ✔
(iv) positive AND the number of moles/molecules of gas increases «from 1 to 4» ✔
(v) ΔH and ΔS are both positive, so ΔG «= ΔH − TΔS» is negative only when TΔS exceeds ΔH ✔
the reaction is therefore not spontaneous at low temperature and becomes spontaneous above a certain temperature / raising the temperature makes ΔG more negative ✔
Schemat punktowania
(i) [3]: M1 six C–C bonds broken (net); M2 three C=C bonds formed; M3 +234 kJ. Award [3] for a correct final answer with working; a full inventory that also breaks and re-forms the C–H bonds is acceptable if it reaches +234 kJ. (ii) [2]: M1 correct expression with but-1-ene (+0.1) and propane (−105) taken from section 13; M2 +248.0 kJ (accept +248). (iii) [1]. (iv) [1]: sign and reason both required. (v) [2]: M1 signs of ΔH and ΔS with the condition ΔG < 0; M2 spontaneous at high temperature.
Komentarz. A full bond inventory gives the same result: 14 C–C and 32 C–H broken, 8 C–C, 3 C=C and 32 C–H formed; the C–H terms cancel, leaving six C–C broken and three C=C formed.