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Zadanie 2c

6 pktobliczenioweśrednie (szac.)Paper 2, listopad 2023, TZ2

Wzorowane na: Chemistry HL Paper 2, November 2023, TZ2, pytanie 2(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Informacja do zadań 2a-2g.

Methanol, a possible fuel and chemical feedstock, can be produced by the hydrogenation of carbon dioxide according to the equilibrium

CO2(g) + 3 H2(g) ⇌ CH3OH(g) + H2O(g)

Polecenie

Bond enthalpies give an estimate of the enthalpy change of a reaction. Assume that all species are gases.

(i) Use bond enthalpies from section 12 of the data booklet to determine the standard enthalpy change, ΔH⊖, of the forward reaction.

(ii) Each bond enthalpy is known to within ±0.1 %. Determine the percentage uncertainty in the value of ΔH⊖ that you calculated in (i).

(iii) Most tabulated bond enthalpies are averages over many compounds. Identify one of the bond enthalpies you used in (i) that is an exact value, not an average, and give a reason for your choice.

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Odpowiedź z klucza

(i) «bond breaking» 2 C=O + 3 H−H «= 2(804) + 3(436)» = 2916 «kJ» ✔

«bond forming» 3 C−H + C−O + 3 O−H «= 3(414) + 358 + 3(463)» = 2989 «kJ» ✔

«ΔH⊖ = 2916 − 2989 =» −73 «kJ mol−1» ✔

(ii) ALTERNATIVE 1

sum of absolute uncertainties «= 0.001 × (2916 + 2989) =» 5.905 «kJ mol−1» ✔

percentage uncertainty «=5.90573×100=8.09== \dfrac{5.905}{73} \times 100 = 8.09 =» 8 «%» ✔

ALTERNATIVE 2

sum of absolute uncertainties, each type of bond counted once «= 0.804 + 0.436 + 0.414 + 0.358 + 0.463 =» 2.475 «kJ mol−1» ✔

percentage uncertainty «=2.47573×100=3.39== \dfrac{2.475}{73} \times 100 = 3.39 =» 3 «%» ✔

(iii) H–H AND it occurs only in H2 «so there is nothing to average» ✔

Schemat punktowania

(i) [3]: M1 bonds broken, M2 bonds formed, M3 the enthalpy change with its sign. Award [3] for a correct final answer. (ii) [2]: award [2] for a correct final answer from either alternative. The percentage is taken relative to the size of ΔH⊖ (73), ignoring its sign. ECF from (i). (iii) [1]: do not accept C–H, C–O or O–H (averages over many compounds); C=O is not accepted either, because its single booklet value is applied to carbonyl groups in many different compounds.

Komentarz

The second alternative counts each type of bond once, as the IB markscheme for the original question also accepted; the first alternative is the one that follows the propagation of uncertainties.

Umiejętności w zadaniu