Zadanie 2c
Wzorowane na: Chemistry HL Paper 2, November 2023, TZ1, pytanie 2(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 2a-2g.
Hydrogen iodide forms from its elements in the gas phase according to the equilibrium
H2(g) + I2(g) ⇌ 2 HI(g)
Polecenie
Approximate enthalpy changes of reactions can be found from bond enthalpies.
(i) Determine the enthalpy change, ΔH⊖, of this reaction, in kJ mol−1, using section 12 of the data booklet.
(ii) Each bond enthalpy is uncertain by 0.1 %. Determine the percentage uncertainty that this produces in the calculated enthalpy change of the reaction.
(iii) Most bond enthalpies are average values. Identify which of the bond enthalpies used in (i) are exact values, and give a reason for your choice.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) «bond breaking» H–H + I–I / 436 + 151 / 587 «kJ» ✔
«bond forming» 2(H–I) / 2 × 298 / 596 «kJ» ✔
«ΔH⊖ = 587 − 596» = −9 «kJ mol−1» ✔
(ii) sum of absolute uncertainties «= 0.436 + 0.151 + 2(0.298)» = 1.183 «kJ mol−1» ✔
percentage uncertainty «» = 13 «%» ✔
(iii) H–H AND I–I AND H–I ✔
«each of the three bonds occurs in a diatomic molecule only, H2, I2 and HI, so no averaging over different molecules is needed»
Schemat punktowania
(i) M1: bonds broken; M2: bonds formed; M3: enthalpy change with the correct sign. Award [3] for a correct final answer. (ii) M1: sum of the absolute uncertainties, with the H–I term counted twice; M2: percentage uncertainty relative to the enthalpy change. Award [2] for a correct final answer; accept 13.1 %. ECF from (i). (iii) [1] for all three bonds, H–H, I–I and H–I, with the reason that each occurs in a diatomic molecule / needs no averaging. Do not award the mark if one of the three is left out.
Komentarz
The percentage uncertainty is large because the enthalpy change is the small difference of two large numbers.