Zadanie 6a, Paper 2, maj 2025, TZ1, Chemia IB HL
Wzorowane na: Chemistry HL Paper 2, May 2025, TZ1, pytanie 6(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
The rate of the reaction between acidified potassium manganate(VII) and sodium ethanedioate, Na2C2O4, can be measured by timing the colour changes. When ethanedioate ions are in excess, an unstable red-brown manganese(III) complex forms before the colour fades.
Polecenie
In the first step, manganese(III) ions formed from the manganate(VII) are complexed by ethanedioate ions.
[Mn(H2O)6]3+(aq) + 2 C2O42−(aq) → [Mn(C2O4)2(H2O)2]−(aq) + 4 H2O(l)
(i) Deduce the oxidation state of carbon in [Mn(C2O4)2(H2O)2]−.
(ii) Explain why this manganese(III) complex ion is coloured rather than colourless.
(iii) Adding a small amount of manganese(II) sulfate, MnSO4(s), at the start makes the reaction much faster, because Mn2+ ions act as a catalyst. Explain how a catalyst brings about this increase in the rate of reaction.
(iv) The red-brown colour disappears as the complex decomposes: the ethanedioate ion, C2O42−, is oxidized to CO2, and Mn3+ is reduced to Mn2+. Deduce the half-equation for the oxidation and the overall redox equation of this second step.
Oxidation half-equation:
Overall redox equation:
(v) Sketch, on labelled axes of potential energy against reaction coordinate, an energy profile for this two-step reaction. Label the reactants, the intermediate and the products, and mark the activation energy, Ea, of each step and the overall enthalpy change, ΔH. Take the overall reaction to be exothermic.
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Pokaż rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) +3 / III «Mn is +3 and H2O is neutral, so each C2O4 ligand carries a 2− charge: 2x + 4(−2) = −2» ✔
(ii) Any three of:
Mn3+ has a partially filled d-sublevel «3d4» ✔
«the ligands cause» the d-orbitals «to» split into two sets of different energy ✔
light is absorbed as an electron is promoted from the lower to the higher set of d-orbitals «d-d transition» ✔
the energy gap corresponds to a wavelength in the visible region of the spectrum ✔
the colour observed «red-brown» is complementary to the colour of the light absorbed ✔
(iii) «the catalyst provides» an alternative reaction pathway / a different mechanism «involving Mn2+» ✔
with a lower activation energy «so a greater proportion of collisions are successful» ✔
(iv) Oxidation half-equation: C2O42− → 2 CO2 + 2 e− ✔
Overall redox equation: 2 Mn3+ + C2O42− → 2 Mn2+ + 2 CO2 ✔
(v) products drawn lower in energy than the reactants ✔
intermediate drawn at an energy minimum between two maxima ✔
Ea marked for both steps «from the reactants and from the intermediate up to the respective maximum» ✔
ΔH marked from the reactants down to the products ✔
Schemat punktowania
(i) [1]; do not accept 3 or 3+. (ii) [3 max] from the five points; the complementary-colour point scores only if absorption of light is mentioned. (iii) M1: alternative pathway or mechanism; M2: lower activation energy. (iv) M1: half-equation with the electrons on the right; M2: overall equation balanced for charge with 2Mn3+, ECF from M1. (v) [1] per feature; any relative peak heights are accepted and single-headed or double-headed arrows both score.
Komentarz. For (i), read the manganese oxidation state from the reactant [Mn(H2O)6]3+ first; the overall 1- charge then fixes each ethanedioate ligand at 2-.