Zadanie 2a
Wzorowane na: Chemistry HL Paper 2, May 2023, TZ2, pytanie 2(a). 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-2d.
Atmospheric pollutants include oxides of sulfur, which cause acid deposition, and ozone, which is formed in photochemical smog.
Polecenie
In air, sulfur dioxide can be oxidized to sulfur trioxide. The reaction is reversible, and at high temperature it reaches equilibrium in a closed container.
2 SO2(g) + O2(g) ⇌ 2 SO3(g)
(i) At 600 °C the equilibrium constant, K, for this reaction is 105. Outline what this indicates about the extent of the reaction.
(ii) Calculate the standard Gibbs energy change, ΔG⊖, in kJ mol−1, for this equilibrium at 600 °C. Use sections 1 and 2 of the data booklet.
(iii) Calculate the value of K at 600 °C for the equilibrium 2 SO3(g) ⇌ 2 SO2(g) + O2(g).
(iv) Calculate the standard enthalpy change, in kJ mol−1, for the forward reaction, using the data in the table.
(v) Calculate the standard entropy change, in J K−1 mol−1, for the forward reaction, using the data in the table.
| Substance | ΔHf⊖ / kJ mol−1 | S⊖ / J K−1 mol−1 |
|---|---|---|
| SO2(g) | −296.8 | +248.2 |
| SO3(g) | −395.7 | +256.8 |
| O2(g) | 0 | +205.1 |
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) the equilibrium lies to the right / products are favoured «at equilibrium»
OR
the equilibrium mixture contains a greater concentration of products than of reactants ✔
(ii) ΔG⊖ «= −RTln K = −8.31 × 873 × ln 105» = −33.8 «kJ mol−1» ✔
(iii) «» = 9.52 × 10−3 ✔
(iv) «ΔH⊖ = 2(−395.7) − 2(−296.8)» = −197.8 «kJ mol−1» ✔
(v) «ΔS⊖ = 2(256.8) − [2(248.2) + 205.1]» = −187.9 «J K−1 mol−1» ✔
Schemat punktowania
[1] for each of (i) to (v). (i) must refer to the position of equilibrium (products favoured); a statement that the reaction goes to completion scores zero. (ii) Accept -33 800 J mol-1 if the unit is stated. (iv) and (v) Do not penalize the unit if the numerical value is correct.
Komentarz
The data are consistent: ΔH − TΔS at 873 K is −33.8 kJ mol-1, which gives K = 105. The negative ΔS reflects three moles of gas becoming two.