Zadanie 2d
Wzorowane na: Chemistry HL Paper 2, May 2023, TZ1, pytanie 2(d). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Polecenie
Sulfur trioxide is manufactured industrially by the oxidation of sulfur dioxide.
2 SO2(g) + O2(g) ⇌ 2 SO3(g) ΔH⊖ = −198.0 kJ mol−1
(i) Outline what is meant by dynamic equilibrium.
(ii) Deduce the Kc expression for this reaction.
(iii) Determine the entropy change, ΔS⊖, for the forward reaction to four significant figures, using the data given.
| Substance | S⊖ / J K−1 mol−1 |
|---|---|
| SO2(g) | 248.2 |
| O2(g) | 205.0 |
| SO3(g) | 256.8 |
(iv) Calculate the temperature, in K, below which this reaction becomes spontaneous. Use section 1 of the data booklet. (If you were unable to obtain an answer for part (iii), use −170.0 J K−1 mol−1, but this is not the correct value.)
(v) The value of Kc for this reaction is 3.71 × 103 at 500 °C. Suggest, with a reason, how raising the temperature affects the value of Kc.
(vi) Calculate the standard Gibbs energy change, ΔG⊖, in kJ mol−1, for this reaction at 500 °C. Use sections 1 and 2 of the data booklet.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) «in a closed system» the rate of the forward reaction equals the rate of the reverse reaction ✔
(ii) ✔
(iii) «ΔS⊖ = ΣS⊖(products) − ΣS⊖(reactants)»
2 × 256.8 − (2 × 248.2 + 205.0) ✔
= −187.8 J K−1 mol−1 ✔
(iv) «ΔG⊖ = ΔH⊖ − TΔS⊖; the reaction is spontaneous when ΔG⊖ < 0»
ΔS⊖ = −0.1878 kJ K−1 mol−1 AND ΔH⊖ = −198.0 kJ mol−1 «same units» ✔
«0 = −198.0 − T × (−0.1878)» T = 1054 «K» ✔
Alternative with the given −170.0 J K−1 mol−1: «T = 198.0/0.1700» = 1165 «K»
(v) «the reaction is» exothermic AND Kc decreases «as the equilibrium shifts to the left» ✔
(vi) «ΔG⊖ = −RTln Kc» = −(8.31 × 773 × ln 3710)/1000 ✔
= −52.8 kJ mol−1 ✔
OR «ΔG⊖ = ΔH⊖ − TΔS⊖» = −198.0 − 773 × (−0.1878) = −52.8 kJ mol−1 ✔✔
Schemat punktowania
(i) [1]. (ii) [1]: square brackets required. (iii) [2]: award [2] for the correct final answer to four significant figures; one mark for the correct subtraction set-up. (iv) [2]: M1 for conversion of ΔH and ΔS to common units; award [2] for the correct final answer. (v) [1]: both the exothermic character and the decrease of Kc are needed. (vi) [2]: award [2] for the correct final answer by either route; temperature must be in kelvin; accept 773.15 K.
Komentarz
The two routes in (vi) agree because the given K was constructed from the same data. In (iv) 'below' is correct because both ΔH⊖ and ΔS⊖ are negative.