Zadanie 4d
Wzorowane na: Chemistry HL Paper 2, May 2022, TZ2, pytanie 4(d). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 4a-4d.
Nitrogen dioxide is a brown gas formed in vehicle engines. Its reactions illustrate how rates and equilibria are studied.
Polecenie
Consider the formation of gaseous nitrogen dioxide from liquid dinitrogen tetroxide.
N2O4(l) → 2 NO2(g) ΔH⊖ = +85.9 kJ mol−1
| Substance | S⊖ / J K−1 mol−1 |
|---|---|
| N2O4(l) | 209.2 |
| NO2(g) | 240.1 |
(i) Calculate the entropy change of the reaction, ΔS⊖, in J K−1 mol−1.
(ii) Predict, giving a reason, how the value of ΔS⊖ would be affected if N2O4(g) were used as the reactant instead of N2O4(l).
(iii) Calculate the Gibbs energy change, ΔG⊖, in kJ mol−1, for the reaction at 298 K. Use section 1 of the data booklet.
(iv) Calculate the equilibrium constant, K, for the reaction at 298 K. Use your answer to (iii) and sections 1 and 2 of the data booklet. If you could not answer (iii), use ΔG⊖ = 4.0 kJ mol−1; this is not the correct value.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) «ΔS⊖ = 2 × 240.1 − 209.2 =» +271.0 «J K−1 mol−1» ✔
(ii) ΔS⊖ lower / less positive AND a gas has a higher entropy than a liquid
OR
ΔS⊖ lower / less positive AND the increase in the number of moles of gas is smaller (1 to 2 instead of 0 to 2) ✔
(iii) «ΔG⊖ = ΔH⊖ − TΔS⊖ = 85.9 − (298 × 0.2710) =» +5.1 «kJ mol−1» ✔
(iv) «» −2.06 ✔
«K =» 0.13 ✔
Schemat punktowania
(i) [1]. (ii) [1] for the direction with a correct reason. (iii) [1]; the units of ΔS⊖ must be converted to kJ, accept 5.14. (iv) M1: ln K (−2.06 to −2.08); M2: K = 0.13. Award [2] for 0.13 with no working. Award [2] for 0.20 obtained from the fallback value 4.0 kJ mol−1. Award [1 max] if ΔG⊖ is used in kJ mol−1 together with R in J K−1 mol−1 (giving K ≈ 1.00), or if the sign of ln K is wrong.