Zadanie 6a
Wzorowane na: Chemistry SL Paper 2, May 2023, TZ1, pytanie 6(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Polecenie
Hydrogen sulfide, H2S, is removed from natural gas in the Claus process. Part of the hydrogen sulfide is burned in air to form sulfur dioxide, which then reacts with the remaining hydrogen sulfide to form sulfur.
(i) Determine the enthalpy change, in kJ mol−1, for the reaction 2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(g), using the equations below.
2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g) ΔH⊖ = −1036.0 kJ mol−1
S(s) + O2(g) → SO2(g) ΔH⊖ = −297.0 kJ mol−1
(ii) Formulate equations showing how sulfur dioxide and carbon dioxide each make rainwater acidic.
(iii) Explain the polarity of the S-Cl bond in sulfur dichloride, SCl2. Use section 9 of the data booklet.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) «reverse the second equation and multiply it by 3, then add it to the first equation:» ΔH⊖ = −1036.0 − 3 × (−297.0) = −145.0 kJ mol−1 ✔
(ii) SO2(g) + H2O(l) → H2SO3(aq) AND CO2(g) + H2O(l) ⇌ H2CO3(aq) ✔
(iii) «electronegativity» Cl = 3.2 and S = 2.6, so Cl is the more electronegative atom / there is a difference of 0.6 ✔
the bonding pair is drawn towards Cl, so Cl is partially negative (δ−) and S is partially positive (δ+) ✔
Schemat punktowania
(i) [1] for −145.0 kJ mol−1 with the sign; accept −145. (ii) [1]: both equations needed; accept a single arrow for the carbon dioxide equation and accept equations that show H+ (or H3O+) with HSO3− and HCO3−. State symbols are ignored. (iii) [2]: M1 the electronegativity values or difference from section 9; M2 the direction of polarity with partial charges. Accept a suitable diagram showing the S-Cl dipole.
Komentarz
Adding the second equation without reversing and multiplying it gives -1333.0, a common slip; the sulfur must cancel, which needs three moles of the second equation reversed.