Zadanie 1c, Paper 1B, specimen 2025, Chemia IB SL
Wzorowane na: Chemistry SL Paper 1B, specimen 2025, pytanie 1(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
Hot tubs and spa pools are often sanitized with bromine compounds rather than with chlorine. The active sanitizer is hypobromous acid, HOBr, which forms when bromine reacts with water.
Br2(aq) + H2O(l) ⇌ HOBr(aq) + HBr(aq)
Polecenie
A student compiles data on the solubility of bromine in water at different temperatures from several websites and textbooks.
| Source | Temperature / °C | Solubility of bromine |
|---|---|---|
| A | 5 | 3.90 g per 100 cm3 of water |
| B | 15 | 1.15 cm3 of liquid bromine per 100 cm3 of water |
| C | 25 | 34.0 g per dm3 of water |
| D | 30 | 3.35 g per 100 g of water |
| E | 40 | 1.02 cm3 of liquid bromine per 100 cm3 of water |
| F | 40 | 3.30 g per 100 cm3 of water |
(i) Identify one problem that arises when the values from these sources are compared with one another.
(ii) Each solubility is then expressed in mol dm−3.
| Source | Temperature / °C | Solubility of bromine | Solubility of bromine / mol dm−3 |
|---|---|---|---|
| A | 5 | 3.90 g per 100 cm3 of water | 0.244 |
| B | 15 | 1.15 cm3 per 100 cm3 of water | 0.22 |
| C | 25 | 34.0 g per dm3 of water | 0.213 |
| D | 30 | 3.35 g per 100 g of water | 0.210 |
| E | 40 | 1.02 cm3 per 100 cm3 of water | |
| F | 40 | 3.30 g per 100 cm3 of water | 0.207 |
Calculate the missing value for source E and so complete the table. Take the density of liquid bromine to be 3.10 g cm−3.
(iii) Suggest an explanation for the trend in the solubility of bromine as the temperature rises.
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Odpowiedź z klucza
(i) Any one of:
different units / different ways of expressing solubility ✔
OR
some sources give the amount of water by mass, others by volume ✔
OR
different precisions / numbers of significant figures ✔
OR
uncertainties «of the values» not given ✔
OR
«for sources B and E» the density of bromine is needed but not stated ✔
(ii) «mass of bromine =» 1.02 «cm3» × 3.10 «g cm−3» = 3.16 «g» ✔
«solubility =» 3.16 «g» / 159.80 «g mol−1» / 0.100 «dm3» = 0.198 «mol dm−3» ✔
(iii) «solubility decreases as temperature increases, so» dissolving «bromine in water» is exothermic «and the equilibrium shifts towards undissolved bromine at higher temperature» ✔
OR
«at higher temperature» the kinetic/thermal energy of the molecules overcomes the intermolecular forces between bromine and water «so bromine escapes from the solution» ✔
OR
the negative entropy change of dissolving becomes more significant at higher temperature ✔
Schemat punktowania
(i) [1 max] for any one valid problem. (ii) M1: mass of bromine from volume and density; M2: conversion to mol dm−3 using M(Br2) = 159.80 g mol−1 and 0.100 dm3 of water; award [2] for a correct final answer (0.198, accept 0.20). ECF from M1. (iii) [1] for a valid explanation consistent with a decrease in solubility with temperature.
Komentarz. In (ii) a common slip is to divide by 100 instead of 0.100, giving 0.00198 mol dm−3; another is to use M = 79.90 g mol−1 for Br2.