Zadanie 5a
Wzorowane na: Chemistry SL Paper 2, May 2024, TZ2, pytanie 5(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 5a-5b.
A student studied the kinetics of the reaction between a purple dye, D+, and aqueous sodium hydroxide, represented by the equation below.
D+(aq) + OH−(aq) → DOH(aq)
The ion D+ is intensely purple and DOH is colourless.
In a preliminary experiment the student timed how long the purple colour of the mixture described in the table took to vanish, and estimated the rate of reaction from the expression below.
estimated rate = [D+]/ time for colour to disappear
| Initial [D+] / mol dm−3 | Initial [OH−] / mol dm−3 | Time for colour to disappear / s | Estimated rate / mol dm−3 s−1 |
|---|---|---|---|
| 2.50 × 10−6 | 4.00 × 10−2 | 140 | 1.79 × 10−8 |
Polecenie
Outline why the estimated rate calculated from the preliminary experiment is not the initial rate of the reaction.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
the rate falls during the 140 s because the reactants are used up «their concentrations decrease» ✔
so the estimated rate is an average over the whole time, lower than the rate at the start «OWTTE»
Schemat punktowania
[1]: the rate decreases with time as concentrations fall, or the estimated value is an average rate. Accept "the rate is not constant".
Komentarz
The preliminary figure (1.79 × 10^-8) is lower than the initial rate measured for the same mixture in experiment 1 of 5b (2.20 × 10^-8), as expected for an average over a falling rate.