Zadanie 2a

9 pktobliczenioweśrednie (szac.)Paper 2, maj 2026, TZ2

Wzorowane na: Chemistry HL Paper 2, May 2026, TZ2, pytanie 2(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.

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Informacja do zadań 2a-2c.

Iodine and selenium are two non-metals that occur in trace amounts in seawater.

Polecenie

Iodine crystals, I2(s), were sealed in a flask and left until an equilibrium was reached with violet vapour above the crystals.

I2(s) ⇌ I2(g)

(i) Name the change of state from I2(s) to I2(g) and state the type of force between I2 molecules that is overcome.

(ii) The flask was then warmed and the violet colour of the vapour became deeper. Explain why.

(iii) The pressure in the flask was higher at the higher temperature. Explain this, in terms of the particles, giving two reasons.

(iv) The sealed 250 cm3 flask was kept at 35 °C. When equilibrium had been reached, the vapour contained 3.05 × 10−3 g of iodine. Calculate the pressure, in kPa, that the iodine vapour exerts.

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(i) Change of state: sublimation ✔

Type of force: London / dispersion forces ✔

(ii) the change from I2(s) to I2(g) is endothermic «absorbs energy» ✔

the equilibrium shifts to the right / more vapour is formed «so the colour is deeper» ✔

(iii) Any two of:

more gas molecules / moles of gas in the flask OR the concentration of the gas increases ✔

the molecules have greater «average» kinetic energy OR the molecules move faster ✔

the molecules collide with the walls of the flask with greater force ✔

the molecules collide with the walls more frequently ✔

(iv) V = 2.50 × 10−4 m3 AND T = 308 K ✔

«n=mM=3.05×10−32×126.90=n = \dfrac{m}{M} = \dfrac{3.05 \times 10^{-3}}{2 \times 126.90} =» 1.20 × 10−5 «mol» ✔

«P=nRTV=1.20×10−5×8.31×3082.50×10−4=P = \dfrac{nRT}{V} = \dfrac{1.20 \times 10^{-5} \times 8.31 \times 308}{2.50 \times 10^{-4}} =» 123 «Pa» / 0.123 «kPa» ✔

Schemat punktowania

(i) [2] M1 sublimation; M2 London / dispersion forces; do not accept van der Waals forces or intermolecular forces for M2. (ii) [2] M1 endothermic change; M2 equilibrium shifts to the vapour side. (iii) [2 max] one mark for each of two different reasons from the key. (iv) [3] M1 volume in m3 AND temperature in K; M2 amount of iodine, using the molar mass of I2; M3 pressure. Award [3] for a correct final answer; ECF from M1 and M2.

Komentarz

In (iv) the molar mass is that of the I2 molecule, 253.80 g mol−1, not that of the iodine atom; using 126.90 doubles the amount and gives about 0.246 kPa.

Umiejętności w zadaniu