Q26.
(a) Chloride (ions) are smaller (than bromide ions)
Must state or imply ions.
Allow chloride has greater charge density (than bromide).
Penalise chlorine ions once only (max 2 / 3).
1
So the force of attraction between chloride ions and water is stronger
This can be implied from M1 and M3 but do not allow intermolecular forces.
1
Chloride ions attract the δ+ on H of water / electron deficient H on water
Allow attraction between ions and polar / dipole water.
Penalise H+ (ions) and mention of hydrogen bonding for M3
Ignore any reference to electronegativity.
Note: If water not mentioned can score M1 only.
1
(b) ΔHsolution = ΔHL + ΔHhyd K+ ions + ΔHhyd Br − ions / = 670 − 322 − 335
Allow ΔHsolution= ΔHL + ΣΔHhyd
1
= (+)13 (kJ mol−1)
Ignore units even if incorrect.
+13 scores M1 and M2
−13 scores 0
−16 scores M2 only (transcription error).
1
(c) (i) The entropy change is positive / entropy increases
ΔS is negative loses M1 and M3
1
Because 1 mol (solid) → 2 mol (aqueous ions) / no of particles increases
Allow the aqueous ions are more disordered (than the solid).
Mention of atoms / molecules loses M2
1
Therefore TΔS > ΔH
1
(ii) Amount of KCl = 5/Mr = 5/74.6 = 0.067(0) mol
If moles of KCl not worked out can score M3, M4 only (answer to M4 likely to be 205.7 K)
1
Heat absorbed = 17.2 × 0.0670 = 1.153 kJ
Process mark for M1 × 17.2
1
Heat absorbed = mass × sp ht × ΔT
(1.153 × 1000) = 20 × 4.18 × ΔT
If calculation uses 25 g not 20, lose M3 only (M4 = 11.04, M5 = 287)
1
ΔT = 1.153 × 1000 / (20 × 4.18) = 13.8 K
If 1000 not used, can only score M1, M2, M3
M4 is for a correct ΔT
Note that 311.8 K scores 4 (M1, M2, M3, M4).
1
T = 298 − 13.8 = 284(.2) K
If final temperature is negative, M5 = 0
Allow no units for final temp, penalise wrong units.
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