A-LevelChemistryChemical energetics9701/42 · Feb/Mar 2024 · Q126 Marks

Potassium iodide, KI, is used as a reagent in both inorganic and organic chemistry.

(a)KI forms an ionic lattice that is soluble in water.
(a)(i)Define enthalpy change of solution, ΔΗsol.[1]
(a)(ii)KI(s) has a high solubility in water although its enthalpy change of solution is endothermic. Explain how this high solubility is possible.[2]
(b)Table 1.1 gives some data about the halide ions, Cl-, Br¯ and I¯, and their potassium salts. Table 1.1 | halide ion | enthalpy change of hydration, ∆Hhyd /kJ mol-1 | lattice energy of potassium halide, ∆Hlatt/kJ mol-1 | |:-----------|:----------------------------------------------|:-----------------------------------------------| | Cl- | -364 | -701 | | Br- | -335 | -670 | | I- | -293 | -629 |
(b)(i)Explain the trend in the enthalpy change of hydration of the halide ions.[2]
(b)(ii)The ∆Hsol values of these potassium halides are almost constant. Use the ∆Hhyd and ∆Hlatt data in Table 1.1 to suggest why.[1]
(b)(iii)The enthalpy change of solution of KI(s) is +21.0kJmol−1. Use this information and the data in Table 1.1 to calculate the enthalpy change of hydration of the potassium ion, K+(g). ∆Hhyd of K+(g) =[1]
(b)(iv)Solid PbI2 forms when KI(aq) is mixed with Pb2+(aq) ions. The solubility product, Ksp, of PbI2 is 7.1 × 10-9mol³dm-9 at 25 °C. Calculate the solubility, in moldm³, of PbI2(s). solubility of PbI2(s) =[2]
(b)(v)The ionic radius of Pb2+ is 0.120nm compared to 0.133nm for K+. Suggest how the ∆Hlatt of PbI2(s) differs from ΔΗatt of KI(s). Explain your answer.[2]
(c)KI slowly oxidises in air, forming I2. reaction 1 4KI(s) + 2CO2(g) + O2(g) → 2K2CO3(s) + 2I2(s) ∆H° = −203.4 kJ mol-1 Table 1.2 shows some data relevant to this question. Table 1.2 | substance | standard entropy, S°/JK-1 mol-1 | |:----------|:----------------------------------| | CO2(g) | 213.6 | | I2(s) | 116.1 | | K2CO3(s) | 155.5 | | KI(s) | 106.3 | | O2(g) | 205.2 |
(c)(i)Calculate the standard entropy change, ∆Sº, of reaction 1. ∆S° =[2]
(c)(ii)Use your answer to (c)(i) to show that reaction 1 is spontaneous at 298 K.[2]
(c)(iii)The Group 1 carbonates are much more thermally stable than the Group 2 carbonates. State and explain the trend in the thermal stability of the Group 2 carbonates.[2]
(d)A student electrolyses a solution of KI(aq) for 8 minutes using a direct current. The half-equation for the reaction that occurs at the anode is given. 2I-(aq) → I2(aq) + 2e-
(d)(i)Write a half-equation for the reaction that occurs at the cathode. Include state symbols.[1]
(d)(ii)After the electrolysis, the I2(aq) produced requires 21.35 cm³ of 0.100moldm-3 Na2S2O3(aq) to react completely. I2(aq) + 2Na2S2O3(aq) → 2NaI(aq) + Na2S4O6(aq) Calculate the average current used in 8 minutes during the electrolysis. current =[3]
(e)KI is used as a source of I- ions in organic synthesis. One example of this is shown in the synthetic route in Fig. 1.1. [Figure 1.1]
(e)(i)Identify the reagents required for steps 1 and 2. step 1 step 2[2]
(e)(ii)Step 3 occurs in two stages. stage I NaNO2 and HCl undergo an acid-base reaction to produce HNO2. stage II HNO2 reacts with C, C6H5NH2, to produce D, C6H5N2+. Complete the equations for stage I and for stage II. stage I NaNO2 + HCl → stage II[2]
(e)(iii)The I- from KI reacts with D in step 4. The mechanism is shown in Fig. 1.1. Suggest the name for this mechanism.[1]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2024). The mark or mark type is shown in brackets.

[1] (enthalpy change when) one mole of a substance / solute AND dissolves in water / turns into an aqueous solution (to form a solution of infinite dilution)
[1] there is a (large) increase in entropy OR ∆S is positive OR T∆S is positive
[1] so ∆G is negative / T∆S outweighs ∆H
[1] anionic charge density decreases (down the group / Cl- to I-)
[1] (so hydration enthalpies become less negative / less exothermic because) less attraction of ion to water/the dipole-ion force weakens
[1] the difference between ∆Hlatt and ∆Hhyd remains roughly constant OR ∆Hlatt and ∆Hhyd become less exothermic by a similar amount
[1] ∆Hhyd(K+(g)) = −629 + 21.0 –(−293) = −315 kJ mol-1
[1] solubility of PbI2 = 7.1 × 10–9 = x(2x)2 ∴ x = ³√(¼ × 7.1 × 10-9)
[1] solubility of PbI2 = 1.21 × 10-3 mol dm-3 min 2sf
[2] Pb2+ has a greater charge
• Pb2+ is smaller (than K+) OR (Pb2+) smaller ionic radius
• greater attraction between Pb2+ and I-
OR ionic bond between Pb2+ and I– is stronger
OR lattice energy is more exothermic / more negative
mark as ✓✓
[1] ∆S = 2(155.5) + 2(116.1) – 4(106.3) – 2(213.6) – 205.2
[1] ∆S = −514.4 (J K-1 mol-1) min 3sf ECF
[1] ∆G = ∆H – T∆S AND use of 298 K for T OR clear working to represent this
[1] ∆G = −203.4 – 298(–514.4 / 1000) ∆G = −50.1 OR –50.2 (kJ mol−1) OR ∆G = −50 108.8 J min 3sf ECF from (c)(i) (negative so spontaneous)
[1] increases (in thermal stability down the group)
[1] AND (cat)ionic radius / ion size increases (down the group)
[1] less polarisation of anion / C—O bond / distortion of carbonate ion / CO32- OR C-O is less weakened / stronger (down the group)
[1] 2H+(aq) + 2e- → H2(g) state symbols required
[1] M1 moles of S2O32- = 21.35 / 1000 × 0.100 = 2.135 × 10-3
[1] M2 calc of Q = 2.135 × 10-3 × 0.5 × 2 × 96500 = 206.02 / 205.68 ECF
[1] M3 = answer I = 206 / (8 × 60) = 0.428 OR 0.429 A min 2sf ECF
[1] reaction 1 concentrated HNO3 AND concentrated H2SO4 (concentrated seen once)
[1] reaction 2 Sn AND (concentrated) HCl
[1] step 1 NaNO2 + HCl → HNO2 + NaCl
[1] step 2 C6H5NH2 + HNO2 + H+ → C6H5N2+ + 2H2O
[1] nucleophilic (aromatic) substitution