A-LevelChemistryGroup 179701/42 · Feb/Mar 2021 · Q318 Marks

Iodates are compounds that contain the IO3¯ anion.

(a)The IO3 anion is shown. [Figure 3.1] Explain, with reference to the qualitative model of electron-pair repulsion, why the IO3 anion has a pyramidal shape.[1]
(b)The reaction of iodine and hot aqueous sodium hydroxide is similar to that of chlorine and hot aqueous sodium hydroxide. Sodium iodate, NaIO3, is formed as one of the products. Suggest an equation for the reaction of iodine and hot aqueous sodium hydroxide.[1]
(c)(i)The decomposition of hydrogen peroxide, H2O2, is catalysed by acidified IO3¯. H2O2 reduces acidified IO3¯ as shown. 5H2O2 + 2H+ + 2IO3¯ → I₂ + 5O2 + 6H2O This reaction is followed by the oxidation of I₂ by H2O2. half-equation E°/V H2O2 + 2H+ + 2e- ⇌ 2H₂O +1.77 IO3¯ + 6H+ + 5e¯ ⇌ ½I₂ + 3H2O +1.19 O2 + 2H+ + 2e¯ ⇌ H2O2 +0.68 Use the data to show that the separate reactions of H2O2 with IO3¯ and with I2 are both feasible under standard conditions. In your answer, give the equation for the reaction of H2O2 with I2.[3]
(c)(ii)Write the overall equation for the decomposition of H2O2 catalysed by acidified IO3¯.[1]
(d)(i)A student collects some data for the reaction of H2O2 with acidified IO3¯, as shown in the table. experiment [H2O2] /mol dm-3 [IO3-] /mol dm-3 [H+] /mol dm-3 initial rate of reaction /mol dm-³s-1 1 0.0500 0.0700 0.025 1.47 x 10-5 2 0.100 0.0700 0.050 2.94 x 10-5 3 0.100 0.140 0.025 5.88 × 10-5 4 0.150 0.140 0.025 8.82 x 10-5 Use the data to determine the order of reaction with respect to [H2O2], [IO3¯] and [H+]. Show your reasoning. order with respect to [H2O2] = .......................... order with respect to [IO3¯] = .......................... order with respect to [H+] = ..........................[3]
(d)(ii)Use your answer to (d)(i) to write the rate equation for this reaction. rate = ..........................[1]
(d)(iii)Calculate the value of the rate constant, k, using data from experiment 4 and your answer to (d)(ii). Give the units of k. k = .......................... units = ..........................[2]
(e)(i)Pb(IO3)2 is only sparingly soluble in water at 25°C. The solubility product, Ksp, of Pb(IO3)2 is 3.69 × 10-13 mol³ dm¯º at 25°C. Write an expression for the solubility product of Pb(IO3)2. Ksp = ..........................[1]
(e)(ii)Calculate the solubility, in moldm¯³, of Pb(IO3)2 at 25 °C. solubility = .......................... mol dm-³[2]
(f)(i)NH4IO3 is an unstable compound that readily decomposes when warmed. The decomposition reaction is shown. NH4IO3(s) → ½N2(g) + ½O2(g) + ½I2(g) + 2H2O(l) ΔΗ = –154.6 kJ mol-1 Use the data in the table to calculate the entropy change of reaction, ΔS, of the decomposition of NH4IO3(s). compound S/JK-1 mol-1 NH4IO3(s) 42 N2(g) 192 O2(g) 205 I2(g) 261 H2O(I) 70 ΔS = .......................... JK-1 mol-1[2]
(f)(ii)This reaction is feasible at all temperatures. Explain why, using the data in (f) and your answer to (f)(i).[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 2021). The mark or mark type is shown in brackets.

[1] 3 bonding-pair centres and one lone pair (on iodine)
[1] 3I2 + 6NaOH → NaIO3 + 5NaI + 3H2O
[3] M1: E_cell for H2O2 with IO3⁻ = E°(IO3⁻/I₂) - E°(O₂/H₂O₂) = +1.19 V - (+0.68 V) = +0.51 V, hence feasible.
[3] M2: E_cell for H2O2 with I₂ = E°(H₂O₂/H₂O) - E°(IO₃⁻/I₂) = +1.77 V - (+1.19 V) = +0.58 V, hence feasible.
[3] M3: Equation: 5H₂O₂ + I₂ → 4H₂O + 2IO₃⁻ + 2H⁺
[1] 2H₂O₂ → 2H₂O + O₂
[3] order with respect to [H₂O₂] = 1 (reasoning: comparing expts 3 and 4, [H₂O₂] × 1.5, rate × 1.5, so 1st order)
[3] order with respect to [IO₃⁻] = 1 (reasoning: comparing expts 1 and 3, [H₂O₂] × 2, [IO₃⁻] × 2, rate × 4. Since [H₂O₂] is 1st order, [IO₃⁻] must be 1st order for rate to be ×4 total.)
[3] order with respect to [H⁺] = 0 (reasoning: comparing expts 1 and 2, [H₂O₂] × 2, [H⁺] × 2, rate × 2. Since [H₂O₂] is 1st order, [H⁺] must be 0th order for rate to be ×2 total.)
[1] rate = k[H₂O₂][IO₃⁻] (ecf from (d)(i))
[2] M1: k = rate / ([H₂O₂][IO₃⁻]) = 8.82 × 10⁻⁵ mol dm⁻³ s⁻¹ / (0.150 mol dm⁻³ × 0.140 mol dm⁻³) = 4.20 × 10⁻³ (ecf from (d)(ii) if arithmetic is correct)
[2] M2: Units = mol⁻¹ dm³ s⁻¹
[1] Ksp = [Pb²⁺][IO₃⁻]²
[2] M1: Let solubility be x. Ksp = (x)(2x)² = 4x³. So, 4x³ = 3.69 × 10⁻¹³.
[2] M2: x = ³√(3.69 × 10⁻¹³ / 4) = 4.52 × 10⁻⁵ mol dm⁻³ (allow 2 or 3 significant figures, e.g., 4.5 × 10⁻⁵ or 4.52 × 10⁻⁵).
[2] M1: ΔS = [½(192) + ½(205) + ½(261) + 2(70)] - 42
[2] M2: ΔS = (+)427 J K⁻¹ mol⁻¹ (ecf)
[1] The reaction is feasible at all temperatures because ΔG is always negative (due to ΔH < 0 / negative OR exothermic AND ΔS > 0/ positive OR –T∆S < 0 for all T)