Hydrogen peroxide is a liquid at 298K. It is moderately stable under room conditions but will decompose quickly if a catalyst is added. reaction 1 2H2O2(l) → 2H2O (l) + O2 (g)
- (a)(i)Define entropy.[1]
- (a)(ii)Predict the sign of the standard entropy change of reaction 1. Explain your answer. sign explanation[1]
- (b)Some bond energy data are shown in Table 3.1. Table 3.1 type of bond bond energy /kJmol⁻¹ O-O 150 O-H 460 O=O 496 Use the data in Table 3.1 to show that the enthalpy change of the following reaction is -196 kJ mol⁻¹. 2H2O2(g) → 2H2O(g) + O2(g)[1]
- (c)Some standard entropies, Sº, are shown in Table 3.2. Table 3.2 substance Sº /J K⁻¹ mol⁻¹ H2O2(l) +102 H2O(l) +70 The enthalpy change and Gibbs free energy change for the following reaction are shown. 2H2O2(l) → 2H2O(l) + O2(g) ΔHº = –196 kJ mol⁻¹ ΔGº = –238 kJ mol⁻¹ Use the data given to calculate the standard entropy of oxygen, Sº, O2(g). Sº, O2(g) =[3]
- (d)The decomposition of H2O2(aq) is catalysed by aqueous iron(III) chloride and by silver metal. Identify which of these two catalysts is acting as a homogeneous catalyst. Explain your answer. homogeneous catalyst explanation[1]
- (e)The E values for two electrode reactions are given. H2O2 + 2H+ + 2e⁻ ⇌ 2H2O E° = +1.77V Cr3+ + e⁻ ⇌ Cr2+ E° = -0.41V
- (e)(i)An electrochemical cell is constructed with the following half-cells (electrodes): • an acidified solution of H2O2, a platinum wire • Cr2+ mixed with Cr3+, a platinum wire. Identify the positive half-cell and calculate the standard cell potential, E°cell. positive half-cell E°cell =[1]
- (e)(ii)Calculate the value of ΔGº for the cell reaction that occurs, per mole of H2O2. ΔGº =[2]
- (f)The E values for two electrode reactions are given. H2O2 + 2H+ + 2e⁻ ⇌ 2H2O E° = +1.77V Co3+ + e⁻ ⇌ Co2+ Eº = +1.82V An electrochemical cell is constructed with the following half-cells. half-cell 1 an acidified solution of H2O2 under standard conditions, a platinum wire half-cell 2 a solution containing 0.020 moldm⁻³ Co³+ and 2.0 moldm⁻³ Co2+, a platinum wire
- (f)(i)Use the Nernst equation to calculate the value of E, the electrode potential of half-cell 2 under these conditions. E[2]
- (f)(ii)Write an equation for the cell reaction that occurs in this cell under these conditions.[1]
- (g)(i)Define enthalpy change of hydration, ΔHhyd.[1]
- (g)(ii)Aluminium fluoride, AlF3, is an ionic solid. Complete and label the energy cycle to show the relationship between: • the enthalpy change of solution of AlF3, ΔHsol • the lattice energy of AlF3, ΔHlatt • the enthalpy changes of hydration of Al3+ and F⁻, ΔHhyd Include state symbols for all substances and ions.[2]
- (g)(iii)Relevant data for this question are given. ΔHsol AlF3 = -209 kJ mol⁻¹ ΔHhyd Al3+ = -4690 kJ mol⁻¹ ΔHhyd F⁻ = -506 kJ mol⁻¹ Use these data and your energy cycle in (g)(ii) to calculate the ΔHlatt of AlF3. ΔHlatt of AlF3 =[1]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.
- [1] (number of possible) arrangements of particles / energy in a system OR measure / degree of disorder in / of a system
- [1] positive / + AND more gas molecules / particles in products OR more moles / molecules in products / RHS
- [1] ΔH = (2 × 150) – (1 × 496) (= −196) OR ΔH = (2 × 150) + (4 × 460) – (1 × 496) – (4 × 460) (= −196)
- [1] ΔG = ΔH -TΔS seen or used with correct signs
- -238 = -196 – 298ΔS
- ΔS = 42/298
- [1] ΔS = (+)0.141 / 0.1409 (kJ K-1 mol-1) OR (+)141 / 140.9 (J K–1 mol-1) ecf from ΔG = ΔH + TΔS
- [1] 141 = (2 × 70) + S(O2(g)) – (2 × 102) S, O2(g) = 205 / 204.94 (J K−¹ mol-1) ecf
- [1] iron(III) chloride / FeCl3 AND same state / phase as reactants / H2O2
- [1] hydrogen peroxide / H2O2 AND +2.18
- [1] ΔG = -nEcellF
- [1] ΔG = (-2 × 2.18 × 96500) = −420.7 (kJ mol-1) ecf
- [1] Nernst: (E =) E° + (0.059/z)log[ox] / [red] u/c OR (E =) E° + (RT/zF)ln[ox] / [red] OR (E =) 1.82 + (0.059/1)log(0.02/2)
- [1] E = 1.82 + (0.059 / 1)log(0.02 / 2) = (+)1.702 (V) min 2sf OR E = 1.82 + [(8.314 × 298 / 1) × (96 500) ] ln(0.02 /2) = (+)1.702
- [1] H2O2 + 2H+ + 2Co2+ → 2H2O + 2Co3+ ECF for reverse equation from (f)(i) if E > 1.77 V
- [1] enthalpy change when one mole of gaseous ions forms an aqueous solution / dissolves in water
- [1] Substances with state symbols: AlF3(s) AlF3(aq) OR Al3+(aq) + (3)F-(aq) Al3+(g) (3)F¯(g)
- [1] changes identified and three correct arrow directions (one given)
- [1] ΔHlatt = −4690 + (3 × – 506) – (−209) = −5999 (kJ mol-1)
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