Complete the table by placing one tick (✓) in each row to indicate the sign of each type of energy change under standard conditions. [Table with energy change types and columns for always positive, always negative, either negative or positive]
- (a)Complete the table by placing one tick (✓) in each row to indicate the sign of each type of energy change under standard conditions. [Table with energy change types and columns for always positive, always negative, either negative or positive][1]
- (b)Define, in words, the term enthalpy change of solution.[1]
- (c)The following enthalpy changes are given. [Table with enthalpy changes and values for formation of K₃PO₄(s), P(s) + 2O₂(g) + 3e⁻ → PO₄³⁻(aq), K(s) → K⁺(aq) + e⁻] Determine the standard enthalpy change of solution of potassium phosphate, K₃PO₄(s). It may be helpful to draw a labelled energy cycle. ΔHsol =[3]
- (d)Some lattice energy values are shown in the table. [Table with compounds CaBr₂(s) and KBr(s) and their lattice energy values] Suggest an explanation for why ΔHlatt CaBr₂ is more exothermic than ΔHlatt KBr.[2]
- (e)(i)For a particular gas phase reaction the variation in standard Gibbs free energy change, ΔGᵒ, with temperature is shown. Assume standard enthalpy change, ΔHᵒ, and standard entropy change, ΔSᵒ, remain constant with temperature. [Figure X.X] Write the equation that relates ΔGᵒ to ΔHᵒ and ΔSᵒ.[1]
- (e)(ii)Use this equation to explain why ΔGᵒ becomes less positive as temperature increases in this reaction.[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 (May/June 2019). The mark or mark type is shown in brackets.
- [1] both
- [1] (energy change) when 1 mole of solute is dissolved in an infinite amount of water to form a dilute solution
- [1] calculation of ΔHᵒ_sol with −251, -1284 and -2035 only and two correct signs
- [2] calculation of ΔHᵒ_sol with −251, -1284 and -2035 only and correct signs OR calculation of ΔHᵒ_sol with (−251 × 3), −1284 and –2035 only and two correct signs
- [3] ΔHᵒ_sol = (3 × −251) + (−1284) – (−2035) = -2 (kJ mol⁻¹)
- [1] Ca²⁺ have a higher charge / greater charge density ora
- [1] stronger electrostatic forces between Br⁻ and Ca²⁺
- [1] ΔGᵒ = ΔHᵒ − TΔSᵒ
- [1] TΔS is more positive OR -TΔS becomes more negative
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