Structured Questions
Answer the questions below in the spaces provided. Check your answers with the mark schemes and examiner tips.
Question 1: Born-Haber Cycles & Bonding Character
7 marksThe Born-Haber cycle for sodium chloride contains several steps.
(a) Name the enthalpy changes represented by the following processes:
• (i) Na(s) → Na(g)
• (ii) Cl(g) + e⁻ → Cl⁻(g)
• (iii) Na⁺(g) + Cl⁻(g) → NaCl(s) [3]
(b) Explain why the lattice enthalpy of formation of magnesium oxide (MgO) is significantly more exothermic than that of sodium chloride (NaCl). [4]
Show Mark Scheme
(a)
- (i) Enthalpy of atomisation of sodium [1]
- (ii) First electron affinity of chlorine [1]
- (iii) Lattice enthalpy of formation of sodium chloride [1]
(b)
- Mg²⁺ and O²⁻ ions have larger charges (+2 and -2 respectively) compared to Na⁺ and Cl⁻ (+1 and -1) [1]
- Mg²⁺ has a smaller ionic radius than Na⁺ OR O²⁻ has a smaller ionic radius than Cl⁻ [1]
- There is a much stronger electrostatic attraction between the ions in magnesium oxide [1]
- More energy is released when the lattice is formed, making the enthalpy of formation significantly more exothermic [1]
Question 2: Dissolution of Magnesium Chloride
8 marksWhen magnesium chloride dissolves in water, the temperature of the mixture increases.
(a) Define enthalpy of solution. [2]
(b) Calculate the enthalpy of solution of magnesium chloride (MgCl₂) using the following data:
• Lattice dissociation enthalpy of MgCl₂ = +2526 kJ mol⁻¹
• Hydration enthalpy of Mg²⁺(g) = −1920 kJ mol⁻¹
• Hydration enthalpy of Cl⁻(g) = −378 kJ mol⁻¹ [3]
(c) Explain, in terms of the forces between particles, why the hydration of gaseous ions is an exothermic process. [3]
Show Mark Scheme
(a)
- The enthalpy change when one mole of an ionic solid [1]
- is completely dissolved in water to form infinitely dilute aqueous ions under standard conditions [1]
(b)
- Enthalpy of Solution = Lattice dissociation enthalpy + Hydration enthalpy of Mg²⁺ + 2 × Hydration enthalpy of Cl⁻ [1]
- Enthalpy of Solution = +2526 + (−1920) + 2 × (−378) [1]
- Enthalpy of Solution = −150 kJ mol⁻¹ [1]
(c)
- Water molecules are polar, with oxygen being delta negative and hydrogen being delta positive [1]
- Electrostatic attractions (ion-dipole forces) form between the gaseous ions and the polar water molecules [1]
- Forming attractions releases energy, making the process exothermic [1]
Question 3: Entropy & Gibbs Free Energy
8 marksThe thermal decomposition of calcium carbonate is represented by the equation:
CaCO₃(s) → CaO(s) + CO₂(g)
Standard enthalpies of formation (ΔHf) and standard entropies (\(S\)) are:
• CaCO₃(s): ΔHf = −1207 kJ mol⁻¹, \(S = 93\text{ J K}^{-1}\text{mol}^{-1}\)
• CaO(s): ΔHf = −635 kJ mol⁻¹, \(S = 40\text{ J K}^{-1}\text{mol}^{-1}\)
• CO₂(g): ΔHf = −394 kJ mol⁻¹, \(S = 214\text{ J K}^{-1}\text{mol}^{-1}\)
(a) Calculate the enthalpy change (ΔH) for this reaction. [2]
(b) Calculate the entropy change (ΔS) for this reaction. [2]
(c) Calculate the minimum temperature, in Kelvin, at which this reaction becomes feasible. Show your working. [4]
Show Mark Scheme
(a)
- \(\Delta H = \Sigma\Delta H_{\text{f}}\text{(products)} - \Sigma\Delta H_{\text{f}}\text{(reactants)}\) [1]
- ΔH = [(−635) + (−394)] − [−1207] = +178 kJ mol⁻¹ [1]
(b)
- \(\Delta S = \Sigma S\text{(products)} - \Sigma S\text{(reactants)}\) [1]
- ΔS = [40 + 214] − 93 = +161 J K⁻¹mol⁻¹ [1]
(c)
- For a reaction to be feasible, ΔG ≤ 0. Setting ΔG = 0 gives \(T = \frac{\Delta H}{\Delta S}\) [1]
- Convert ΔH to J mol⁻¹: +178 kJ mol⁻¹ = +178,000 J mol⁻¹ [1]
- \(T = \frac{178,000}{161}\) [1]
- \(T =\) 1106 K (or 1105.6 K) [1]