Thermodynamics Exam Practice
Exam-style practice questions with detailed mark schemes and examiner tips.
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)
- Mg^2+ and O^2- ions have larger charges (+2 and -2 respectively) compared to Na^+ and Cl^- (+1 and -1) [1]
- Mg^2+ has a smaller ionic radius than Na^+ OR O^2- 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 Mg^2+(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 Mg^2+ + 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 J K^-1 mol^-1
• CaO(s): ΔHf = -635 kJ mol^-1, S = 40 J K^-1 mol^-1
• CO2(g): ΔHf = -394 kJ mol^-1, S = 214 J K^-1 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)
- ΔH = ΣΔHf(Products) − ΣΔHf(Reactants) [1]
- ΔH = [(-635) + (-394)] − [-1207] = +178 kJ mol^-1 [1]
(b)
- ΔS = ΣS(Products) − ΣS(Reactants) [1]
- ΔS = [40 + 214] − 93 = +161 J K^-1 mol^-1 [1]
(c)
- For a reaction to be feasible, ΔG ≤ 0. Setting ΔG = 0 gives T = ΔH / ΔS [1]
- Convert ΔH to J mol^-1: +178 kJ mol^-1 = +178,000 J mol^-1 [1]
- T = 178,000 / 161 [1]
- T = 1106 K (or 1105.6 K) [1]