A Level Chemistry: Lattice Energy, Entropy and Gibbs Free Energy — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Lattice Energy, Entropy and Gibbs Free Energy revision notes

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Section A

1. Define lattice energy. [2]

2. Explain why the second electron affinity of oxygen is endothermic. [2]

3. Predict, with reasons, whether MgO or NaCl has the more exothermic lattice energy. [3]

4. Predict the sign of ΔS for each and justify: (a) CaCO₃(s) → CaO(s) + CO₂(g); (b) N₂(g) + 3H₂(g) → 2NH₃(g). [4]

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Section B

5. Use the following data to construct a Born–Haber cycle and calculate the lattice energy of potassium chloride.

enthalpy of formation of KCl(s)      -437 kJ mol-1
enthalpy of atomisation of K(s)      +89  kJ mol-1
first ionisation energy of K         +418 kJ mol-1
enthalpy of atomisation of Cl2(g)    +122 kJ mol-1  (per mole of Cl atoms)
first electron affinity of Cl        -349 kJ mol-1

(a) Calculate the lattice energy. [3]

(b) The theoretical lattice energy calculated from a purely ionic model is −702 kJ mol⁻¹. Comment on the difference. [2]

6. For the decomposition of calcium carbonate:

CaCO3(s) -> CaO(s) + CO2(g)     delta-H = +178 kJ mol-1,  delta-S = +161 J K-1 mol-1

(a) Calculate ΔG at 298 K and state whether the reaction is feasible. [3]

(b) Calculate the minimum temperature at which the reaction becomes feasible. [2]

(c) Room temperature (298 K) is far below the temperature found in (b). Using your answers to (a) and (b), explain why the reaction does not proceed at room temperature, however long you wait. [2]

Section C

7. Define enthalpy change of hydration and enthalpy change of solution, and explain why hydration enthalpy is always exothermic while enthalpy of solution can be either sign. [3]

8. Given ΔH_latt(LiCl) = −846 kJ mol⁻¹, ΔH_hyd(Li⁺) = −520 kJ mol⁻¹ and ΔH_hyd(Cl⁻) = −364 kJ mol⁻¹, calculate ΔH_sol(LiCl). [3]

9. Explain how ionic charge and ionic radius affect the magnitude of hydration enthalpy. [2]

10. A reaction has ΔH negative and ΔS negative. State whether it is feasible at all temperatures, never feasible, or feasible only within a certain temperature range — and justify your answer. [2]

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Answers

1. The enthalpy change when one mole of an ionic compound [1] is formed from its gaseous ions [1].

2. An electron is being added to an already negatively charged ion (O⁻) [1], so energy must be supplied to overcome the electrostatic repulsion [1].

3. MgO [1]. Its ions carry double the charge (2+ and 2−, so the charge product is four times greater) [1] and the ionic radii are smaller [1]. Charge dominates over radius — say so.

4. (a) Positive [1] — a gas is produced from a solid, greatly increasing disorder [1]. (b) Negative [1] — four moles of gas become two, so the number of gas particles decreases [1].

5. (a) ΔH_f = ΔH_at(K) + IE(K) + ΔH_at(Cl) + EA(Cl) + ΔH_latt −437 = 89 + 418 + 122 + (−349) + ΔH_latt [1] −437 = 280 + ΔH_latt [1] ΔH_latt = −717 kJ mol⁻¹ [1].

(b) The experimental value is more exothermic than the theoretical [1], indicating a degree of covalent character — the cation polarises the anion’s electron cloud [1].

6. (a) ΔS must be converted: 161 J = 0.161 kJ [1]. ΔG = 178 − (298 × 0.161) = 178 − 47.98 = +130 kJ mol⁻¹ [1]. ΔG is positive, so the reaction is not feasible at 298 K [1].

(b) At the point of feasibility ΔG = 0, so T = ΔH ÷ ΔS [1] = 178 ÷ 0.161 = 1106 K [1].

(c) At 298 K, ΔG is positive (part (a)) — the reaction is simply not thermodynamically feasible at this temperature, so it cannot occur spontaneously no matter how long you wait [1]. Only once T reaches 1106 K (part (b)) does the −TΔS term grow large enough to make ΔG negative and the reaction feasible [1].

7. Hydration enthalpy is the enthalpy change when one mole of a gaseous ion dissolves in water to form one mole of aqueous ions [1]. Enthalpy of solution is the enthalpy change when one mole of an ionic solid dissolves to form an infinitely dilute solution [1]. Hydration is always exothermic because it is dominated by the attraction between ions and polar water molecules, but solution is the balance between breaking the lattice (endothermic) and hydrating the ions (exothermic), so its sign depends on which term is larger [1].

8. ΔH_sol = −ΔH_latt + ΔH_hyd(cation) + ΔH_hyd(anion) [1] = −(−846) + (−520) + (−364) [1] = 846 − 884 = −38 kJ mol⁻¹ [1].

9. Hydration enthalpy becomes more exothermic as ionic charge increases and as ionic radius decreases [1], because a smaller, more highly charged ion has a more concentrated charge density, attracting the surrounding water dipoles more strongly [1].

10. Feasible only below a certain temperature [1]. With ΔH negative, the −TΔS term becomes increasingly positive (since ΔS is negative) as T rises, so ΔG eventually turns positive at high temperature even though it starts negative at low temperature [1].

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Where marks are usually lost

• Omitting “one mole” or “gaseous” from the lattice energy definition.
• Mixing J and kJ in ΔG = ΔH − TΔS — the single most common arithmetic error.
• Forgetting the ×2 for atomisation of a diatomic molecule, or the second ionisation energy for a 2+ ion.
• Saying a feasible reaction will definitely be observed.