Pearson Edexcel Edexcel 2019-style practice: Use the following data to calculate the lattice energy of sodium chloride by constructing a Born-Haber cycle: H_f = -411; atomisation of Na = +107; first ionisation of Na = +496; atomisation of Cl = +122; electron affinity of Cl = -349 (all kJ mol^-1).

Apply Hess's law around the cycle: the formation route equals the elemental route plus the lattice step. H_f = H_at(Na) + IE_1(Na) + H_at(Cl) + EA(Cl) + H_latt (1, cycle set up). Rearrange: H_latt = H_f - [ H_at(Na) + IE_1 + H_at(Cl) + EA] (1). = -411 - (107 + 496 + 122 - 349) (1) = -411 - 376 (1) = -787\ kJ mol^-1 (1). Markers reward a correct cycle, correct signs (electron affinity is exothermic), and the final exothermic lattice energy.

Pearson Edexcel Edexcel 2021-style practice: For the decomposition CaCO_3(s) → CaO(s) + CO_2(g), H = +178\ kJ mol^-1 and S = +161\ J K^-1\,mol^-1. (a) Calculate G at 298\ K and state whether the reaction is feasible. (b) Calculate the temperature above which the reaction becomes feasible.

Use G = H - T S, converting entropy to kJ. (a) S = +0.161\ kJ K^-1\,mol^-1. G = 178 - 298 × 0.161 = 178 - 48.0 = +130\ kJ mol^-1 (1). Since G > 0, the reaction is not feasible at 298\ K (1). (b) At the feasibility limit G = 0, so T = H S = 1780.161 = 1106\ K (1). The reaction becomes feasible above about 1106\ K (around 830\ ^C) (1).

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Why do some reactions happen spontaneously while others do not?

Born-Haber cycles and lattice energy, enthalpies of solution, hydration and atomisation, entropy, and Gibbs free energy as the criterion for feasibility.

An Edexcel 9CH0 Topic 12 answer covering Born-Haber cycles and lattice energy, enthalpies of solution and hydration, entropy, and Gibbs free energy as the test of feasibility.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

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What this topic is asking

Edexcel Topic 12 wants you to construct and use Born-Haber cycles to find lattice energies, relate lattice and hydration enthalpies to ionic charge and radius, use enthalpies of solution, calculate entropy changes, and use Gibbs free energy to decide feasibility, including the temperature at which a reaction becomes feasible.

The answer

Born-Haber cycles

To build the cycle, draw the elements at the bottom, form gaseous atoms (atomisation), then gaseous ions (ionisation and electron affinity), and finally bring the gaseous ions together to form the solid lattice (lattice energy). By Hess's law, the direct formation enthalpy equals the sum of all the steps, so any one unknown can be found by difference.

Enthalpies of solution and hydration

When an ionic solid dissolves, the lattice is broken into gaseous ions and the gaseous ions are then hydrated:

A salt dissolves readily when the hydration enthalpies released are enough to compensate for the lattice energy that must be put in to separate the ions.

Entropy and Gibbs free energy

Always convert $\Delta S$ from $\text{J K}^{-1}\,\text{mol}^{-1}$ to $\text{kJ K}^{-1}\,\text{mol}^{-1}$ before combining it with $\Delta H$ . A reaction with a positive $\Delta H$ (endothermic) can still become feasible at high temperature if $\Delta S$ is positive, because the $-T\Delta S$ term grows more negative as $T$ rises. The temperature at which feasibility just begins is found by setting $\Delta G = 0$ , giving $T = \Delta H / \Delta S$ .

Finding the temperature at which a reaction becomes feasible

For $\text{MgCO}_3(\text{s}) \rightarrow \text{MgO}(\text{s}) + \text{CO}_2(\text{g})$ , $\Delta H = +117\ \text{kJ mol}^{-1}$ and $\Delta S = +175\ \text{J K}^{-1}\,\text{mol}^{-1}$ . Find $\Delta G$ at $298\ \text{K}$ and the temperature above which it becomes feasible.

step 1: Convert the entropy units

$\Delta S = 175\ \text{J K}^{-1}\,\text{mol}^{-1} = 0.175\ \text{kJ K}^{-1}\,\text{mol}^{-1}$ .

step 2: Calculate $\Delta G$ at 298 K

$\Delta G = \Delta H - T\Delta S = 117 - 298 \times 0.175 = 117 - 52.2 = +64.8\ \text{kJ mol}^{-1}$ .

step 3: Interpret

$\Delta G > 0$ , so the decomposition is not feasible at $298\ \text{K}$ .

step 4: Find the feasibility temperature

Set $\Delta G = 0$ : $T = \dfrac{\Delta H}{\Delta S} = \dfrac{117}{0.175} = 669\ \text{K}$ . Above about $669\ \text{K}$ (around $396\ ^\circ\text{C}$ ) the reaction becomes feasible.

Examples in context

Example 1. Why limestone is heated in a kiln. The decomposition of calcium carbonate to lime is endothermic with a positive entropy change (a gas is released). At room temperature $\Delta G$ is positive, so nothing happens, but heating the kiln above about $1100\ \text{K}$ makes the $-T\Delta S$ term dominate, $\Delta G$ becomes negative, and decomposition proceeds. This is the industrial reason kilns run so hot and is a direct application of the $T = \Delta H/\Delta S$ feasibility calculation.

Example 2. Dissolving and the role of entropy. Some salts, such as ammonium nitrate, dissolve even though the process is endothermic (the pack gets cold). The endothermic enthalpy of solution would forbid it on enthalpy alone, but the large positive entropy change as the ordered lattice disperses into solution makes $\Delta G$ negative. This is why instant cold packs work and shows that feasibility depends on both $\Delta H$ and $\Delta S$ , not enthalpy alone.

Try this

Q1. State the equation for Gibbs free energy and the condition for feasibility. [2 marks]

Cue. $\Delta G = \Delta H - T\Delta S$ ; feasible when $\Delta G \leq 0$ .

Q2. Explain why the lattice energy of magnesium oxide is more exothermic than that of sodium chloride. [2 marks]

Cue. $Mg^{2+}$ and $O^{2-}$ have higher charges and smaller radii than $Na^+$ and $Cl^-$ , so stronger electrostatic attraction.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20195 marksUse the following data to calculate the lattice energy of sodium chloride by constructing a Born-Haber cycle:

\Delta H_f = -411

; atomisation of Na

= +107

; first ionisation of Na

= +496

; atomisation of Cl

= +122

; electron affinity of Cl

= -349

(all

\text{kJ mol}^{-1}

Show worked answer →

Apply Hess's law around the cycle: the formation route equals the elemental route plus the lattice step.

$\Delta H_f = \Delta H_{at}(\text{Na}) + IE_1(\text{Na}) + \Delta H_{at}(\text{Cl}) + EA(\text{Cl}) + \Delta H_{latt}$ (1, cycle set up).

Rearrange: $\Delta H_{latt} = \Delta H_f - [\Delta H_{at}(\text{Na}) + IE_1 + \Delta H_{at}(\text{Cl}) + EA]$ (1).

$= -411 - (107 + 496 + 122 - 349)$ (1) $= -411 - 376$ (1) $= -787\ \text{kJ mol}^{-1}$ (1).

Markers reward a correct cycle, correct signs (electron affinity is exothermic), and the final exothermic lattice energy.

Edexcel 20214 marksFor the decomposition

\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})

\Delta H = +178\ \text{kJ mol}^{-1}

and

\Delta S = +161\ \text{J K}^{-1}\,\text{mol}^{-1}

. (a) Calculate

\Delta G

298\ \text{K}

and state whether the reaction is feasible. (b) Calculate the temperature above which the reaction becomes feasible.

Show worked answer →

Use $\Delta G = \Delta H - T\Delta S$ , converting entropy to $\text{kJ}$ .

(a) $\Delta S = +0.161\ \text{kJ K}^{-1}\,\text{mol}^{-1}$ . $\Delta G = 178 - 298 \times 0.161 = 178 - 48.0 = +130\ \text{kJ mol}^{-1}$ (1). Since $\Delta G > 0$ , the reaction is not feasible at $298\ \text{K}$ (1).

(b) At the feasibility limit $\Delta G = 0$ , so $T = \dfrac{\Delta H}{\Delta S} = \dfrac{178}{0.161} = 1106\ \text{K}$ (1). The reaction becomes feasible above about $1106\ \text{K}$ (around $830\ ^\circ\text{C}$ ) (1).

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Sources & how we know this

Pearson Edexcel A-Level Chemistry (9CH0) specification — Pearson Edexcel (2015)