What is enthalpy of solution?

When an ionic solid dissolves, two enthalpy terms compete: the lattice must be broken apart (the reverse of the exothermic lattice enthalpy, so endothermic) and the gaseous ions are then hydrated (exothermic enthalpy of hydration). The enthalpy of solution is the sum of these, H_sol = - H_latt + H_hyd. If hydration releases more energy than is needed to break the lattice, dissolving is exothermic; if not, it is endothermic and may only occur because of a favourable entropy increase as ordered solid disperses into solution.

State the condition on G for a reaction to be feasible. [1 mark]

Predict the sign of S for CaCO_3(s) → CaO(s) + CO_2(g). [1 mark]

Explain why a reaction with G < 0 may still not occur at room temperature. [1 mark]

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What makes a reaction feasible?

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

A focused answer to WJEC A-Level Chemistry Unit 3, covering Born-Haber cycles and lattice enthalpy, enthalpy of solution, entropy changes, and Gibbs free energy as the criterion for reaction feasibility.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

WJEC wants you to construct Born-Haber cycles to find lattice enthalpy, use enthalpy of solution, understand entropy as a measure of disorder, and use Gibbs free energy to decide whether a reaction is feasible.

The answer

Lattice enthalpy and Born-Haber cycles

Lattice enthalpy is more exothermic for ions of higher charge and smaller radius, because the electrostatic attraction is stronger.

Entropy

Gibbs free energy

Lattice enthalpy from a Born-Haber cycle

For magnesium oxide, given $\Delta H_f = -602$ , atomisation of Mg $+148$ , first and second ionisation of Mg $+738$ and $+1451$ , atomisation of O $+249$ , first and second electron affinity of O $+(-141)$ and $+798$ , all kJ mol $^{-1}$ , find the lattice enthalpy.

step 1 Sum the upward steps

$148 + 738 + 1451 + 249 + (-141) + 798 = 3243$ kJ mol $^{-1}$ .

step 2 Apply Hess's law

$\Delta H_f = (\text{sum of steps}) + \Delta H_{latt}$ .

step 3 Rearrange

$\Delta H_{latt} = \Delta H_f - 3243 = -602 - 3243 = -3845$ kJ mol $^{-1}$ .

step 4 Interpret

The very exothermic value reflects the high charges ( $\text{Mg}^{2+}$ , $\text{O}^{2-}$ ) and small ions.

Enthalpy of solution

When an ionic solid dissolves, two enthalpy terms compete: the lattice must be broken apart (the reverse of the exothermic lattice enthalpy, so endothermic) and the gaseous ions are then hydrated (exothermic enthalpy of hydration). The enthalpy of solution is the sum of these, $\Delta H_{sol} = -\Delta H_{latt} + \Sigma \Delta H_{hyd}$ . If hydration releases more energy than is needed to break the lattice, dissolving is exothermic; if not, it is endothermic and may only occur because of a favourable entropy increase as ordered solid disperses into solution.

Feasibility versus rate

A key distinction in this topic is between thermodynamic feasibility and kinetic rate. Gibbs free energy tells you only whether a reaction can happen ( $\Delta G \le 0$ ), not how fast. The combustion of diamond to carbon dioxide has a strongly negative $\Delta G$ , yet diamonds do not burn at room temperature because the activation energy is enormous. So a reaction can be feasible but immeasurably slow, which is why feasibility predictions must always be qualified with the reminder that a high activation energy can block a thermodynamically favourable reaction.

Examples in context

Why some salts dissolve. Whether a salt dissolves depends on the balance of lattice enthalpy and hydration enthalpy, the enthalpy of solution, and on the entropy gain of dispersing ions, a direct use of Born-Haber and Gibbs ideas. Thermal decomposition. Heating drives endothermic, entropy-increasing decompositions (like carbonates) feasible above a threshold temperature where $-T\Delta S$ outweighs $\Delta H$ .

Try this

Q1. State the sign of lattice enthalpy and explain why. [1 mark]

Cue. Negative (exothermic), because energy is released when gaseous ions attract to form a lattice.

Q2. State the condition on $\Delta G$ for a reaction to be feasible. [1 mark]

Cue. $\Delta G \le 0$ .

Q3. Predict the sign of $\Delta S$ for $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$ . [1 mark]

Cue. Positive, because a gas is produced, increasing disorder.

Q4. State the two enthalpy terms that combine to give the enthalpy of solution. [1 mark]

Cue. The lattice enthalpy (breaking the lattice) and the enthalpy of hydration of the ions.

Q5. Explain why a reaction with $\Delta G < 0$ may still not occur at room temperature. [1 mark]

Cue. A high activation energy can make the reaction immeasurably slow despite being feasible.

Exam-style practice questions

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

WJEC 20205 marksUse a Born-Haber cycle to calculate the lattice enthalpy of sodium chloride from: enthalpy of formation -411, atomisation of Na +107, first ionisation of Na +496, atomisation of Cl +122, electron affinity of Cl -349 kJ mol-1.

Show worked answer →

By Hess's law around the cycle, the enthalpy of formation equals the sum of the atomisation, ionisation and electron-affinity steps plus the lattice enthalpy (formation of the lattice from gaseous ions).

$\Delta H_f = \Delta H_{at}(\text{Na}) + IE_1 + \Delta H_{at}(\text{Cl}) + EA(\text{Cl}) + \Delta H_{latt}$ .

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

$\Delta H_{latt} = -411 - [107 + 496 + 122 + (-349)] = -411 - 376 = -787$ kJ mol-1.

Markers reward the cycle, correct signs, the rearrangement, and a negative lattice enthalpy.

WJEC 20194 marksA reaction has delta-H = +30.0 kJ mol-1 and delta-S = +120 J K-1 mol-1. Determine the temperature above which the reaction becomes feasible.

Show worked answer →

A reaction is feasible when $\Delta G \le 0$ , where $\Delta G = \Delta H - T\Delta S$ .

At the threshold $\Delta G = 0$ , so $T = \dfrac{\Delta H}{\Delta S}$ .

Convert $\Delta S$ to kJ: $120$ J K-1 mol-1 $= 0.120$ kJ K-1 mol-1.

$T = \dfrac{30.0}{0.120} = 250$ K. Above $250$ K the $-T\Delta S$ term outweighs $\Delta H$ , so the reaction is feasible.

Markers reward $\Delta G = \Delta H - T\Delta S$ , setting $\Delta G = 0$ , converting units, and the temperature of $250$ K.

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Sources & how we know this

WJEC A-level Chemistry specification — WJEC (2015)