Enthalpy of Solution and Hydration — Formula and Worked Examples
Why does ammonium chloride make a beaker feel ice-cold while calcium chloride makes it warm? Both are white solids dissolving in water, yet one absorbs heat and the other releases it. The answer is a simple energy balance between two competing steps, and once you can set that balance up you can predict the sign of the heat change for almost any salt. This is a standard Class 12 thermodynamics question and it also appears in IIT-JAM and GATE papers.
The two steps inside dissolving
When an ionic solid dissolves in water, two things must happen:
- The lattice must break apart. The oppositely charged ions in the crystal are pulling on each other strongly, so pulling them into the gas phase costs energy. This is the lattice dissociation enthalpy, and it is always positive (endothermic).
- The free ions must be hydrated. Water molecules surround each ion — the negative oxygen end points at cations, the positive hydrogen end at anions — and new ion–dipole attractions form. Forming attractions releases energy, so hydration enthalpy is always negative (exothermic).
The sum is over every ion the formula unit produces. CaCl₂ gives one Ca²⁺ and two Cl⁻, so the chloride hydration enthalpy is counted twice.
The sign convention trap — two definitions of lattice enthalpy
Textbooks genuinely differ here, and mixing the two definitions is the single biggest source of wrong answers.
| Name used | Process it describes | Sign |
|---|---|---|
| Lattice dissociation enthalpy | NaCl(s) → Na⁺(g) + Cl⁻(g) | Positive (endothermic) |
| Lattice formation enthalpy | Na⁺(g) + Cl⁻(g) → NaCl(s) | Negative (exothermic) |
NCERT defines lattice enthalpy as the enthalpy change when one mole of solid is formed from its gaseous ions, so in NCERT it is a negative number. Many other books tabulate the dissociation value, which is positive. They are the same magnitude with opposite signs. The formula box above uses the dissociation value. If your data table gives the formation value, change its sign before adding, or use ΔHsolution = Σ ΔHhydration − ΔHlattice formation. Always read the heading of the table you are given.
Worked example 1 — sodium chloride
Data: lattice dissociation enthalpy of NaCl = +788 kJ/mol; ΔHhyd(Na⁺) = −406 kJ/mol; ΔHhyd(Cl⁻) = −378 kJ/mol.
Sum of hydration enthalpies = (−406) + (−378) = −784 kJ/mol
ΔHsolution = (+788) + (−784) = +4 kJ/mol
A small positive value: dissolving table salt in water is very slightly endothermic, so the solution cools by a barely noticeable amount. The measured value is about +3.9 kJ/mol, which agrees well.
Worked example 2 — potassium chloride
Data: lattice dissociation enthalpy of KCl = +718 kJ/mol; ΔHhyd(K⁺) = −322 kJ/mol; ΔHhyd(Cl⁻) = −378 kJ/mol.
Sum of hydration enthalpies = (−322) + (−378) = −700 kJ/mol
ΔHsolution = (+718) + (−700) = +18 kJ/mol
KCl absorbs noticeably more heat than NaCl. The reason is visible in the numbers: K⁺ is a bigger ion than Na⁺, so its hydration enthalpy is 84 kJ/mol less negative, while the lattice enthalpy only falls by 70 kJ/mol. Hydration loses the race.
Worked example 3 — calcium chloride, the exothermic case
Data: lattice dissociation enthalpy of CaCl₂ = +2258 kJ/mol; ΔHhyd(Ca²⁺) = −1577 kJ/mol; ΔHhyd(Cl⁻) = −378 kJ/mol.
Careful — there are two chloride ions:
Sum of hydration = (−1577) + 2 × (−378) = −1577 − 756 = −2333 kJ/mol
ΔHsolution = (+2258) + (−2333) = −75 kJ/mol
Strongly exothermic, which is why anhydrous CaCl₂ is used in self-heating packs and why the beaker gets hot. The doubly charged Ca²⁺ is hydrated far more strongly than any 1+ ion, and that is what tips the balance.
A warning about data: hydration and lattice enthalpies are not measured directly, they are derived from thermodynamic cycles, and different tables quote values that differ by tens of kJ/mol. In an exam, always use the numbers printed in the question, not remembered ones.
Why hydration enthalpies follow a pattern
Hydration enthalpy becomes more negative as charge density rises — that is, as the charge goes up or the ion gets smaller.
| Ion | Charge | Relative size | ΔHhyd (kJ/mol, typical) |
|---|---|---|---|
| Li⁺ | 1+ | smallest of the three | −520 |
| Na⁺ | 1+ | medium | −406 |
| K⁺ | 1+ | largest | −322 |
| Mg²⁺ | 2+ | small | −1921 |
| Ca²⁺ | 2+ | larger | −1577 |
Notice that the jump from 1+ to 2+ is much larger than any change within a group. Doubling the charge roughly quadruples the ion–dipole attraction energy, so 2+ and 3+ ions are hydrated enormously more strongly than 1+ ions of similar size.
Worked example 4 — measuring ΔHsolution in the lab
You do not need lattice data to get this number. A polystyrene-cup calorimeter is enough.
Experiment: 5.00 g of NH₄Cl is stirred into 100.0 g of water. The temperature falls from 25.0 °C to 21.8 °C. Take the specific heat capacity of the solution as 4.18 J g⁻¹ K⁻¹.
Step 1 — heat exchanged. Mass of solution = 100.0 + 5.00 = 105.0 g;
ΔT = 21.8 − 25.0 = −3.2 K.
q = m c ΔT = 105.0 × 4.18 × (−3.2) = −1404.5 J
The solution lost 1404.5 J, so the dissolving process absorbed +1404.5 J.
Step 2 — moles of NH₄Cl.
M = 14.007 + 4 × 1.008 + 35.45 = 14.007 + 4.032 + 35.45 = 53.489 g/mol
n = 5.00 ÷ 53.489 = 0.09347 mol
Step 3 — express per mole.
ΔHsolution = +1404.5 J ÷ 0.09347 mol = 15 025 J/mol =
+15.0 kJ/mol
Book values for NH₄Cl sit close to +15 kJ/mol, so a simple cup calorimeter gets remarkably near. Note that the mass used is the mass of the whole solution, not just the water.
If it absorbs heat, why does it dissolve at all?
Enthalpy is only half the story. Spontaneity is decided by Gibbs free energy:
Breaking an ordered crystal into freely moving hydrated ions raises the disorder, so ΔSsolution is usually positive. For NH₄Cl, taking ΔS ≈ +75 J K⁻¹ mol⁻¹ at 298 K:
TΔS = 298 × 75 = 22 350 J/mol = +22.35 kJ/mol
ΔG = (+15.0) − (+22.35) = −7.35 kJ/mol
ΔG is negative, so dissolving is spontaneous even though it cools the beaker. The entropy term simply wins.
This also explains why warming the water helps most endothermic dissolutions: a larger T makes the −TΔS term more negative.
Common mistakes that cost marks
- Mixing the two lattice-enthalpy conventions. Adding a negative lattice formation enthalpy to negative hydration enthalpies gives a hugely negative, obviously wrong answer. Check the sign before you add.
- Counting an ion once when the formula gives two. CaCl₂, MgBr₂ and Na₂SO₄ all need one hydration term multiplied.
- Using the mass of solvent only in q = mcΔT. The thermometer sits in the whole solution, so use the total mass.
- Forgetting to divide by moles. q is in joules for the sample; ΔHsolution is per mole. Missing this step is the most common calorimetry error of all.
- Saying "endothermic means it will not dissolve". Solubility is decided by ΔG, not by ΔH alone.
- Writing "hydration" for a non-aqueous solvent. In general the word is solvation; hydration is the special case where the solvent is water.
Where this appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 12 | Calculate ΔHsolution from given lattice and hydration data; explain why a salt cools water |
| Class 11–12 practical | Determine the enthalpy of solution of a salt by calorimetry |
| JEE/NEET | Predict the sign of ΔHsolution; compare solubility trends down a group |
| IIT-JAM / CUET-PG | Born–Haber style cycles combined with hydration enthalpies |
| GATE / CSIR-NET | Charge-density arguments for hydration trends; ΔG of dissolution |
Finish the spontaneity step in seconds. Once you have ΔHsolution, the Gibbs Free Energy tool takes ΔH, ΔS and T and returns ΔG = ΔH − TΔS with the unit handling done for you — exactly the step from the NH₄Cl example above.
Open the Gibbs Free Energy (ΔG = ΔH − TΔS) Calculator →Thermodynamics is where most Class 12 students lose easy marks — the physics is fine, the sign conventions are not. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with separate batches for board and competitive preparation: abcchemistry.in.