Enthalpy of Solution / Dissolution Calculator
Calculate the heat released or absorbed when a solute dissolves in water.
Find the temperature change in a coffee-cup calorimeter setup.
Enthalpy of solution (ΔH_soln) is the heat change when one mole of a solute dissolves in a solvent.
Heat released or absorbed by dissolution:
q = n × ΔH_soln
Where n = moles of solute = mass / molar mass.
Temperature change of the solution:
ΔT = −q / (m_solution × Cp_solution)
The minus sign matters and it trips people constantly. Here q is the heat of the dissolving system, so the solution gains whatever the system loses. An exothermic dissolution has a negative q, and −q is therefore positive: the water warms up. Drop the sign and you get an answer that says a beaker of dissolving sodium hydroxide turns to ice, which is memorably wrong to anyone who has ever made up a batch.
For dilute aqueous solutions, Cp ≈ 4.18 J/g·°C (same as water). m_solution = mass of solvent + mass of solute.
Common ΔH_soln values at 25°C:
| Compound | ΔH_soln (kJ/mol) | Type |
|---|---|---|
| NaCl (table salt) | +3.9 | Endothermic (slightly absorbs heat) |
| NaOH | −44.5 | Exothermic (dissolves vigorously, gets HOT) |
| KOH | −57.1 | Exothermic |
| NH₄NO₃ | +25.7 | Endothermic (used in instant cold packs) |
| NH₄Cl | +14.8 | Endothermic |
| LiCl | −37.0 | Exothermic |
| CaCl₂ | −81.3 | Exothermic (road de-icer, hand warmers) |
| KNO₃ | +34.9 | Endothermic |
| MgSO₄ (anhydrous) | −91.2 | Exothermic |
| MgSO₄·7H₂O (Epsom salt) | +16.1 | Endothermic |
| Glucose | +10.6 | Endothermic |
| Urea | +15.4 | Endothermic |
Practical applications:
- Instant cold packs: ammonium nitrate dissolving in water (endothermic, absorbs heat)
- Chemical hand warmers: calcium chloride dissolving in water (exothermic). The click-to-activate sodium acetate warmers are a different mechanism entirely: they release heat by crystallizing out of a supersaturated solution, not by dissolving
- Safety note: NaOH dissolution is highly exothermic — always add solid to water, never add water to solid
Why lattice energy and hydration fight each other
Every dissolution is two competing steps. Pulling the crystal apart against its lattice energy costs energy, always. Surrounding the freed ions with water molecules releases energy, always. The enthalpy of solution is just the difference, which is why it can land on either side of zero and why the numbers are so much smaller than either underlying quantity. Sodium chloride is the classic near-tie: about +788 kJ/mol to break the lattice, about −784 to hydrate the ions, leaving a barely-endothermic +3.9. Dissolve a spoonful of table salt and the water gets imperceptibly cooler.
That near-cancellation also explains why hydrates behave so differently from their anhydrous forms. Anhydrous magnesium sulfate has bare ions waiting to be hydrated, so dissolving it releases a lot of heat. Epsom salt arrives with its water of crystallization already attached, most of the hydration energy is spent, and what remains is mildly endothermic. Same compound on the label, opposite thermal behaviour in the beaker.
How we build and check this calculator
This calculator runs entirely in your browser, so the numbers you enter stay on your device. The math behind it is written by hand and tested against worked examples and standard references before the page goes live.
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