Enthalpy Calculator

Enter your solution's mass, how much its temperature changed, and the moles of limiting reactant (or its grams plus molar mass) and this page computes the heat of your experiment and the enthalpy change in kJ/mol. The sign is reasoned out loud on every result, because the minus in the formula is where most of the points go missing.

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How this enthalpy calculator works

This is the coffee cup calorimetry calculation, the one behind nearly every "calculate the enthalpy" assignment: run a reaction in a nested pair of foam cups, watch the thermometer, and turn that temperature change into a number that describes the reaction itself. Two equations do all the work. First, q = mcΔT finds the heat the solution gained or lost, using the whole solution's mass and, unless you override it, a specific heat of 4.18 J/(g C), because a dilute aqueous solution is mostly water. Second, ΔH = −q ÷ n converts that heat into the reaction's enthalpy change per mole of limiting reactant.

That minus sign is the entire reason this page narrates its steps. More points are lost to it than to any arithmetic, so the sign is never silently applied here: every result reasons it out loud, from which way the temperature moved to which way the energy went.

The formula

q = m × c × ΔT
ΔH = −q ÷ n

q is the heat absorbed by the solution in joules, m the mass of the whole solution in grams, c its specific heat in J/(g C), and ΔT its temperature change, with the sign kept. n is the moles of limiting reactant, and ΔH is the enthalpy change in J/mol, which this page also reports in the kJ/mol that tables use. The minus sign changes the point of view: q describes the water, ΔH describes the chemistry, and energy that arrived in one left the other.

Worked example

Mix 50.0 mL of 1.0 M HCl with 50.0 mL of 1.0 M NaOH in a foam cup. The combined solution is close to 100.0 g, and the thermometer climbs 6.68 C. Each solution brought 0.0500 mol, they react one to one, so n = 0.0500 mol.

q = 100.0 × 4.18 × 6.68 = 2,792.24 J absorbed by the solution.

The solution warmed, so the reaction released that heat: ΔH = −2,792.24 ÷ 0.0500 = −55,844.8 J/mol = −55.84 kJ/mol, an exothermic reaction.

The tables put strong acid plus strong base at about −57 kJ/mol, and the gap is worth understanding rather than worrying about: some of the reaction's heat warmed the cup, the thermometer, and the room instead of the water, and every escaped joule shrinks the measured answer. Running a few percent low is what a foam cup does, and it means your experiment worked, not that it failed.

The sign is the chemistry

Here is the reasoning this page renders on every result, because it is the step that decides the answer's direction. Your thermometer watches the water, not the reaction. If the water warmed, that heat had to come from somewhere, and it came out of the chemistry: the reaction released energy, so from the reaction's point of view the change is a loss and ΔH is negative. Exothermic. If the water cooled, the chemistry was pulling heat in, and ΔH is positive. Endothermic. The minus sign in ΔH = −q ÷ n is exactly that switch of viewpoint, from the water's books to the reaction's books, and nothing deeper.

Positive enthalpy changes are not exotic. An instant cold pack is a bag of ammonium nitrate waiting to dissolve, and that dissolution runs at about +26 kJ/mol, pulling heat out of the water and your bruised ankle alike. If your solution went cold, this calculator will hand you a positive ΔH and say endothermic, and both of you will be right.

Hess's law: routes add, because enthalpy is a state function

Enthalpy depends only on where you start and where you end, never on the route between them, the same way the altitude gained on a hike does not care which trail you took. That property, being a state function, is what makes Hess's law work: if a reaction can be written as the sum of steps whose enthalpies you know, its enthalpy is the sum of theirs. Flip an equation and its sign flips with it; double an equation and its enthalpy doubles.

Burning carbon to carbon monoxide refuses to run cleanly on a bench, because some CO always burns on to CO₂. Hess's law routes around the mess with two reactions that do burn cleanly:

C + O₂ → CO₂, ΔH = −393.5 kJ
CO + ½O₂ → CO₂, ΔH = −283.0 kJ

Flip the second equation so CO₂ becomes a reactant, which flips its sign to +283.0, then add. The CO₂ cancels, leaving C + ½O₂ → CO with ΔH = −393.5 + 283.0 = −110.5 kJ/mol. A number nobody can measure directly, built from two that anybody can.

Formation enthalpies: the same law, pre-packaged

Tables of standard formation enthalpies are Hess's law with the bookkeeping done for you: every compound's ΔH_f is its enthalpy of assembly from elements, so any reaction's enthalpy is the sum for the products minus the sum for the reactants, each times its coefficient. For burning methane, CH₄ + 2O₂ → CO₂ + 2H₂O(l), the standard textbook figures give (−393.5 + 2 × −285.8) − (−74.6 + 0) = −890.5 kJ/mol. Elements in their standard states, like the O₂, count as zero by definition, since assembling an element from itself is no assembly at all.

This page deliberately embeds no formation enthalpy table. Those tables run to thousands of compounds, keeping one honest is a reference work's job, and this calculator's job is the calorimetry case: your cup, your thermometer, your number. The worked figures above are quoted so you can see the bookkeeping, and the NIST WebBook is the place to look values up.

There is a third route you will meet: bond enthalpies, adding up the energy of bonds broken minus bonds formed. Treat its answers as estimates, honestly rougher than the other two, because tabulated bond energies are averages over many molecules and no bond in your particular molecule is exactly average.

q is your beaker; ΔH is portable

The heat q is a fact about one afternoon: this much solution, this thermometer, these joules. Double the volumes and q doubles, while the reaction has not changed at all. Dividing by n strips your experiment's size out of the answer, and attaching the sign convention strips out the viewpoint, leaving a number that belongs to the reaction itself. That is why tables list ΔH in kJ/mol and never q: per mole is what makes your 100 g cup comparable with every other cup, lab, and textbook on earth. The division by n is the whole difference between a measurement and a result.

Specific heat or enthalpy: which page do you need

This page and our specific heat calculator are a deliberate pair, and the split is clean. Specific heat answers how much energy does it take to warm this stuff: it is a property of a material, measured in J/(g C), and q = mcΔT is its equation run forward. Enthalpy answers how much heat did this reaction move, per mole: it is a property of a change, measured in kJ/mol, with a sign that carries the direction. The calorimetry experiment is the bridge between them: the specific heat page's equation measures q, and this page divides it by moles and handles the sign.

The n usually arrives by way of a molar mass, so our molar mass calculator is the natural first stop when you have grams instead of moles, and this form will do that division for you if you flip the toggle. For reactions involving gases, our gas law calculator turns pressures and volumes into the moles this page needs.

Sources

Where the numbers on this page come from. We go to the body that publishes the figure, not to another calculator. See how we verify.

Standard enthalpies of formation, NIST Chemistry WebBook. The formation enthalpies quoted in the worked examples (CO2 -393.5, H2O liquid -285.8, CH4 -74.6, CO -110.5 kJ/mol) are the standard textbook figures, checked September 2026. This page deliberately embeds no formation enthalpy table: those tables run to thousands of compounds, and the calculator itself is for the calorimetry case.
Heat of neutralization, Standard general chemistry references. The strong acid plus strong base neutralization enthalpy is about -57 kJ/mol of water formed. The worked example's -55.84 kJ/mol is deliberately a little smaller, because a student calorimeter loses some heat to the cup and the room, and the page says so.

Frequently asked questions

How do you calculate enthalpy?

For a reaction in solution, measure the heat with q = m x c x (temperature change), using the solution's mass and 4.18 J/(g C) for dilute aqueous solutions, then divide by the moles of limiting reactant and flip the sign: enthalpy change = -q / n. A solution that warmed means the reaction released heat, so the enthalpy change comes out negative.

Why is the enthalpy change negative when the solution gets hot?

Because the heat your thermometer sees had to come from somewhere, and it came out of the chemistry. The enthalpy change describes the reaction, not the water: energy leaving the reaction is a loss from the reaction's point of view, so its sign is negative, and we call it exothermic. The minus sign in the formula is exactly that change of viewpoint, nothing deeper.

What is Hess's law?

Enthalpy is a state function: it depends only on where you start and where you end, not on the route. So if a reaction can be written as the sum of steps with known enthalpies, its enthalpy is the sum of theirs. Flip an equation and its sign flips with it; double an equation and its enthalpy doubles. It is the accounting rule that lets chemists measure reactions that refuse to run cleanly on a bench.

What is the difference between enthalpy and heat?

The heat q is what your particular experiment moved, in joules, and it depends on how much you happened to mix. The enthalpy change is q converted to a per-mole basis with a sign convention attached, in kJ/mol, and it belongs to the reaction itself. Your beaker's q is not portable; the enthalpy change is, which is why tables list it and not q.

What does per mole mean here?

Per mole of the limiting reactant as the balanced equation is written, which is why the same measurement gives one number per mole of HCl and would give a different one per mole of something with a coefficient of 2. Dividing your measured heat by the moles that actually reacted is what turns your experiment into a number anyone else can use.

Can an enthalpy change be positive?

Yes, and you have held one: an instant cold pack works because dissolving ammonium nitrate has an enthalpy change of about +26 kJ/mol, so it pulls heat out of its surroundings and the pack goes cold. A positive enthalpy change, endothermic, just means the chemistry absorbed heat instead of releasing it. Your solution cooling is the same signal in this calculator.

Why is my measured enthalpy smaller than the textbook value?

Almost certainly the calorimeter, not your arithmetic. A nested pair of foam cups still lets some heat warm the cup, the thermometer, and the room, and every joule that escapes your measurement shrinks the answer. Textbook values come from instruments built to trap all of it. Running a few percent low is normal and expected; the fix is better insulation and catching the peak temperature quickly, not different math.