Reaction Enthalpy Calculator (Hess's Law)
ΔH° from 414 formation enthalpies, from bond enthalpies, or by combining equations.
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Result
Values used
Heat per mole and per gram of each substance
How it was worked out
Look up a substance in the data table
| Substance | ΔfH° kJ/mol | ΔfG° kJ/mol | S° J/(mol·K) | Source |
|---|
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About the Reaction Enthalpy Calculator (Hess's Law)
Work out the enthalpy change of a reaction, ΔrH°, three ways. From enthalpies of formation: type the equation (it is balanced for you) and every substance is looked up in a table of 414 entries — elements, compounds, ions in water and common organic compounds — with the state and form chosen for you (graphite, calcite, liquid water) and editable, so ΔrH° = Σν·ΔfH°(products) − Σν·ΔfH°(reactants) follows at once, with the heat per mole and per gram of every substance and for any amount you weigh out. From bond enthalpies: for gases, the bonds broken minus the bonds formed, with the bonds of common molecules filled in. By Hess’s law: combine given thermochemical equations into a target one; the multipliers (×2, reversed, ×½) are solved exactly.
The values come from the CODATA Key Values for Thermodynamics, the NIST-JANAF tables, the NBS tables of chemical thermodynamic properties and the NIST Chemistry WebBook, each marked with its source; the bond enthalpies are worked out from those same formation enthalpies. Type your textbook’s value into any row to match its answer.
Without a pass the free preview shows the reaction enthalpy ΔrH° in full, with the first rows of each table and the first step of the working; the other figures are hidden; with a Pro pass, or after unlocking this result, you can copy the full answer or download it as CSV.
How to use it
- Choose Formation enthalpies, Bond enthalpies or Combine equations.
- Type the equation, for example
CH4 + O2 -> CO2 + H2O(l); add (s), (l), (g) or (aq) where the state matters — water as (l) or (g) changes the answer. - Check each substance: pick another state or form from its list, or type your own ΔfH° (or the bonds of a molecule) to replace the table value.
- Read ΔrH° in the dark box: negative is exothermic, positive endothermic. Add an amount (grams or moles of one substance) to get the heat for it.
- For Hess’s law, type the target equation and the given equations with their ΔH; the calculator finds how to combine them.
- Copy the result or download it as CSV — with a Pro pass, or after unlocking this result.
Examples
CH4 + O2 -> CO2 + H2O(l)
ΔrH° = −890.3 kJ per mole of CH₄ (−55.49 kJ/g); with H₂O(g), −802.3 kJ
CODATA values for CO₂ and H₂O, NIST-JANAF for CH₄.
C3H8 + O2 -> CO2 + H2O(l)
ΔrH° = −2219.2 kJ/mol (−50.32 kJ/g)
CaCO3 -> CaO + CO2
ΔrH° = +178.5 kJ/mol: endothermic
NH4NO3(s) -> NH4^+(aq) + NO3^-(aq)
ΔrH° = +25.45 kJ/mol: dissolving ammonium nitrate absorbs heat
H^+(aq) + OH^-(aq) -> H2O(l)
ΔrH° = −55.8 kJ/mol
CH4 + 2O2 -> CO2 + 2H2O (bond mode)
2650.0 kJ broken − 3462.6 kJ formed = −812.6 kJ, against −802.3 kJ from formation enthalpies of the gases
Target C(s) + ½O₂ → CO; given C + O₂ → CO₂ (−393.51 kJ) and CO + ½O₂ → CO₂ (−282.98 kJ)
First as given, second reversed: ΔH = −110.53 kJ
Common uses
- Thermochemistry homework: ΔH from a table, from bond enthalpies or by Hess’s law, with every step.
- Comparing fuels by the heat they release per gram.
- Checking whether a reaction in the lab will warm or cool the mixture, and by how much for the amount used.
- Preparing the ΔH° input for a Gibbs energy or equilibrium calculation.
The three methods
- Enthalpies of formation. ΔfH° is the enthalpy change when one mole of a substance forms from its elements in their standard states, so for any reaction ΔrH° = Σ ν·ΔfH°(products) − Σ ν·ΔfH°(reactants). An element in its standard state — O₂(g), graphite, Fe(s) — has ΔfH° = 0 by definition.
- Bond enthalpies. In the gas phase, breaking bonds takes energy and forming them releases it: ΔrH ≈ Σ D(bonds broken) − Σ D(bonds formed). Mean bond enthalpies are averages, so the result is an estimate, usually within about 10–30 kJ/mol; the calculator shows the formation-enthalpy value beside it when the gases are in the table.
- Hess’s law. Enthalpy is a state function, so the ΔH of a reaction is the sum of the ΔH of any steps that add up to it. Reversing an equation changes the sign of its ΔH; multiplying it multiplies ΔH.
About the data
- 414 entries at 298.15 K and 1 bar: ΔfH° (kJ/mol), ΔfG° (kJ/mol) and S° (J/(mol·K)), each with its source.
- CODATA Key Values for Thermodynamics: internationally agreed values for about 150 key substances and states, including the main ions in water. CODATA gives ΔfH° and S°; ΔfG° is worked out from them.
- NIST-JANAF Thermochemical Tables: many gases, oxides, halides, hydroxides and carbonates.
- NBS Tables of Chemical Thermodynamic Properties: further salts, ions in water and C₁–C₂ organic compounds; every entry with all three values was checked against ΔfG° = ΔfH° − TΔfS°.
- NIST Chemistry WebBook: larger organic compounds (propane, butane, octane, benzene, acetone, sucrose …).
- An ion in water uses the convention ΔfH°(H⁺, aq) = 0. A dissolved strong electrolyte without its own entry, such as NaCl(aq), is the sum of its ions.
- Tables differ slightly (CH₄: −74.87 kJ/mol here, −74.8 or −74.6 in other tables). Type your textbook’s value to match its answer.
How the bond enthalpies were worked out
Each mean bond enthalpy comes from one small molecule’s enthalpy of atomization, ΔatH = Σ n·ΔfH°(atom, g) − ΔfH°(molecule, g), all from the same data: H–H from H₂ (436.0 kJ/mol), O–H from water (463.5), N–H from ammonia (390.9), C=O from CO₂ (804.3), and so on. C–H and C–C are solved from ethane and propane together (413.3 and 346.0), so the alkanes come out right. Bonds in other molecules are not exactly these means — a C=O in an aldehyde or ketone is weaker than in CO₂, for example — so edit a value when your problem gives its own.
Sources
- CODATA, Key Values for Thermodynamics (Cox, Wagman and Medvedev)
- NIST, NIST-JANAF Thermochemical Tables (Chase)
- Wagman et al., The NBS Tables of Chemical Thermodynamic Properties, J. Phys. Chem. Ref. Data 11, Supplement 2
- NIST, Chemistry WebBook (SRD 69)
Limitations
- The values are for 298.15 K and 1 bar. At other temperatures ΔH changes with the heat capacities, which this calculator does not include.
- Substances that are not in the table need their ΔfH° typed in; isomers are matched by how the formula is written (C₂H₅OH is ethanol), so check the name shown.
- Bond enthalpies are means and apply to gases; liquids and solids need their enthalpies of vaporization or sublimation as well.
- Ions in water are at infinite dilution; real solutions give slightly different heats.
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Frequently asked questions
What do I get without a pass?
Without a pass, Reaction Enthalpy Calculator (Hess's Law) shows the reaction enthalpy ΔrH° in full, with the first rows of each table and the first step of the working; the other figures are hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.
What is the standard enthalpy of formation?
ΔfH° is the enthalpy change when one mole of a compound forms from its elements, each in its standard state, at 1 bar (here at 298.15 K). It is zero for an element in its standard state, such as O₂(g) or graphite.
How do I calculate ΔH of a reaction from enthalpies of formation?
Multiply each substance’s ΔfH° by its coefficient, add the products, add the reactants, and subtract: ΔrH° = Σν·ΔfH°(products) − Σν·ΔfH°(reactants). For CH₄ + 2O₂ → CO₂ + 2H₂O(l): [−393.51 + 2(−285.83)] − [−74.87 + 0] = −890.3 kJ.
Why does the state of water change the answer?
Liquid water is 44.0 kJ/mol lower in enthalpy than water vapour (−285.83 against −241.83 kJ/mol). With H₂O(l) you get the higher heating value of a fuel, with H₂O(g) the lower one.
Why is the bond-enthalpy answer different from the table answer?
Bond enthalpies are averages over many molecules, while each real bond depends on its neighbours. The estimate is usually within 10–30 kJ/mol of the value from formation enthalpies, which is the accurate one.
How does Hess’s law combine equations?
Find multiples of the given equations that add up to the target: reverse an equation to put a substance on the other side (its ΔH changes sign), and multiply it to get the right amount (its ΔH is multiplied too). The calculator solves the multipliers exactly and shows each step.
Is a negative ΔH exothermic?
Yes. A negative ΔrH° means the products have less enthalpy than the reactants: heat is released to the surroundings (exothermic). A positive ΔrH° means heat is absorbed (endothermic).