About the Enthalpy Change Calculator
This enthalpy change calculator finds the standard heat of reaction, ΔH°rxn, in two ways. In heats-of-formation mode you enter each reactant and product with its stoichiometric coefficient and standard enthalpy of formation (ΔHf°), and it applies ΔH = Σ n·ΔHf(products) − Σ n·ΔHf(reactants). In Hess’s law mode you enter the ΔH of up to four known reactions and how each is scaled or reversed, and it adds them up.
It is built for general and physical chemistry students, and for anyone estimating how much heat a combustion or synthesis releases. Enter how many moles of the first reactant are consumed to convert the molar ΔH into actual heat in kJ — for example, the energy released by burning a given amount of fuel.
The defaults use standard-state values at 25 °C and 1 bar from the NIST Chemistry WebBook (methane −74.87, CO₂ −393.52, liquid water −285.83 kJ/mol). Watch the physical state: liquid water and water vapour differ by about 44 kJ/mol, which changes the answer noticeably. Elements in their standard state, such as O₂(g), always have ΔHf° = 0.
How to use the enthalpy change calculator
- 1Choose heats of formation or Hess’s law.
- 2For formation: enter each reactant and product, its coefficient from the balanced equation and its ΔHf° from a data table.
- 3For Hess’s law: enter each known reaction’s ΔH and the multiplier you apply (−1 to reverse).
- 4Optionally enter how many moles of the first reactant are consumed.
- 5Read ΔH and whether the reaction is exothermic or endothermic.
Formula and method
Enthalpy is a state function, so the heat of a reaction depends only on where it starts and ends. Imagine breaking every reactant into its elements (the reverse of formation, −ΔHf°) and then building the products from those elements (+ΔHf°): adding the two gives ΔH°rxn = Σ n·ΔHf°(products) − Σ n·ΔHf°(reactants), with n the coefficients of the balanced equation.
Hess’s law uses the same idea with whole reactions: if a target reaction is the sum of known steps, its ΔH is the sum of their ΔH values, each multiplied by the factor applied to that step (negative when a step is reversed). ΔH is per mole of reaction as written, so the heat for a real amount is ΔH × (moles of reactant 1 ÷ its coefficient).
- ΔHf°
- Standard enthalpy of formation, kJ/mol (0 for elements in their standard state)
- n
- Stoichiometric coefficient in the balanced equation
- mᵢ
- Multiplier applied to step i in Hess’s law (−1 = reversed)
- q
- Heat for the actual amount, kJ (negative = released)
Worked examples
Combustion of methane
Products: −393.52 + 2 × (−285.83) = −965.18 kJ. Reactants: −74.87 + 2 × 0 = −74.87 kJ. ΔH = −965.18 − (−74.87) = −890.31 kJ per mole of methane burned.
Haber process: N₂ + 3H₂ → 2NH₃
Both elements have ΔHf° = 0, so ΔH = 2 × (−45.94) = −91.88 kJ per mole of N₂. Reacting 2.5 mol of N₂ releases 2.5 × 91.88 = 229.7 kJ.
Hess’s law: enthalpy of C + ½O₂ → CO
Take C + O₂ → CO₂ (−393.5 kJ) as written and reverse CO + ½O₂ → CO₂ (−283.0 kJ → +283.0 kJ). Adding them cancels CO₂ and gives −393.5 + 283.0 = −110.5 kJ.
Frequently asked questions
How do you calculate enthalpy change from heats of formation?+
Multiply each species’ ΔHf° by its coefficient, add up the products and the reactants separately, then subtract: ΔH = Σ products − Σ reactants. Elements in their standard state count as zero.
What does a negative ΔH mean?+
A negative ΔH means the reaction is exothermic: the products have lower enthalpy than the reactants and heat is released to the surroundings. A positive ΔH is endothermic and absorbs heat.
What is Hess’s law?+
Hess’s law says the total enthalpy change of a reaction is the same whatever route is taken, so you can add the ΔH values of steps that sum to your target reaction, flipping signs for reversed steps and scaling by any multipliers.
Why is the enthalpy of formation of O₂ zero?+
ΔHf° is defined as the enthalpy change to form one mole of a substance from its elements in their standard states. Forming O₂(g) from O₂(g) involves no change, so its value is zero by definition.
Does the state of water matter?+
Yes. ΔHf° of liquid water is −285.83 kJ/mol but water vapour is −241.83 kJ/mol. Using vapour for methane combustion gives about −802 kJ (the lower heating value) instead of −890 kJ.