Enthalpy of Reaction Calculator

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How to Use the Enthalpy of Reaction Calculator

1. 1. Write the balanced equation with states. For example CH4(g) + 2 O2(g) -> CO2(g) + 2 H2O(l). Put spaces around +, use (g), (l), (s) or (aq), and write ions as Na+ or SO4^2-. You can also press “+ reactant” or “+ product” in the table instead of typing.
2. 2. Fix anything flagged. A species that is not in the table is named, and a species that exists in several states, like water, asks for the state. If you know its ΔfH°, add it yourself with @, as in C8H18(l)@-250.1.
3. 3. Read ΔH°rxn. You get both sums, each species' contribution, and the working. The tool warns if the equation is not balanced, because the answer scales with the coefficients you typed.
4. 4. Ask for a real quantity. Choose a species and an amount in mol or grams to get the heat released or absorbed. Switch to Hess's law mode to add up known reactions instead.

The Formula

ΔH°rxn = Σ n·ΔH°f(products) − Σ m·ΔH°f(reactants)

ΔH°f = 0 for an element in its standard state

ΔH(net) = Σ (multiplier × ±ΔH of each step)

The standard enthalpy of formation ΔH°f is the enthalpy change when one mole of a compound forms from its elements, each in its most stable form at 1 bar and, in these tables, 298.15 K. Because the elements themselves need no change to form, their ΔH°f is zero: O₂(g), H₂(g), Br₂(l), graphite and so on. The coefficient n multiplies each value, so the answer is the enthalpy change per mole of reaction as written; double every coefficient and ΔH°rxn doubles.

Worked Example: Burning Methane

For CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l), with ΔH°f values of −74.6 for CH₄(g), −393.5 for CO₂(g) and −285.8 kJ/mol for H₂O(l):

• Products: 1 × (−393.5) + 2 × (−285.8) = −965.1 kJ
• Reactants: 1 × (−74.6) + 2 × 0 = −74.6 kJ
• ΔH°rxn = −965.1 − (−74.6) = −890.5 kJ per mole of CH₄ burned

The sign says the reaction is exothermic. Change H₂O(l) to H₂O(g), at −241.8 kJ/mol, and the answer becomes −802.5 kJ: the 88.0 kJ difference is the heat of condensing two moles of water vapour at 44.0 kJ each. This is why the state symbols are not decoration — drop them, and the tool will ask for the one you mean when a formula has more than one entry.

For a real quantity, pick CH₄(g) in “Express the heat for” and enter 10 g. The molar mass is 16.043 g/mol, so 10 g is 0.6233 mol and the heat released is 0.6233 × 890.5 = 555.1 kJ. A second example: glucose, C₆H₁₂O₆(s) + 6 O₂(g) → 6 CO₂(g) + 6 H₂O(l), gives ΔH°rxn = 6(−393.5) + 6(−285.8) − (−1273.3) = −2802.5 kJ, or about 155.6 kJ released per 10 g.

Hess's Law: Combining Known Reactions

Enthalpy is a state function, so the enthalpy change for a reaction is the same however it is reached, and the ΔH of a target reaction is the sum of the ΔH of steps that add up to it. Three rules cover every case: reverse a reaction and flip the sign of its ΔH; multiply a reaction and multiply its ΔH by the same number; add the results, cancelling species that appear on both sides.

To find ΔH for C(s) + ½O₂(g) → CO(g), which cannot be measured cleanly because some CO₂ always forms, combine C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ, with the reverse of CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ. Reversing flips −283.0 to +283.0, and the sum is −393.5 + 283.0 = −110.5 kJ, which matches the tabulated ΔH°f of CO. In Hess's law mode, enter each step's ΔH, tick “Reverse” on the second, and the tool does the arithmetic. It adds the numbers you give it; checking that the steps really add up to the target equation is up to you.

Frequently Asked Questions

What is the formula for the enthalpy of reaction from enthalpies of formation?

ΔH°rxn = Σ n·ΔH°f(products) − Σ m·ΔH°f(reactants), where n and m are the balanced-equation coefficients. Multiply each ΔH°f by its coefficient, add the product side and the reactant side separately, and subtract reactants from products.

Why is the enthalpy of formation of an element zero?

Formation is defined as making a compound from its elements in their standard states, so an element in its standard state needs no change to form. O₂(g), H₂(g), Br₂(l) and graphite are zero. The same element in another form is not: diamond is +1.9 kJ/mol and ozone O₃(g) is +142.7 kJ/mol.

Do I multiply ΔHf by the coefficients?

Yes. Each ΔH°f in the table is for one mole, so a coefficient of 2 means twice the value. The final ΔH°rxn is then per mole of reaction as written, and halving every coefficient halves the answer.

What does a negative ΔH mean?

A negative ΔH means the reaction releases heat to its surroundings at constant pressure; it is exothermic. A positive ΔH means it absorbs heat; it is endothermic. The sign says nothing about speed, and on its own it does not decide spontaneity, for which you need ΔG.

Why does the state of water change the answer?

Liquid and gaseous water have different enthalpies of formation, −285.8 and −241.8 kJ/mol. Forming H₂O(l) releases the extra 44.0 kJ/mol that condensation gives out, so the same combustion is more exothermic when the water ends up as a liquid.

Why does my answer differ slightly from my textbook's?

Handbooks tabulate slightly different values, and some round to the nearest 0.1 or 1 kJ/mol. Check which table your course uses and, if marks depend on it, type your book's values using @ for each species.