Enthalpy of reaction calculator

How do you calculate the enthalpy of a reaction from enthalpies of formation?

ΔH°rxn = Σ n × ΔH°f(products) − Σ n × ΔH°f(reactants), where n is each substance’s coefficient in the balanced equation and ΔH°f is its standard enthalpy of formation; an element in its standard state has ΔH°f = 0 (OpenStax Chemistry 2e, section 5.3). For the combustion of ethanol, C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), the products give 2 × (−393.51) + 3 × (−285.83) = −1,644.51 kJ, the reactants −277.6 kJ, and ΔH°rxn = −1,644.51 − (−277.6) = −1,366.91 kJ per mole of reaction as written, close to the −1,366.8 kJ OpenStax gives in Table 5.2. Type any reaction; if the coefficients don’t balance it, the calculator balances it first.

How do I know if a reaction is exothermic or endothermic?

From the sign: a negative ΔH means heat is released to the surroundings (exothermic), a positive ΔH means heat is absorbed (endothermic) (OpenStax Chemistry 2e, section 5.3). The size depends on the physical states, so they must be written: H₂(g) + ½O₂(g) → H₂O(l) releases 285.83 kJ, but forming gaseous water releases only 241.82 kJ, the 44.01 kJ difference being the heat of vaporization of one mole of water. That is why the calculator wants (g), (l), (s) or (aq) after every formula.

How much heat is released by a given mass of fuel?

Scale ΔH°rxn by moles: heat = ΔH°rxn × (moles of the substance ÷ its coefficient). Pick the substance, type the grams or moles, and the calculator does the conversion with IUPAC atomic weights. For the pre-filled example, 100 g of ethanol (46.07 g/mol) is 2.171 mol, so burning it releases 2.171 × 1,366.91 = about 2,967 kJ. OpenStax Example 5.10 does the same with the tabulated combustion enthalpy of isooctane. The result is the heat at constant pressure, ΔH, not the heat measured in a bomb calorimeter.

What if a substance isn’t in the table?

The data are the 330 substances in OpenStax Chemistry 2e Appendix G that carry a state symbol, at 298.15 K; the appendix is a selection, so it has no Mg(s), MgO(s) or Ni(s), for example. A missing substance is named, and an input box appears for its ΔH°f in kJ/mol, which you can take from a data book. Remember that an element in its standard state is 0. If you leave out a state and the table has several (water is both l and g), you are asked to add one; for substances with several forms the first listed is used (C is graphite, type C(diamond) for diamond). Ions are written with the charge, like Ag+(aq) and Cl-(aq).

What is Hess’s law and how do I use the steps tab?

If a reaction can be written as the sum of several steps, its enthalpy change is the sum of the steps’ (OpenStax Chemistry 2e, section 5.3). Reversing a step changes the sign of its ΔH, and multiplying its coefficients by a number multiplies ΔH by it. The Hess’s law tab takes each step’s ΔH, a multiplier and the direction. OpenStax Example 5.14: ClF + F₂ → ClF₃ is ½ × (+214.0 kJ) for the first step, ½ × (−49.4 kJ) reversed for the second (+24.7 kJ) and (+236.2 kJ) reversed for the third (−236.2 kJ), giving ΔH° = +107.0 + 24.7 − 236.2 = −104.5 kJ. To turn ΔH into a verdict on spontaneity add the entropy change in the Gibbs free energy calculator.

Why can my answer differ from the textbook’s?

Rounding and a few slips in the book. OpenStax Example 5.15, 3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g), prints −136.80 kJ using −206.64 kJ/mol for nitric acid (a row of Appendix G with no state), while its own second method gets −138.4 kJ; with HNO₃(aq) = −207.4 kJ/mol, as listed, the full-precision answer is −138.3 kJ, which is what this calculator returns. Likewise Table 5.2 lists the combustion of isooctane as −5,465.5 kJ/mol while Example 5.10 uses −5,460.

Sources: OpenStax Chemistry 2e, section 5.3 “Enthalpy” (equation for ΔH°rxn, Hess’s law, Examples 5.10 to 5.15), Appendix G (Table G1), and the chapter 5 exercises and answer key. Related: Gibbs free energy calculator (ΔH goes into ΔG = ΔH − TΔS), stoichiometry calculator, specific heat calculator and chemical equation balancer.