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🔥Enthalpy of Reaction Calculator - ΔH, ΔS, ΔG, K and Hess's Law

The standard enthalpy change of a reaction is products minus reactants: ΔrH° = Σν·ΔfH°(products) − Σν·ΔfH°(reactants), and ΔrS° and ΔrG° are found the same way. For the combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔrH° = [−393.51 + 2(−285.83)] − [−74.81 + 0] = −890.36 kJ/mol, ΔrS° = −243.0 J/(mol·K) and ΔrG° = −817.9 kJ/mol. For CaCO₃(s) → CaO(s) + CO₂(g), ΔH° ≈ +178.3 kJ/mol and ΔS° ≈ +160.6 J/(mol·K); because both are positive, setting ΔG = ΔH − TΔS = 0 gives a crossover temperature of about 1110 K (837 °C), above which limestone decomposition becomes spontaneous. This calculator includes about 190 substances with ΔfH°, ΔfG° and S° at 298.15 K (elements, oxides, acids, bases, salts, organic compounds and common aqueous ions, from the NBS/NIST tables and the CRC Handbook). Type an equation and it is balanced automatically, each species is matched by its state symbol (s), (l), (g) or (aq), and missing values can be entered by hand. Results include ΔrG° both from ΔfG° and from ΔH − TΔS, ΔG and K = exp(−ΔG/RT) at any temperature, the four-case spontaneity sign table and a ΔG versus T chart. Separate tabs handle Hess's law (combine reactions with factors, solved automatically) and bond-enthalpy estimates.

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How to use Reaction enthalpy & Gibbs energy

  1. 1Type a chemical equation (state symbols and fractions such as 1/2 O2 are fine) or pick an example; adjust states or thermodynamic values if needed
  2. 2Read ΔrH°, ΔrS°, ΔrG°, K, the crossover temperature and the spontaneity verdict; enter a temperature to see ΔG at that T
  3. 3Switch to the Hess's law or bond enthalpy tab to combine known reactions or list the bonds broken and formed

FAQ

Why are there two slightly different values of ΔrG°?
One sums the standard Gibbs energies of formation ΔfG°; the other uses ΔrG° = ΔrH° − TΔrS° at 298.15 K. They are equal in theory, but each tabulated value is rounded independently, so they usually differ by a few tenths of a kJ/mol.
How do I know at what temperature a reaction is spontaneous?
Look at the sign of ΔG = ΔH − TΔS. If ΔH < 0 and ΔS > 0 the reaction is spontaneous at every temperature; if ΔH > 0 and ΔS < 0 it never is. When both have the same sign there is a crossover temperature T = ΔH/ΔS: below it if both are negative, above it if both are positive. CaCO₃ decomposition, for example, becomes spontaneous above about 1110 K. This assumes ΔH and ΔS do not change with temperature.
How do I solve a Hess's law problem with this tool?
On the Hess's law tab enter each known reaction with its ΔH, type the target reaction and press Solve factors. The tool finds the factor for each reaction (negative means reversed), shows the net reaction, checks it against the target and combines the ΔH values with the same factors — for example, the combustion enthalpies of C and CO give ΔH = −110.5 kJ/mol for C(s) + ½O₂(g) → CO(g).
Why does the bond enthalpy estimate differ from the table value?
Average bond enthalpies are averages over many molecules and apply only to gases, so they give estimates. Methane combustion to water vapour comes out at about −808 kJ/mol from bond enthalpies versus −802.3 kJ/mol from enthalpies of formation. Use enthalpies of formation when you need an accurate answer.