The formula
- Adding stepwise reactions
ΔH_reaction = ΣΔH_steps- Reversing a step
reverse the equation and change the sign of ΔH- Scaling a step
multiply the equation by a factor and multiply ΔH by the same factor- From standard enthalpies of formation
ΔH°_reaction = Σn·ΔH°_f(products) - Σn·ΔH°_f(reactants)
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
ΔH_reaction | Enthalpy change of the target reaction | kJ (or kJ/mol as written) |
ΔH_steps | Enthalpy change of each known step, after any reversing or scaling | kJ |
ΔH°_f | Standard enthalpy of formation, for making one mole of a substance from its elements in their standard states | kJ/mol |
n | Stoichiometric coefficient of that substance in the balanced equation | mol |
° | Standard state marker: 1 bar, and the stated temperature, usually 298 K | dimensionless |
When it applies
- The target reaction is hard or impossible to run cleanly on its own, but it can be assembled from combustion or formation reactions that are tabulated.
- Every intermediate species has to cancel between the steps. If something is left over on the wrong side, the steps are not yet arranged correctly.
- Physical states have to match to cancel. H₂O(l) and H₂O(g) are different species here, and swapping them changes the answer by the heat of vaporization.
- The formation route needs ΔH°_f for every reactant and product. An element in its standard state has ΔH°_f of zero, which is why those terms drop out.
Worked example
Problem. Find ΔH for C(s) + ½O₂(g) → CO(g), which cannot be measured directly because some CO₂ always forms. Use C(s) + O₂(g) → CO₂(g), ΔH = -393.5 kJ, and CO(g) + ½O₂(g) → CO₂(g), ΔH = -283.0 kJ.
- The target has C(s) on the left, and so does the first reaction, so keep that one as written: ΔH = -393.5 kJ.
- The target has CO(g) on the right, but the second reaction has it on the left, so reverse it: CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ.
- Add the two: C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g). CO₂ cancels completely and ½O₂ cancels from the O₂ on the left, leaving C(s) + ½O₂(g) → CO(g), which is the target.
- Add the enthalpies the same way: ΔH = -393.5 + 283.0 = -110.5 kJ.
Answer. -110.5 kJ, an exothermic reaction, obtained without ever running it on its own.
Common mistakes
- Reversing a reaction and leaving the sign of ΔH alone. Reversing always flips the sign, and this single slip is the most common way a Hess's law answer goes wrong.
- Scaling the equation but not the enthalpy. Doubling the coefficients doubles ΔH, because enthalpy is an extensive quantity.
- Letting a species cancel across different physical states. Liquid water on one side does not cancel gaseous water on the other.
- Subtracting reactants from products in the wrong order on the formation route. It is products minus reactants, and the reverse gives an answer correct in size but wrong in sign.
Related formulas
- Nernst equation:
E_cell = E°_cell - (RT / nF)·ln Q - Gibbs free energy equation:
G = H - TS