Heat of Reaction & Hess's Law

Compute ΔH°rxn from formation enthalpies and use Hess's Law to trust the answer even when the pathway changes.

Material & Energy BalancesChemical Engineering Year 1Free preview
⏱️ About 18 min

You can't measure the heat of every reaction directly — so how do engineers compute it reliably from tabulated data?

💡
The big idea: Reaction enthalpy is a state function: ΔH°rxn can be computed from standard heats of formation, independent of the path (Hess's Law).
🎯 By the end, you'll be able to
  • Define standard heat of reaction and the reference state convention (25°C, 1 bar)
  • Compute ΔH°rxn from standard heats of formation
  • Explain Hess's Law (path independence) and why it works
  • Apply sign convention correctly (negative = exothermic)
📎 Helpful to know first

Standard Heat of Reaction from Formation Enthalpies

The standard heat of reaction ΔH°rxn is the enthalpy change when reactants in their standard states form products in their standard states (typically 25°C and 1 bar).

A powerful way to compute it is with standard heats of formation ΔH°f, tabulated for many species.

\[ \Delta H^\circ_{\text{rxn}} = \sum_i \nu_i\, \Delta H^\circ_{f,i}\big|_{\text{products}} \;-\; \sum_i \nu_i\, \Delta H^\circ_{f,i}\big|_{\text{reactants}} \]
Compute ΔH°rxn by summing ν·ΔH°f over products minus reactants (ν are stoichiometric coefficients in the written balanced reaction).

Hess's Law (Path Independence)

Hess's Law states that the enthalpy change depends only on initial and final states, not on the reaction path. This works because enthalpy is a state function.

Practically: you can add/subtract intermediate reactions (or use formation reactions) and the ΔH values add accordingly.

⚠️ Pitfall: Sign convention (negative = exothermic)

If ΔH°rxn < 0, the reaction releases heat (exothermic). If ΔH°rxn > 0, it absorbs heat (endothermic). Keep the sign with you into the energy balance — don't flip it accidentally.

📝 Worked example: Compute ΔH°rxn at 25°C for the combustion of methane to CO₂(g) and H₂O(g): CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). Use standard reference values: ΔH°f[CH₄(g)] = −74.8 kJ/mol, ΔH°f[CO₂(g)] = −393.5 kJ/mol, ΔH°f[H₂O(g)] = −241.8 kJ/mol, ΔH°f[O₂(g)] = 0.
  1. Write the products sum (ν·ΔH°f):
  2. Σ products = (1)(−393.5) + (2)(−241.8) kJ/mol = −393.5 − 483.6 = −877.1 kJ/mol
  3. Write the reactants sum (ν·ΔH°f):
  4. Σ reactants = (1)(−74.8) + (2)(0) kJ/mol = −74.8 kJ/mol
  5. Compute ΔH°rxn:
  6. ΔH°rxn = Σ products − Σ reactants = (−877.1) − (−74.8) = −877.1 + 74.8 = −802.3 kJ/mol
✓ ΔH°rxn = −802.3 kJ/mol CH₄ (to CO₂(g) + 2H₂O(g) at 25°C)
✏️ Practice: For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), using the same standard reference values, what is ΔH°rxn (kJ per mol CH₄)?
kJ/mol
Solution
  1. Products: (−393.5) + 2(−241.8) = −877.1 kJ/mol
  2. Reactants: (−74.8) + 2(0) = −74.8 kJ/mol
  3. ΔH°rxn = −877.1 − (−74.8) = −802.3 kJ/mol

Check your understanding

1. Hess's Law is valid because:
Enthalpy depends only on state, not path, so enthalpy changes add along any sequence connecting the same endpoints.
2. For an exothermic reaction at standard conditions, ΔH°rxn is:
By convention, ΔH < 0 means the system releases heat to the surroundings (exothermic).
🔎 Related lessons
Heat of Combustion · soon
✅ Key takeaways
  • ΔH°rxn can be computed from ΔH°f values: products minus reactants
  • Hess's Law: ΔH depends only on initial and final states, so reaction-path details don't matter for ΔH
  • Elements in their standard states have ΔH°f = 0 (e.g., O₂(g))
  • Negative ΔH°rxn indicates exothermic heat release
➡️ Combustion is a special (and extremely common) reaction class. Next we connect heats of combustion to real fuel ratings, and explain why HHV and LHV differ by the latent heat of the water produced.
Want to test yourself on this? Try the Chemical Aptitude test →