Hess's law

Hess's law states that the change of enthalpy in a chemical reaction is the same regardless of whether the reaction takes place in one step or several steps, provided the initial and final states of the reactants and products are the same. Enthalpy is an extensive property, meaning that its value is proportional to the system size.[3] Because of this, the enthalpy change is proportional to the number of moles participating in a given reaction.

In other words, if a chemical change takes place by several different routes, the overall enthalpy change is the same, regardless of the route by which the chemical change occurs (provided the initial and final condition are the same). If this were not true, then one could violate the first law of thermodynamics.

Hess's law allows the enthalpy change (ΔH) for a reaction to be calculated even when it cannot be measured directly. This is accomplished by performing basic algebraic operations based on the chemical equations of reactions using previously determined values for the enthalpies of formation.

Combination of chemical equations leads to a net or overall equation. If the enthalpy changes are known for all the equations in the sequence, their sum will be the enthalpy change for the net equation. If the net enthalpy change is negative (${\displaystyle \Delta H_{\text{net}}<0}$), the reaction is exothermic and is more likely to be spontaneous; positive ΔH values correspond to endothermic reactions. (Entropy also plays an important role in determining spontaneity, as some reactions with a positive enthalpy change are nevertheless spontaneous due to an entropy increase in the reaction system.)

Hess's law states that enthalpy changes are additive. Thus the value of the standard enthalpy of reaction can be calculated from standard enthalpies of formation of products and reactants as follows:

${\displaystyle \Delta H_{\text{reaction}}^{\ominus }=\sum _{i}a_{i}\Delta _{\text{f}}H_{products}^{\ominus }-\sum _{i}b_{i}\Delta _{\text{f}}H_{reactants}^{\ominus }}$

Here, the first sum is over all products and the second over all reactants, ${\displaystyle a_{i}}$ and ${\displaystyle b_{i}}$ are the stoichiometric coefficients of products and reactants respectively, ${\displaystyle \Delta _{\text{f}}H_{products}^{\ominus }}$ and${\displaystyle \Delta _{\text{f}}H_{reactants}^{\ominus }}$ are the standard enthalpies of formation of products and reactants respectively, and the ^o superscript indicates standard state values. This may be considered as the sum of two (real or fictitious) reactions:

Reactants → Elements (in their standard states)

${\displaystyle \Delta H_{\text{RE}}^{\ominus }=-\sum _{i}b_{i}\Delta _{\text{f}}H_{reactants}^{\ominus }}$

and Elements → Products

${\displaystyle \Delta H_{\text{EP}}^{\ominus }=\sum _{i}a_{i}\Delta _{\text{f}}H_{products}^{\ominus }}$

1. Given:
   1. C_graphite + O₂ → CO₂(g) ( ΔH = −393.5 kJ/mol) (direct step)
   2. C_graphite + 1/2 O₂ → CO(g) (ΔH = −110.5 kJ/mol)
   3. CO(g) +1/2 O₂ → CO₂(g) (ΔH = −283.0 kJ/mol)

   Reaction (a) is the sum of reactions (b) and (c), for which the total ΔH = −393.5 kJ/mol, which is equal to ΔH in (a).

2. Given:
   • B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g) (ΔH = 2035 kJ/mol)
   • H₂O(l) → H₂O(g) (ΔH = 44 kJ/mol)
   • H₂(g) + 1/2 O₂(g) → H₂O(l) (ΔH = −286 kJ/mol)
   • 2B(s) + 3H₂(g) → B₂H₆(g) (ΔH = 36 kJ/mol)

   Find the ΔH_f of:

   • 2B(s) + 3/2 O₂(g) → B₂O₃(s)

   After multiplying the equations (and their enthalpy changes) by appropriate factors and reversing the direction when necessary, the result is:

   • B₂H₆(g) + 3O₂(g) → B₂O₃(s) + 3H₂O(g) (ΔH = 2035 × (−1) = −2035 kJ/mol)
   • 3H₂O(g) → 3H₂O(l) (ΔH = 44 × (−3) = −132 kJ/mol)
   • 3H₂O(l) → 3H₂(g) + (3/2) O₂(g) (ΔH = −286 × (−3) = 858 kJ/mol)
   • 2B(s) + 3H₂(g) → B₂H₆(g) (ΔH = 36 kJ/mol)

   Adding these equations and canceling out the common terms on both sides, we obtain

   • 2B(s) + 3/2 O₂(g) → B₂O₃(s) (ΔH = −1273 kJ/mol)

The concepts of Hess's law can be expanded to include changes in entropy and in Gibbs free energy, since these are also state functions. The Bordwell thermodynamic cycle is an example of such an extension that takes advantage of easily measured equilibria and redox potentials to determine experimentally inaccessible Gibbs free energy values. Combining ΔG^o values from Bordwell thermodynamic cycles and ΔH^o values found with Hess's law can be helpful in determining entropy values that have not been measured directly and therefore need to be calculated through alternative paths.

For the free energy:

${\displaystyle \Delta G_{\text{reaction}}^{\ominus }=\sum \nu _{\text{p}}\Delta G_{\mathrm {f} \,({\text{p}})}^{\ominus }-\sum \nu _{\text{r}}\Delta G_{\mathrm {f} \,({\text{r}})}^{\ominus }.}$

For entropy, the situation is a little different. Because entropy can be measured as an absolute value, not relative to those of the elements in their reference states (as with ΔH^o and ΔG^o), there is no need to use the entropy of formation; one simply uses the absolute entropies for products and reactants:

${\displaystyle \Delta S_{\text{reaction}}^{\ominus }=\sum \nu _{\text{p}}S_{({\text{p}})}^{\ominus }-\sum \nu _{\text{r}}S_{({\text{r}})}^{\ominus }.}$

Hess's law is useful in the determination of enthalpies of the following:[1]

1. Heats of formation of unstable intermediates like CO_(g) and NO_(g).
2. Heat changes in phase transitions and allotropic transitions.
3. Lattice energies of ionic substances by constructing Born–Haber cycles if the electron affinity to form the anion is known, or
4. Electron affinities using a Born–Haber cycle with a theoretical lattice energy.

• Thermochemistry
• Thermodynamics

• Leicester, Henry M. (1951). "Germain Henri Hess and the Foundations of Thermochemistry". The Journal of Chemical Education. 28 (11): 581–583. Bibcode:1951JChEd..28..581L. doi:10.1021/ed028p581.

• Hess' paper (1840) on which his law is based (at ChemTeam site)
• a Hess’ Law experiment