Let me first briefly describe the laws as I understand them.

Hess' law states that the enthalpy of reaction can be obtained by taking the sum of the enthalpies of formation for the compounds involved in the reaction, weighted by their stoichiometric coefficient. This is often shortened to $\Delta_\mathrm{r} H = \sum \Delta_\mathrm{fm} H_\mathrm{products} - \sum \Delta_\mathrm{fm} H_\mathrm{reactants}$. A similar law exists for entropy. To get the enthalpy of a compound, one usually looks up the tabulated standard state enthalpy, and then integrates the compound's heat capacity $C_p$ to get the enthalpy at the temperature of interest. A similar procedure is used for computing the entropy. Although I have not seen it done, I guess there is nothing in the way of using such laws in a partial reaction (i.e. equilibrium): by summing the enthalpies of the various compounds at equilibrium and subtracting the enthalpies of the compounds you started with, you should get the "enthalpy of equilibrium".

What assumptions underlie these laws? For Hess' law, I would think "enthalpy of mixing is zero" is one of them, but I've yet to see it stated.

An admittedly wage bonus question: I am here talking about $H$ and $S$, but what about similar laws (if they exist) for other potentials (Gibbs energy $G$, Helmholtz energy $A$ and internal energy $U$)?


1 Answer 1


The typically unstated assumption in applying Hess's law is that you are going from an initial state of pure reactants (say, in separate containers) to a final state of pure products (also in separate containers). In the case of ideal gas reactions, the enthalpy of mixing is zero and enthalpy is independent of pressure. So, for ideal gas situations, Hess's law can be extended and applied directly to starting and ending states in which the reactants and products are mixed (and even simultaneously present) and the partial pressures are all different from the standard pressure of 1 atm.

Of course, since U, S, and G are functions of state, the strict version of the law, applying only to pure reactants and products, still holds. However, for ideal gas mixtures, partial molar entropy and partial molar free energy are functions of mole fraction, so this needs to be taken into account for mixtures.

  • $\begingroup$ Thanks. Follow up: As far as I understand, measuring heats of reaction is how you determine enthalpies of formation in the first place. But if you're neglecting the contribution from enthalpy of mixing, wouldn't you get inconsistent results by using different reactions to determine a compound's enthalpy of formation? Or are tabulated enthalpies of formation based on the assumption that enthalpy of mixing is negligible compared to the enthalpy of reaction? $\endgroup$
    – user22112
    Oct 21, 2015 at 15:28
  • 1
    $\begingroup$ In the table of formation enthalpies, no mixing is involved. The tabulated values are for going from the pure elements to the pure compound. Any mixing that you may be thinking of is accounted for by including process steps to separate the various materials after reaction (conceptually using semipermeable membranes, say) to arrive at a pure product compound separate from any reactants that may be remaining. This is all inherently included in what we call the heat of formation. For an ideal gas, of course, no separation is necessary. $\endgroup$ Oct 21, 2015 at 17:44

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