The knowledge of standard heat of formation of various substances can be used to calculate the heat of reactions under standard conditions. The standard heat of any reaction is equal to the difference between the standard heat of formation of the products and the reactants

Why is that so? I'd prefer physical derivation for this: $$\Delta H^\circ = \sum\Delta H^\circ_{products} - \sum\Delta H^\circ_{reactants}$$

  • $\begingroup$ Technically, I prefer the form where the equation essentially states that the standard enthalpy change of formation is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. Now, the physical (as in measurable) verification becomes problematic when the energy released includes things like light. Note: Hess's Law applies to heat. $\endgroup$
    – AJKOER
    Oct 6, 2020 at 13:01

1 Answer 1


It is the consequence of the Hess's law and that is the consequence of the energy conservation law. Overall it says the enthalpy change does not depend on the path, but only on the start and final point.

Comparing to the real reaction path, you can formally get there also

  • via the path of conversion reactants to pure elements in their standard form
  • and then via forming products from those elements.

If the result were not the same, you could create perpetuum mobile switching both states via 2 different paths, gaining the nonzero net energy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.