The standard enthalpy of reaction is given as $$\Delta_r H = \sum\nu\; H_m(\text{products})- \sum\nu\;H_m(\text{reactants}).$$

Peter Atkins in his book Elements of Physical Chemistry then writes regarding this equation:

The problem with this eqn, is that we have no way of knowing the absolute enthalpies of the substances . To avoid this problem, we can imagine the reaction as taking place by an indirect route, in which the reactants are first broken down into the elements. [...]

With the introduction of standard enthalpies of formation, we can write $$\Delta_r H = \sum\nu\; \Delta_f H(\text{products})- \sum\nu\;\Delta_f H(\text{reactants}).$$ The first term on the right is the enthalpy of formation of all the products from their elements; the second term on the right is the enthalpy of formation of all the reactants from their elements..the fact that the enthalpy is a state function means that a reaction enthalpy calculated this way is identical to the value that would be calculated from the former eqn, if absolute enthalpies were available.

I'm not understanding why we can't calculate the absolute enthalpies. After all, enthalpy of a substance at a certain pressure (which is constant as we are talking about standard state & that is 1 bar) is given by the defining equation:

$$H= U +pV$$.

Yes, can't we calculate the internal energy taking a certain state as the zero of the energy? Why isn't it possible to calculate the absolute enthalpy?

  • 1
    $\begingroup$ If you "take a certain state as the zero of the energy", then it is no longer an absolute internal energy, it is a relative internal energy. But I see what you mean. After all entropies are also defined to be 0 at absolute zero and we call those "absolute". I imagine it has more to do with the internal energy of a molecule not being easy to quantify. It is probably possible, but not worth the effort, which makes the values "not available". $\endgroup$ Commented Nov 7, 2015 at 12:08
  • $\begingroup$ @orthocresol: I do count it as an answer but let the question take some time more to see what others think. $\endgroup$
    – user5764
    Commented Nov 7, 2015 at 12:23
  • $\begingroup$ Nah, it doesn't deserve to be an answer. $\endgroup$ Commented Nov 7, 2015 at 12:24
  • $\begingroup$ @orthocresol: It's your opinion but your argument has logic, after all. $\endgroup$
    – user5764
    Commented Nov 7, 2015 at 12:25
  • 1
    $\begingroup$ Related: Why it is not possible to find absolute value of internal energy? $\endgroup$
    – user7951
    Commented Nov 7, 2015 at 12:58


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