I'm having trouble (not alone, apparently) understanding the concept of Gibbs free energy, but I'm beginning to. What I still don't understand, however, is how Gibbs free energy of formation differs from heat- or enthalpy of formation. Is it more precise, somehow? given that entropy is accounted for in Gibbs free energy?

Thanks in advance


1 Answer 1


Gibbs free energy is not more "precise." The two are different but complementary. Enthalpy and Gibbs Free Energy indicate different things. Enthalpy can tell you about the relative stabilities of the products and reactants. Gibbs free energy however can tell you about whether a reaction is spontaneous (whether a reaction will occur) under a set of specified conditions.

A lot of people will confuse exothermic - bond improving - reactions with spontaneous reactions. While it is true that a bond improving reaction is in itself quite favorable - after all - all chemical processes tend toward greater stability and lower potential energy - we must remember that there is another dimension to the problem of reaching maximum stability - namely, entropy (a measure of disorder).

Recall that all physical processes tend toward greater entropy (greater disorder) as well. Just as one's bedroom becomes more and more chaotic despite the pleas from Mom to clean one's room, physical processes tend toward chaos (disorder) as well. Gibbs free energy takes entropy into account with enthalpy and can tell us whether a process will actually go.

Consider this reaction:

$\ce{2NO_2(g) -> N_2O_4(g)}\ $

Consider only the forward reaction. The forward reaction is spontaneous under standard state conditions and 298 K - i.e. we have a negative Gibbs free energy value. Don't believe me? Nitrogen dioxide is brown. Nitrogen tetraoxide however is colorless. Put some brown nitrogen dioxide into a flask and if you observe it, you'll notice that the brown color fades somewhat as the colorless nitrogen tetraoxide is made.

The enthalpy of reaction is definitely negative - the reaction is exothermic because the reaction is a dimerization reaction - we are forming a bond here (specifically, a N-N bond) by using two nitrogen dioxide molecules (which are actually radicals, and remember that radicals are highly unstable). Nitrogen tetraoxide however is not a radical.

The entropy of reaction however is negative as well - i.e. we have an unfavorable change in entropy. Remember that $ΔS^o = S_{final} - S_{initial}$. In this case, it is clear that $S_{final} < S_{initial}$ because on the right hand side, we have one mole of gas. On the left hand (reactant) side, we have two moles of gas. There is definitely a lot more chaos (disorder) with 2 moles of gas molecules bumping around and whizzing around every single way imaginable. So we've gone from a disordered system on the left to a somewhat more ordered system on the right.

We would describe this reaction as (at 298 K and std. state conditions):

1) Spontaneous 2) Exothermic 3) Entropy lowering 4) Exergonic

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    $\begingroup$ spontaneous = exergonic. What is a bond improving reaction? Do you mean bond formation? During an exothermic reaction bonds are usually formed and broken or non of it at all. ...all chemical processes tend toward greater stability and lower potential energy... is not entirely correct. All processes are tending to a state of equilibrium and of all equilibria the lowest in energy will be populated most. $\endgroup$ May 13, 2014 at 6:49
  • $\begingroup$ Right, and equilibrium is the state of the lowest free energy. A bond improving reaction can be a bond forming reaction. A bond improving reaction also may not involve the formation of bonds at all but simply bond improvement. $\endgroup$
    – Dissenter
    May 13, 2014 at 12:48
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    $\begingroup$ I have never heard this term - how can a bond be improved? $\endgroup$ May 13, 2014 at 13:20
  • $\begingroup$ Not a bond but the aggregate of bonds can be improved in a molecule. $\endgroup$
    – Dissenter
    May 13, 2014 at 13:40
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    $\begingroup$ That would be yet again bond formation or solvatation or dispersion or some other kind of interaction. A bond cannot be improved. Improvement is also a very soft term relying on a reference state which might be chosen arbitrarily. $\endgroup$ May 13, 2014 at 13:50

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