I was doing a couple of problems for homework:

Calculate $K_\mathrm{sp}$ of $\ce{AgI}$ at $55.0\ \mathrm{^\circ C}$

Calculate $K_\mathrm{b}$ of $\ce{NH3}$ at $36.0\ \mathrm{^\circ C}$

I have to use $\Delta G^\circ= -RT\ln K$ and $\Delta G= \Delta H-T\,\Delta S$

When I did this $\Delta G^\circ$ is positive ($89.59\ \mathrm{kJ/mol}$ and $28.037831\ \mathrm{kJ/mol}$ respectively), yet $K_\mathrm{sp}$ for $\ce{AgI}$ is $5.5\times10^{-15}$ and $K_\mathrm{b}$ for $\ce{NH3}$ is $1.8\times10^{-5}$, indicating that there are some products and the reactions do occur. Plus, $1.0\ \mathrm M$ $\ce{NH3}$ in solution has a $\mathrm{pH}$ of $11.6$ so it must react a little.

According to the second law of thermodynamics, if $\Delta G$ is positive, the reaction is not spontaneous, right? But clearly, they, in fact, are to a certain extent. What is going on?

  • 1
    $\begingroup$ For a Q vs K argument (instead of $\Delta_r G$ vs $\Delta_r G^\circ$) see chemistry.stackexchange.com/questions/26251/… $\endgroup$
    – Karsten
    Commented May 19, 2019 at 10:37
  • $\begingroup$ Individual molecules know nothing of Gibbs, or temperatures, or entropy. They just collide and perhaps react, by exchanging energy. As long as molecules do what the conservation of energy and impulse dictates them (within the realm of structures that are possible by QM), the laws of thermodynamics emerge naturally for a macroscopic sample. $\endgroup$
    – Karl
    Commented May 20, 2019 at 19:57

3 Answers 3


$\Delta G^\circ_m$ is the difference in molar Gibbs free energy between the reagents and products in their standard states (in the case of $\ce{AgI(s)}$, the standard state for the reagent is the pure solid and that for the product is the solute in a $\pu{1 molal}$ ideal solution, all at $p^\circ =\pu{ 1 bar}$).

If $\Delta G^\circ_m>0$ you cannot convert the reactants completely into products without doing some accompanying work (for instance, applying an electrical voltage, in the case of an electrochemical cell).

If you want to see where the reaction is headed, for instance, if you start with only reactants (say $\ce{AgI(s)}$), then you should compute the following:

$$\Delta G_m=\Delta G^\circ_m+RT\log(Q)$$

For a reaction $\ce{A<->B}$ this is illustrated in the following plot (generated assuming an equilibrium constant $K_{eq}=10$, and $\Delta G$ in arbitrary units):

enter image description here

Here $\zeta$ represents a progress coordinate ($\zeta=Q/(1+Q)= \chi _B$).

Note that some authors place a "tilde" following $\Delta G_m$ to emphasize that it is a partial molar quantity, that is, it measures the change in free energy with reaction progress. At the start of the reaction, when there is no solute dissolved, Q=0 and $\Delta G_m=- \infty$. In other words, the driving force for the reaction to occur is theoretically unlimited. At equilibrium that force goes to zero. Past the equilibrium point, the reverse reaction becomes more favorable.

See "What is the difference between ∆G and ∆G°?" for an elaborate answer addressing a similar question.


The reaction Gibbs energy says what is $\Delta G$ for the conversion of reagents to products.

But it does not (directly) say what is $G$ value between these 2 states.

The $G$ for the transition from reagents to products is not a linear function, but it has the minimal value at the reaction equilibrium.

If there is reaction $\ce{A <=> B}$ with $\Delta G^\circ_\mathrm{r}=0$ and $K=1$, then the $G_\mathrm{min}$ will be at $x_\mathrm{a}=x_\mathrm{b}=0.5$

If the above reaction has $\Delta G^\circ_\mathrm{r}$ slightly positive, then $K\lt 1$.

The reaction will not magically stop happening, but the equilibrium with $G_\mathrm{min}$ will be slightly shifted toward reactants.


It is only a "rule of thumb" that a reaction is spontaneous only if the standard change in free energy is negative. In reality, a reaction can be considered spontaneous if, starting with stoichiometric proportions of reactants, the amounts of reactants remaining at final equilibrium is small (i.e., the equilibrium constant of the reaction is high, $\Delta G^0$ is negative, and the reaction goes nearly to completion).

Conversely, a reaction is considered non-spontaneous if, starting with stoichiometric proportions of reactants, the amounts of products produced when final equilibrium is achieved is small (i.e., the equilibrium constant of the reaction is low, $\Delta G^0$ is positive, and the reaction does not approach completion).

It was a stupid idea in the first place for teachers to introduce the concept that the sign of $\Delta G^0$ can be used as an on-off switch for determining whether a reaction is spontaneous or not; it has caused an endless amount of confusion to students. In reality, all reactions are spontaneous to some extent.


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