Let’s assume we have a reaction medium containing only simple molecules such as $\ce{H2O}$, $\ce{CH4}$, $\ce{NH4+}$ (and maybe some more).

Now we consider the following equation ;

$\Delta G_\mathrm r = \Delta G_\mathrm r^{\circ} + RT \ln Q_\mathrm r$

were $\Delta G_\mathrm r$ is the Gibbs energy change associated to a chemical reaction occurring in this reaction medium, $\Delta G_\mathrm r^{\circ}$ the standard Gibbs energy change of this reaction and $Q_\mathrm r$ the mass action ratio of the reaction in the reaction medium at a given time.

Let’s consider the reaction consisting of the synthesis of a very complex molecule from the simple molecules of the reaction medium (for example, a protein). Whatever the complex molecule is, it is so complex that we are sure that it is absent from the reaction medium, i.e. its concentration is exactly 0. Also, the standard Gibbs energy change of this reaction must be very high (positive), but not infinite.

If the concentration of the complex molecule starts at 0; then $Q_\mathrm r$ is 0; then $RT \ln Q_\mathrm r$ is $-\infty$ ; as $\Delta G_\mathrm r^{\circ}$ is finite, $\Delta G_\mathrm r$ is $-\infty$. So in these conditions, the synthesis of at least one copy of our very complex molecule is infinitely exergonic. This reasoning could be applied to any molecule whose concentration in the reaction medium is 0 at a given time and whose synthesis from the reaction medium is theoretically possible. In the thermodynamics courses I attended, a negative Gibbs energy change was associated with the “favourableness” of a reaction, so those reactions would be “infinitely favourable”.

Of course, this is absurd. But why?

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    $\begingroup$ The equation you use is only valid under equilibrium conditions. So, at the start of the reaction when there is no product present in the medium, the equation you quoted does not apply. Only after equilibrium has been established you can use it. And in an equilibrium there is always at least some tiny amount of product in the mixture otherwise there wouldn't be a reaction. Thus, no infinite $\Delta G$ can arise. $\endgroup$ – Philipp Aug 14 '15 at 11:08

Within the context of the equation in question, $\Delta G$ refers to the change in free energy between two specific thermodynamic equilibrium states:

State 1: Stoichiometric quantities of pure reactants in separate containers at temperature T and individual specified pressures

State 2: Corresponding stoichiometric quantities of pure products in separate containers at temperature T and individual specified pressures

In the case of $\Delta G^0$, the specified pressures are all 1 atmosphere. This means that Q = 1, and $\Delta G =\Delta G^0$.

If the specified pressures in states 1 and 2 happen to correspond to the partial pressures of an equilibrium mixture of reactants and products, $\Delta G$ will be equal to zero. In this case, if one had an equilibrium mixture of the reactants and products, the partial pressures in the equilibrium mixture could also be held in equilibrium through individual semipermeable membranes with the gases in the separate containers of reactants and products (with no tendency for the reactants or products to enter or leave the reaction mixture through the semipermeable membranes).


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