I know how to derive the expression for the equilibrium constant starting from $\mathrm dG=0$, $$\ln(K)=-\frac{\Delta G ^\circ}{RT}$$

However, if we are not at equilibrium, $\mathrm dG$ is not necessarily zero. In this case: $$\frac{\partial G}{\partial \epsilon} = \sum \mu_i \nu_i = \Delta G ^\circ + RT \ln \left( \prod a_i ^{\nu_i} \right)$$

Where $\nu_i$ are the stoichiometric coefficients, $\mu_i$ are the chemical potentials, $\Delta G^\circ$ is the Gibbs free energy of reaction at standard state and $a_i$ are the activities.

Hence, $$\frac{\partial G}{\partial \epsilon} = RT \ln \left(\frac{Q}{K} \right)$$

Where $Q=\prod a_i ^{\nu_i}$ is the reaction quotient.

So the question is, if we want to kown if a reaction is spontaneous, do we look at the ratio $Q/K$? If this is correct, if $Q > K$ then $\frac{\partial G}{\partial \epsilon} > 0$ and the reaction would be non-spontaneous, whereas if $Q < K$ then $\frac{\partial G}{\partial \epsilon} < 0$ and the reaction is spontaneous.

If so, if we start with reactants only, $Q=0$ (since the activites of the products are zero because there are none) and the reaction would be spontaneous, and we would always get at least a little bit of products.

However, various pages in internet say that if $\Delta G ^\circ < 0$ then the reaction is spontaneous, and it is not if $\Delta G ^\circ>0$.

So how do you know, then, if you start with only reactants, if a reaction is going to happen?


I think you are confusing two related concepts here. Spontaneity of a reaction is determined by the net change in free energy, $\Delta G$ - notice the capital Delta.

Which direction you will move along a reaction coordinate is determined by $dG$ - note the little $d$.

$\Delta G$ is the overall change in free energy when comparing products to reactants. It tells you whether the reactants are "uphill" or "downhill" of the products.

$dG$ is the slope of the free energy surface at your current position on the reaction coordinate. In the equation you gave, that's given by Q/K. K basically tells you what the equilibrium ratio between reactants and products is, and Q tells you what the current ratio is. The ratio of Q/K is an indication of how far to the left or right you are of equilibrium.

This picture might help explain it a little better:

Difference between net change in free energy and derivative of free energy

Image the free energy surface as a gravitational potential energy surface (we can get away with this because free energy is the thermodynamic potential for systems at constant T & P). The ball represents the reacting system, and the x-axis is a reaction coordinate.

$dG$ (the differential) tells us the slope at a given point - it tells us where the system is going to go right now. $\Delta G$ tells us the overall change - it tells us where the system is going to eventually end up.

When you look up standard state $\Delta G$ values for a reaction, you are determining whether an overall reaction is spontaneous or not - you are figuring out which side of the equilibrium has more stuff - reactants or products.

When you look at Q, you are figuring out which direction things need to go to get back to equilibrium, based on where you currently are.


Your analysis is correct. It is silly to say that a reaction is or is not spontaneous based soley on $\Delta G^{\circ}$. If you have only reactants present and no products present, then the reaction will always be spontaneous, irrespective of $\Delta G^{\circ}$.

  • $\begingroup$ So you are saying that the reaction $\ce{O2 -> 2O}$ is spontaneous when all you have is $\ce{O2}$? That doesn't seem right... $\endgroup$ – thomij Jun 25 '15 at 4:45
  • $\begingroup$ thomji, it is right. But the equilibrium constant might be so small that at "room" temperature, the expected amount of O atoms would be less than 1, even in a system the size of the atmosphere. $\endgroup$ – Curt F. Jun 25 '15 at 6:37
  • $\begingroup$ There is a difference between saying a reaction can happen and saying it is spontaneous. $\endgroup$ – thomij Jun 25 '15 at 20:45
  • $\begingroup$ I think that @thomij is correct. The reaction O$_2$ + N$_2$=2NO , for example, has a $\Delta G^0$ of +180 kJ/mol. In a thermodynamic sense then O$_2$ + N$_2$ are stable as there is no process that can lead to a diminution in free energy. $\endgroup$ – porphyrin Jul 30 '16 at 9:02

IMHO, I think you had the right answer: "whereas if Q < K then ∂G∂ϵ<0 and the reaction is spontaneous."

So if you have only reactants, Q = 0. You then look up K for this temperature. Since K > 0 (even if it's very close to it), ΔG<0 and so the reaction will occur spontaneously. When products builds up, you will reach a point where Q = K, and at that point ΔG will be equal to zero. (ΔG depends on the relative concentration of reactants\products and changes as products build up from ΔG < 0 to ΔG = 0 at equilibrium).


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