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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?

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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.

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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 $\Delta G^{\circ}$.

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  • $\begingroup$ 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
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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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