[lakealer96]

Hi!
I'm going crazy on the relation between ΔG° and ΔG:

ΔG = ΔG° + RTlnQ


Then, consider this graph.


The green arrows are the ΔG and they show the "distance" from the equilibrium point.
The red arrow instead is the standard free energy. 

Now, according  to this equation, Q=1 implies that ΔG = ΔG°. The question is: at the middle of the graph, when Q = 1, ΔG seems to be extremelly lower than the ΔG° shown in the graph. I mean, if we drew a arrow when Q=1, it would be shorter than ΔG°.   How is that possible? Am I wrong?
Surely, there is a mistake I'm making, but I really can find it!
Thanks and sorry if my English is bad, I'm not English!

[Corribus]

I'm not 100% sure what you're asking. But, here is a link to a thread where the differences between ΔG and ΔG° are discussed.

http://www.chemicalforums.com/index.php?topic=66737.msg241353#msg241353

There is nothing generally special about the situation where Q = 1. The comparisons that are most meaningful are Q compared to K and ΔG compared to 0. ΔG° is simply a reference value that specifies where the equilibrium point is for a specific reaction.

In the case of the reaction shown in your post (conversion of n-butane to iso-butane), because iso-butane is more thermodynamically stable than n-butane, at equilibrium there will be more iso-butane in the product mixture than n-butane. This is true regardless of whether you start with a flask of pure n-butane or a flask of pure iso-butane. For either of those scenarios, calculating ΔG will specify the "direction of motion" of the reaction toward the equilibrium point. If you start with pure n-butane, you will form some iso-butane. If you start with pure iso-butane, you will form some n-butane. The ΔG° value, which can be calculated from an equilibrium constant or alternatively from standard thermodynamic enthalpy/entropy values, is an important component of this calculation because it calibrates the calculation appropriately. That is, it specifies where equilibrium is, so that you can determine which way you need to move to get to equilibrium from any random set of reaction conditions.  It is generally easy to predict the direction of a reaction like n-butane to iso-butane when you start with a flask of pure reagent. However, if I started you with a flask containing 50% n-butane and 50% iso-butane, how would you know whether more n-butane is formed or iso-butane is formed? This is why you need ΔG°, which specifies what the equilibrium position is.

[mjc123]

Let's think a bit about what "ΔG" is. It's not that obvious. The green arrows on the diagram are misleading, they are not "ΔG" in the equation. (Think what would happen to ΔG at extent of reaction = 0 or 1.) What is "ΔG", the quantity that is equal to zero at equilibrium? It is, for a given composition, represented by Q, the free energy change of converting a mole of A (n-butane) to B (isobutane) at that composition. (It is not, e.g., the difference between starting materials and mixture at Q; or, as your diagram might suggest, between mixture at Q and equilibrium mixture.) Imagine that you had a huge vessel, with a million moles of mixture at a given composition; converting 1 mole of A to B would have negligible effect on the overall composition. The free energy change would be "ΔG". Scale this system down to 1 mole of mixture and ΔG, per mole of A converted, is equal to dG/dα, where α is the extent of reaction. In other words, it is equal to the slope of the green G-α curve. This slope is zero at the equilibrium point, where G is a minimum. If the curve is properly drawn, at the composition where Q = 1 the slope of the green curve should be equal to the slope of the red line (= ΔG°/1).