- #1

- 739

- 3

So for 1) the graph of Gibbs energy against the progress of reaction would look like this: http://postimg.org/image/phycri0vr/

So the difference between 100% product and 100% reactant is the ΔG of the reaction which makes sense.

But for 2) the graph of Gibbs energy against the progress of reaction would look like this: http://postimg.org/image/grmh7oknz/

So now the ΔG° would be for 1 mole of reactant to form 1 mole of product and also, the actual difference is just the G value of the products minus the G value of the reactants. However, during equilibrium the ΔG is now 0. However, if we just take the difference between the G value of the equilibrium point, E and the 1 mole of reactant, that value is non-zero so why is it always known for the ΔG at equilibrium to be 0?

Lastly, I also read that no reaction is fully irreversible so in fact all reactions has the graph on 2). However, in that case if all reaction has an eventual ΔG of 0 then how can any work be done? And we always say that using the formula ΔG°=ΔH°-TΔS° would give us the maximum work being done. But since all reactions would have an actual ΔG of 0, how can there any work be done?